A Workbook in Topology

S. Kumaresan
School of Math. and Stat.
University of Hyderabad
Hyderabad 500046
kumaresa@gmail.com

Preface

This is a summary of courses on General Topology, offered by me at the Department of

Mathematics, University of Mumbai during the academic year 2004-2005, at the Department
of Mathematics and Statistics, University of Hyderabad in Jan–April 2012 and at the School
of Mathematics and Statistics, University of Hyderabad in Jan–April 2015. There are minimal
number of proofs in this set of notes. This may be used as a workbook by students of topology.
Because of its brevity, its emphasis on the most useful concepts in the practice of modern
topology and wealth of concrete examples, the student will be adequately prepared to do
mathematics which require a deeper knowledge of topology rather a nodding acquaintance to
a large number of dated concepts.

Its merit, if any, lies in the choice of topics, their development and the emphasis on
concrete and geometric examples and exercises. I plan to add a bit more material and a lot
of pictures so that it could serve as a skeleton of a course in General Topology. Later I plan
to develop this into a text-book. (So, please do not plagiarize!)

This set may be used in conjunction with many articles of mine on Topology. They are
the topics which a student who wishes to specialize in Topology as practiced now need to
know and they take them further into some of the topics dealt with the main text. These are
appended at the end of the workbook as appendices.

Topology of Metric Spaces, 2nd edition, by S. Kumaresan is published by Narosa. The

books Topology by Munkres and Topology by Armstrong are available in Indian edition. These
three books may be used to fill in the details of my outline. My book is strongly recommended
for pictures, geometric insights and developing a taste for topology.

I would appreciate receiving your comments and views.

1
Contents

1 Basic Notions 5

2 Continuity 13

3 Limit and Cluster Points 23

4 First and Second Countable Spaces 28

5 Dense Subsets in a Topological Space 32

6 Homeomorphisms 36

7 New Topologies from the Old 40

8 Compact Spaces 59

9 Connected and Path-connected Spaces 70

10 Locally P-Spaces 79

11 One Point Compactification 82

12 Baire Category Theorems 84

13 Completely Regular and Normal Spaces 88

14 Homotopy 94

15 Covering Space 98

A Concepts Summary 101

B Finite Sets 102

C Cardinality and Countability 104

D Subspace Topology 106

2
E Generating Topologies — A Unified View of
Subspace, Product and Quotient Topologies 108

F Quotient Topology 113

G Tychonoff ’s Theorem 121

H Compact Spaces 122

H.1 Heine-Borel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
H.2 Characterization of Compact Metric Spaces . . . . . . . . . . . . . . . . . . . 124
H.3 Tychonoff’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
H.4 Continuous Functions on Compact Spaces . . . . . . . . . . . . . . . . . . . . 128
H.5 Characterization of Compact Metrizable Spaces
via Metrics and Continuous Functions . . . . . . . . . . . . . . . . . . . . . . 129

I Connected and Path Connected spaces 132

I.1 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
I.2 Path Connected spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
I.3 Examples & Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

J Proper Maps 138

K Existence of Continuous Functions 141

K.1 Case of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
K.2 Normal Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

L Topological Groups — via Problems 146

M Discrete subgroups of Rn 150

N Non-contractibility of the circle and

Brouwer Fixed Point Theorem 151

O Maps into Punctured Plane 154

P Rm is not homeomorphic to Rn if m 6= n 160

P.1 Barycentric Subdivision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

3
P.2 Sperner’s Lemma and its Corollaries . . . . . . . . . . . . . . . . . . . . . . . 162
P.3 dim T n = n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

4
1 Basic Notions

1.1 Finite sets. Let X be a set. We say that it is finite if X = ∅ or if there exists a bijective
map of X into an initial segment In := {k ∈ N : 1 ≤ k ≤ n} of N.
Using induction/well-ordering principle, one can show that if X has a bijection with Im
and In , then m = n. The unique n is called the number of elements in X. (For a proof,
see my article on Finite sets.) The number of elements in the emptyset is 0.
1.2 Countable and uncountable sets. We say that a set X is countable if either X = ∅
or if either of the equivalent conditions are satisfied:
(a) There is a one-one map f : X → N.
(b) There exists an onto map g : N → X.
Applications: Countability of N × N, Q+ , Q, countable union of countable sets, finite
product of countable sets. (See my article on Countable and Uncountable sets, also
Munkres.)
1.3 Uncountability of 2N : Cantor’s theorem: there exists no onto map from X to P (X).
We prove this by contradiction. Assume f : P (X) → X be onto. Consider the set
S := {x ∈ X : x ∈ / f (x)}. Since f is onto there exists a ∈ X such that f (a) ∈ S. Now
exactly one of the following must happen: (i) a ∈ S or (ii) a ∈
/ S. If a ∈ S, by the very
definition of S, a ∈
/ f (a) = S, contradiction. Similarly (ii) cannot happen. Hence we
conclude that no such f exists.
1.4 Metric Spaces: In most of a first course in real analysis, we just needed the notion of
a distance between two real numbers to define the concept of convergent sequences or
the concept of continuous functions. Motivated by this we define a metric or a distance
function on a (nonempty) set X as a function d : X × X → R which satisfies the following
properties:
(a) For all x, y ∈ X, we have d(x, y) ≥ 0 and d(x, y) = 0 iff x = y.
(b) d(x, y) = d(y, x) for all x, y ∈ X.
(c) For all x, y, z ∈ X, we have the traingle inequality:
d(x, z) ≤ d(x, y) + d(y, z).
1.5 Metrics in R2 : L1 and L∞ metrics, called the sum and max metrics:
n
X
d1 (x, y) := |xk − yk |
k=1
dmax (x, y) ≡ d∞ (x, y) := max{|xk − yk | : 1 ≤ k ≤ n}.

These distances generalize to function spaces. See Item 8.

1.6 Normed linear spaces. A norm on a vector space V over R (or over C) is a function
k k : V → R satisfying the following conditions:
(i) For x ∈ V , kxk ≥ 0 and kxk = 0 iff x = 0.
(ii) For x ∈ V and λ ∈ R (or λ ∈ C if X is vector space over C) , we have kλxk = kλk kxk.
(iii) For x, y ∈ V , we have the triangle inequality kx + y k ≤ kxk + ky k.

5
 1.7 Examples of normed linear spaces:

(a) Finite dimensional normed linear spaces: On Rn , we have the following norms:
n
X
kxk1 := |xk | and kxk∞ := max{|xk | : 1 ≤ k ≤ n}.
k=1

That these are norms is easily verified.

2 1/2 .
Pn

Another norm is the standard/Euclidean norm: kxk2 := k=1 |xk | We need
Cauchy-Schwarz inequality to verify that this is a norm.

1.8 Function spaces.

(a) Let X be any nonempty set. Let B(X, R) denote the real vector space of all bounded
real valued functions on X. Then kf k∞ := sup{|f (x)| : x ∈ X} is a norm on
B(X, R).
(b) Let X = [0, 1]. Let V := C(X, R) the vector space of all continuous real valued
R1
functions on X. Then kf k1 := 0 |f (t)| dt defines a norm on V .
(c) Since C([0, 1], R) ⊂ B([0, 1], R), we have another norm on V , namely, kf k∞ .

1.9 `1 , the space of sequences whose associated series are absolutely summable is defined as
follows: ( )
X
`1 := (zn ) : zn ∈ R; |zn | is convergent.
n

is a norm on `1 .
P
Then kz k = k(zn )k := n |zn |

1.10 Open balls. Let (X, d) be a metric space. Fix a ∈ X and r > 0. The open ball B(a, r)
and the closed ball B[a, r] centred at a and radius r are defined by

B(a, r) := {x ∈ X : d(x, a) < r} & B[a, r] := {x ∈ X : d(x, a) ≤ r}.

We now look at some examples.

(a) in R: B(p, r) = (p − r, p + r).

(b) B(0, 1) in R2 with k k1 , k k2 and k k∞ . Picture!

Look at the pictures. How do we arrive at them? The “boundary” of B(0, 1) is

identified. In the case of k k1 , the boundary is defined by |x| + |y| = 1. Hence
B(0, 1) in this space is the ‘region’ enclosed by lines x + y = 1, −(x + y) = 1,
x − y = 1 and y − x = 1. In the case of k k∞ , the bounding lines are x = 1, −x = 1,
y = 1 and −y = 1.
(c) in Z with the induced metric. Identify all open balls. Answer: Any set of 2n + 1
consecutive integers.
(d) Relations between B(x, r) and B(y, s).
If x = y and r < s, then B(x, r) ⊆ B(x, s). Equality can occur. Consider the
discrete metric and r = 1/2 and s = 3/4.
If d is discrete, and if r > 1 and s > 1, then for any x, y, we have B(x, r) = B(y, s).
(e) Visualizing the open balls in C[0, 1] under k k∞ .

6
 (f) In an normed linear space , B(x, r) = x + rB(0, 1).

1.11 Open sets in a metric space. A subset U of a metric space is said to be open or
d-open if for each x ∈ U , there exists rx > 0 such that B(x, rx ) ⊂ U . We now look at
lots of examples to build our intuition. In each of the examples, draw pictures of the sets
and see whether you can enclose each of the points x in an open ball B(x, rx ) contained
in the given set. In most of the cases, the geometry will lead you to the ‘best possible’
radius rx . This will develop your intuition to ‘identify’ the open sets “instantly”. Pictures!

(a) in R: various examples such as open intervals, union of open intervals and non-
examples such as Z, Q, R \ Q,
(b) {(x, y) ∈ R2 : x > 0, y > 0} in R2 .
(c) {(x, y) ∈ R2 : x ≥ 0, y > 0} in R2 .
(d) {(x, y) ∈ R2 : x2 + y 2 > 1} in R2 .
(e) {(x, y) ∈ R2 : x2 + y 2 < 1} in R2 .
(f) {(x, y) ∈ R2 : x2 + y 2 ≤ 1} in R2 .
(g) Various conic sections in R2 .
(h) In a normed linear space V , if any vector subspace W is open, then W = V . Appli-
cation: Is C[0, 1] open in BF [0, 1], the set of bounded functions?
(i) Is R \ Z open in R?
(j) The open ball B(x, r) is open in any metric space.
Draw picture. If y ∈ B(x, r), we need to find s > 0 such that B(y, s) ⊂ B(x, r). Let
us find such an s. Let z ∈ B(y, s). We need to show z ∈ B(x, r). That is we need an
estimate for d(z, x). The obvious estimate is d(z, x) ≤ d(z, y) + d(y, x) < s + d(y, x).
If we can show s + d(y, x) < r, we are through. This suggest we choose 0 < s <
r − d(x, y).
(k) {y ∈ X : d(x, y) > r} is open. Hint: Modify the idea of the last sub-item.
(l) What are the open sets in a finite metric space?
(m) Can {h ∈ C[0, 1] : f (x) < h(x) < g(x)} for some f, g ∈ C[0, 1] be an open ball? Is it
an open set?
(n) Is the open unit ball in (C[0, 1], k k∞ ) open in (C[0, 1], k k1 )?
(o) If U is an open subset in an normed linear space , (X, k k), then
i. x + U is open for any x ∈ X
ii. A + U is open for any A ⊂ X
iii. αU is open for any nonzero scaler α.
(p) Is the set U := {f ∈ C[0, 1] : f (1/2) 6= 0} open in (C[0, 1], k k∞ )?
(q) Any open subgroup G of R is R.
For, 0 ∈ G and hence (−ε, ε) ⊂ G for some ε > 0. Since G is group, for all
x, y ∈ (−ε, ε) we have x+y ∈ G, that is, (−2ε, 2ε) ⊂ G. By induction, (−nε, nε) ⊂ G
for n ∈ N.. Now let x ∈ R be nonzero. By Archimedean property, there exists N ∈ N
such that N ε > |x|. Hence x ∈ (−N ε, N ε). It follows that R = ∪n∈N (−nε, nε) ⊂ G.

7
 (r) A subset U of a metric space is open iff it is the union of a family of open balls.
For, if x ∈ U , there exists rx > 0 such that B(x, rx ) ⊂ U . We then have an indexed
family {B(x, rx ) : x ∈ U } of open balls. Clearly, U = ∪x∈U B(x, rx ).
(s) A subset U ⊂ R is open iff it is the union of a countable family of pair-wise disjoint
open intervals. (See Lemma 1.2.42 on Page 23 of my book on Metric spaces.)

1.12 The class T of open subsets of a metric space (X, d) have the following properties:

(a) ∅, X ∈ T .
(b) If {Ui : i ∈ I} is any collection of elements in T , then U := ∪i∈I Ui ∈ T .
(c) If Uk , 1 ≤ k ≤ n are in T , then U1 ∩ U2 ∩ · · · ∩ Un ∈ T .

1.13 Topology: Definition and Examples. A topology on a set X is a collection T of

subsets of X which satisfies the three conditions (a)–(c) of the last item. Elements of T
are called open sets, to be precise T -open.

(a) Metric topology: Let (X, d) be a metric space. Then the collection of (d-) open
subsets is a topology on the metric space. This topology is called the metric topology
on the metric space.
(b) Discrete topology: Here T = P (X), the power set of X. Thus, every subset is open.
(c) The topology on a finite metric space is discrete.
(d) Indiscrete topology: U is open iff U = ∅ or U = X, that is, T = {∅, X}.
(e) Co-finite topology: U is open iff U = ∅ or X \ U is finite, that is ,

T = {U ⊂ X : Either U = ∅ or X \ U is finite.}

(f) Co-countable topology: U is open iff U = ∅ or X \ U is countable, that is,

T = {U ⊂ X : Either U = ∅ or X \ U is countable.}

(g) VIP topology: Fix p ∈ X. U is open iff U = ∅ or p ∈ U .

(h) Outcast topology: Fix p ∈ X. U is open iff U = X or p ∈
/ U.
/ U or U c is finite.
(i) Outcast + co-finite topology: U is open iff either p ∈

1.14 A topology on Z. Let B be the set of arithmetic progressions in Z. Any element B ∈ B

is of the form a + Zb for some nonzero b. Note that B is nothing other than the set of
cosets of all additive (non-trivial) subgroups of Z. An example: 2 + 5Z. We define a
topology T on Z as follows: a subset U ⊂ Z is open iff for each x ∈ U , a coset of the
from x + Zb ⊂ U . Clearly, ∅, Z ∈ T . If {Ui } is a collection of sets in T and x ∈ ∪i Ui ,
then x ∈ Uj for some j and hence there is a b 6= 0 such that x ∈ x + Zb ⊂ Uj ⊂ ∪i Ui . If
x ∈ U ∩ V for some U, V ∈ T , then there exist b, c such that x + Zb ⊂ U and x + Zc ⊂ V .
Clearly, x + Z lcm (b, c) ⊂ U ∩ V . Hence T is a topology on Z.
(1) Observe that any element of B can be written in the form r + Zb where b > 0 and
0 ≤ r < b − 1. Hence in view of Z = ∪0≤r<b−1 r + Zb, we see that any element of B and
its complement are both open!
(2) Another observation is that no nonempty finite set can be open.

8
 As an application of these observations, we now give a topological proof of Euclid’s
theorem on the infinitude of primes in Z. We prove this by contradiction. Assume that
p1 , . . . , pn are the set of all primes. Now the only integers that are not divisible by any
prime are ±1. Hence
Z \ {±1} = ∪nk=1 Zpk = ∪k Uk , say.
Let us take the complements on both sides of the above equality. The complement of left
side is ∩k Ukc , a finite intersection of open sets (in view of Observation 1) and hence is
open. Hence the left side, a finite set is open, a contradiction to observation 2).

1.15 We now present two interesting examples of topological spaces. We leave the detailed
verifications as exercise for the readers. As we learn new concepts, we shall keep revisiting
them often.

(a) Let X = R. Let Td denote the standard topology on R. Let T be the collection of
all subsets of the form G := U \ A where U ∈ Td and A is a countable subset of R.
Show that T is a topology on R. In the sequel we shall always denote elements of
T as G = U \ A etc.
(b) We say that a subset F ⊂ N is small if k∈F k1 is convergent. The empty set is
P
defined to be small. The following are easy to see.
i.Any finite subset F ⊂ N is small.
ii.If S is small and T ⊂ S, then T is small.
iii.If Fk is small for 1 ≤ k ≤ N , then F := ∪N k=1 Fk is small.
iv. If S is an infinite subset of N, there exists T ⊂ S such that
T is an infinite small set.
To see this, observe that for each k ∈ N there exists nk ∈ S such that nk > 2k .
v. N is not small.
Let X := N ∪ {0}. We consider the following collection

T := {U ⊂ X : either U ⊂ N or 0 ∈ U and X \ U is small}.

Show that T is a topology on X.

1.16 Basis of a topological space and basis for a topology on a set.

Basis for a topological space. Let (X, T ) be a topological space. A subset B ⊂ T of open
sets is said to be a basis for T if every element in T is a union of elements from B. In
other words, B is a basis for T if for any U ∈ T and x ∈ U , there exists B ∈ B such that
x ∈ B ⊂ U . The typical example of a basis is the set of all open balls for the topology
on a metric space.

1.17 Examples of bases:

(a) {B(x, r) : x ∈ X, r > 0} is a basis for the metric topology on any metric space. The
indexing set is X × (0, ∞).
(b) {B(x, 1/n) : x ∈ X, n ∈ N} is a basis for the metric topology on any metric space.
The indexing set is X × N.

9
 (c) When X = R, we can do better than the last two bases. Consider B := {(a, b) :
a, b ∈ Q} is a basis for the standard topology on R. Note that this basis is countable,
as it is indexed by Q × Q+ . (Why? (a, b) = B(c, r) where c = (a + b)/2 ∈ Q and
r = (b − a)/2 ∈ Q+ .)
(d) A basis for the VIP topology is {p} ∪ {{p, q} : q ∈ X, q 6= p}.
(e) A basis for outcast topology is {X} ∪ {{q} : q ∈ X, q 6= p}.
(f) B := {{x} : x ∈ X} is a basis for the discrete topology on a set X.
(g) B := {X} is a basis for the indiscrete topology on a set X.
(h) Note that (0, 1) = (0, 1/2) ∪ (1/4, 1) = ∪n≥n (0, (n − 1)/n). Hence there is no unique-
ness while expressing an open set as a union of some elements from the basis.

1.18 The second notion is a basis for a topology on a set X. The question here is: given a set
X and a subset B ⊂ P (X) of subsets of X, does there exist a topology T on X for which
B is a basis? Suppose such a topology T exists. Then X ∈ T so that a first requirement
is (1) ∪B∈B B = X. Also, since any B ∈ B must be in T , B1 ∩ B2 ∈ T for any B1 , B2 ∈ B.
Hence the second condition: (2) for any B1 , B2 ∈ B and x ∈ B1 ∩ B2 , there exists B ∈ B
such that x ∈ B ⊂ B1 ∩ B2 . If these two conditions are satisfied, we define a topology T
on X as follows:

T := {U ⊂ X : ∀x ∈ U, ∃B ∈ B such that x ∈ B ⊂ U }.

It is easy to verify that T is a topology on X and that B is a basis for this topology.

1.19 Order Topology: partial and total orders, dictionary order on products, C is totally
ordered but is not an ordered field. Intervals of the form (a, b) and rays of the form
(−∞, a) and (b, ∞). Examples in R2 : the rays (−∞, (1, 2)), ((−1, 1), ∞) and the intervals
((−1, 1), (3, −2)) and ((0, 0), (0, 10)). Basis for order topology. What is the order topology Pictures!

on R, on Z, on N and on a finite totally ordered set? Details!

(a) Let R ⊂ X × X be a subset with the following properties:

(i) Given x, y ∈ X, either x = y or (x, y) or (y, x) ∈ R.
(ii) For each x ∈ X, we have (x, x) ∈ / R.
(iii) If (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.
Such an R is called a simple or linear order on X. If (x, y) ∈ R, we usually denote
it by x < y. We called (X, R) or (X, <) a totally ordered set. Note that (i) is the
law of trichotomy.
(b) A standard example is the standard < relation on R. Another example is defined
by (x, y) ∈ R if either x2 < y 2 or if x2 = y 2 , then x < y.
Question: How do you describe R as a subset of R × R?
(c) If X and Y are totally ordered sets, the dictionary order on X × Y is defined as
follows: (x1 , y1 ) < (x2 , y2 ) if x1 < x2 or if x1 = x2 , then y1 < y2 .
(d) Let X be a totally ordered set. For any a, b ∈ X with a < b, we define intervals
[a, b], (a, b), [a, b) and (a, b] as in R. How do we define (−∞, b) or (a, ∞)?
Draw pictures of the following intervals in R × R with lexicographic order: (i) the
ray (−∞, (1, 2)), (ii) the ray ((−1, 1), ∞), (iii) the intervals ((−1, 1), (3, −2)) and
((0, 0), (0, 10)).

10
 (e) Let B be the collection of all open intervals of the form (a, b) in X along with [m, b)
if there exists a minimum m ∈ X and/or (a, M ] if there exists a maximum M ∈ X.
It is easy to see that B is a basis for a topology T on X, called the order topology
on X.
(f) What is the order topology on R, on Z, on N and on a finite totally ordered set?
What is the order topology on [0, 1)?
(g) Is an open disk, say B((0, 0), 1) open in the order topology on R2 ? Is the open
interval ((0, 0), (0, 10)) open in R2 with the standard topology?

1.20 Lower Limit Topology: Consider B := {[a, b) : a, b ∈ R, a < b}. It is easy to see that B
satisfies both the conditions laid out in Item 1.18 on Page 10. The topology associated
with this basis is known as the lower limit topology on R, denoted by TL . The space
(R, TL ) is denoted by R` .
When is a subset U ⊂ R open in TL ? If for x ∈ U , we can find [a, b) ∈ B such that
x ∈ [a, b) ⊂ U . A picture will immediately lead you to a ‘better’ condition: for x ∈ U ,
we can find b > x such that [x, b) ⊂ U . In particular, any interval (a, b) ∈ TL . Hence the
lower limit topology is finer than the standard topology on R. In fact, it is strictly finer,
since [a, b) is open In TL but not in the standard topology.
Note that no countable sub-collection of {[a, b) : a, b ∈ R, a < b} will serve as a basis for
the lower limit topology. For, if {[an , bn ) : n ∈ N} is one such, then choose a ∈ R such
that a 6= an for n ∈ N. Then the open set [a, a + 1) cannot be written as a union of
any such elements. Why? For, a has to be in one of them, say, [ak , bk ) Since ak 6= a, it
follows that ak < a < bk so that the union will have elements from [ak , a) which are not
in [a, a + 1).
Question: How about the collection B = {(a, b] : a, b ∈ R, a < b}? Is it a basis for some
topology on R? If so,what will you call it?

1.21 The class of all topologies on a given set is a partially ordered set: if T1 and T2 are
topologies on X, we define T1 ≤ T2 iff T1 ⊂ T2 , as subsets of P (X). The indiscrete
topology is the smallest element and the discrete topology is the largest element of the
class of topologies on X.
The union of topologies on X need not be topology. Let X = {a, b, c} be a three element
set. Let T1 := {∅, {a}, X} and T2 := {∅, {b}, X}. These are two topologies on X but their
union is not a topology.
However, the intersection of a (nonempty) family of topologies on X is again a topology,
as can be easily verified.
Compare this with analogous results from algebra: intersections of subgroups of a group
is again a group, intersection of vector subspaces of a vector spaces a vector subspace,
intersection of ideas in a ring is again an ideal and so on. Associated with this phenomenon
is the concept of subgroup (a vector subspace, an ideal, or a submodule) generated by
subset S in a group (in a vector space, in a ring, or in a module over a ring).
These motivate us to define the following: if A is an arbitrary collection of subsets of a
set X, there exists a unique smallest topology on X which contains A and is called the
topology generated by A. We shall later see a practical way of looking at this topology.
See Item 10.2 on Page 40. For the time being, let us work out two examples.

11
 • Let X be a nonempty set with at least three elements. Let S be the collection of
all two element subsets of X. What is the smallest topology T containing S? Fix
a ∈ X. We can find two distinct elements, say, x, y ∈ X none of which is a. Then
{a, x} and {a, y} lie in S and hence in T . It follows that {a} ∈ T . Thus,every
singleton subset is in T and hence T is the discrete topology on X.
Question: What is T if X has only two elements?
• Let S consist of single element A ⊂ X. Then T = {∅, A, X}.

1.22 Let X be a set and Tc and Tf be respectively co-countable and co-finite topologies on X.
Then the co-countable topology is finer than the co-finite topology.
They are the same iff X is finite. If X is finite, then the two topologies are the same.
To see the converse, we need a result form set theory: If X is an infinite set, then there
exists a set A such that X \ A is infinite and countable.
In Item 1.15a on Page 9, T is finer than Td . The lower limit topology T` (Item 1.20 on
Page 11) is finer than the standard topology on R.
Note that any topology on X is finer than the indiscrete topology on X and the discrete
topology on X is finer than any topology on X.

1.23 We can use bases to say something about the topologies on a set.

Theorem 1. Let X be any set. Let Bi be a basis for some topology Ti on X, for i = 1, 2.
Then T1 ≤ T2 iff the following holds: if B1 ∈ B1 , then B1 ∈ T2 . In particular, T1 = T2 iff
every B1 ∈ B1 is in T2 and every B2 ∈ B2 is in T1 .

We may use this to show that the order topology TO on R × R is (strictly) finer than the
usual topology Tstd on R2 . For if a point p = (a, b) ∈ B, an open ball, then there exist
ε > 0 such that {a} × (b − ε, b + ε)× ⊂ B. (The ε can be explicitly determined!) But the
set {a} × (b − ε, b + ε) is a basic open set in TO . Further, this basic open set is not open
in the standard topology. Thus, TO is strictly finer than Tstd .

12
2 Continuity

2.1 Continuity: Let (X, TX ) and (Y, TY ) be topological spaces. Let f : (X, TX ) → (Y, TY )
be a map and x0 ∈ X. We say that f is continuous at x0 if for any given open set V
containing f (x0 ), there exists an open set U containing x0 such that f (U ) ⊂ V . This
definition is an abstraction of the standard ε-δ definition of continuity, say, of functions
f : R → R. In this context, V = (f (x0 ) − ε, f (x0 ) + ε) and U = (x0 − δ, x0 + δ). In fact,
we have the following theorem:
Theorem 2. Let f : (X, d) → (Y, d) be a map between metric spaces. Let x0 ∈ X. Let
TX and TY be the topologies on X and Y induced buy their respective metrics. Then
f : (X, TX ) → (Y, TY ) is continuous at x0 iff for every ε > 0 there exists δ > 0 such that
whenever d(x, x0 ) < δ, we have d(f (x), f (x0 )) < ε.

Proof. Let us assume that f : (X, TX ) → (Y, TY ) is continuous at x0 . Let ε > 0 be given.
Then V := B(f (x0 ), ε) is an open set containing f (x0 ). Hence there exists an open set
U 3 x0 such that for all x ∈ U we have f (x) ∈ V . Since U is open there exists δ > 0
such that B(x0 , δ) ⊂ U . Hence it follows d(x, x0 ) < δ =⇒ d(f (x), f (x0 )) < ε, that is,
f : (X, d) → (Y, d) is continuous at x0 . Pictures!

The converse is similar. Let f : (X, d) → (Y, d) is continuous at x0 . Assume that an

open V 3 f (x0 ) is given. Then we can find ε > 0 such that B(f (x0 ), ε) ⊂ V . For this
ε > 0 by the definition of continuity in metric space context, there exists δ > 0 such that
d(x, x0 ) < δ =⇒ d(f (x), f (x0 )) < ε. If we let U := B(x0 , δ), then U is open, x0 ∈ U ,
and for x ∈ U , we have f (x) ∈ B(f (x0 < ε) ⊂ V .

2.2 Let f : X → Y be any map between two sets. Let B ⊂ Y . The set f −1 (B) := {x ∈ X :
f (x) ∈ B} is called the inverse image of B under f . The following are well-known facts: Details!

(a) If {Bi : i ∈ I} is a family of subsets of Y , then

i. f −1 (∪i∈I Bi ) = ∪i∈I f −1 (Bi ).
ii. f −1 (∩i∈I Bi ) = ∩i∈I f −1 (Bi ).
(b) For any set B ⊂ Y , we have X \ f −1 (B) = f −1 (Y \ B).
(c) Let f : X →
Y and g : Y → Z be maps. Let W ⊂ Z. Then (g ◦ f )−1 (W ) =
f −1 g −1 (W ) .

Thus, “the inverse images behave well under set-theoretic operations and compositions.“

2.3 Let X and Y be topological spaces. Then a map f : X → Y is said to be continuous on

X iff it is continuous at each point x ∈ X.
Let X and Y be topological spaces and f : X → Y be continuous at each point x ∈ X.
Let V ⊂ Y be any open subset of Y . Let U := f −1 (V ) = {x ∈ X : f (x) ∈ V }. Let a ∈ U .
By the definition of U , f (a) ∈ V . Since f is continuous at a and V 3 f (a) is an open set,
there exists an open set Ua 3 a such that for all x ∈ Ua , we have f (x) ∈ V . This implies
Ua ⊂ U . Since a ∈ U was arbitrary, what we have shown is that for each a ∈ U , there
exists an open set Ua such that a ∈ Ua and Ua ⊂ U . In particular, U = ∪a∈U Ua is open.
(The argument of this paragraph teaches an algorithm: in the case of a topological space
if we want to show that a set U is open , we need to find an open set Ua for each a ∈ U

13
 such that a ∈ Ua and Ua ⊂ U . Compare this with the algorithm to show a subset of a
metric space is open, Item 1.11r on Page 8.)
We have thus shown that f −1 (V ) is open in X for each open subset V ⊂ Y of Y .
Is the converse true? That is, if f −1 (V ) is open in X for each open subset V ⊂ Y of
Y , is f continuous on X? This is easy. Let a ∈ X and V 3 f (a) be open in Y . Then
U := f −1 (V ) is open by hypothesis . Clearly a ∈ U . Also, for each x ∈ U , f (x) ∈ V , that
is, f is continuous at a. Since a is arbitrary, it follows that f is continuous on X.
We have thus arrived at the following result.

Theorem 3. Let X and Y be topological spaces. Then a map f : X → Y is continuous

on X iff for every open subset V ⊂ Y , the inverse image f −1 (V ) is open in X.

2.4 The theorem of the last item leads us to the following result:
Let T1 and T2 be two topologies on the same set X. Then T1 ≤ T2 iff the identity
map I : (X, T2 ) → (X, T1 ) is continuous. In particular, T1 = T2 iff the identity maps
I : (X, T1 ) → (X, T2 ) and I : (X, T2 ) → (X, T1 ) are continuous. (This is same as saying
that the identity map is a homeomorphism, a concept to be defined in Item 6.3 on
Page 36.)

2.5 We looked at the following examples:

(a) Any constant map from a topological space to another is continuous.

(b) The identity map from (X, TX ) to itself is continuous.
(c) If T1 and T2 are topologies on a set X, then the identity map I : (X, T1 ) → (X, T2 )
is continuous iff T1 is finer than T2 .
(d) Let (X, TX ) be a topological space with the property that any map f : (X, TX ) →
(Y, TY ) is continuous. Then TX is discrete and conversely.
(e) Let (Y, TY ) be a topological space with the property that any map f : (X, TX ) →
(Y, TY ) is continuous. Then TY is indiscrete and conversely.
(f) The identity map from X with co-countable topology to X with co-finite topology
is continuous. The other way map is continuous iff X is finite.
(g) Let X be a set with at least two elements and p ∈ X. Let V (resp. O) denote the
VIP topology (resp. the outcast topology) on X with respect to p. Then
i. The identity map I : (X, V ) → (X, O) is not continuous. However it is continu-
ous at x = 0 and at no other point.
ii. The identity map I : (X, O) → (X, V ) is not continuous at any point.

2.6 The identity map from R with the lower limit topology is continuous to R with the usual
topology.

2.7 Let k kk , k = 1, 2, be two norms on a vector space V . Then they are equivalent iff the
identity map I : (V, k k1 ) → (V, k k2 ) and I : (V, k k2 ) → (V, k k1 ) are continuous.

2.8 Let X be an uncountable set with co-countable topology Tc . Then the only continuous
functions f : (X, Tc ) → R are constants.

14
 2.9 Find the set of points of continuity of all real valued functions on the following spaces.
(i) R with VIP topology with 0 as the VIP.
(ii) R with outcast topology with 0 as the outcast.
(iii) N with the topology T := {∅, N} ∪ {In : n ∈ N} where In = {1, 2, . . . , n}.

2.10 On any metric space X, we have lots of real valued continuous functions: f (x) := d(x, p)
for any fixed p ∈ X. In particular, given p 6= q in X, there exists a real valued continuous
function f on X such that f (p) 6= f (q).

2.11 Let A be a nonempty subset of a metric space X. We defined dA (x) ≡ d(x, A) :=

inf{d(x, a) : a ∈ A}. Identify as much as possible dA for the subsets below and draw their
graphs. Picture!

(a) X = R and A = [−1, 1].

(b) X = R and A = Q.
(c) X = R and A = Z.
(d) X = R2 and A is the x-axis.
(e) X = R2 and A = {(x, y) : x2 + y 2 = 1}.
(f) W is a vector subspace of Rn . Hint: If Rn = W ⊕ W ⊥ , and if x = w + w0 , then
dW (x) = kw0 k = kx − pW (x)k, where pW : Rn → W is the orthogonal projection.

2.12 We claim that for any nonempty subset A of a metric space X, the function dA : X → R
is continuous.
dA (x) ≤ d(x, a) ≤ d(x, y) + d(y, a), for a ∈ A. Hence dA (x) is a lower bound for the set
{d(x, y) + d(y, a) : a ∈ A}. But then inf{d(x, y) + d(y, a) : a ∈ A} = d(x, y) + dA (y).

2.13 The function x 7→ kxk is continuous on an normed linear space (V, k k). Note that
kxk = kx − y + y k ≤ kx − y k + ky k so that kxk − ky k ≤ kx − y k. Interchanging x and
y we get
| kxk − ky k | ≤ kx − y k .
This establishes the (uniform) continuity of the norm function. Note that this has the
continuity of modulus/absolute value as a special case.

2.14 The functions πj : x 7→ xj , the coordinate projections are continuous on Rn (with respect
to any of the norms k ki , i = 1, 2, ∞):

|πj (x) − πj (a)| = |xj − aj | ≤ kx − ak , 1 ≤ j ≤ n.

2.15 Composite of continuous functions is continuous: Let X, Y, Z be topological spaces. Let

f : X → Y be continuous at p ∈ X and g : Y → Z be continuous at q := f (p) ∈ Y . Then
g ◦ f : X → Z is continuous at p.
Let W ⊂ Z be an open set such that (g ◦ f )(p) = g(q) ∈ W . Since g is continuous
at q, there exists an open V 3 q such that g(V ) ⊂ W . Since f is continuous at p and
V 3 f (p) =, there exists an open U 3 x such that f (U ) ⊂ V . Clearly, (g ◦ f )(U ) ⊂ W .

2.16 Let f : X → R be a continuous function. Then |f | : X → Y defined by |f |(x) := |f (x)| is

continuous on X. For, it is the composite of two continuous functions |f | = | | ◦ f , where
| | : R → R is the modulus function, | |(x) := |x|.

15
2.17 Let X be a topological space. Let Rn be given the metric topology arising form the
standard Euclidean metric. Let f : X → Rn . Then we can write f (x) = (f1 (x), . . . , fn (x)).
Note that fj (x) = πj ◦ f where πj is the projection as in Item 2.14 on Page 15.
We claim that f is continuous iff each fj : X → R, 1 ≤ j ≤ n, is continuous. Assume that
f is continuous. Since fj = πj ◦ f , it follows from Items 2.14–2.15 on Page 15 that fj is
continuous.
Now the converse. Fix a ∈ X. Let V ⊂ Rn be open containing f (a). Let ε > 0 be such
that B(f (a), ε) ⊂ V . By continuity of fj at a, there exists an open set Uj ⊂ X such that
√
a ∈ Uj and fj (Uj ) ⊂ B(fj (a), ε/ n), 1 ≤ j ≤ n. Then U := ∩nj=1 Uj is an open set which
contains a and is such that f (x) ∈ B(f (a), ε) for all x ∈ U :
n
X
d(f (x), f (a))2 = (fj (x) − fj (a))2 < n(ε2 /n) = ε2 .
j=1

Hence for x ∈ U , we have f (x) ∈ B(f (a), ε) ⊂ V , that is, f is continuous at a.

2.18 Let f, g : X → R be continuous functions. Consider R2 with k k being one of the three
norms: k k1 , k k2 , k kmax . Then the function ϕ : X → R2 given by ϕ(x) = (f (x), g(x))
is continuous.
This is a special case of the last item.

2.19 The functions R2 → R given by α : (x, y) 7→ x + y and µ : (x, y) 7→ xy are continuous.

To establish the continuity of these function we use Theorem 2 in Item 1.
Let (a, b) ∈ R2 . Let ε > 0 be given. Assume δ > 0 serves. We estimate

|α(x, y) − α(a, b)| = |(x + y) − (a + b)| = |(x − a) + (y − b)|

≤ |x − a| + |y − b|
≤ d((x, y), (a, b)) + d((x, y), (a, b)).

If d((x, y), (a, b)) < δ, the above estimate suggests that we take 2δ < ε.
Let ε > 0 be given. Assume δ > 0 serves. We may assume that 0 < δ < 1. If
d((x, y), (a, b)) < δ, then |x − a| < δ < 1 and |y − b| < δ < 1. Hence |y| ≤ |y − b| + |b| <
1 + |b|. We now estimate

|µ(x, y) − µ(a, b)| = |xy − ab| = |xy − ay + ay − ab| ≤ |y||x − a| + |a||y − b|

≤ (1 + |b|)|x − a| + |a||y − b|
< M 2δ,
ε
where M = max{1 + |b|, |a|}. If we choose δ < 2M , as well as δ < 1, the estimates above
establish |µ(x, y) − µ(a, b)| < ε.

2.20 If f, g are continuous functions from a topological space to R and if a, b ∈ R, then the
functions af + bg and f g are continuous. Hint: Use Items 15–19.
Thus the set C(X, R) of all real valued continuous functions on a topological space is a
vector space over R. It is also a commutative ring with identity, in fact, an algebra over
R.

16
2.21 Given two real numbers a, b we wish to find a “formula” for max{a, b} and min{a, b}.
Given a, b, their mid point is (a + b)/2. To reach the maximum of these two, we need to
move to the right for half of the distance between them, that is, we need to add |a − b|/2
to their mid point. Similar analysis can be done for minimum. Hence we arrive at the
following formulas:
(a + b) + |a − b| (a + b) − |a − b|
max{a, b} = and min{a, b} = .
2 2

2.22 If f, g : X → R are two continuous functions on a topological space X, then max{f, g}

and min{f, g} are continuous. This follows from Items 21, 16 and 20.

2.23 Any polynomial function f : Rn → R is continuous. This follows from Item 14 and 22.
Examples of polynomial functions on R2 and R3 are p(x, y) = 3x2 +y 2 −xy 2 +6x−7y +10,
q(x, y, z) = z 10 − 9y 2 + 17xyz 3 + 2012 etc.

2.24 The map ρ : R∗ → R∗ given by ρ(x) = 1/x is continuous. Look at the estimate:

|x − y| 2|x − y|
|ρ(x) − ρ(y)| ≤ ≤ ,
|xy| |x2 |

if we restrict y in such a way that |x − y| < |x|/2.

2.25 Let f : X → R be continuous and assume that f (x) 6= 0 for all x ∈ X. Then 1/f : X → R
is continuous. For, 1/f is the composition ρ ◦ f , where ρ is as in the last item.

2.26 Any linear map from Rn with any one of our three standard norms to any normed linear
space is continuous. In particular, any linear map from Rm to Rn is continuous.
More generally, any linear map T : Rn → X, where X is any normed linear space is
(uniformly) continuous.
n
P let {ei :n1 ≤ i ≤ n} be the standard basis of R . Then for any x = (x1 , . . . , xn ) =
For
i xi ei ∈ R we have
n
! n
X X
kT xk = T xi e i ≤ |xi | kT ei k


i=1 i=1
n
X
≤ M kxk , where M := max{kT ei k : 1 ≤ i ≤ n}
i=1
= M n kxk .

Note that |xi | ≤ kxk where k k could be either k k1 , k k2 or k kmax . Hence kT x − T y k =

kT (x − y)k ≤ M n kx − y k so that T is Lipschitz and hence uniformly continuous.

2.27 Let Mm×n (R) denote the set of all m × n matrices with real entries. We identify it with
Rmn using an obvious linear isomorphism:

X = (xij ) 7→ (x11 , . . . , x1n , x21 , . . . , x2n , . . . , xm1 , . . . , xmn ).

We use any one of the standard norms on Mm×n (R). We let M (n, R) := Mn×n (R). Then
we have

17
 (a) The ‘transpose’ map X 7→ X T from M (n, R) to itself is continuous. For, the map is
(x11 , x12 , . . . , xn1 , . . . , xnn ) → (x11 , x21 , . . . , x1n , . . . , xnn ). The coordinate maps are
fij (X) = xji and hence are continuous. (See Item 17.)
(b) The ‘trace’ map X 7→ Tr(X) is continuous from M (n, R) to R. Observe that it is a
linear map.
(c) The determinant map det : M (n, R) → R, defined by X 7→ det(X), is a“polynomial

a b
function” and hence is continuous. When n = 2 and the matrix is X = , then
c d
det(X) = ad − bc. For general n, recall the P
formula for the determinant (Laplace
expansion) as an alternating sum, det(X) := σ∈Sn sign(σ)x1σ(1) · · · xnσ(n) .

2.28 One can use functions whose continuity are known to assert that certain subsets are open.
This is a very useful observation.

(a) Since polynomial functions from Rn to R are continuous

i. The subsets {(x, y) ∈ R2 : xy 6= 0}, {(x, y) ∈ R2 : x2 + y 2 6= 1} and {(x, y) ∈
R2 : xy 6= 1} are all open.
ii. The subset {(x, y) ∈ R2 : x3 − 34x2 y − 28xy 2 − y 3 + 7xy − 19y + 125 6= 0} is
open in R2 .
iii. R3 \ P , where P := {(x, y, z) : ax + by + cz = d} is a plane, is open in R3 .
iv. The rectangle R := (a, b) × (c, d) is open in R2 : R = p−1 −1
1 (a, b) ∩ p2 (c, d), where
p1 (x, y) = x etc.
(b) The set {f ∈ C[0, 1] : f (1/2) 6= 0} in X := (C[0, 1], k k∞ ) is open. Hint: Consider
T : X → R given by T (f ) := f (1/2).
(c) Let W be a vector subspace of Rn . Then Rn \ W is open in Rn . Hint: Write
Rn = W ⊕ W ⊥ and let u1 , . . . , uk be an orthonormal basis of W ⊥ . Then x ∈ Rn lies
in W iff hx, ui i = 0 for all 1 ≤ i ≤ k.
Alternately, consider the orthogonal projection π : Rn → W ⊥ . Then Rn \ W =
π −1 (W ⊥ \ {0}).
(d) GL(n, R), the set of all invertible matrices is open in M (n, R).
(e) The set of symmetric matrices, being a vector subspace, cannot be open in M (n, R).
Hint: See Item 11h.
Same holds true for the set of skew symmetric matrices.
(f) Let Y be an ordered set with the order topology. Let f, g : X → Y be continuous.
Then the set L := {x ∈ X : f (x) < g(x)} is an open set. Hint: Fix a ∈ L. Either
there exists α ∈ Y such that f (a) < α < g(a) or none. In the former consider the
open sets V1 := (−∞, α), V2 := (α, ∞) in the former case and V1 = (−∞, g(a)) and
V2 = (f (a), ∞). Consider U := f −1 (V1 ) ∩ g −1 (V2 ). Then a ∈ U . For each x ∈ U , we
have f (x) < g(x).

2.29 To check continuity, it suffices to show that the inverse images of basic elements in the
codomain are open in the domain:

Lemma 4. Let (Xi , Ti ) be topological spaces i = 1, 2 and let B2 be a basis for T2 . Then
f : (X1 , T1 ) → (X2 , T2 ) is continuous iff f −1 (B2 ) ∈ T1 for all B2 ∈ B2 .

18
 Item 6 is an immediate consequence of this.

2.30 Consider M (n, R) the set of all n × n real matrices. Then the map Repetition: Item 27

ϕ : A 7→ (a11 , . . . , a1n , a21 , . . . , an1 , . . . , ann )

2
is a linear isomorphism of M (n, R) onto Rn . We usePthis to transfer the Euclidean norm
2
on Rn to M (n, R) as follows: kAk2 := kϕ(A)k2 = ni,j=1 |aij |2 .
Show that the map M (n, R) × M (n, R) → M (n, R) given by µ(X, Y ) = XY , the matrix
product is continuous.

2.31 Let X and Y be normed linear spaces. A linear map T : X → Y is continuous at 0 ∈ X

iff there exists a positive constant C such that kT xk ≤ C kxk for all x ∈ X. Hint: Use
ε-δ definition of continuity at 0.
Deduce that a linear map between two normed linear space ’s is continuous iff it is
continuous at 0.

2.32 When do two norms k kj , j = 1, 2 generate the same topology on a vector space X?
They do iff the identity maps I : (X, k k1 ) → (X, k k2 ) and I : (X, k k2 ) → (X, k k1 ) are
continuous. (Why?) By the last item, this means that we can find positive constants
C1 and C2 such that C1 kxk1 ≤ kxk2 ≤ C2 kxk1 for all x ∈ X. We thus arrive at the
following result.
Two norms k kj , j = 1, 2 generate the same topology on a vector space X iff positive
constants C1 and C2 such that C1 kxk1 ≤ kxk2 ≤ C2 kxk1 for all x ∈ X. We then say
that the two norms k k1 and k k2 are equivalent.

2.33 In Rn , the three norms k k1 , k k2 and k k∞ are equivalent. This follows from Item 26.
It follows also from the observation:
1 1
kxk1 ≤ √ kxk2 ≤ kxk∞ ≤ kxk2 ≤ kxk1 .
n n

Later, we shall show that all norms on Rn induce the same topology, that is, they are all
equivalent.

2.34 Closed Sets: Let (X, T ) be a topological space. A set F ⊂ X is called a closed set (or
said to be closed) in X if X \ F is open in X. Let C be the class of all closed subsets in
X. The following are more or less immediate:

(a) ∅, X ∈ C.
(b) If {Fi : i ∈ I} is a family of closed sets, then their intersection ∩i∈I Fi is again closed.
(c) If F1 and F2 are closed, then so is F1 ∪ F2 .

2.35 Examples of Closed Sets:

(a) ∅ and X are both open and closed in any topological space.
(b) Z is closed in R.
(c) There exist sets which are neither open nor closed: [0, 1), Q, R \ Q in R with usual
topology,

19
(d) Any finite subset of a metric space is closed.
(e) Any closed ball B[x, r] in a metric space is closed. Hence any closed interval [a, b] is
closed in R.
(f) Any sphere S(x, r) := {y ∈ X : d(x, y) = r} in a metric space is closed.
(g) The set {1/n : n ∈ N} ∪ {0} is closed in R.
(h) The set (−∞, 0) ∪ [1, ∞) is closed in R with lower limit topology but not closed in
R with the usual topology.
(i) The only subsets of R which are both open and closed are ∅ and R.
Let A be both open and closed in R. Assume that A is not empty. We need to prove
A = R. Let a ∈ A. Since a is open there exists r > 0 such that (a − r, a + r) ⊂ A.
Consider
E := {c ∈ R : c > a, (a − ε, c) ⊂ A}.
Then a + ε ∈ E . If sup E = ∞, then it follows that (a − ε, ∞) ⊂ A. Assume
sup E = α ∈ R. Now either α ∈ A or α ∈ / A.
If α ∈ A, since A is open there exists δ > 0 such that (α − δ, α + δ) ⊂ A. Since
α−δ < α = sup E, there exists c ∈ E such that (a−ε, c) ⊂ A. Clearly, (a−ε, α+δ) =
(a − ε, c) ∪ (α − δ, α + δ) ⊂ A. Hence α + (δ/2) ∈ E, contradiction to α = sup E.
If α ∈/ A, then α ∈ R \ A, an open set. Hence there exists δ > 0 such that
(α − δ, α + δ) ⊂ R \ A. As earlier, there exists c ∈ E such that α − δ < c. Hence
the interval (α − δ, c) lies in both A and its complement, a contradiction. Thus
we conclude that sup E = ∞ so that (a − ε, ∞) ⊂ A. Similarly, we can conclude
(−∞, a + ε) ⊂ A and hence A = R.
(j) The set [0, 1) is neither closed nor open in R.
(k) Any subset of a discrete space is open as well as closed.
(l) Any subset A ⊂ R∗ is closed in R with VIP topology with 0 as the VIP.
(m) What are the sets which are both open and closed in R with VIP topology with 0
as the VIP?
(n) Any subset of R containing 0 is closed in R with the outcast topology with 0 as the
outcast.
(o) What are the sets which are both open and closed in R with the outcast topology
with 0 as the outcast?
(p) Any vector subspace of Rn is closed. So are its translates.
Let V be a vector subspace of Rn . Let Rn = V ⊕V ⊥ be the orthogonal decomposition.
Then x ∈ Rn lies in V iff v · u ≡ hx, ui = 0 for all u ∈ V ⊥ . The map fu : Rn → R
given by fu (x) := x · u is linear and hence by Item 26, it is continuous. Hence the
kernel fu−1 (0) is a closed subset of Rn . Since V = ∩u∈V ⊥ fu−1 (0) is the intersection
of closed sets, V is closed.
(q) The set of n × n symmetric matrices and the set of n × n skew-symmetric matrices
are closed in M (n, R).
(r) The set GL(n, R) is not closed in M (n, R).
(s) The set of singular matrices in M (n, R) is closed.
(t) The set {f ∈ C[0, 1] : f (1/2) = 0} in X := (C[0, 1], k k∞ ) is closed.

20
 (u) The sets Q and R \ Q are neither closed nor open in R.

2.36 We have the following characterization of continuity in terms of closed sets.

Theorem 5. Let f : X → Y be a map between topological spaces. Then f is continuous

iff f −1 (B) is closed in X for every closed set B ⊂ Y .

2.37 As we did earlier in the case of continuity and open sets, we may use the above theorem
to assert that certain subsets are closed.

(a) The set {(x, y) ∈ R2 : xy = 0}, {(x, y) ∈ R2 : xy = 1}, {(x, y) ∈ R2 : x2 + y 2 = 1}

are closed in R2 .
(b) The closed rectangle R := [a, b] × [c, d] is closed in R2 .
(c) The unit n-dimensional sphere S n := {x ∈ Rn+1 : kxk = 1} is closed in Rn+1 .
(d) The set SL(n, R) of matrices A ∈ M (n, R) with determinant 1 is closed in M (n, R).
(e) The subset of matrices whose trace is 0 is closed in M (n, R). (Also follows from
Item 35p.)
(f) The set O(n) of orthogonal matricesPis closed in M (n, R). Hint: The maps M (n, R) →
R given by A 7→ Ri (A) · Rj (A) ≡ nk=1 aik ajk are continuous. Here Ri (A) denotes
the i-th row of A.
Or, use the fact that the map F : A 7→ (A, AT ) composed with (A, B) → AB is
continuous. Then O(n, R) is the inverse image F −1 (I).
(g) The set of singular matrices in M (n, R) is closed.
(h) The set of nilpotent matrices in M (n, R) is closed.
(i) Keep the notation of Item 28f. Then the set {x ∈ X : f (x) ≤ g(x)} is closed. .

2.38 Let A be a subset of a topological space. The characteristic function χA of A is defined

by (
1 if x ∈ A
χA (x) := .
0 if x ∈
/A
What can you conclude about A if the function χA is continuous on X?

2.39 Go back to Item 21. If you understand the principle in work in it, you would have
foreseen what follows. For any set A of a topological space (X, T ), the smallest closed set
containing A exists. It is denoted by A and called the closure of A in X. (Compare this
with the existence of the smallest topology containing a family {Ai : i ∈ I} of subsets of
a set X.) Note that A ⊂ A.

2.40 Examples of closures:

(a) The closure of (a, b) ⊂ R is [a, b].

(b) The closure of Q in R is R.
(c) The closure of an open ball B(x, r) in Rn is the closed ball B[x, r]. In a general metric
space, this need not be true. Consider B(x, 1) and B[x, 1] in a discrete metric space
with at least two points.

21
 (d) Let R be given the VIP topology with 0 as the VIP. Then the closure of A = {0} is
R. The closure of R \ Q is itself. The closure of {a} is itself if a 6= 0.
(e) Investigate the case of R with outcast topology.

2.41 Let (X, T ) be a topological space and A ⊂ X. Then x ∈/ A iff there exists an open set
U 3 x with U ∩ A = ∅. Hence, x ∈ A iff for every open set U 3 x, we have U ∩ A 6= ∅.
This suggests the following definition.

22
3 Limit and Cluster Points

3.1 x ∈ X is said to be a limit point of A if for every open set U 3 x, we have U ∩ A 6= ∅.

This is NOT the standard definition and hence should not be confused with the no-
tion of cluster or an accumulation point which we shall see below. We shall follow our
nomenclature only.

3.2 Consider the lower limit topology TL on R. Let A = [a, b). Is b in the closure of A? That
is, is b a limit point of [a, b)?

3.3 Consider R2 with order topology. Let Q := {(x, y) ∈ R2 : x > 0 & y > 0} be the first
quadrant. What is Q? Points of the other three quadrants are not in the closure. Any
point (a, 0) with a > 0 is in Q while (0, 0) is not.

3.4 Every point of A is a limit point of A.

3.5 x ∈ A iff x is a limit point of A. (This is true because of our definition of a limit point.
See Item 12.)
For, let x ∈ A and U 3 x be open. If U ∩ A = ∅, then A ⊂ X \ U , a closed set and hence
A ⊂ X \ U . But x ∈ A and x ∈/ X \ U , a contradiction. Hence x is a limit point of A.
Conversely, if x is a limit point of A and x ∈
/ A, then x ∈ U := X \ A, an open set. But
U ∩ A ⊂ U ∩ A = ∅. Hence x is a not a limit point of A, a contradiction.

3.6 Let (X, d) be a metric space, A ⊂ X. Then x ∈ X is a limit point of A iff there exists a
sequence (an ) in A such that an → x.

3.7 With the notation as in the last item, x ∈ A or x is a limit point of A iff dA (x) = 0.

3.8 In any normed linear space (X, k k), the closure of an open ball B(p, r) is B[p, r]. Thus,
q ∈ X is a limit point of B(p, r) iff d(p, q) ≤ r. In particular, B(p, r) = B[p, r].
If q ∈ B[p, r], consider the line segment (1 − t)p + tq, 0 ≤ t ≤ 1. Draw picture. All
points with 0 ≤ t < 1 are in B(p, r). From this line segment, you can find a sequence
pk ∈ B(p, r) which converges to q. Or, consider B(q, ε) for ε > 0. Then for any 0 < t < 1,
we have
d(q, (1 − t)p + tq) = k(1 − t)(q − p)k = (1 − t)r < ε,
if t is near to 1. Thus, any open set containing q contains points of B(p, r) other than q,

3.9 The set theoretic results about the closure operation:

(a) If A ⊂ B, then A ⊂ B.
(b) A ∪ B = A ∪ B.
(c) A ∩ B ⊂ A ∩ B. Strict containment can occur.
(d) ∪i∈I Ai ⊂ ∪i∈I A. Strict containment can occur.

(a) follows from the fact that any closed set that contains B will contain A. In particular,
the smallest closed set B that contains B will contain A. Hence A, the smallest closed
set containing A, will be contained in B.

23
 (b). Since LHS is the smallest closed set containing A ∪ B, and since A ∪ B is a closed
set containing A ∪ B, it follows that A ∪ B ⊂ A ∪ B. Let x ∈ A ∪ B. Assume WLOG
that x ∈ B. Then for any open set U 3 x, we have ∅ = 6 U ∩ B ⊂ U ∩ (A ∪ B). That is, x
is a limit point A ∪ B and hence x ∈ A ∪ B.
(c) Since A ∩ B is a closed set containing A ∩ B, it follows that A ∩ B ⊂ A ∩ B. An
instance of the strict containment, consider A = Q and B = R \ Q in R.
(d) Consider Q = ∪x∈Q {x}.

3.10 x ∈ X is a cluster or an accumulation point of A iff for every open set U 3 x, the set
(U \ {x}) ∩ A 6= ∅, that is, any open set U ∈ x contains a point of A other than x.

3.11 Intuitively, A accumulates or clusters around x. (They are like celebrities of A!) Obvi-
ously, any cluster point of A is a limit point of A, but not conversely. The notion of a
cluster point is much stronger and more stringent than that of a limit point.

3.12 Let (X, T ) be any topological space and A ⊂ X. Then A is the union of A and the cluster
points of A. (Compare and contrast this with Item 5.)

3.13 Every point of A = Z ⊂ R is a limit point of A but there exists no cluster point of A in
R.
Since Z = Z, this examples also shows that ‘limit point’ cannot be replaced by ‘cluster
point’ in Item 5.

3.14 Consider R with VIP topology with 0 as the VIP. Then any nonzero real number is a
cluster point of A = {0}. Zero is obviously a limit point of A but not a cluster point of
A.

3.15 The last example also shows that the following can occur. x may be a cluster point of A,
but there may exist open sets U 3 x with U ∩ A is finite!

3.16 Any point in any ball (open or closed) in an normed linear space is a cluster point of the
ball. The idea in Item 8 proves this.

3.17 Analyze the situation in a metric space. In a metric space, if x is a cluster point of A,
then every open set U 3 x will contain infinitely many points of A. The proof suggested
the following definition.

3.18 A topological space X is said to be Hausdorff iff for every pair x, y ∈ X of distinct points,
there exist open set U, V such that x ∈ U and y ∈ V and U ∩ V = ∅. That is, any two
distinct points can be “separated by open sets.”
We also say that a topology T on a set X is Hausdorff if the space (X, T ) is Hausdorff.

3.19 Let (X, T ) be a Hausdorff (topological) space and A ⊂ X. Then x ∈ X is a cluster point
of A iff for every open set U 3 x, the set U ∩ A is infinite.
We prove this by contradiction. Let x be a cluster point of A and assume that there exists
an open set U such that x ∈ U and U ∩ A is finite. Let (U \ {x}) ∩ A = {a1 , . . . , an }.
Since X is Hausdorff, for each j, 1 ≤ j ≤ n, the exists an open set Uj 3 x and Vj 3 aj
such that Uj ∩ Vj = ∅, 1 ≤ j ≤ n. Then U = ∩nj=1 Uj is an open set such that U ∩ A is at
most {x}.

24
3.20 A finite set in a Hausdorff space cannot have a cluster point. (Hausdorff condition is
required. Look at R with VIP topology with zero as the VIP and A = {0}.) If a subset
A of a Hausdorff space X has a cluster point, then A is infinite.
But there exists an infinite set in a Hausdorff space which has no cluster point. Look at
Z in R.

3.21 Let us now look at some examples of Hausdorff spaces.

(a) Any metric space is Hausdorff. For if x1 , x2 ∈ (X, d) are distinct, then d(x1 , x2 ) > 0.
Let r = d(x1 , x2 )/2. Then B(xj , r) is an open set containing xj and B(x1 , r) ∩
B(x2 , r) = ∅. For, x is a common point, then

d(x1 , x2 ) ≤ d(x1 , x) + d(x, x2 ) < r + r < d(x1 , x2 ),

a contradiction. In particular, Rn with the standard metric and normed linear spaces
are Hausdorff.
(b) Any discrete topology is Hausdorff.
(c) The indiscrete topology on a set X with at least two elements is not Hausdorff.
(d) (R, TV ) with 0 VIP is not Hausdorff.
(e) If we have T1 ≤ T2 and T1 is Hausdorff, so is T2 . As special cases, we have the
following.
i. The order topology on R2 with dictionary order is Hausdorff.
ii. The lower limit topology on R is Hausdorff.
(f) Let f : X → Y be a 1-1 continuous function. If Y is Hausdorff, so is X. Let x1 , x2
be distinct elements of X. Then f (x1 ) and f (x2 ) are distinct elements of Y and
hence there exist disjoint open sets Vj 3 f (xj ). Consider Uj := f −1 (Vj ), j = 1, 2.

3.22 We now give an example of a Hausdorff space in which two disjoint closed sets cannot be
separated by open sets.
Let X = R. For any fixed p ∈ R and m ∈ N, let Bp,m := {p + km : k ∈ Z+ }. Let T
be the set of all subsets U ⊂ R such that for any p ∈ U , there exists m ∈ N such that
Bp,m ⊂ U . Then it is easy to check that T is a topology on R.
We claim that it is Hausdorff. Consider p 6= q in R. If p − q is not an integer, then
Bp,m ∩ Bq,m = ∅ for any m ∈ N. For, otherwise, if z is a common element, then z =
p + km = q + kn. It follows that p − q = k(n − m), an integer — a contradiction.
If p − q = m ∈ Z,say, then the basic open sets Bp,2m abd Bq,2m separate p and q. (Verify
this.)
Fix p ∈ R and m ∈ N. We claim that each element of {p−km : k ∈ N} is a cluster point of
Bp,m . Let q = p − km. Consider a basic open set Bq,n 3 q. The element q + mkn ∈ Bq,n .
Since
q + kmn = p − km + kmn = p + mk(n − 1) ∈ Bp,m ,
the claim follows.
Consider now the two disjoint sets F1 := {1} and F2 := {x ∈ R : x ≤ 0}. F1 is closed
since the space is Hausdorff. F2 is also closed, since its complement is open. For, note
that for any p > 0 and m ∈ N, Up,m ⊂ (0, ∞).

25
 We claim that they cannot be separated by open sets. Assume the contrary. Let U1 ⊃ F1
and U2 ⊃ F2 be open sets separating them. Then there exists a basic open set B1,m ⊂ U1 .
Now 1 − 2m is a cluster point of B1,m . But no point of F2 can be cluster point of U1 since
F2 ⊂ U2 and U2 ∩ F1 ⊂ U2 ∩ U1 = ∅.
Thus we have an example of a Hausdorff space in which two distinct points can be
separated by open sets but not any two disjoint closed sets.

3.23 This examples is from Munkres. Consider R with the topology TK whose basic open sets
are open intervals (a, b) and open intervals (a, b) \ K where K := {1/n : n ∈ N}. Then
{0} and K are disjoint closed subsets which cannot be separated by open sets.

3.24 We say that a sequence (xn ) in a topological space (X, T ) converges to a point x ∈ X, if
for every open set U 3 x, there exists n0 ∈ N such that xn ∈ U for all n ≥ n0 . The point
x is called the limit of the sequence and (xn ) is said to be convergent.

3.25 If (X, T ) is a Hausdorff (topological) space, then any convergent sequence has a unique
limit.
This need not be true in a general space. For instance, if we consider R with indiscrete
topology, any sequence is convergent to any point of R!

3.26 Consider the sequence (1/n) in R with co-finite topology. Then 1/n → x, for any x ∈ R!
(Co-finite topology on R is not the discrete topology.)

3.27 In any Hausdorff space, any finite set is closed. Details!

This need not be true in an arbitrary topological space. For instance, consider the indis-
crete topology on R. Or, the set {x, 0}, x 6= 0, in R with VIP topology with VIP=0.
Hence conclude: The topology of any finite Hausdorff is discrete. (See also Item 11l.)

3.28 Examples of Convergent sequences:

(a) The only convergent sequences in any discrete space are eventually constant se-
quences.
For, let xn → x. Then {x} 3 x is open so there exists N ∈ N such that ≥ N =⇒
xk ∈ {x}. Thus xk = x for k ≥ N .
(b) In the normed linear space (B(X, R), k k∞ ), a sequence (fn ) converges to f ∈
B(X, R) iff fn converges to f uniformly on X.
Assume that fn → f uniformly on X. Let ε > 0 be given. Choose N such that for
all k ≥ N , and x ∈ X, we have |f (x) − fk (x)| < ε/2. Hence supx∈X |f (x) − fk (x)| ≤
ε/2 < ε. That is, kfk − f k∞ < ε for k ≥ N and hence fk converge to f in the norm.
Other way implication is easier.
(c) A sequence (xk ) in Rn converges to x ∈ Rn iff xkj → xj as k → ∞ for 1 ≤ j ≤ n.

3.29 Let f : X → Y be continuous. Let (xn ) be a sequence in X such that xn → x. Then

f (xn ) → f (x) in Y .
Let V ⊂ Y be an open set containing f (x). Since f is continuous at x, there exists an
open set U 3 x such that f (z) ∈ V fo rany z ∈ U . Since xn → x there exists N ∈ N such
that for all k ≥ N , we ave xk ∈ U and hence f (xk ) ∈ V .

26
3.30 We analyzed the proof of Item 6 and arrived at the following conclusion:

Let (X, T ) be a space with the following property: For every x ∈ X, there
exists a countable collection of open sets {Un,x : n ∈ N} such that
(a) For every open set U 3 x, there exists n such that x ∈ Un,x ⊂ U
(b) ∩n Un,x = {x}.
Then, x ∈ X is a limit point of A ⊂ X iff there exists a sequence (an ) in A
such that an → x.

27
 4 First and Second Countable Spaces

4.1 The foregoing item led us to the following concepts.

4.2 Let (X, T ) be a topological space and p ∈ X. Then by a local base at p, we mean a
family {Up,i : i ∈ I} of open sets containing p with the property that if U is an open set
containing p, then there exists i ∈ I such that x ∈ Up,i ⊂ U .
A typical example to keep in mind: {B(p, r) : r > 0} is a local base at p in a metric space
X.

4.3 A space is said to be first countable if there exists a countable local base at every point
p ∈ X.

4.4 Observe that if (X, T ) is first countable, then we may assume that a local base {Up,n :
n ∈ N} at p is decreasing sequence. For, if {Vp,n } is a local base at p, consider Up,n :=
Vp,1 ∩ · · · ∩ Vp,n .

4.5 We look at some examples:

(a) In R with standard topology, {(p − n1 , p + n1 ) : n ∈ N} is a local base at p. Hence

R is first countable. More generally, {B(p, 1/n) : n ∈ N} is a local base at p in any
metric space. Hence any metric is first countable.
(b) If R is endowed with the discrete topology, then a local base at x can be taken as
{x}. Hence R with discrete topology is first countable.
(c) Consider R with VIP topology. (Convention: VIP is always 0.) Then the set {p, 0}
is a local base at any p ∈ R. (If p = 0, then the set {p, 0} = {0}!) Hence R with
VIP topology is first countable.
(d) Any indiscrete topology is first countable.
(e) R with the lower limit topology TL is first countable. At any a ∈ R, consider
{[a, a + n1 ) : n ∈ N}.

4.6 Let (X, T ) be a Hausdorff, first countable space. Let {Up,n : n ∈ N} be a countable local
base. Then ∩n Un,p = {p}. (We do not need the full power of Hausdorff condition. We
could have achieved the same result with less stringent hypothesis, but we shall not worry
about this!)

4.7 In view of Item 30 and Item 6, we have the following.

Theorem 6. Let (X, T ) be first countable and Hausdorff. Then x is a limit point of A
iff there exists a sequence (an ) in such that an → x.

4.8 We say that a topological space (X, T ) is second countable if there exists a countable
basis for T .

4.9 Clearly, any second countable space is first countable.

4.10 Examples and non-examples:

(a) R with the standard topology is second countable. (See Item 17c.)

28
 (b) A discrete space X is second countable iff the set X is countable.
(c) R with VIP topology is first countable but not second countable. Why? Consider
the basis {{x, 0} : x ∈ R}. If {Bn : n ∈ N} is a countable basis, then for x ∈ {x, 0}
there will be n(x) ∈ N such that x ∈ Bn(x) ⊂ {x, 0}. Since Bn(x) will always contain
{0, x}, it follows that Bn(x) = {0, x}. But the family {{x, 0} : x ∈ R} is uncountable
where as {Bn : n ∈ N} is countable.
(d) The outcast topology on R is first countable but not second countable. For the
second countability part, argue with {{x, 0} : x 6= 0} and consider f : R \ {0} → N.
(e) The lower limit topology on R is not 2nd countable. Assume the contrary and let
{Bn : n ∈ N} be a countable basis. For each basic open set [a, a + 1) there exists
n(a) ∈ N such that a ∈ Bn(a) ⊂ [a, a + 1). Observe that inf Bn(a) = a = inf[a, a + 1).
Hence the map f : R → N given by f (a) = n(a) is one-one. That is, R is countable,
an absurdity.
Thus (R, TL ) is a first countable, separable space which is not second countable.
(f) Any indiscrete space is second countable.

4.11 Think over this: What will be the counter part (in terms of open sets) of the smallest
closed set containing A? It is the largest open set contained in A. It is called the interior
of A and is denoted by Int (A).

4.12 Examples of interior of a set:

(a) The interior of an open set is itself.

(b) The interior of [a, b] ⊂ R is (a, b).
(c) The interior of Q ⊂ R is the empty set. What is Int (R \ Q)?
(d) The interior of a proper vector subspace of Rn is empty. Does this generalize to any
normed linear space ?
(e) The interior of a closed ball B[p, r] in any normed linear space is the open ball
B(p, r). In a general metric space, such a result is not true.
(f) Let (X, T ) be a discrete space. Then Int (A) = A for any A ⊂ X.
(g) Let (X, T ) be an indiscrete space. Then Int (A) = ∅ for any A ⊂ X, A 6= X.
(h) Consider R with the VIP topology (VIP is 0). The interior of R∗ is the empty set.
What is Int (Q) and Int (R \ Q) in this topology? More generally, if 0 ∈ A, then
Int (A) = A and if 0 ∈
/ A, then Int (A) = ∅.
(i) Consider R with the outcast topology (outcast is 0). The interior of any set A is
A \ {0}.

4.13 A is open iff A = Int (A).

4.14 Set theoretic results about the interior operation:

(a) If A ⊂ B, then Int (A) ⊂ Int (B).

(b) Int (A) ∪ Int (B) ⊂ Int (A ∪ B).
(c) Int (A ∩ B) = Int (A) ∩ Int (B).
(d) ∪i∈I Int (Ai ) ⊂ Int (∪i∈I Ai ).

29
4.15 Let X be a (metric) space and A ⊂ X. A point x ∈ X is said to be a boundary point of
A in X if every open set that contains x intersects both A and X \ A non-trivially. The
boundary of A in X is the set of boundary points of A in X. We denote it by ∂A.

4.16 Find the boundaries of each of the following sets:

(a) A1 = (a, b] ⊂ R with the standard topology.

(b) A2 = R \ {0} ⊂ R with the standard topology.
(c) A = Q ⊂ R with the standard topology.
(d) ∂∅ = ∅ = ∂X for any topological space X.
(e) The boundary of an open or closed ball in Rn is the sphere: ∂B(x, r) = ∂B[x, r] =
S(x, r) := {y ∈ Rn : d(x, y) = r}. Is this true in an normed linear space ? in an
arbitrary metric space?
(f) In R with VIP topology and R with outcast topology, find ∂A, where A = {0}, {x}, Q
and R \ Q. (x is a nonzero real number.)
(g) Let B be an open ball in Rn . Find the boundary of B minus a finite number of
points.
(h) Let A := {z ∈ C : z = reit , r ∈ [0, 1], t ∈ (0, 2π)}. (Draw a picture.) Find the
boundary of A.

4.17 A few more examples to sharpen our geometric intuition.

(a) Consider A = R × {0} ⊂ R2 . What is the boundary of A in R2 ?

(b) A = U1 ∪ U2 ∪ U3 is the subset of R2 where U1 := {x2 + y 2 < 1, y > 0}, U2 := {−1 ≤
x ≤ 1, y = 0} and U3 := {x2 + y 2 = 1, y < 0}.
(c) A = {(x, y) : x2 + y 2 = 1}.

4.18 Show that for any subset A of a topological space (X, T ), ∂A = A ∩ X \ A. (This is the
standard definition.)

4.19 While trying to prove the equivalence of the definition of continuity at a point (of a
function between two metric spaces) with the sequential definition, we established the
following.

Theorem 7. Let X and Y be arbitrary topological spaces and p ∈ X. Let f : X → Y be

a map.
1. If f is continuous at p, then for every sequence (xn ) in X converging to p, we have
f (xn ) → f (p).
2. Assume that X is first countable and Hausdorff. Assume further that f has the
property that for every sequence (xn ) converging to p, the sequence (f (xn )) converges to
f (p) in Y . Then f is continuous at p.

Proof. 1. Let V ⊂ Y be open with f (p) ∈ V . By continuity of f at p, there exists an

open set U 3 p such that for x ∈ U , we have f (x) ∈ V . Since xn → p, for this U , there
exists N ∈ N such that k ≥ N =⇒ xk ∈ U . Hence for k ≥ N , we see that f (xk ) ∈ V ,
that is, f (xk ) → f (x).

30
2. We prove this by contradiction. Assume that (Bn ) is a local base at p such that
Bn+1 ⊂ Bn and ∩n Bn = {p}. Since f is not continuous at p, there exists an open set
V 3 f (p) such that given any open set U 3 p, there exists x ∈ U such that f (x) ∈/ V . In
particular, for each n ∈ N, there exists xn ∈ Bn such that f (xn ) ∈
/ V . Clearly, xn → p.
For, let W 3 p be an open set. Then there exists N such that p ∈ BN ⊂ W . If k ≥ N ,
then xk ∈ Bk ⊂ BN ⊂ W . Thus, we conclude xn → p. Now by hypothesis, f (xn ) → f (p).
If we apply the definition of convergence to the set V , we find that f (xn ) ∈
/ V for any
n.

31
5 Dense Subsets in a Topological Space

5.1 A subset D ⊂ X of a topological space is dense in X if for every nonempty open set
U ⊂ X, we have D ∩ U 6= ∅, that is U intersects D non-trivially.
5.2 Examples of dense sets:
(a) Q is dense in R. Is R \ Q dense in R? Can you think of a countable dense subset in
R2 ? in Rn ?
(b) In R, with the lower limit topology, the sets Q and R \ Q are dense.
(c) The set A := {x ∈ `1 : xn = 0 for all n ≥ N for some N } is dense in `1 .
(d) The set Dn of all sequences x = (xm ) ∈ `1 whose terms are rational and xk = 0 for
k > n. Let D := ∪n∈N Dn . Then D is a countable dense subset of `1 .
(e) The only dense subset of a discrete space X is X itself.
(f) In an indiscrete space, any nonempty subset is dense.
(g) The set {0} is dense in R with the VIP topology. The set R \ Q is not dense.
(h) The set R \ {0} is dense in R with the outcast topology. This space cannot have a
countable dense set.
√ √
(i) S := {n+m 2 : n, m ∈ Z} is dense in R. (Did you notice that Z and 2Z are closed
and S is a sum of two closed sets? If the result is true, then S cannot be closed in
R. Why? If S is closed and dense, then S = R, but S is countable! Hence we have
an example of two closed sets in R whose sum is not closed.) See Lemma 2.5.7/Page
52 of my book on Metric spaces..
(j) Is Q2 dense in R2 with the order topology?
(k) Weierstrass approximation theorem says that the vector subspace of polynomials in
the normed linear space (C[0, 1], k k∞ ) is dense. (This should be a topic for Student
Seminar!)
(l) A dyadic rational is a real number of the form m/2n where m is an integer and
n ∈ N. Let D denote the set of dyadic rationals. Then D is dense in R. Consider
an open interval of the form (a − ε, a + ε). Choose n so that 1/2n < ε. If there
is no dyadic rational in this interval, then there exists an odd integer m such that
m/2n < a − ε and (m + 2)/2n > a + ε. (Why?) But then

2/2n = 2−n ((m + 2) − m) > a + ε − (a − ε) = 2ε, a contradiction.

5.3 D ⊂ X is dense in a space (X, d) iff every point of X is a limit point of D.

Let D be dense in X. Let x ∈ X and U 3 x be open. Then U ∩ D 6= ∅. Thus x is a limit
point of D.
Conversely, if every x ∈ X is a limit point of D, we claim that D is dense in X. For,
if not, there exists a nonempty open set U such that U ∩ D = ∅. Since U is nonempty,
choose x ∈ U . Then x is not a limit point of D as U 3 x is open but U ∩ D = ∅.
5.4 D ⊂ X is dense in the space X iff its closure D = X. (This is the standard definition.)
Recall (from Item 5) that the closure of any set A is the set of limit points of A. The
result now follows from the last item.

32
 5.5 In a metric space (X, d), a set A is dense in X iff for every x ∈ X and ε > 0, there
exists an a ∈ A such that d(x, a) < ε. (Thus, A is dense in X, if we can “approximate”
any point x ∈ X to “any level of approximation” by an element of A. See Item 2k to
understand this vague remark. Also recall that Q is dense in R, which means that any
real number can be approximated to any level of accuracy by a rational number.)

5.6 Let (X, d) be a metric space. Assume that the only dense subset is X itself. Can we say
something about the topology? Hint: What are the maximal proper subsets of X?

5.7 Let A, B be two dense subsets of a space X. Is A ∪ B dense? Is A ∩ B dense?

5.8 If A, B are open dense subsets of a space X, is A ∩ B dense in X?

5.9 Give an example of a proper open dense subset of R.

5.10 Continuation of the last item. If we write an open set U = ∪J ˙ k , as the disjoint union of
P
open intervals (Item 11s), then we say that the “measure” or “length” of U is k `(Jk ),
the sum of lengths of the intervals Jk . Given ε > 0, can you find an open dense subset of
R whose length is less than or equal to ε?

5.11 Let D be dense in (X, T1 ). Is D (necessarily) dense in (X, T2 ) where T2 is finer (respec-
tively, coarser) than T1 ?

5.12 Let X, Y be topological spaces. Assume that A is dense in X and f : X → Y is continuous

and onto. Then f (A) is dense in Y .

5.13 The set of matrices in M (n, C) with distinct eigenvalues is dense. In particular, the set of
all diagonalizable matrices in M (n, C) is dense. This exercise requires a good background
in Linear Algebra.
It is well-known fact in linear algebra that any A ∈ M (n, C) can be brought to upper
triangular form, say T , via conjugation by a unitary matrix U such that T = U AU −1 .
The eigenvalues are the diagonal entries, say, dj . We can find very small εj ’s so that
dj + εj ’s are all distinct. We thus get a new upper triangular matrix, say T1 whose entries
are the same as that of T except dj is replaced by dj + εj . Again it is well known that
T1 is diagonalizable. The matrix A1 := U T1 U −1 has distinct eigenvalues and hence is
diagonalizable. It is close to A if εj ’s are small. This follows from the observation kAk
on M (n, C) comes from the inner product (X, Y ) 7→ Tr XY ∗ and the inner product is
invariant under conjugation by unitary matrices. The reader is encouraged to work out
the details and submit it as an assignment to the instructor. Details!

5.14 A topological space is separable if there exists a countable dense subset.

5.15 Examples and non-examples of separable spaces:

(a) Rn is separable. Consider {x = (x1 , . . . , xn ) ∈ Rn : xj ∈ Q}.

(b) A discrete space X is separable iff X is countable.
(c) `1 is separable.
(d) R with VIP topology is separable.
(e) R with outcast topology is not separable.

33
 (f) Any second countable space is separable. If {Bn } is any countable basis, choose one
element, say, xn ∈ Bn . Then D := {xn } is a countable dense set.
(g) Let X be infinite with co-finite topology and let A be any infinite subset of X. Then
any x ∈ X is a limit point of A. In particular, X with co-finite topology is separable.

(h) Is R2 with the order topology separable? (Recall the geometric description of basic
open sets in this space. See Item 19.)
No. For, consider the uncountable collection of pair-wise disjoint basic open sets of
the form {Bx := {x} × R : x ∈ R}. If D is a countable dense set, then for each
x ∈ R, there exists y(x) ∈ D ∩ Bx . The map f : R → D given by x 7→ y(x) is one-one
and hence R is countable.

5.16 Let X be uncountable with co-finite topology. Then X is not first countable but separable
by Item 15g.

5.17 Let X be uncountable with co-countable topology. No countable set can have a cluster
limit point and hence X is not separable.

5.18 Let `∞ denote the set of all bounded real sequences. It is a normed linear space with
respect to the norm kxk∞ := sup{|xn | : n ∈ N}. The space (`∞ , k k∞ ) is not separable.
Hint: Consider the uncountable subset {x : N → {0, 1}} of `∞ .

5.19 A metric space is separable iff it is second countable.

Let X be a separable metric space with D := {an : n ∈ N} as a countable dense set. Then
the collection {B(an , 1/k) : n, k ∈ N} is a countable basis for the metric topology. Let
U be an open set and x ∈ U . Then there exists r > 0 such that B(x, r) ⊂ U . Choose k
1 1
such that 1/k < r. Let aan ∈ D ∩ B(x, 2k ). We claim that x ∈ B(an , 2k ) ⊂ B(x, k1 ) ⊂ U .
1
If y ∈ B(an , 2k ), then observe that

1 1 1
d(y, x) ≤ d(y, an ) + d(an , x) < + = < r.
2k 2k k

5.20 R` , the space R with lower limit topology is first countable, separable but not second
countable. This is Item 10e
and may be
deleted.
5.21 Let f, g : X → Y be continuous and Y be Hausdorff. Then the set A := {x ∈ X : f (x) =
g(x)} is closed in X.
We show that B := X \ A is open. Let b ∈ B. Then f (b) 6= g(b) and hence there exist
open sets V1 3 f (b) and V2 3 g(b) with V1 ∩ V2 = ∅. By continuity of f and g at b,
there exist open sets U1 3 b and U2 3 b such that f (U1 ) ⊂ V1 and g(U2 ) ⊂ V2 . Then
Ub := U1 ∩U2 is an open set containing b and we have for x ∈ Ub , f (x) ∈ V1 and g(x) ∈ V2 .
Hence f (x) 6= g(x) for x ∈ Ub . That is, Ub ⊂ B. Hence B = ∪b∈B Ub is open.

5.22 Let the hypothesis be as in the last item. Assume that D is dense in X and that f (x) =
g(x) for all x ∈ D. Then f (x) = g(x) for all x ∈ X.
Let the notation be as in the last item. Then A is a closed set containing D. Since D is
dense D = X ⊂ A. That is, A = X.

34
An alternative proof is by contradiction. Let A be the set on which f and g agree.
Then A is dense in X. Let a ∈ X be such that f (a) 6= g(a). Since Y is Hausdorff,
there exist open sets V1 3 f (a) and V2 3 g(a) with V1 ∩ V2 = ∅. Let U1 := f −1 (V1 )
and U2 := g −1 (V2 ). Then Ui are open (why?) with a as a common element. Since
U := U1 ∩ U2 3 a is a nonempty open set, and A is dense in X, there exists b ∈ U ∩ A.
Since b ∈ A, f (b) = g(b). Since b ∈ U = U1 ∩ U2 , we have f (b) ∈ V1 and g(b) ∈ V2 . It
follows that f (b) = g(b) ∈ V1 ∩ V2 = ∅, a contradiction.

35
6 Homeomorphisms

6.1 Let X, Y be sets. Suppose f : X → Y is a bijection. Assume further that one of the sets
has an extra mathematical structure such as a group, vector space, metric or a topology.
Then we can transfer the structure to the other set using the bijection. We look at some
specific instances.

(a) Let X be a group. Then we define y1 · y2 to be f (x1 · x2 ) where f (xi ) = yi , i = 1, 2.

It turns out that Y is group and that f : X → Y , by virtue of the very definition of
group law on Y , is a group homomorphism (and hence an isomorphism.)
(b) Let Y be a metric space. Then we set d(x1 , x2 ) := d(f (x1 ), f (x2 )). Then the metric
space (X, d) is isometric to (Y, d).
(c) Let X be a topological space. Let TX be the topology on X. We then define a
topology TY on Y be declaring that V ∈ TY iff there exists U ∈ TX such that
V = f (U ). Then the map f : (X, TX ) → (Y, TY ) is a homeomorphism, a term not
yet defined!

6.2 To illustrate this principle, we use the bijection t 7→ et from X := R to Y := (0, ∞) to

make Y into a vector space over R. Given y1 , y2 ∈ Y , we look at their (unique) pre-images
xj = log yj , carry out the vector addition in X, obtain x1 +x2 = log y1 +log y2 = log(y1 y2 )
and map it by the bijection. The result is y1 y2 . Similarly, the scalar multiple of y by
α ∈ R is α log y 7→ eα log y = y α . Thus the vector addition of y1 and y2 is y1 y2 and the
scalar multiple α · y is y α . The ‘additive identity’ is 1. Note that the map t 7→ et is a
linear isomorphism.

6.3 A map f : X → Y between two topological spaces is a homeomorphism if (i) f is bijective,

(ii) f is continuous and (iii) f −1 : Y → X is continuous.
This is the analogue of isomorphisms in Algebra. Note that there also one requires a
bijective map f such that f and its inverse f −1 preserve the ‘algebraic structures’ such as
group, ring, vector space structures. It turns out in the context of algebra, if f preserves
the structure, then f −1 does automatically.
We say that two topological spaces X and Y are homeomorphic if there exists a homeo-
morphism f : X → Y .

6.4 The relation of being homeomorphic is an equivalence relation among topological spaces.

6.5 Examples of homeomorphisms.

(a) Any f : R → R of the form f (x) = ax + b for a nonzero a ∈ R is a homeomorphism.

(b) f : R → R given by f (x) = x3 is a homeomorphism.
(c) Any linear isomorphism of Rn is a homeomorphism.
More generally, any linear isomorphism from (Rn , k k) to (Rn , k k0 ), where k k and
k k0 are any of the norms k k1 , k k2 and k kmax , is a homeomorphism.
In particular, the identity map is a homeomorphism. As a corollary, we conclude
that the topologies induced by these norms are the same:

Tk k1 = Tk k2 = Tk kmax .

36
 (d) Let us now look at some homeomorphisms of a normed linear space. Let (X, k k)
be a normed linear space. Then the maps (a) x 7→ λx for 0 6= λ ∈ R, (b) x 7→ x + v,
where v ∈ X is fixed are homeomorphisms.
(e) Consider M (n, R). Then the maps (a) X 7→ X t , (b) X 7→ X + A for fixed A ∈
M (n, R) and (c) X 7→ AX for a fixed nonsingular matrix A are homeomorphisms.
(f) Any two discrete spaces are homeomorphic iff they have the same cardinality.
(g) If two metric spaces are isometric, then they are homeomorphic.
(h) In the examples of this item, the subsets are given the metric topology from the
induced metric.
(i) Let X be a set with at least two elements.. Let T1 and T2 be respectively the
indiscrete and discrete topology on X. Let f be the identity map on X. Then
(i) f : (X, T1 ) → (X, T2 ) is a bijection but not continuous and hence is not a
homeomorphism.
(ii) f : (X, T2 ) → (X, T1 ) is a continuous bijection but not a homeomorphism.
(j) Keep the notation of the last example. Then any bijection of (X, Ti ) to itself is a
homeomorphism, where i = 1, 2.
(k) A few more examples.
i. [a, b] ' [0, 1]. More generally, [a, b] ' [c, d].
ii. (−1, 1) ' R.
iii. (0, 1] ' [1, ∞).
iv. [0, 1) ' (0, 1].
v. Can Q be homeomorphic to Z?
vi. Is N ' Z?
(l) A bijective continuous map need not be a homeomorphism. Examples and a non-
example:
i. R with discrete topology and R with indiscrete topology.
ii. f : [0, 2π) → S 1 ⊂ C given by f (t) = eit . (A more instructive exercise.)
iii. Any bijective continuous map of a finite topological space X to itself is a home-
omorphism.
(m) The spaces (R, VIP) and (R, Outcast) are not homeomorphic.
We shall see later a lot of examples of homeomorphisms.

6.6 Open and closed maps. A map f : X → Y is said to be open if f (U ) is open in Y for
every U open in X. A closed map is defined similarly.

6.7 A continuous map need not be an open map. Similarly, it need not be a closed map.
Example: The identity map from (X, T2 ) to (X, T1 ) is continuous and is neither open nor
closed. Notation as in Item 5i.
A closed (respectively open) map need not be continuous.
Example: The identity map from (X, T1 ) to (X, T2 ) is not continuous and is an open map
as well as a closed map.

37
 6.8 A bijective continuous map is a homeomorphism iff it is an open map.
The key observation is the following. Let f : X → Y be any map. For any B ⊂ Y , we
have the inverse image B under f , namely f −1 (B) ⊂ X. If f happens to be a bijection
and if g : Y → X is its inverse then g(B) = f −1 (B). We prove this.
Let x ∈ g(B). Then there exists y ∈ B such that x = g(y). Now, x ∈ f −1 (B) iff f (x) ∈ B.
This is the case iff f (g(y)) ∈ B. Since f ◦ g is the identity of Y , we see that f (x) = y ∈ B.
Thus g(B) ⊂ f −1 (B).
The reverse inclusion is similar. Let x ∈ f −1 (B). Then y := f (x) ∈ B. What is g(y)?
We have g(y) = g(f (x)) = x since g ◦ f is the identity on X. Hence x = g(y) ∈ g(B).
Now let us prove the stated result. Let f : X → Y be a homeomorphism and U ⊂ X be
open. We observe that f (U ) = g −1 (U ). Since g is continuous, it follows that g −1 (U ) =
f (U ) is open. The proof of f (U ) = g −1 (U ) is similar to that of f −1 (B) = g(B). Replace
f by g and B by U in the argument and we get the result.
Let f be a continuous bijection which is also an open map. Let g be the inverse of f .
To establish the continuity of g, let U ⊂ X be open. Is g −1 (U ) open in Y ? We know
g −1 (U ) = f (U ) by the observation made in the beginning. Since f is open, f (U ) = g −1 (U )
is open.

6.9 Application: The map f : R → R given by f (x) = x3 is a homeomorphism.

We have already seen that any polynomial function on R is continuous. The function f
is strictly increasing and hence is one-one. (To see that f is increasing, either apply the
derivative test or observe that x3 − y 3 = (x − y)(x2 + xy + y 2 ) and the second factor on
the right side is always non-negative.) It is onto due to the intermediate value theorem.
Given α ∈ R, we are on the look-out for an x ∈ R with x3 = α. Let N ∈ N be such that
|x| < N . When restricted to [−N, N ], the function g(x) = x3 − α takes values of opposite
signs.

6.10 A bijective continuous map is a homeomorphism iff it is a closed map.

This again depends on the following set theoretic observation: Let f : X → Y be any
map. Let B ⊂ Y . Then f −1 (Y \ B) = X \ f −1 (B). Proof is easy. Let x ∈ f −1 (Y \ B).
That is, f (x) ∈
/ B. Hence x ∈ / f −1 (B). This shows that f −1 (Y \ B) ⊂ X \ f −1 (B). To
prove the reverse inclusion, let x ∈ X \ f −1 (B). Note that this implies f (x) ∈
/ B. We
−1
need to show that x ∈ f (Y \ B), that is, f (x) ∈ / B, which is true.
To prove the stated result, let f : X → Y be a homeomorphism and K ⊂ X be closed.
Note that f (X \ K) = g −1 (X \ K) = Y \ f (K). Since f is a homeomorphism and X \ K
is open, we see that f (X \ K) = Y \ f (K) is open. That is, f (K) is closed. We leave the
converse to the reader.
Application: Item 22b. We have to wait for this.

6.11 Keep the notation of Item 5i. Let X be a set with at least two elements. Can some
bijection f : (X, T1 ) → (X, T2 ) be a homeomorphism? How about a bijection g : (X, T2 ) →
(X, T1 )?

6.12 We say that property of a topological space is a topological property if every space Y
homeomorphic to X also has the property. Examples:

38
(a) The space being Hausdorff is a topological property.
(b) The space being first countable is a topological property.
(c) The space being second countable is a topological property.
(d) The space being separable is a topological property. (Item 12 is of use here.)
(e) Existence of a nonempty, proper subset which is both open and closed is a topological
property.
(f) Let us say that a topological space X has BCP if every continuous real valued
function is bounded. For example all closed and bounded intervals have this property.
Is BCP a topological property? Hint: If ϕ : X → Y is a homeomorphism there is
a “natural map” ϕ∗ : C(Y, R) → C(X, R) where C(X, R) stands for the set of real
valued continuous functions on X etc.
The “adjoint” map ϕ∗ is defined by ϕ∗ (g) := g ◦ ϕ. If ϕ is a bijection, then ϕ∗ is also
a bijection. If ϕ is a homeomorphism, then ϕ∗ (g) ∈ C(X, R) for any g ∈ C(Y, R).
(g) Two metric spaces can be homeomorphic, but one of them could be bounded while
the other is not. Hence ‘being bounded’ is not a topological property among metric
spaces.
(h) Similarly, completeness is not a topological property among the metric spaces.

We shall see later a lot of examples of topological properties.

The study of topology is mainly understanding topological properties and using

them to assert whether given two spaces are homeomorphic or not.

39
7 New Topologies from the Old

7.1 We now look at some natural questions which lead us to the generation of new topologies.

7.2 Given a set X and a collection S of subsets of X, how to we describe the open sets in the
smallest topology, say, TS that contains S? (We assume, as this is the case that occurs
in practice, that for every x ∈ X, there exists S ∈ S such that x ∈ S.) We do this in two
steps.

(a) We wanted a base for some topology on X which will also contain S. Clearly,
B := {S1 ∩ · · · ∩ Sn : Sj ∈ S, n ∈ N} is a base for some topology and S ⊂ B.
(b) The topology TB := {U ⊂ X : ∀x ∈ U, ∃B ∈ B such that x ∈ B ⊂ U } is then the
smallest topology that contains S.
(c) Thus, we can rid of the intermediate B and define the topology directly in terms of
S. We say U ∈ TS iff for every x ∈ U , there exists n ∈ N such that we can find Sj ,
1 ≤ j ≤ n with x ∈ S1 ∩ · · · ∩ Sn ⊂ U . One can again show directly that this is the
smallest topology containing S.
(d) S is called a subbase and TS is the topology generated by S.

7.3 Let us look at some concrete examples:

(a) Consider S := {(−∞, a) : a ∈ R} ∪ {(b, ∞) : b ∈ R}. The topology generated by S

on R is the usual topology.
(b) Consider R and S := {{0, x} : x 6= 0, x ∈ R}. What is the topology on R?
(c) Let X be a set with at least 3 elements. Let S be the family of two-element subsets
of X. The topology generated by S is the discrete topology.
(d) What is the topology on R2 , if we take the subbase consisting of all straight lines in
R2 ?
(e) What is the topology on R2 , if we take the subbase consisting of all straight lines
parallel to the x-axis in R2 ? Which of the following sets are open in this topology?
(i) the open unit disk, B(0, 1), (ii) the open vertical band {(x, y) ∈ R2 : 0 < x < 1},
(iii) the open horizontal band {(x, y) ∈ R2 : 0 < y < 1}, (iv) any subset which is
bounded in the Euclidean metric.
(f) What is the topology on R2 , if we take the subbase consisting of all circles in R2 ?
(g) What is the topology on R2 , if we take the subbase consisting of all circles, with
centre at the origin, in R2 ?
(h) Consider S = {X} as a subbase on X. What topology do we get on X?

7.4 Let f : X → Y be any map between two sets. Assume that one of them is a topological
space. What we wish to do is to endow the other set with an optimal topology in such a
way that f : X → Y becomes a continuous map between the spaces.

(a) Let Y be a topological space. Then if we endow X with the discrete topology, then
the problem is solved! But this topology has no bearing on Y and/or on the map f !
So what we require is the smallest topology on X making f continuous.

40
 (b) Let X to be a topological space. Considerations similar to the last item suggest us
that we require the largest topology on Y making f continuous.

7.5 These problems arise in a very natural way.

(a) Let X be a subset a topological space Y . Then we have an obvious or natural map
i : X → Y , the inclusion of X in Y , that is, the restriction of the identity on Y to
X.
(b) Let X be any topological space and ∼ an equivalence relation on X. Then as Y , we
take the quotient set X/ ∼, that is, the set of equivalence classes. Once again, we
have a natural map π : X → Y , where π(x) is the equivalence class of x.

7.6 More general situations may also arise. Let X be a set and Yi be topological spaces,
indexed by a set I. Assume that we are given certain maps fi : X → Yi for each i ∈ I.
We again ask for a single smallest topology on X making all the maps fi continuous. Or
the other way around, we have maps fj : Xj → Y where Xj ’s topological spaces.
Typical instances of this phenomenon are:

(a) Let {Xj : U j ∈ I} be an indexed family of (pairwise disjoint) topological spaces.

Let X := j∈I Xj , the disjoint union of Xj ’s. We have natural inclusion maps
ιj : Xj → X. We wish to endow X with the largest topology with respect to which
all ιj ’s are continuous.
Q
(b) Let {Xi : i ∈ I} be an indexed family of topological spaces. Let X := i∈I Xi . We
have obvious maps πi (x) = xi , the i-th projection. We wish to equip X with the
smallest topology such that each of the projections becomes continuous.
(c) Let E be a set and let X := F be a family of functions from E to R. Consider
the evaluation maps εx (f ) := f (x) for each x ∈ E. Thus, we have a family of
maps εx : X → R and we want the smallest topology which will make all these maps
continuous.

7.7 Let us deal with various cases. Let X be a set and Y be a topological space and f : X → Y
be a map. Any topology on X which makes f continuous must contain the set U :=
{f −1 (V ) : V ∈ TY }. It turns out this collection is already a topology and hence is the
smallest topology on X, as required. (We were lucky this time!)

7.8 Let us look at the concrete case in Item 5a. Then the topology on X is given by

TX := {i−1 (V ) : V ∈ TY } = {V ∩ A : V ∈ TY }.

The topology TX is called the subspace topology on Y and any U ∈ TX is said to be open
in X. We say that F ⊂ X is closed in X if its complement, X \ F , in X is open in X.

7.9 The following are immediate from the definition of subspace topology and

7.10 Assume that Y is ordered and is with order topology. Let f, g : X → Y be continuous.
Let h(x) := min{f (x), g(x)}. Then h is continuous. Hint: Item 28f.
Is ϕ := max{f, g} continuous?

41
7.11 Did you notice that all the examples above use the second version of the gluing lemma?
There is a beautiful application of the ‘open version’ of the gluing lemma in my book
“Topology of Metric Spaces”. See Lemma 3.2.52 in the book.

7.12 (a) If B is a basis for the topology TY on Y , then BX := {B ∩ X : B ∈ B} is a basis for

the subspace topology on X. Details!

(b) If Bx is a local basis at x ∈ Y for the topology TY on Y , then Bx,X := {B ∩ X : B ∈

Bx } is a local basis at x ∈ X for the subspace topology on X.

7.13 Let us look at some examples to develop our intuition about subspace topology. Use the
last item to identify a local basis at each point of the subset A.

(a) Consider A = [0, 1] ⊂ R. Then the sets [0, 1/2), (1/2, 1] and (1/2, 3/4) are open in
in A.
(b) Consider Y := {(x, y) : xy = 0} ⊂ R2 be the two axes. Then the basic open sets
near (0,0) are crosses (of two line segments along the x and y-axes.) At other points,
just intervals around them.
(c) Let A := {1/n : n ∈ N} ∪ {0}. Then the basic open sets are the singletons {1/n} for
n ∈ N and {1/n : n ≥ n0 } ∪ {0}. The latter are basic opens sets near 0 in A.
(d) Let S := {(x, y) ∈ R2 : x2 + y 2 = 1} ⊂ R2 be the unit circle in R2 . The basic open
sets in S are open arcs of the circle.
√ √
(e) Consider A = Q ⊂ R. Then the set {r ∈ Q : − 2 ≤ r ≤ 2} is both open and
closed in Q.
(f) Let X be a metric space and ∅ = 6 A ⊂ X. Then we have two topologies on A: (i) one
comes from the induced metric, call it dA , on A and (ii) the other is the subspace
topology. They are the same.
Let TdA denote the metric topology on A and TA denote subspace topology on A.
The local base at a ∈ A for TA is {B(A,dA ) (a, r) : r > 0} and the one for TA is
{B(X,d) (a, r) ∩ A : r > 0}. But, B(A,dA ) (a, r) = B(X,d) (a, r) ∩ A for each r > 0.
Observe the following:

B(A,dA ) (a, r) := {x ∈ A : dA (x, a) < r} = {x ∈ A : d(a, x) < r} = B(X,d) (a, r) ∩ A.

Hence the local bases are the same at each point a ∈ A for both the topologies.
(g) Let A := [0, 1] × [0, 1]. Then A has the order topology as well as the subspace
topology as a subset of R2 with order topology. They are not the same. (Contrast
this with the last item.)
Consider the set V := {(0, y) : 1/2 < y ≤ 1}. Then V is is open in the subspace
topology but not in the order topology on the ordered set A. Draw a picture of A
and use the definitions of subspace topology and order topology. V is open in the
subspace topology, since it is the intersection A with basic open set in R2 with an
interval (in the order topology): V = A ∩ (a, b) where a = (0, 1/2) and b = (0, 2).
Let, if possible, A be open in the order topology on A. Then there exists an open
interval (c, d) such that that point p = (0, 1) ∈ (c, d). Let c = (x1 , y1 ) and d =
(x2 , y2 ). Then x1 , x2 ≥ 0 and y1 , y2 ≤ 1. Now, (x1 , y1 ) < (0, 1) in the dictionary
order. We conclude that x1 = 0 and y1 < 1. Similarly, x2 > 0 and y2 ≥ 0. But an
element of the form (x2 /2, y) with y ≥ 0 lies in the basic open set but not be in V .

42
 (h) Consider R with VIP topology and A = R∗ . Then the subspace topology on R∗ is
the discrete topology. The subspace topology on Q is the VIP topology on Q. (Do
you understand this statement?)
(i) Investigate the subspace topology on Q considered as a subset of R with outcast
topology.
(j) Let X be a Hausdorff space, A ⊂ X be endowed with the subspace topology. Then
A is Hausdorff.
(k) Let [a, b] ⊂ R` . Then {b} is open in [a, b] with the subspace topology inherited from
the lower limit topology on R.

7.14 Let A be nonempty and open in X. Then U ⊂ A is open in A iff it is open in X.

7.15 Let B ⊂ A ⊂ X. Let (X, TX ) be a topological space. Let TA denote the subspace topology
on A. Let x ∈ A. Then x ∈ A is a limit point of B in A iff x is a limit point of B in X.
Let x be a limit point of B in A. Let U ∈ TX such that x ∈ U . Since U ∩ A ∈ TA is an
open set containing x, (U ∩ A) ∩ B 6= ∅. But, (U ∩ A) ∩ B = U ∩ B. Thus, x is a limit
point of B in X.
Conversely, let x ∈ A be a limit point of B in X. Let V be open in A with x ∈ V . We
need to show that V ∩ B 6= ∅. There exists U ∈ TX such that V = U ∩ A. Then x ∈ U
and since x is a limit point of B in X, we have U ∩ B 6= ∅. Since B = B ∩ A, it follows
that x ∈ (U ∩ A) ∩ B 6= ∅, that is, V ∩ B 6= ∅, or x is a limit point of B in A.

7.16 Let A ⊂ X. Then F ⊂ A is closed in A iff there exists a closed set C in X such that
F = A ∩ C.
You may use the last item to prove this. Let F = A ∩ C where C is a closed subset of X.
To show that F is closed in A, it suffices to show that if x ∈ A is a limit point of F in A,
then x ∈ F . By the last item, this is true. Hence F is closed in A.
To prove the converse, if F is closed in A, where do we find C ⊂ X closed in X with
F = A ∩ C? Obvious choice is F , the closure of F in X. Do we have F = F ∩ A? Clearly,
we have F = F ∩ A ⊂ F ∩ A. To prove the reverse inclusion, let x ∈ F ∩ A. Then, x is a
limit point of F in A and hence by the last item, it is a limit point of F in the subspace
topology. Since F is closed in the subspace topology, it contains all its limit points in A,
in particular x.
Or, we may proceed directly as follows. (This is essentially set-theoretic exercise and so
the reader should try on his own.)
Let F = A ∩ C. We show that the complement of F in A is open. Let U = X \ C.
Then U is open in X. We claim that U ∩ A = A \ F . To show (X \ C) ∩ A ⊂ A \ F , let
x ∈ (X \ C) ∩ A. If x ∈ F , then x ∈ F = C ∩ A and hence x ∈ C, a contradiction. For
the reverse inclusion, let x ∈ A \ F . We need to show that x ∈ X \ C. If false, then x ∈ C
and hence x ∈ A ∩ C = F , that is x ∈ F , a contradiction.
Assume that F is closed in A. Then A \ F is open in A. Let U be open in X such that
A \ F = U ∩ A. Let C := X \ U . Then C is closed in X. We claim that C ∩ A = F .
To show that C ∩ A ⊂ F , let x ∈ C ∩ A. Suppose x ∈ / F , then x ∈ (A \ F ) = A ∩ U .
Therefore, x ∈ U or U ∈
/ C, a contradiction. To prove the reverse inclusion, let x ∈ F . If
x∈/ C, then x ∈ U and hence x ∈ U ∩ A = (A \ F ), that is x ∈/ F , a contradiction.

43
 As a specific example, the set of Item 13e is open as well as closed in Q. (Contrast this
with Item 35i.)

7.17 We shall put to use some of the concepts we learned to gain a different perspective of the
limit of a sequence.
Consider the set X := {1/n : n ∈ N} ∪ {0} ⊂ R with the subspace topology inherited
from R. (Refer to Item 13c.) Let f : X → R be a function. Note that any such f is
continuous at any point of the from 1/n. When is it continuous at 0? Let an := f (1/n)
and a := f (0). We claim that f is continuous at 0 iff the sequence (an ) converges to a.

Let f be continuous at 0. Consider V := (a − ε, a + ε) 3 0 an open set containing

a = f (0). Since f is continuous at 0, there exists a basic (relatively) open set U :=
{1/k : k > N } ∪ {0} such that for all k ∈ U , we have f (k) ∈ (a − ε, a + ε). That is, for
k > N , we have |ak − a| < ε. This proves that (an ) converges to a. Converse is along
the same lines.

Consider X = N ∪ {∞}, where ∞ is just a symbol representing an element not in N. (We

could have used ? in place of ∞!) Let ϕ : X → Y := {1/n : n ∈ N} ∪ {0} be defined
by ϕ(n) = 1/n and ϕ(∞) = 0. Then ϕ is a bijection. We endow Y with the subspace
topology as a subset of R. Using the bijection ϕ, we transfer the topology on Y to X.
The local basic open set in Y at n are {n}, and at ∞ are {k : k ≥ N } for some N ∈ N.
(See Item 13c.)
Now if f : X → R is a function, when is it continuous at ∞? If we restrict f to N ⊂ X,
we can think of f as a real sequence, say, (an ), in R. Do you see any relation between the
continuity of f at ∞ and the convergence of (an )? If we replace R by a topological space
Z, do the results (concerning the convergence of sequences in Z) continue to be true?
We shall return to this example later when we talk of one point compactifications.

7.18 Let f : X → Y be a continuous map between two topological spaces. Let A ⊂ X be a

subset endowed with the subspace topology. Then the restriction fA of f to A gives rise
to a map fA : A → Y . Is it continuous?
If V ⊂ Y is open, then fA−1 (V ) = f −1 (V ) ∩ A is open in A, since f −1 (V ) is open in X.

7.19 A question ‘dual’ to the one in the last item: Let f : X → Y be continuous. Assume that
f (X) ⊂ B ⊂ Y . We then have an induced map g : X → B defined by g(x) = f (x) for
x ∈ X. Let B be given the subspace topology. Is g continuous? The answer is ‘Yes.’
Let W ⊂ B be open in B. Then there exists V , an open subset of Y such that W = V ∩B.
It is easy to check that g −1 (W ) = f −1 (V ).
For, we have

g −1 (W ) := {x ∈ X : g(x) ∈ W } = {x ∈ X : f (x) ∈ W } ⊂ f −1 (V ).

On the other hand, if x ∈ f −1 (V ), then f (x) ∈ V ∩ B, that is, g(x) ∈ W . Hence

x ∈ g −1 (W ).

Since f is continuous, f −1 (V ) is open in X and hence g −1 (W ). (We shall return to this

later when we talk of universal mapping properties. See Item 49.)

44
 Is the converse true? If g is continuous, it f continuous? Let V ⊂ Y be open. Since
V ∩ B is open in B, we have g −1 (V ) is open in X. What is g −1 (B)?
g −1 (V ) = {x ∈ X : g(x) ∈ V } = {x ∈ X : f (x) ∈ V } = f −1 (V ).
Since g is continuous, it follows that g −1 (V ) = f −1 (V ) is open for any open set V ⊂ Y .
That is, f is continuous.
Items 18–19 are most often used when we deal with subspaces without explicit
mention.
7.20 At this stage we are curious about the following questions.
(a) Let X and Y be topological spaces. Assume that {Ai ∈ I} is a family of subsets
of X such that ∪i∈I Ai = X. Further assume that for each i, we have a continuous
function fi : A → Y . (Here Ai ’s are given the subspace topology.) Can we get
‘glue’ them together to get a continuous function f : X → Y in such a way that the
restriction f |Ai = fi for i ∈ I?
A necessary condition is that fi (x) = fj (x) for each x ∈ Ai ∩ Aj , i, j ∈ I. This will
ensure that we get a function f from the set X to Y whose restrictions to Ai are as
required.
(b) Let X and Y be topological spaces. Let f : X → Y be a map. Assume that
{Ai ∈ I} is a family of subsets of X such that ∪i∈I Ai = X and that f |Ai : Ai → Y
is continuous. Can we conclude f is continuous?
7.21 Let us investigate the situation. Let V ⊂ Y be open. Observe that
f −1 (V ) = f −1 (V ) ∩ X = ∪i∈I f −1 (V ) ∩ Ai .

Each term in the union, f −1 (V ) ∩ Ai = fi−1 (V ) is open in Ai . If we can ensure that each
of these open in X then f −1 (V ) is open in X. We know a sufficient condition which will
ensure this, namely, we demand each Ai is open.
Let V ⊂ Y be closed. Since each f −1 (V ) ∩ Ai = fi−1 (V ) is closed in Ai , to ensure f −1 (V )
is closed, we may demand that each Ai is closed. But then f −1 (V ) is a union of closed
sets and it is closed if we further assume that I is finite. We have thus arrived at
7.22 Gluing Lemma:
Lemma 8. Let X, Y be topological spaces and let f : X → Y be any map. Assume
that {Ai : i ∈ I} is a family of subsets of X whose union is X. Assume further that
fi := f |Ai : Ai → Y is continuous for each i ∈ I. Then
1. f is continuous if each Ai is open.
2. f is continuous if each Ai is closed and I is finite.

7.23 Typical applications of the gluing lemma.

(a) The absolute-value function | | : R → R is continuous.
(b) Let f, g : [0, 1] → X be continuous. Assume that f (1) = g(0). Define
(
f (2t) 0 ≤ t ≤ 1/2
h(t) :=
g(2t − 1) 1/2 ≤ t ≤ 1.
Then h is continuous.

45
(c) Let f : Rn → Rn be defined as follows:
(
x if kxk ≤ 1
f (x) = x
k x k2
if kxk > 1.

Then f is continuous.
(d) Assume that Y is ordered and is with order topology. Let f, g : X → Y be continuous.
Let h(x) := min{f (x), g(x)}. Then h is continuous. Hint: Item 28f.
Is ϕ := max{f, g} continuous? If you wish, you may assume Y = R to gain insights.
Note that the application in this series is a special case of this general result.
(e) Did you notice that all the examples above use the second version of the gluing
lemma? There is a beautiful application of the ‘open version’ of the gluing lemma
in my book “Topology of Metric Spaces”. See Lemma 3.2.52 in the book.

46
7.24 Let us consider the general case in Item 6: Let X be a set and Yi be topological spaces,
indexed by a set I. Assume that we are given certain maps fi : X → Yi for each i ∈ I.
We again ask for a single smallest topology on X making all the maps fi continuous.
We want the smallest topology T that contains all sets of of the form fi−1 (Vi ) where Vi
is open in Xi and i ∈ I. That is T is the smallest topology containing the family of sets
S := {fi−1 (Vi ) : Vi ∈ Ti ; i ∈ I}, where Ti is the topology on Xi .
There is no reason to believe that fi−1 (Vi ) ∩ fj−1 (Vj ) must be again of the form fr−1 (Vr )
for some r ∈ I. Hence S may not be topology on X.

7.25 We now want to look at the concrete case in Item 6b. As a preliminary, we review the
concept of Cartesian product.
Q
Let {Xi : i ∈ I} be an indexed family of sets. Then the Cartesian product X := i∈I Xi
is defined by Y
Xi := {x : I → ∪i∈I Xi : x(i) ∈ Xi for each i ∈ I}.
i∈I
Q
(a) We usually write x ∈ i∈I Xi as Q x = (xi ), where xi := x(i). We shall call xi as the
i-th coordinate of x. Let πj : j∈I Xi → Xj denote the map πj (x) = x(j) = xj .
This is called the j-th projection of X onto the j-th factor Xj .
(b) As a convention, if I = {1, 2, . . . , n}, we identify X with X1 × · · · × Xn , that is,
with the set of “ordered n-tuples” (x1 , . . . , xn ). Similarly, if I = N, we identify
X with X1 × X2 × · · · × Xn × · · · , that is the set of ordered infinite tuples x 7→
(x1 , x2 , . . . , xn , . . .). Details!

(c) If Vj ⊂ Xj , then πj−1 (Vj ) = i∈I Ui where Ui = Xi for i 6= j and Uj = Vj . In

Q

particular, π1−1 (V1 ) = V1 × X2 where X = X1 × X2 etc.

7.26 We now give a few examples to instill some confidence to work with the concept of
Cartesian products.
Q
(a) Let I = R and Xt := R for t ∈ I. We claim that X := t∈I Xt is the set of all
functions f : R → R.. Details!
Q
(b) Let I = N and Xn := R. How do we visualize n∈I Xn ? It is the set of vertical lines
passing through the points (n, 0) in the x-axis of R2 .
Note that the product set is the set of all real sequences. This can be seen as in the
previous example.
(c) Let I := [0, ∞) and Xt := [0, t] for t ∈ R. How do we visualize the product? It is
the region {(x, y) ∈ R2 : 0 ≤ y ≤ x}.
Does the element (t2 )t∈I lie in the product? Note that if t = 2, then the x(t) = 4 ∈
/
[0, t] ≡ [0, 2].
Q
7.27 What we requite on X := i∈I Xi to make the projections πi (i ∈ I) continuous is the
smallest topology that contains

S := {πi−1 (Vi ) : Vi ∈ Ti , i ∈ I}.

This is the question we have already answered in Item 2. What is πi−1 (Vi )? An element
x = (xi ) ∈ i∈I Xi lies in πi−1 (Vi ) iff πi (x) = xi ∈ Vi . What about xj ’s for j 6= i? No
Q

47
 condition sis imposed on theseQcomponents, which means they could be any element of
Xj for j 6= i. Thus, πi−1 (Vi ) = j∈I Vj where Vj = Vi if j = i and Vj = Xj for j 6= i.
Can you visualize π1−1 ((2, 3)) where π1 : R × R → R is the projection onto the first factor?
Arguing along similar lines, we find that
Y
πi−1 (Vi ) ∩ πj−1 (Vj ) = {x = (xr ) ∈ Xr : xi ∈ Vi and xj ∈ Vj }
r∈I
Y
= Ur where Ur = Vr if r ∈ {i, j} and Ur = Xr otherwise.
r∈I

7.28 We apply the process of Item 2.toQ the problem posed in Item 27. Thus we arrive at the
definition of product topology on i∈I Xi as follows.
As a subbase for a topology on X, we take the set
( )
Y
S := Ui : where Ui = Xi for all but finitely many i and Ui is open in Xi .
i∈I

The basis for the product topology on X is finite intersections of elements from S.
In particular, G ⊂ X is open iff for every x ∈ G, there exists S1 , . . . , Sn ∈ S such that
x ∈ S1 ∩ · · · ∩ Sn ⊂ G.

Thus, G ⊂ X is open in the product topology iff for a given x ∈ G, there exists
Q
a finite subset F ⊂ I and open subsets Uj ⊂ Xj forQ j ∈ F such that x ∈ i Vi
/ F , Vj = Uj for j ∈ F and x ∈ i∈I Vi ⊂ G.
where Vi = Xi for i ∈

Q
7.29 Let ∅ 6= Ui Xi , be open in Xi for i ∈ I. Then U = i∈I Ui could never be open in X
unless I is finite.
Assume I is infinite. Let x = (xi ) ∈ U . If U were Qopen, then there exists a finite set
F ⊂ I, open sets Vj for j ∈ F such that x ∈ W = i Wi ⊂ U where Wi = Xi for i ∈ /F
and Wj = Vj for j ∈ F . Choose an r ∈ I \ F . Since Ur 6= Xr , there exists yr ∈ Xr \ Ur .
Consider z = (zi ) where zi = xi for i 6= r and zr = yr . Then z ∈ W but z ∈
/ U.
7.30 If I is finite, say, I = {1, 2, . . . , n}, then the basic open sets are of the form U1 × · · · × Un
where Ui is an arbitrary open set in Xi for each 1 ≤ i ≤ n.
7.31 Warning: If, at first, we defined finite products of topological spaces with basis as in the
last item, we would
Q be tempted to use the following collection as a basis for a topology
on the product i∈I Xi :
( )
Y
B := Ui : where Ui is an arbitrary open set in Xi .
i∈I

The topology given rise to by this basis is called the box topology. Evidently, this is finer
than the product topology.
The product topology on X is the smallest topology which makes all the pro-
jection maps πi continuous. We shall always use this topology on the product
sets.

48
7.32 We shall see how to visualize the subbasic and basic open sets of the product topology.
This will allow us to gain some geometric intuition. Pictures!

(a) Consider X × Y . We visualize this the first quadrant in R2 where X and Y are
represented by [0, ∞). Then any subbasic open set if of the form U × Y or X × V
where U ⊂ X and V ⊂ Y are open. We visualize this by a vertical strip of the from
(a, b) × [0, ∞) or as a horizontal strip of the from [0, ∞) × (c, d). Hence any basic
open set is represented by a rectangle of the from (a, b) × (c, d).
This can be extended to a finite product.
Q
(b) We now consider a countable product, say X = n∈N Xn . We visualize X as vertical
half-lines erected at (n, 0) ∈ R2 : A basic open set is therefore of the form half-lines at
all points except at finitely many n1 , . . . nk and at nj an interval of the form (aj , bj ).
Q
(c) Consider X := t∈R R. The product set can be identified with the set of functions
f : R → R. Each function can be represented by its graph in R × R. Fix a finite set
of points {t1 , . . . , tn } and a finite set of intervals (aj , bj ), 1 ≤ j ≤ n. Then the basic
open set corresponding to this data is R at all t ∈ / {tk : 1 ≤ k ≤ n} and (ak , bk )
if t = tk , 1 ≤ k ≤ n. Thus the elements in this basic set are functions such that
f (tk ) ∈ (ak , bk ). We can visualize this by means of their graphs.

7.33 To have a feeling for the product topology, we look at the following results/questions:

(a) The product of Hausdorff spaces is Hausdorff.

Q
Easy. if x = (xi ) and y = (yi ) are in X = i Xi are distinct elements, then there
exists j ∈ I such that xj 6= yj . Since Xj is Hausdorff, there exist Uj andQVj open
in XjQwith xj ∈ Uj , yj ∈ Vj and Uj ∩ Vj = ∅. Consider the open set U = Ui and
V = Vi where Ui = Xi = Vi for i 6= j and at j the disjoint open sets Uj and Vj .
Then U and V separate x and y.
(b) A sequence (xk ) in the product space is convergent to an element x iff it converges
coordinate-wise, that is, iff πi (xk ) → πi (x) for each i ∈ I.
To avoid confusion with indices, we use the Greek alphabet to denote elements of
the index set I.
Q
Let xk ∈ X = α∈I Xα converge to x. Fix β ∈ I. Let xkβ := πβ (xk Q ) and xβ :=
πβ (x). Let Uβ 3 xβ be open. Consider the subbasic open set U = Uα where
Uα = Xα for α 6= β and Uα = Uβ for α = β. Then x ∈ U and since xk → x, there
exists N ∈ N such that xk ∈ U for k ≥ N . It follows that xkβ ∈ Uβ for k ≥ N .
Hence xkβ → xβ .
Note that this can also be proved using Item 29. Apply it to the sequence (xn ) and
to the function f = πα .
Q
To prove the converse, let U 3 x be a basic open set, say, of the form U = Uα
where Uα = Xα for α ∈ / F ⊂ I, a finite subset and Uβ are open subsets in Xβ ,
β ∈ F . Since xkα → xα , it follows that there exists Nβ such that for k ≥ Nβ , we
have xkβ ∈ Uβ for β ∈ F . Let N = max{Nβ : β ∈ F }. We claim that xk ∈ U for
k ≥ N . For, any α ∈ / F , πα (xk ) = xkα ∈ Uα = Xα and for β ∈ F , since k ≥ N ,
πβ (xk ) = xkβ ∈ Uβ .
Q
Thus, the convergence in i∈I Xi is ”coordinate-wise convergence.”

49
 Q
(c) Let Di be dense in Xi for each i. Then D := i∈I Di is dense in X. Details!

Using our standard notation, it suffices to prove that D ∩ πF−1 (VF ) is nonempty for
a any finite subset of I and Vi open sets in Xi for i ∈ F . Since Vi is nonempty open
set in Xi and Di is dense in Xi , there exists zi ∈ Di ∩ Vi for i ∈ F . For j ∈ I \ F ,
we select an element yj ∈ Dj . Then the element (xr ) ∈ X, where xr = yr for r ∈ /F
−1
and yr = zr for r ∈ F lies in D ∩ πF (VF ).
Q Q
(d) Let Ai ⊂ Xi and A := i∈I AQ i . Then A = i∈I Ai . In particular, if each Ai is
closed, then the product A := i∈I Ai is closed in the product space X. Contrast
this with Item 29.
The proofs of this runs almost like the earlier three items. We need to show that
x = (xi ) is a limit point of A iff each xi is a limit point of Ai . Let x be a limit points
of A. Let Ui 3 xi . We need to show that Ui ∩Ai 6= ∅. Let U := πi−1 (Ui ). Then U 3 x
and is an open set. Hence there exists z ∈ U ∩A. Clearly, π(z) ∈ Ui ∩πi (A) = Ui ∩Ai .
Conversely, let xi be a limit point of Ai for i ∈ I. Let x = (xi ) and U 3 x be an open
set. We need to show that U ∩ A 6= ∅. Since x ∈ U and U is open in the product
topology, there exists a finite set F ⊂ I and a finite collection Vi of open sets in Xi
for i ∈ F such that x ∈ πF−1 (VF ) ⊂ U . Note that xi ∈ Vi for i ∈ F . Since xi is a
limit point of Ai in Xi , there exists zi ∈ Vi ∩ Ai for i ∈ F . Let y = (yj ) be defined
as yj ∈ Aj is arbitrary and yj = zj if j ∈ F . Then y ∈ πF−1 (VF ) ∩ A.
Q
(e) Let Xi be a discrete space for each i. When is i∈I Xi is discrete?
(f) Let X, Y be metric spaces. We have a product metric on the product X × Y given
by δ((x1 , y1 ), (x2 , y2 )) := max{d(x1 , x2 ), d(y1 , y2 )}. Thus we have two topologies on
X × Y , namely, the topology induced by the metric δ and the product topology (got
out of the metric topologies on X and Y ). We claim that these two topologies are
the same. Details!

It suffices to show that the identity map of X × Y is a homoeomophism of these two

spaces. Let I : (X × Y, Tδ ) → X × Y, TX × TY ) be the identiy map.
Later, we shall see an easy proof. Reference?

Investigate whether the converses (wherever applicable) are true.

Optional: Investigate how many of them are true if we equip X with the box topology.
Note that if D is dense in (X, T2 ) and if T1 is another topology on X with T1 weaker than
T2 , then D is dense in (X, T1 ).

7.34 This is a continuation the theme of Item 33.

Q
Let f : Y → i∈I Xi be a map from a topological space Y to the topological space
(a) Q
i∈I Xi with product topology. Then f is continuous iff each fi , is continuous,
where fi = πi ◦ f for i ∈ I,
If f is continuous, then fi is the composition of two continuous functions and hence
is continuous.
Assume that fi are continuous. To prove that f is continuous, let U be an open
subset of X. Since U is a union of basic open sets, it is enough to show that f −1 (B)
is open for any basic open set. But B is of the form ∩j∈F πj−1 (Uj ) where F ⊂ I is
finite, and Uj is open subset of Xj , j ∈ F . Since taking inverse images behaves well

50
 with set-theoretic operations, it suffices to show that f −1 (πj−1 (Uj )) is open in Y for
any j ∈ I and any Vj open in Xj . But

f −1 (πj−1 (Uj )) = (πj ◦ f )−1 (uj ) = fj−1 (Uj ), (1)

which is open in Y by the continuity of fj . (We used Item 2c in (1).)

An application: Let X = R2 . Let TP be the product topology on R × R and T be
the standard topology. Then the identity map is a homeomorphism of (R × R, TP )
onto (R2 , T ). Hence TP = T . Details!

(b) Let f : X → Y be continuous. Let Graph(f ) := {(x, f (x)) : x ∈ X} be the graph

of f . Let Graph(f ) ⊂ X × Y be endowed with the subspace topology. Then X is
homeomorphic to Graph(f ).
Consider ϕ : X → Graph(f ) given by ϕ(x) = (x, f (x)). Then ϕ is continuous as a
map from X to X × Y by the last sub-item. By Item 19, ϕ : X → Graph(f ) is also
continuous. Clearly ϕ is a bijection and its inverse ϕ−1 : Graph(f ) → X is given by
ϕ((x, f (x)) = f (x). That is, ϕ−1 is the restriction of the projection of X ×Y onto its
first factor. By the definition of product topology, the projection map is continuous
on X × Y . Its restriction to Graph(f ) is continuous by Item 18.
(c) The map x 7→ (x, y0 ) of X into X × Y is a homeomorphism of X with X × {y0 }
with the subspace topology inherited from X × Y .
Argue as in the last sub-item.
(d) Let X, Y be topological spaces. Let A ⊂ X and B ⊂ Y . Let TA denote the subspace
topology on A induced from the topology on X etc. Let TA × TB (respectively,
TX × TY ) denote the product topology on A × B, (respectively, the product topology
on X × Y ). Let TA×B denote the subspace topology on A × B considered as a subset
of X × Y . Then TA × TB = TA×B .
Easy if you know how to set up a notation which keeps your head clear. Let W ∈
TA × TB . Let x = (a, b) ∈ W . There exist Ua ∈ TA and Vb ∈ TB such that
x = (a, b) ∈ Ua × Vb ⊂ W . Since Ua ∈ TA , there is a U ∈ TX such that Ua = U ∩ A.
Similarly, Vb = V ∩ B. Hence

x = (a, b) ∈ (U ∩ A) × (V ∩ B) = (U × V ) ∩ (A × B) ⊂ W.

Since (U × V ) ∩ (A × B) ∈ TA×B , it follows that W ∈ TA×B .

Reverse inclusion is proved similarly. Let W ∈ TA×B . Then there exists W 0 ∈
TX × TY such that W = W 0 ∩ (A × B). Let x = (a, b) ∈ W so that x ∈ W 0 . We can
find U 0 ∈ TX and V 0 ∈ TY such that x = (a, b) ∈ U 0 × V 0 ⊂ W 0 . Hence it follows
that

x = (a, b) ∈ (U 0 ∩ A) × (V 0 ∩ B) = (U 0 × V 0 ) ∩ (A × B) ⊂ W 0 ∩ (A × B).

Since (U 0 ∩ A) × (V 0 ∩ B) is a basic open set in TA × TB , it follows that W ∈ TA × TB .

(e) Let ∆(X) denote the diagonal {(x, x) : x ∈ X × X} ⊂ X × X. Then X is Hausdorff
iff ∆(X) is closed in X × X. (We leave this as a very easy exercise.)
Q
7.35 A frequently used observation. The projection maps πi : i∈I Xi → Xi are open.
Q
Observe that for any set A := Ai , with Ai ’s nonempty we have πi (A) = Ai .

51
 Q
Let xj ∈ πj (A). Then there exists x ∈ A such that πj (x) = xj . Since x ∈ Ai , we
have xj ∈ Aj . Let aj ∈ Aj . Since Ai ’s are non-empty, there exist xi ∈ Ai , i 6= j.
Consider (zi ) where zi = xi for 6= j and zj = aj . Then z ∈ A and we have πj (z) = aj .
Q
Let U ⊂ Xi be open. Let Gi := πi (U ) and ai ∈ Gi . Let a ∈ U be any pointQ such that
−1
πi (a) = ai . Since a ∈ U and U is open there
Qexists a basic open set πF (VF ) = i Vi such
that a ∈ πF−1 (VF ) ⊂ U . We have Vi = πi ( i Vi ) ⊂ πi (U ). Hence ai ∈ Vi ⊂ Gi . Thus Gi
is open .
Q
7.36 The projection maps πi : i∈I Xi → Xi are not closed. For example, the projection of
a rectangular hyperbola on to the x-axis is R∗ . Let F := {(x, y) ∈ R × R : xy = 1}.
Then F is closed (why?) in R2 with the standard topology (which is same as the product
topology, why?). The projection of F on the first component, namely, π1 (F ) = R∗ is
open in R and not closed in R.
7.37 Let X and Y be X
Q sets. The set Y of functions from X to Y can be considered Q as the
product space x∈X Yx where Yx = Y for x ∈ X via the map ϕ(f ) = (f (x)) ∈ x∈X Yx .
Thus, what the last line of Item 33b says is that the if we use ϕ to transfer the topology
to Y X , then a sequence of functions (fn ) in Y X converges to a function f ∈ Y X iff
fn (x) → f (x) for each x ∈ X, that is, convergence in Y X is pointwise convergence.
Because of this, product topology is known as topology of pointwise convergence.
7.38 Is
Q the product of first/second countable spaces first/second countable? We show that
t∈R R is not first countable. Note that R is second countable. The key idea comes from
the last item. It may be worthwhile to review Item 32c.
The product space is the set of functions from R to R and the convergence is pointwise
convergence. How does local base at (the constant function) 0 look like? Fix a finite
subset F ⊂ R and k ∈ N. Then a typical element of the local base is of the form

UF,k := {f : R → R : f (t) ∈ (−1/k, 1/k), t ∈ F }.

How many such basic open sets are there? As many as in the set F × N where F is the
set of finite subsets of R, that is, the cardinality of F × N.
Q It is intuitively clear and we
expect that this space has no countable local base at 0 ∈ t∈R) R.
We would like to translate these ideas into a rigorous argument. An obvious method of
attack is to prove this by contradiction. Consider the set

E := {f ∈ RR : f (t) = 0 or 1 & {t ∈ R : f (t) = 1} is finite .}

We claim the constant function 0 is a limit point of E. For, let UF,k be a basic open set
containing 0. Then the function f (t) = 1 for t ∈
/ F and f (t) = 0 for t ∈ F lies in UF,k ∩ E.
If the product topology were first countable, then there exists a sequence (fn ) in E such
that fn → f in the product topology. Let Fn := {t : fn (t) = 0}. Let A := ∪n Fn . Then
A is countable. Observe that for t ∈/ A, fn (t) = 1 and therefore f (t) = lim fn (t) = 1 for
t∈/ A. This is a contradiction
7.39 Contrast Item 42b with the following. Let E be any set and let B(E, R) denote the set
of all bounded real valued functions on E. If we endow this vector space with the norm
kf k∞ := supx∈E |f (x)|, then fn → f in this normed linear space iff fn → f uniformly on
E. (This is Item 28b.)

52
7.40 Refer to Item 38. Each of the factors in RR is a metric space. But the topology on RR is
not first countable and hence there cannot be any metric d on the product RR which will
induce the product topology.

7.41 In most
Q of the examples above, we looked at subsets of the product set X which are of the
form i∈I Ai , where Ai ⊂ Xi . You should be aware that not all subsets of X are of this
form. For example, S := {(x, y) ∈ R × R : x2 + y 2 = 1}, D := {(x, x) ∈ R2 : x ∈ R × R}
are not a product of subsets of R.
For, if D = A × B, then (1, 1), (2, 2) ∈ D = A × B. Hence 1, 2 ∈ A, 1, 2 ∈ B and hence
(1, 2) ∈ A × B = D!

7.42 It is equally important to recognize product spaces in disguise. The following are very
typical of this situation.

(a) Define a topology on the set S of all real sequences such that a sequence (xk )
in S converges to x ∈ S iff the xkn → xn as n → ∞ for all k where xk =
(xk1 , xk2 , . . . , xkn , . . .). (Convergence = Coordinate-wise convergence).
(b) Let X denote the set of all real valued functions on R. Define a topology on X such
that a sequence (fn ) of functions in X converge to a function f ∈ X iff fn (x) → f (x)
for all x ∈ R. (Convergence = point-wise convergence of functions.)
Q
(c) Let I = N and Xi = {0, 1} for i ∈ N. Then the product space X := i∈N Xi “is
isomorphic to” the Cantor set. We have to introduce concepts and develop some
more theory to explain this satisfactorily.

7.43 A problem similar to Item 33a: Let X be any set and F be a collection of real valued
functions on X with the property that for any pair of distinct points x, y ∈ X, there
exists f ∈ F such that f (x) 6= f (y). Then the smallest topology on X which makes all
the functions in F continuous is Hausdorff.

53
7.44 Let (X, TX ) be a topological space and Y , a set. Let f : X → Y be a map. What is the
largest topology on Y which makes f continuous?
Let T be any topology which makes f : (X, TX ) → (Y, T ) continuous. Then for any
V ∈ T , we should have f −1 (V ) ∈ TX . This suggests that we consider the collection
TY := {V ⊂ Y : f −1 (V ) ∈ TX }. It is easy to see that this is the largest topology on Y
ensuring the continuity of f .

7.45 An easy example: With the notation above, assume that TX is discrete. Then TY is the
discrete topology on Y .

7.46 Let X be a topological space and ∼ is an equivalence relation on X. Let Y := X/ ∼ be

the quotient set, that is, the set of all equivalence classes. Let π : X → Y be the quotient
map π(x) := [x], the equivalence class of x. The largest topology on Y with respect to
which π is continuous is called the quotient topology on Y . It is given by

V : π −1 (V ) is open in X .


7.47 Two of the standard ways in which an equivalence relation is prescribed on a topological
space are:
(i) when a group acts on the set X underlying the topology and
(ii) when we have a map from X onto a set Y .
In the first case we say p ∼ q iff there exists a group element g such that g · p = q. This
is same as requiring that p and q lie in the same orbit of the group.
In the second case, we say p ∼ q iff f (p) = f (q).

7.48 It is a good idea at this juncture to read my article “Generation Topologies — A Unified
View of Subspace, Product and Quotient Topologies”.

7.49 Now that we have created these new objects, how do we work with them? The answer is
provided by the so-called universal mapping properties.

Theorem 9 (Universal Mapping Property).

(1.) Let f : X → Y be a map from a set X to a topological space Y . Let X be given the
topology generated by f . Then a function h : Z → X is continuous iff f ◦ h : Z → Y is
continuous.
(2.) Let g : X → Y be a map from a topological space X to a set Y . Let Y be endowed
with the topology generated by g. Then a map h : Y → Z is continuous iff the map
h ◦ g : X → Z is continuous.
(3.) Let πi : X → Xi be maps from the set X to topological spaces Xi for i ∈ I. Let X be
given the topology generated by πi ’s. Then a map h : Y → X is continuous iff the maps
πi ◦ h : Z → Xi are continuous.

Proof. Let us prove (1) as a sample, as the proofs are all similar and easy. To prove the
nontrivial part, let us assume that the map f ◦ h is continuous. Let U ⊂ X be open. We

54
 need to show that h−1 (U ) is open in Z. By the very definition of the topology on X,
there exists an open set V ⊂ Y such that U = f −1 (V ). Now

h−1 (U ) = h−1 (f −1 (V )) = (f ◦ h)−1 (V ),

which is open by the continuity of f ◦ h.

To prove (2), let W be open in Z. We need to show that h−1 (W ) is open in Y . By the
continuity of h ◦ g, the set

(h ◦ g)−1 (W ) = g −1 (h−1 (W ))

is open in X. By the definition of topology on Y , the subset h−1 (W ) is open.

To prove (3), we observe that it is enough to show that h−1 (U ) is open for

U ∈ S := {U : πi−1 (Vi ) for an open Vi ⊂ Xi for some i ∈ I}.

(Prove this. Refer to Item 29.) If U = πi−1 (V ), we have h−1 (U ) := h−1 πi−1 (U ) =

(πi ◦ h)−1 (U ) is open, by the continuity of πi ◦ h.

7.50 The most important thing to observe in the theorem is that the problem of establishing
continuity of a map either from or to a newly constructed space is reduced to showing
the continuity of a ‘natural composite map’ between the ‘known spaces’. Go back to the
statement and understand this remark. Also, go through the next few item.

7.51 Let us explicate the theorem in the concrete situations.

• If f : X → Y is the inclusion map (that is, the restriction of the identity of Y to X)

of a subset X into a topological space Y , then (1) of the theorem says that a map
h : Z → X is continuous iff we think of h as a map form the space Z to Y (taking
values only in X) is continuous.
• Let X be a topological space, ∼ be an equivalence relation on X and Y = X/ ∼ be
the quotient set with the quotient map π : X → Y . Then a map h : Y → Z form the
quotient space Y to a space Z is continuous iff the map h ◦ π : X → Z is continuous.
Q
• Let X := i∈I Xi is the Cartesian product of topological spaces with the product
topology. Then a map h : Z → X can be written as h(z) = (hi (z)) where the
coordinate maps hi (z) := πi ◦ h(z). Thus h : Z → X is continuous iff the coordinate
maps hi are continuous.

7.52 The universal mapping property is the most important to deal with the newly constructed
topologies. For instance, in the case of quotient space Y = X/ ∼, giving a map h : Y → Z
is the same as giving a map h̃ : X → Z which is constant on the equivalence classes. Hence
the continuity of h is the same as that of h̃. As a concrete example, let X = R and x ∼ y
iff x−y ∈ Z. Let S := {z ∈ C : |z| = 1} be with the subspace topology. We have a natural
map h̃ : X → S given by h̃(t) := e2πit . Then h̃ gives rise to map h : Y → S which is a
bijection from well-known properties of the exponential map. Also, h is continuous since
h̃ is so. Since any continuous map from a compact space to a Hausdorff space is a closed
map, h is a homeomorphism. Thus Y is the circle in C! For more such applications, see
my article on Quotient spaces.

55
7.53 Let X = R2 and we let the additive group Z act on X as follows: n · (x, y) = (x + n, y).
Then (x1 , y1 ) ∼ (x2 , y2 ) iff x1 − x2 ∈ Z. Consider the set E : −{(x, y) ∈ R2 : 0 ≤ x ≤
1, y ∈ R}. Note that any (x, y) ∈ R2 is equivalent to a unique point (a, y) ∈ E if x ∈ /Z
and (x, y) ∼ (0, y) ∼ (1, y) if x ∈ Z. Thus if we identify the vertical lines x = 0 and
x = 1, we may be tempted to believe that the quotient space is homeomorphic to the
cylinder S := {(x, y, z) ∈ R3 : x2 + y 2 = 1}. Can we at least get a continuous bijection
of Q := X/ ∼ onto S? Consider the map f ([x, y]) := (cos 2πx, sin 2πx, y). It is easy to
check that this is well defined bijection. To check continuity, we use UMP. The natural
composite map is R2 → S given by (x, y) 7→ (cos 2πx, sin 2πx, y), which is continuous.

7.54 Real Projective Spaces: An important class of spaces which arise as quotient spaces is
the class of projective spaces.
We shall construct them in three different ways. Let X := Rn+1 \ {0} be the set of
nonzero vectors in Rn+1 with the subspace topology. Let the group R∗ act on it via
(t, x) 7→ tx. Note that any group action gives rise to an equivalence relation, which in
this case translates as follows: x ∼ y iff there exists t ∈ R∗ such that y = tx. The orbits
of the group action are the equivalence classes. Then the n-dimensional projective space
(over R), also called n-dimensional real projective space Pn (R) is the set of orbits with
the quotient topology.
There is a second description of Pn (R). Observe that any orbit of the last paragraph
intersects the n-dimensional sphere S n := {x ∈ Rn+1 : kxk = 1} at exactly two points. Details!

Thus the equivalence relation on X := Rn+1 \ {0} induces an equivalence relation on

n n
Y := S : Given two points u, v ∈ S , we say that u ∼ v iff v = ±u. This equivalence is
again induced by the Z2 action on S n : 0.u = u and 1.u = −u. (It may be better to think
of Z2 as the multiplicative group of square roots of unity to ‘appreciate’ this action!) The
quotient space of Y is also denoted by Pn (R). There is an obvious bijection between the
quotient set X and that of Y which turns out to be a homeomorphism. This will be
proved later.
A third description is Pn (R) is the collection of all lines passing though the origin in
Rn+1 . This does not arise by quotient construction. (Why?) However, we have an
obvious bijection of this set with the quotient set of X of the first description. A line
through the origin is determined as soon as we know a nonzero vector on it and any two
such different by a (necessarily nonzero) scalar multiple. Now using this bijection, we
transfer the topology on the X/ ∼ to the set of lines passing through the origin. (See
Item 1.)

7.55 We have an obvious map ϕ : S n → Pn (R) defined by x 7→ [x]. This map is an onto
continuous map. For, it is the composite of the maps x 7→ x from S n → Rn+1 \ {0}
followed by the quotient map x 7→ [x]. Note that the equivalence relation induced by f
is the one we defined on S n . Hence this map induces a homeomorphism of the quotient
of S n relative to the equivalence onto Pn (R).
See how we established the continuity of S n → Pn (R). It cannot be done using UMP.

56
7.56 More examples of homeomorphisms. Recall that a map f : X → Y between two
topological spaces is a homeomorphism if (i) f is bijective, (ii) f is continuous and (iii)
f −1 : Y → X is continuous.
(a) Any f : R → R of the form f (x) = ax for a nonzero a ∈ R is a homeomorphism.
(b) Fix v ∈ Rn . Then x 7→ x + v is a homeomorphism of Rn .
(c) Fix a ∈ R nonzero. Then the map x 7→ ax is a homeomorphism of Rn . (This and
the last item can be generalized to any normed linear space.)
(d) f : R → R given by f (x) = x3 is a homeomorphism.
(e) Any linear isomorphism of Rn is a homeomorphism. (In general this is not true for
a normed linear space, as such a linear isomorphism need not even be continuous!)
(f) [a, b] ' [0, 1]. More generally, [a, b] ' [c, d].
(g) (−1, 1) ' R.
(h) (0, 1] ' [1, ∞).
(i) [0, 1) ' (0, 1].
(j) Any two discrete spaces are homeomorphic iff they have the same cardinality.
(k) Can Q be homeomorphic to Z with the subspace topologies (induced from R)?
(l) Is N ' Z with the subspace topologies (induced from R)?
(m) If two metric spaces are isometric, then they are homeomorphic.
(n) B(0, 1) ' Rn .
(o) S n \ {en+1 } ' Rn . (Refer to Example 3.3.5 on Page 73 of my book, “Topology of
Metric Spaces”, 2nd edition.)
(p) f : X → Y continuous. Then the graph of f with the subspace topology of X × Y
is homeomorphic to X. Applications:
i. R is homeomorphic to the parabola y = x2 .
ii. R∗ is homeomorphic to the hyperbola xy = 1.
(q) The product space [−1, 1] × S 1 is homeomorphic to a cylinder.
(r) The annulus {p ∈ R2 : 1 ≤ kpk ≤ 2} is homeomorphic to the cylinder {(x, y, z) ∈
R3 : x2 + y 2 = 1, 1 ≤ z ≤ 2}.
(s) Let f : X → Y be a homeomorphism and let A ⊂ X. Then f induces homeomor-
phism between A and f (A) (and between X \ A and f (X \ A)).
This is a very useful fact. Typical ways of applying this are:
i. [0, 1) is not homeomorphic to (0, 1).
ii. R is not homeomorphic to R2 .
Both these results need connectedness at least in disguise, but can be proved at this
stage using the intermediate value theorem.
For example, let, if possible, ϕ : [0, 1) → (0, 1) be a homeomorphism. Then we have
a homeomorphism, again denoted by ϕ : (0, 1) → (0, 1) \ {ϕ(0)}. Then f : (0, 1) \
{ϕ(0)} → R defined as f (x) = x is a continuous function. Let c, d ∈ (0, 1) be
such that c < ϕ(0) < d. Let ϕ(a) = c and ϕ(b) = d. Then a, b ∈ (0, 1). The
function f ◦ ϕ : (0, 1) → R is continuous on the interval (0, 1) such that f ◦ ϕ(a) = c,
f ◦ ϕ(b) = d. But it misses the point ϕ(0) ∈ (c, d). This contradicts the intermediate
value theorem.

57
 (t) Homeomorphism between conic sections:
i. A circle is homeomorphic to an ellipse.
ii. A parabola is homeomorphic to a line.
iii. A (rectangular) hyperbola is homeomorphic to R∗ .
iv. A pair of intersecting lines is not homeomorphic to any of the other conic sec-
tions. More generally, a circle, a parabola, a hyperbola and a pair of intersecting
lines are mutually non-homeomorphic. (We shall see a proof of this later. Mean-
while you may try to prove along this along the lines of a proof of Item 56(s)i.)
(u) Any k-dimensional subspace W of Rn with subspace topology is homeomorphic to
Rk . Let {wi : 1 ≤ i ≤ k} be a basis of W . Then the map f : Rk → Rn given by
f (a1 , . . . , ak ) := a1 w1 + · · · + ak wk is continuous. For, write wi := wi1 e1 + · · · + win en
with respect to the standard basis of Rn . What are the components fi , (1 ≤ i ≤ n),
of f ? Why is the inverse continuous? Details!

7.57 Let X and Y be metric spaces. A map f : X → Y is said to be distance preserving if for
any x1 , x2 ∈ X, we have d(f (x1 ), f (x2 )) = d(x1 , x2 ). Any distance preserving map f is a
homeomorphism of X onto f (X).
A distance preserving map need to be onto. For instance, consider X = Y = [0, ∞) with
the standard metric. Then f (x) := x = 1 maps X onto [1, ∞) .

7.58 An isometry of between metric spaces is an onto distance preserving map.

Any orthogonal/unitary linear transformation of a finite dimensional (real/complex) inner
product space is an isometry.

7.59 In any normed linear space , any two open balls are homeomorphic. Recall that B(x, r) =
x + rB(0, 1) and B(y, s) = y + sB(0, 1).
The map u 7→ x + ru is a homeomorphism of B(0, 1) onto B(x, r).

7.60 In any normed linear space , any open ball is homeomorphic to the entire space. Enough
to show that B(0, 1) is homeomorphic to the normed linear space X. Any nonzero x is of
the from x = tu, 0 ≤ t < 1. Can we map [0, 1) to [0, ∞) homeomorphically? If yes, then
the finite radial line tu, 0 ≤ t < 1 will be mapped to the line segment emanating from 0
in the direction of u.

7.61 In Rn , we have B∞ [0, 1] ' B2 [0, 1]. For a complete proof, see Example 3.3.6 on Page 73
of my book on Metric Spaces.

7.62 Rm ' Rn iff m = n. More generally, if U ⊂ Rm and V ⊂ Rn are non-empty open sets
which are homeomorphic, then m = n. This is a highly nontrivial result, known as the
invariance of domain, We shall not prove this in our course!

7.63 Another most important way of proving that a map is a homeomorphism is to use the
following result which you might have seen in TYBSc.
A bijective continuous map from a compact metric space to another metric space is a
closed map and hence is a homeomorphism.
We shall see a more general result later in Item 22b.

58
8 Compact Spaces

8.1 Let X be a topological space and A ⊂ X. We say that a collection {Ui : i ∈ I} of subsets
of X is an open cover of A if (i) each Ui is an open subset of X and (ii) A ⊆ ∪i∈I Ui .
Given an open cover {Ui : i ∈ I} of A, by a subcover of A, we mean a subfamily
{Ui : i ∈ J} for some subset J ⊂ I such that {Ui : i ∈ J} is an open cover of A.
We say that the subcover is proper if J is a proper subset of I.
We say that the given open cover admits a finite subcover, if J (in the notation above) is
a finite set.
For example, {(a, b) : a, b ∈ R, a < b} is an open cover of R. The collection {(a, b) : a, b ∈
Q, a < b} is a subcover of R.
Let X be a space with discrete topology. The family {A, X \ A} is an open cover for X
where A is a nonempty proper subset of X with no proper subcover.
Let X be a space with discrete topology. Let ∅ 6= A ( X. Consider {A} ∪ {{x} : x ∈
X \ A}. Then this collection is an open cover of X with no proper subcover.

8.2 Let X = (0, 1) ⊂ R be with the metric/subspace topology. Let Un := (0, n−1
n ). Then
{Un : n ∈ N} is an open cover of (0, 1). (Why?) Can you think of a proper subcover?
Can such a subcover be finite?

8.3 Examples of open covers:

(a) “Non trivial” open covers of R:

i. {(−n, n) : n ∈ N}.
ii. {(−∞, n) : n ∈ N}.
iii. {(−r, 2r) : r ∈ Q+ }.
Do they admit finite subcovers? proper subcover?
(b) Nontrivial open covers of (−1, 1).
(c) In any metric space, {B(x, rx ) : x ∈ X} is an open cover where rx > 0 is pre-assigned
for x ∈ X. Such a cover arises “naturally” in the following way: Let f : X → R be a
continuous function. Let ε > 0 be given. Given x ∈ X, by the continuity of f at x,
there exists rx > 0 such that for all y with d(x, y) < rx , we have |f (x) − f (y)| < ε.
The collection {B(x, rx ) : x ∈ X} is an open cover of X.
(d) Given a Hausdorff space with at least two elements, think of a nontrivial open cover.
(e) Can you say something specific about any open cover of R with outcast topology?
(f) Give an open cover of R with VIP topology which has no proper subcover.
(g) Give a non-trivial open cover of R with lower limit topology. For example, {[a, b) :
a < b}. Think of a non-trivial open cover which does not admit any proper subcover.
(h) Open covers of S n :
i. Rn+1 \ {0}. (This is a trivial open cover!)
ii. U = Rn+1 \ {N } and V := Rn+1 \ {S}, where N, S are north and south poles of
the sphere S n respectively.
iii. Ui± := {x ∈ Rn+1 : xi ≶ 0}, 1 ≤ i ≤ n + 1.

59
 (i) Open cover for a discrete space X. Let A ⊂ X. Look at {A} ∪ {{x} : x ∈
/ A}. This
is an open cover of X. What if A = ∅?
(j) Open cover for an uncountable space with co-countable topology. If X = R with
co-countable topology and U = R \ N, I would think of defining Un := U ∪ {n} or
Vn := U ∪ {k : 1 ≤ k ≤ n}. Then {Un : n ∈ N} and {Vn : n ∈ N} are open covers of
R. Do they admit any proper subcovers of R?
What will you do for an countable set X with co-countable topology?
(k) Open cover for a set with co-finite topology.
8.4 A subset A of a topological space X is said to be compact if given any open cover {Vi :
i ∈ I} of A where each Vi is open in A, we can find a finite subcover. We say that X is a
compact space if X is a compact subset of X.
8.5 Given an open cover {Ui : i ∈ I} of A by means of open subsets of X, then we have a
“natural” open cover {Vi : i ∈ I} of a subset A ⊂ X by means of subsets of A which are
open in A and conversely. (Note the indices. “Naturality” does not mean that given Vi ’s,
the Ui ’s are unique!)
The significance of this observation is that when dealing with compactness of a subset
K ⊂ X we may either work an open cover of K by means of open subsets in X or by sets
open in K. See Items 8, 11, 11 where this observation is exploited.
8.6 Examples of compact sets.
(a) A finite subset of any space is compact. In particular, the empty set is compact.
(b) An indiscrete space is compact.
(c) A discrete space is compact iff it is finite.
(d) R, Q and Z are not compact.
(e) The intervals of the form (a, b), [a, b), (a, b], any infinite interval are not compact.
(f) R with lower limit topology TL is not compact. Go through the proof. Can you
make a general principle of which this is a special case?
Let T1 and T2 are two topologies on the same set X. Assume that T1 ≤ T2 . Then
if (X, T1 ) is not compact, then (X, T2 ) is not compact and (X, T2 ) is compact, so is
(X, T1 ).
This is reminiscent of the comparison test for infinite series of positive terms.
(g) Any open ball in Rn (or in any normed linear space ) is not compact.
(h) Rn is not compact.
(i) Any closed and bounded interval [a, b] ⊂ R is compact.
Let {Ui : i ∈ I} be a collection of open sets in R such that [a, b] ⊂ ∪i Ui . Consider
the set
E := {x ∈ [a, b] : ∃ a finite set Fx ⊂ I such that [a, x] ⊂ ∪j∈Fx Uj }.
If a ∈ Ui , then there exists ε > 0 such that x ∈ (a − ε, a + ε) ⊂ Ui and also a + ε < b.
Then a + ε/2 ∈ E. Let c = l.u.b. E. (Why does it exist?) We claim c ∈ [a.b], c = b
and c = b ∈ E. If c < b, then c ∈ Uj . By the argument above, there exists ε > 0
such that E 3 c + ε/2 < b, a contradiction. Hence c = b. Repeat the same argument
to conclude b ∈ E.

60
 (j) R with VIP topology is not compact.
(k) R with outcast topology is compact.
(l) Any set with co-finite topology is compact.
(m) An uncountable set with co-countable topology is not compact. See Item 3j.
(n) A finite union of compact sets is compact.
(o) The intersection of two compact sets need not be compact. See, however, Item 8.
Consider Z with the discrete topology. Let {±∞} be two distinct elements not in
Z. Let X = Z ∪ {±∞}. We say a subset U ⊂ X is open if either (i) U ⊂ Z or if
either of ±∞ lies in U ⊂ X, then both the elements lie in U and X \ U is finite.
It is easy to verify that this defines a topology on X. The sets A := Z ∪ {∞} and
B := Z ∪ {−∞} are compact but their intersection Z is not compact. Note that
neither A nor B is closed.

8.7 A closed subset K of a compact space X is compact.

Let {Ui : i ∈ I} be an open cover of K. To exploit the compactness of X, we need an
open cover of X. Clearly, if we add the open set X \ K to the given open cover of K, we
end up with an open cover, say, U of X. Let U0 be a finite subcover of X. It is possible
U0 contains X \ K. In any case, U0 \ {X \ K} is a finite subcover of K.

8.8 Let K ⊂ X be a compact subset of X. Is K closed in X?

That is, is X \ K open? The only way of doing this it for each x ∈ X, to find an open set
Ux 3 x such that Ux ∩ K = ∅. We also need to exploit the compactness of K. That is,
we need to find an open cover of K (via open sets of X), members of which do not have
x. This suggests that we may require X to be Hausdorff. (All these were arrived at by
students!) We have the following result.
In a Hausdorff space a compact subset is closed and hence the intersection of compact
sets is compact in a Hausdorff space.
Let X be Hausdorff and K ⊂ X be compact. We shall show that the complement
K c := X \ K is open. Given p ∈ K c , we need to show that there exists an open set Up 3 p
with Up ⊂ K c . We need to exploit Hausdorffness of X and the compactness of K. This
means we need to generate an open cover of K. For any q ∈ K, we have disjoint open
sets Vq 3 q and Upq 3 p. Hence {Vq : q ∈ K} is an open cover of K and hence there exists
a finite set {q1 , . . . , qn } of K such that K ⊂ ∪ni=1 Vqi . Let Up := ∩ni=1 Upqi . Then Up 3 p is
an open set and it lies in K c . Thus, K c = ∪p∈K c Up is open. Picture!

8.9 Note that the a proof in the last item establishes the following result.
Let X be a compact Hausdorff space. Let K ⊂ X be closed and x ∈
/ K. Then there exist
disjoint open sets Ux 3 x and UK ⊃ K.
A space X is said to be regular if K is closed in X and x ∈
/ K, there exist disjoint open
sets Ux 3 x and UK ⊃ K.
Hence a compact Hausdorff space is regular.
Question: In the result about a compact Hausdorff space X, can we replace x by a closed
set L disjoint from K?

61
8.10 Let (X, d) be a metric space. We say that A ⊂ X is bounded if there exist x0 ∈ X and
r > 0 such that A ⊂ B(x0 , r). The following are easily seen results about this concept:

(a) A is bounded iff for every x1 ∈ X, there exists R > 0 such that A ⊂ B(x1 , R).
Easy. Observe
d(a, x1 ) ≤ d(a, x0 ) + d(x0 , x1 ) < r + d(x0 , x1 ).
Hence let R = r + d(x0 , x1 ).
(b) Let (X, k k) be an normed linear space . Show that A ⊂ X is bounded iff there
exists M > 0 such that kxk ≤ M for all x ∈ A. Easy. A ⊂ B(0, M ) for some M > 0
by the last subitem.
(c) Any finite set is bounded.
(d) Any open or closed ball is bounded.
(e) A is bounded iff there exists M > 0 such that d(x, y) ≤ M for all x, y ∈ A.
(f) If A 6= ∅, we set diam (A) := sup{d(x, y) : x, y ∈ A}, which is set to ∞ if the
supremum does not exist. The extended real number diam (A) is called the diameter
of A. A set A is bounded iff either A = ∅ or diam (A) < ∞.
(g) diam (B(x, r)) ≤ 2r and strict inequality can occur.
(h) In an normed linear space , diam (B(x, r)) = 2r. Hint: Go through Item 8.
(i) Any convergent sequence in a metric space is bounded.
(j) Boundedness is not a topological property. Already seen in Item 12g.
(k) Which vector subspaces of an normed linear space are bounded subsets?
(l) The set O(n) of all orthogonal matrices (that is, the set of matrices satisfying AAt =
I = At A) is a bounded subset of M (n, R). Here M (n, R) is considered as an normed
2 P P 2 
linear space as in Ex. 27. Observe that kAk = i j |aij | = n.

(m) The set SL(n, R) of all n × n real matrices with determinant 1 is not bounded in
M (n, R).
(n) The set of all nilpotent matrices in M (n, R) is not a bounded set. Hint: What is he
“canonical form” of a nilpotent matrix?
(o) Let G be a subgroup of the multiplicative group C∗ of the non-zero complex numbers.
Assume that as a subset of C it is bounded. Then |g| = 1 for all g ∈ G.

8.11 In a metric space any compact set is bounded in X.

Let K be a compact subset of a metric space X. Fix a ∈ X. Consider {B(a, n) : n ∈ N}
This is an open cover of X and hence the collection {B(a, n) : n ∈ N} has a finite subcover
of K. Since (B(a, n) is increasing, there exists N ∈ N such that K ⊂ B(a, N ).
Applications:

(a) SL(n, R) is not a compact subset of M (n, R).

(b) The set of symmetric (respectively, the skew-symmetric) matrices is not compact in
M (n, R). So is the set of matrices with trace zero.
(c) The set of nilpotent matrices in M (n, R) is not compact.

62
8.12 In any topological space, any convergent sequence along with its limit is a compact subset.
Let xn → x. Given an open cover {Ui : i ∈ I} of {xn : n ∈ N} ∪ {x}, let x ∈ Uj . Then all
but finitely many xn ∈ Uj .

8.13 If A is a nonempty compact subset of R, then sup A and inf A exist and they belong to
A.
Let β = sup A. Then there exists xn such that β − n1 < xn ≤ β. Hence xn → β and hence
β is a limit point of A. Heine-Borel says that A is closed.

8.14 Assume that f : X → Y is continuous and that X is compact. Then f (X) is compact. In
particular, compactness is a topological property.

8.15 The product X × Y of two spaces is compact iff X and Y are compact.
To understand the proof, draw a picture of X × Y as a closed rectangle, as explained in
Item 32. If {Ui × Vi : i ∈ I} is an open cover by means of basic open sets, then we have an
cover of {x} × Y , a “vertical line”. Since this is compact, we have a finite subcover which
turns out to be an open cover of a “band” fattening the vertical line, that is, a (super)set
of the form Ux × Y , Ux 3 x open. (A more challenging and instructive exercise could be
to carry out this in the case of an open cover of the circle x2 + y 2 = 1 by means of open
disks in R2 .) These Ux ’s cover X and hence they have a finite subcover. Thus we end up
with a finite subcover of X × Y .
Let us now work out the details. WLOG, we may assume that we are given an open cover
by means of basic open sets as in the last paragraph. Since the inclusion map x 7→ (x, y)
is continuous (Why? See Items 34b and 34c.), {x} × Y is compact by Item 14. Hence
there exists a finite subcover, say, {Ui × Vi : i ∈ Fx } for a finite subset Fx ⊂ I. Then
x ∈ Ux = ∩j∈Fx Ui is an open set. Thus the finite subcover {Ui × Vi : i ∈ Fx } is an open
cover of Ux × Y . As x varies over X, we have an open cover {Ux : x ∈ X} of the compact
space X. Let A ⊂ X be finite such that {Ux : x ∈ A} is an open cover of X. Then the
collection {Ui × Vi : i ∈ Fx , x ∈ A} is a finite subcover of X × Y .
(Why? Let (x, y) ∈ X × Y . Since {Ua : a ∈ A} is a finite open cover of X, there exists
a ∈ A such that x ∈ Ua . Hence (x, y) ∈ Ua × Y . Now, {Uj × Vj : j ∈ Fa } is a finite open
cover of Ua × Y , there exists j ∈ Fa such that (x, y) ∈ Uj × Vj , j ∈ Fa , a ∈ A, as claimed.)
P
(How many elements are there in this finite subcover? Answer: x∈A |Fx |.)

8.16 A more general result known as Tykhonoff’s theorem is true, which has very far-reaching
applications in analysis.
Theorem 10Q(Tykhonoff). Let {Xi : i ∈ I} be a family of compact spaces. The the
product space i∈I Xi with product topology is compact.

For a proof, see my article on Compact spaces.

8.17 Any cube [−R, R]n ⊂ Rn is compact. This follows from Items 6i and 15.

8.18 Let K be closed and bounded subset of Rn . Let R > 0 be such that kxk ≤ R for x ∈ K.
Then |xi | ≤ R for x = (x1 , . . . , xn ) ∈ K. Thus, K ⊂ [−R, R]n . Hence by the last result,
[−R, R]n is compact. By Item 7, K is compact. We have thus proved the sufficiency part
of the following

63
 Theorem 11 (Heine-Borel). A subset K ⊂ Rn is compact iff K is closed and bounded.

The necessary part follows from Items 8 and 11.

8.19 Applications of Heine-Borel theorem.

(a) Among the non-degenerate conics in R2 , only circles and ellipses are compact.
(b) The unit sphere S n := {x ∈ Rn+1 : kxk = 1} is compact.
(c) O(n, R), the set of orthogonal matrices is compact subset of M (n, R).
(d) The subgroup SL(n, R) is closed and unbounded. It is not a compact subset of
M (n, R).
(e) The set of nilpotent matrices in M (n, R) is closed and unbounded. It is not a
compact subset of M (n, R).
(f) All norms on Rn are equivalent.
For this beautiful application, we refer the reader to Theorem 4.3.26 on Page 105 of
my book on Metric Spaces (2nd edition).
Application: Any finite dimensional vector subspace of an normed linear space is
always closed. Hints: If two equivalent norms k k1 and k k2 are given on a vector
space X, then (X, k k1 ) is complete iff (X, k k2 ) is complete.

8.20 The set [a, b] ⊂ R` is not compact in the lower limit topology TL on R. Hint: See Item 13k.
Consider the open cover {Un : [a, b − 1/n) : n ∈ N} ∪ {[b, b + 1)}.

8.21 In general, a closed and bounded subset of a metric space need not be compact.
Standard example: If X is an infinite set with the discrete metric, then X is bounded,
closed but not compact.
For another, which will throw an illuminating insight into an example you learnt in the
theory of convergence of functions, see Item 34h.)

8.22 Compact sets and maps:

(a) Assume that f : X → Y is continuous and that X is compact. Then f (X) is compact.
In particular, compactness is a topological property.
Let {Vi : i ∈ I} be an open cover of f (X) with Vi open in Y . By continuity of f ,
each Ui := f −1 (Vi ) is open in X. We claim that {Ui } is an open cover of X. If
x ∈ X, then there exists i ∈ I such that f (x) ∈ Vi , that is, x ∈ f −1 (Vi ) = Ui . Since
X is compact, there exists a a finite subset F ⊂ I such that X = ∪i∈F Ui . We claim
that f (X) ⊂ ∪i∈F Vi . For, if y = f (x) ∈ f (X), then x ∈ Uj for some j ∈ F . Hence
f (x) ∈ Vj , that is, y ∈ ∪i∈F Vi .
(b) Let X be compact and Y be Hausdorff. Then any continuous bijection f : X → Y
is a homeomorphism.
We claim that f is a closed map. Let C ⊂ X be a closed set. Then C is compact by
Item 7. Hence f (C) is a compact subset of Y by sub-item (a). Since Y is Hausdorff
space, and the compact set f (C) is closed in Y by Item 8.
This is a very useful result. Some applications are given below.

64
 i. Typical applications arise in the theory of quotient spaces: The quotient space
[0, 2π]/ ∼ is homeomorphic to S 1 .
ii. Let f be any map (not assumed to be continuous) from a compact Hausdorff
space X to a compact space Y . Assume that the graph of f is closed as a subset
of the product space X × Y . Then f is continuous.
We have a bijection ϕ : X → Graph(f ) given by ϕ(x) = (x, f (x)). If we show
that ϕ is continuous, then as a component of ϕ, the function f must be contin-
uous. To use Item 22b, the requirements that the domain and co-domain are
compact are to be met and we need a continuous bijection.
Since Graph(f ) is closed subset of the compact space X × Y , we see that
Graph(f ) is compact. If we let ψ := ϕ−1 , then ψ(x, f (x)) = x is a continu-
ous bijection from the compact space Graph(f ) to the compact Hausdorff X.
Hence it is a homeomorphism. We conclude its inverse ϕ is also continuous.
This may be called a Closed Graph Theorem, in analogy with a result bearing
the same name in functional analysis: Let X and Y be complete normed linear
spaces. Let T : X → Y be a linear map whose graph is closed in X × Y . Then
T is continuous.
iii. Let X be a set with two distinct topologies T1 and T2 . Assume that T1 ⊂ T2
and further that (X, T2 ) is compact Hausdorff. Then (X, T1 ) is compact but not
Hausdorff.
is
(c) Let Y be compact. Then the projection map π := πX : X × Y → X is a closed map.
Let K ⊂ X × Y be closed. Let C := π(K). We claim X \ C is open. Let a ∈ X \ C.
Note that for any y ∈ Y , we have (a, y) ∈/ K. Since K is closed in X ×Y , there exists
a basic open set of the form Uy ×Vy such that (a, y) ∈ Uy ×Vy and K ∩(Uy ×Vy ) = ∅.
We thus end up with an open cover {Vy : y ∈ Y } of the compact space Y . Hence
there exists a finite subset Fa ⊂ Y such that {Vy : y ∈ Fa } is a finite subcover. Let
U := ∩y∈Fa Uy . Then a ∈ U and U is an open set. We claim that U ∩ C = ∅. If
not there exists x ∈ U ∩ C and hence there exists z ∈ Y such that (x, z) ∈ K. Now
z ∈ Vy for some y ∈ Fa . Since U ⊂ Uy , it follows that (x, z) ∈ Uy × Vy so that
(Uy × Vy ) ∩ K 6= ∅, a contradiction. Hence a ∈ U ⊂ X \ C. This proves C is closed.
(d) Let X be compact and Y be a metric space. Then any continuous map f : X → Y
is bounded.
Let f : X → Y be a continuous function from a compact space X to a metric space Y .
Fix q ∈ Y . Consider Vn := B(q, n). Then Un := f −1 (Vn ) is open. The sequence (Un )
is increasing and ∪n Un = X. Hence X = UN for some N , that is, f (X) ⊂ B(q, N ).
Note that this also follows form Items 14 and 11.
The converse is not true, in general. See Items 8 and 6m. For metric spaces, the
converse is true. For a proof, see my article on Compact Spaces. Details!

(e) Let X be compact. Then any continuous function f : X → R attains its bounds.
Let X be a compact space and f : X → R be continuous. By the last sub-item f (X)
is a bounded subset of R. Let M = sup f (X) and m = inf f (X). If there does not
exists any a ∈ X such that f (a) = M , then Un := {x ∈ X : f (x) < M − n1 } is
open, Un ⊂ Un+1 and ∪n Un = X. (Why?) By compactness, there exists N such
that f (X) = UN . But then sup f (X) ≤ M − N1 , a contradiction. Similar proof
establishes the existence of b ∈ X such that f (b) = m.

65
 This can also be proved using Item 14, Heine-Borel theorem and Item 13.
Applications:
i. Let X be compact and f : X → R be continuous. Assume that f (x) > 0 for all
x ∈ X. Then there is a δ > 0 such that f (x) ≥ δ for all x ∈ X.
ii. Let K be a compact and C a closed subsets of a metric space X such that
K ∩ C = ∅. Then d(K, C) > 0.
iii. Let K be a nonempty compact subset of a normed linear space X. Then there
exists x ∈ K such that ky k ≤ kxk for all y ∈ K.
(f) Let X and Y be metric spaces. Assume that X is compact. Then any continuous
map f : X → Y is uniformly continuous.
Fix ε > 0. For each x ∈ X, let δx correspond to ε/2 and the continuity of f at x.
Then {B(x, δx /2) : x ∈ X} is an open cover of X. Let {B(xk , δk /2) : 1 ≤ k ≤ n} be
a finite subcover where δk = δxk . Let δ := min{δk /2 : 1 ≤ k ≤ n}.
Let s, t ∈ X be such that d(s, t) < δ. If s ∈ B(xk , δk /2), then d(t, xk ) ≤ d(t, s) +
d(s, xk ) < δk . Hence that

d(f (s), f (t)) ≤ d(f (s), f (xk )) + d(f (xk ), t) < ε.

8.23 Given an open cover {Ui : i ∈ I} of a metric space (X, d), we say that a positive number
δ is a Lebesgue number of the cover, if for any subset A ⊂ X whose diameter is less than
δ, there exists i ∈ I such that A ⊂ Ui .
If δ is a Lebesgue number of the cover and 0 < δ 0 ≤ δ, then δ 0 is also a Lebesgue number
of the given open cover.

8.24 In general, an open cover may not have a Lebesgue number. Let X = (0, 1) with the
usual metric. Let Un := (1/n, 1). Then {Un : n ∈ N} is an open cover of X. Does there
exist a Lebesgue number for this cover?
Theorem 12 (Lebesgue Covering Lemma). Let (X, d) be a compact metric space. Let
{Ui : i ∈ I} be an open cover of X. Then a Lebesgue number exists for this cover.

We mimic the argument of Item 22f. For each x ∈ X, if x ∈ Ui , then there exists δx such
that B(x, δx ) ⊂ Ui . Consider the open cover {B(x, δx /2) : x ∈ X} like earlier and arrive
at δ, which does the job.

8.25 Use the last theorem to prove Item 22f. Note that the proofs of Item 22f and Lebesgue
covering lemma are also similar.

8.26 Let X be compact and E ⊂ X be infinite. Then E has a cluster point in X. (This is
known as Bolzano-Weierstrass property.)
We shall prove this by contradiction. Given x ∈ X, x is not a cluster point of E. Hence
there exists Ux 3 x an open set such that Ux ∩ E is either empty or {x}. Now {Ux }x∈X
is an open cover of X and hence there exists a finite set A ⊂ X such that {Ux : x ∈ A}
is a finite subcover. Since

E = E ∩ X = E ∩ (∪x∈A Ux ) = ∪x∈A (E ∩ Ux ) ,

we see that E contains at most |A| elements, a contradiction as A is infinite.

66
8.27 Definition of FIP: A family of subsets {Fi : i ∈ I} of a set X is said to have the finite
intersection property, (FIP, in short), if every finite collection of members of the family
has a nonempty intersection. Examples:

(a) Let X be any set and (Fn ) be a decreasing sequence of nonempty subsets of X. Then
{Fn : n ∈ N} enjoys FIP.
(b) Let X be noncompact. Then there exists an open cover {Ui : i ∈ I} of X which
does not admit a subcover. Consider the family of closed sets {Fi : i ∈ I} where
Fi := X \ Ui . This family of closed sets has F.I.P.

8.28 A topological space is compact iff every family of closed sets with FIP has a nonempty
intersection.
Let X be compact. Let {Ai : i ∈ I} be a family of closed sets with FIP. We are required
to show that ∩i Ai 6= ∅. Assume on the contrary that ∩i Ai = ∅. Let Ui := X \ Ai . Then
{Ui : i ∈ I} is an open cover of X. Since X is compact, there exists a finite set F ⊂ I
such that ∪j∈F Uj = X. By taking complements of this equation, we obtain ∩j∈F Aj = ∅.
This contradicts our hypothesis that {Ai : i ∈ I} enjoys FIP.
Converse is exactly along the same lines. If X has the said property, we need to show
that X is compact. Let {Ui : i ∈ I} be an open cover of X. Assume that it does not
admit a finite subcover. Let Ai := X \ Ui . Then {Ai : i ∈ I} is a family of closed set with
FIP. Hence ∩i Ai 6= ∅ which entails ∪i Ui 6= X!
This characterization is used in the proof of Tykhonoff’s theorem.
Question: Can we use this to prove Tykhonoff’s theorem for the product two compact
spaces?
Let X and Y be compact. Let {Ki : i ∈ I} be a family of closed sets with FIP. Let
Ai : πX (K)i) and Bi := πY (Ki ). Since πX and πY are closed maps (by Item 22c), each
of Ai and Bi is closed. We claim {Ai : i ∈ I} has FIP. For if there exists a finite subset
F ⊂ I such that ∩i∈F Ai = ∅, then ∩i∈F Ki = ∅. (For, otherwise, if (x, y) ∈ ∩i∈F , then
x = πX (x, y) ∈ π(Ki for each i ∈ F . In other words, x ∈ ∩i∈F Ai .) This contradiction
establishes that {Ai : i ∈ I} has FIP. Since X is compact, there exists x ∈ ∩i∈I Ai . By a
similar argument, we conclude the existence of some y ∈ ∩i∈I Bi . We may wish to claim
that (x, y) ∈ Ki for each i ∈ I. This cannot be proved, as it stands.

8.29 Cantor intersection theorem. This is an analogue of the nested interval theorem of real
analysis.
Theorem 13. Let X be any Hausdorff topological space. Let (Kn ) be a decreasing se-
quence of nonempty compact subsets of X. Then ∩n Kn 6= ∅.

Assume the contrary. Let Un := X \ Kn . Then each Un is open. (Why?) (Un ) is an

increasing sequence of open sets whose union is X. Hence {Un : n ∈ N} is an open cover
for K1 and hence there exists N such that K1 ⊂ UN . That is, K1 ⊂ X \ KN . Since
c .
KN 6= ∅, if we select p ∈ KN ⊂ K1 , we arrive at a contradiction p ∈ K1 ⊂ KN

8.30 A subset A of a metric space (X, d) is said to be totally bounded if for any given ε > 0,
there exist a finite number of points x1 , . . . , xn ∈ X such that A ⊂ ∪nk=1 B(xk , ε).
The finite set {xk : 1 ≤ k ≤ n} is usually referred to as an ε-net for A.

67
8.31 Examples, non-examples and properties of totally bounded sets.
(a) Any compact subset of a metric space is totally bounded.
(b) If B is totally bounded and A ⊂ B, then A is totally bounded.
(c) If A is totally bounded, so is its closure A.
If {xk : 1 ≤ k ≤ n} is an ε-net for A, then it is 2ε-net for A.
(d) Any totally bounded subset is bounded. The converse is not true.
Standard example: an infinite set with discrete metric.
A slightly more demanding example: In `2 , the orthonormal set {en : n ∈ N}.
An interesting example: fn (x) = xn , n ∈ N, in (C[0, 1], k k∞ ).
(e) Any bounded subset of R is totally bounded. (This is essentially Archimedean
property.) In fact, any bounded subset of Rn is totally bounded.
One can prove this directly. Or, if A ⊂ Rn is bounded, so is K := A. Hence K is
closed and bounded. By Heine-Borel, K is compact and hence totally bounded. A
being a subset of K is therefore totally bounded by Item 31b.

8.32 Characterization of compact metric spaces.

Theorem 14. Let X be a metric space. Then the following are equivalent.
1. X is compact.
2. X is complete and totally bounded.
3. (Bolzano-Weierstrass property.) Every infinite subset of X has a cluster point in X.
4. (Sequential compactness.) Every sequence in X has a convergent subsequence.

For a proof, we refer the reader to Theorem 4.3.15 on Page 101 of my book on Metric
Spaces (2nd edition).
8.33 Applications of 2nd characterization:
(a) Arzela-Ascoli theorem as a characterization of compact subsets of (C(X), k k∞ ),
where X is a compact metric space. (Perhaps statement only.)
(b) A subset A ⊂ `1 is compact iffP
A is closed, bounded and is such that for every ε > 0,
there exists N ∈ N such that n≥N |xn | < ε for all x ∈ A.
8.34 Applications of (perhaps the most useful) 4th characterization.
(a) Any continuous map from a compact space to a metric space is bounded.
(b) Any continuous real valued function on a compact space attains its bounds.
(c) Let K be a nonempty compact subset of R. Show that sup K, inf K ∈ K. Deduce
the last item from this.
Let α := inf K. Then there exists x ∈ K such that α ≤ xn < α + n1 . Hence xn → α.
Since K is closed, we obtain α ∈ K.
To deduce the last result, take K = f (X).
(d) Let A, B be disjoint compact subsets of a metric space. Then there exist a ∈ A, b ∈ B
such that d(A, B) = d(a, b), and hence d(A, B) > 0.
This result need not be true if the sets are assumed to be closed. Consider A :=
{(x, 0) ∈ R2 : x ∈ R} and B := {(x, y) ∈ R2 : x > 0, y > 0 and xy = 1}.

68
(e) Let K be a compact subset and C a closed set in Rn . If K ∩ C = ∅, then there exist
x ∈ K and y ∈ C such that d(x, y) = d(K, C).
It is easy to see that there exists an x ∈ K such that d(K, C) = d(x, C). To get y,
observe that there exists a sequence (yn ) in C such that d(x, yn ) → d(x, C). You
need to apply Bolzano-Weierstrass theorem to the sequence (yn ).
(f) Let K, C be as in the last item. Then K + C is closed in Rn . Details!

This result need not be true if the sets are assumed to be closed. Consider A :=
{(x, 0) ∈ R2 : x ∈ R} and B := {(x, y) ∈ R2 : x > 0, y > 0 and xy = 1}. Then
A + B = {(x, y) ∈ R2 : y > 0}.
This can be seen geometrically. Can also be proved rigorously! Details!

(g) Let X, Y be compact metric spaces. Then X × Y is compact.

An obvious line of attack needs a careful argument. Observe that 4th character-
ization applied to the sequences (xn ) and (yn ) may produce subsequences of the
form (x2n ) and (y2n−1 ) converging to x and y respectively. This will not help us to
produce a convergent subsequence of (xn .yn )!
If ((xn , yn )) is a sequence in X × Y , by compactness of X, there exists a subsequence
(xnk ) which converges to some x ∈ X. Now consider the sequence (ynk ) in the
compact metric space Y . Assume a subsequence (ynkr ) converges to y ∈ Y . Then
the subsequence (xnkr , ynkr ) converges to (x, y) by Item 33b.
(h) Let X denote the normed linear space of all bounded real valued functions on [0, 1]
under the sup norm k k∞ . Then the closed unit ball in X is closed and bounded
but not compact.
Recall Item 28b and consider the sequence (xn ) in C[0, 1].
If this has a convergent subsequence, say (xnk ), then the convergence is uniform
convergence (by Item 28b) and its limit is continuous. But the pointwise limit of
the original sequence is the discontinuous function: f (x) = 0 for 0 ≤ x < 1 and
f (1) = 1. Details!

69
9 Connected and Path-connected Spaces

9.1 Connected Spaces. Look at

(a) R, an interval,
(b) a circle, a parabola, an ellipse, two intersecting lines, a disk, a circle, a parabola or
an ellipse along with a tangent line at one of its points in R2 ,
(c) a plane, a sphere, a ball in R3 .

All of them seem to be in a “single piece.” Consider now

(a) {−1, 1}, Z, (−1, 0) ∪ (0, 1) in R,

(b) two (distinct) parallel lines, a hyperbola, two disjoint open disks in R2 ,
(c) two distinct parallel planes, the set consisting of the unit ball B(0, 1) along with the
plane x = 2.

All of these seem to have more than one piece.

9.2 A topological space X is said to be connected if the only subsets of X which are both
open and closed are ∅ and X. If there exists a subset ∅ =6 A 6= X which is both open and
closed, then the space is said to be disconnected or not connected.
Clearly, connnectedness is a topological property.
We say that a subset A of a topological space X is connected (or a connected subset of
X), if A is a connected space with the subspace topology.

9.3 If X is not connected, say ∅ 6= A 6= X is both open and closed, then B := X \ A is such
that ∅ 6= B 6= X and it is both open and closed. Hence, X is disconnected iff there exist
. . .. (Complete this sentence.) Thus X has two “pieces” A and B!
One usually calls A or the pair (A, B) as a disconnection of X.

9.4 A topological space X is connected iff it has the following property: If U and V are
nonempty open sets such that X = U ∪ V , then U ∩ V 6= ∅.

9.5 A subset A is connected iff the following condition is satisfied: If U and V are open subsets
of X such that U ∩ A and V ∩ A are nonempty and A ⊂ U ∪ V , then U ∩ V ∩ A 6= ∅.

9.6 We now give some examples. (More examples will follow once we prove a powerful char-
acterization of connected spaces. See Items 7–8.)

(a) R is connected. See Item 35i. Similar proof shows that any interval is connected.
(b) Q and R \ Q are not connected. See Item 13e.
(c) Any discrete space with more than one element is disconnected.
(d) Any indiscrete space is connected.
(e) Is the empty set connected?

9.7 The following theorem is a powerful characterization of connected spaces. The theorem
remain true if we take Z to be any discrete space with at least two elements, for instance,
Z ⊂ R with the subspace topology.

70
 Theorem 15. Consider Z := {±1} ⊂ R with subspace topology. A topological space is
connected iff any continuous map f : X → Z is a constant.

Let X be connected. Let f : X → Z be continuous. If f (a) = 1 and f (b) = −1, let

A := f −1 ({1}) and B := f −1 ({−1}). Then A and B are non-empty, open, disjoint, with
X = A ∪ B. Hence X is not connected.
Conversely, if X is not connected, let (A, B) be a disconnection of X. Define f = 1 on
A and f = −1 on B. If V is an open set in {±1}, then V = ∅, V = {±1}, V = {1} or
V = {−1}. Their inverse images are ∅, X, A and B respectively. Hence f is a continuous
function from X onto {±1}.
When we use this result to deal with connectedness of subsets in a topological space, we
shall make use of Items 18–19.

9.8 Applications of the last theorem.

(a) Any interval is connected. Use intermediate value theorem.

(b) A subset of R is connected iff it is an interval. As one can give a direct proof of this,
we have the intermediate value theorem as a corollary.
(c) Let M (n, R) denote the set of all n × n matrices of real numbers. Then GL(n, R) :=
{A ∈ M (n, R) : det(A) 6= 0} is not connected.
(d) O(n, R) := {A ∈ GL(n, R) : AAt = I} is not connected.
(e) Let X be a topological space. Let A and B be two connected subsets of X such that
A ∩ B 6= ∅. Then A ∪ B is connected. Generalize this.
(f) Let X be a connected topological space and g : X → Y be a continuous map. Then
g(X) is connected. Applications:
i. Any line segment in an normed linear space is connected.
ii. The circle {(x, y) ∈ R2 : x2 + y 2 = 1} is connected. Similarly, the ellipse and
parabola are connected.
iii. SO(2, R) := {A ∈ O(2, R) : det A = 1} is connected.
iv. GL(n, R) is not connected.
v. O(n, R) is not connected.
(g) Let X be such that every pair of points of X lies in a connected subset. Then X is
connected.
In fact, we can weaken the hypothesis. Let p ∈ X be fixed. Assume that for any
x ∈ X, there exists a connected subset Ax ⊂ X such that p, x ∈ Ax . Then X is
connected.
Applications:
i. Any star-shaped subset of a normed linear space is connected. A subset E of a
normed linear space X is said to be star-shaped at p ∈ E if for any q ∈ E, the
line segment [p, q] := {(1 − t)p + tq : 0 ≤ t ≤ 1} ⊂ E.
ii. A subset C of a normed linear space is said to be convex if it is star-shaped at
each of its points. Hence a convex set in a normed linear space is connected.

71
 iii. It is easy to see that a ball B(a, r) in a normed linear space is convex. If
x, y ∈ B(a, r), we have, for t ∈ [0, 1],

d((1 − t)x + ty, a) = k(1 − t)(x − a) + t(y − a)k

≤ (1 − t) kx − ak + t ky − ak
< (1 − t)r + tr = r.

Consequently, any ball (open or closed) in a normed linear space is connected.

iv. R2 \ {0} is connected.
v. R2 \ {(n, 0) : n ∈ Z} is connected.
vi. R with the lower limit topology TL is not connected.
For, the basic open set [a, b) is both open and closed. Its complement is
(−∞, a) ∪ [b, ∞). The set (−∞, a) is open in the standard topology of R and
since TL is finer than the standard topology, it is open in the lower limit topol-
ogy. Alternatively, if x ∈ (−∞, a), then x < a and hence x belongs to the basic
open set [x, a) ⊂ (−∞, a).
Given any x ∈ [b, ∞), the basic open set [x, x + 1) ⊂ [b, ∞). Hence the set [b, ∞)
is open in TL . Thus the complement of [a, b) is the union of two TL open sets
and hence is open in TL .
(h) Let A be a connected subset of a space X. Let A ⊂ B ⊂ A. Then B is connected.
Let f : B → {±1} be continuous. Restricted to A, f is a constant, say 1. Let x ∈ B.
We show that f (x) = 1. Let, if possible, f (x) = −1. Then there exists a set Ux 3 x,
open in B, such that f (Ux ) ⊂ (−3/2, −1/2). Since x is a limit point of A, we can
find a ∈ Ux ∩ A. But then f (a) = 1 ∈/ (−3/2, −1/2).
Application:
• Consider the set L := {(t, 0) : t ∈ [0, 1]}, An := {(1/n, y) : y ∈ [0, 1]} for
n ∈ N and A0 := {(0, y) : y ∈ [0, 1]}. Then E := L ∪ (∪n An ) is connected and
so its closure, E ∪ A0 is connected. Hence the set E ∪ {(0, 1)} is connected.
(X := E ∪ A0 is known as the comb space.)
(i) Let X be the union of open disk in R2 along with the tangent line x = 1. It is
connected.
(j) The open unit disk in R2 along with any subset of its boundary is connected. (This
is geometrically ‘obvious’.)
(k) Let {Ai : i ∈ I} be a collection of connected subsets of a space X with the property
that for all i, j ∈ I we have Ai ∩ Aj 6= ∅. Then A := ∪i Ai is connected.
Applications:
Any star-shaped subset of a normed linear space is connected. In particular, we have
• Any convex subset of a normed linear space is connected.
• Any open/closed ball in any normed linear space is connected.
• Any vector subspace in a normed linear space (in particular Rn ) is connected.
• Any coset of a vector subspace in a normed linear space (or Rn ) is connected.
(l) Let X be a topological space. Assume that {Ai : i ∈ I} is a family of connected
subsets of X. Let L be another connected subset such that L ∩ Ai 6= ∅ for all i ∈ I.
Show that L ∪ (∪i∈I Ai ) is a connected subset of X.

72
 (m) Let X and Y be topological spaces. Then the product space X × Y is connected iff
both X and Y are connected.
Let f : X × Y → {±1} be a continuous function. Fix (a, b) ∈ X × Y . We show that
for any (x, y) ∈ X ×Y , we have f (x, y) = f (a, b). Since {a}×Y is connected (why?),
the restriction of f to this set is a constant. In particular, f (a, y) = f (a, b). Now
the subset X × {y} is connected and hence the restriction of f to it is a constant.
In particular, f (a, y) = f (x, y). Hence f (x, y) = f (a, b). Picture!

Applications:
i. R2 \ {(0, 0)} is connected as it is the product of (0, ∞) × [0, 2π).
ii. A cylinder {(x, y, z) : x2 + y 2 = 1} is the product of circle and R and hence is
connected.
(n) The sphere S n := {x ∈ Rn+1 : kxk = 1} is connected.
The case n = 1 is already seen. Assume n > 1.
Note that S = S+ ∪ S− where S± := {x ∈ Rn+1 : ±xn+1 > 0}, union of two closed
hemi-spheres. The map ϕ : S− → B[0, 1] given by ϕ(x1 , . . . , xn+1 ) = (x1 , . . . , xn ) is
a bijective
q continuous map whose inverse is ψ : B[0, 1] ⊂ Rn → S− given by ψ(u) =
P 2
u, − 1 − uj . Hence it is homeomorphism and S− is connected. Similarly,
S+ is connected. Now the intersection of these two hemi sphere is the equator
{xn+1 = 0}. By Item 8e, the sphere is connected.
Alternatively, connectedness of S n can also seen as follows. Since S n is the image the
polar coordinate map, S n is connected. For instance, ϕ : [−π/2, π/2] × [0, 2π] → S 2
is given by ϕ(u, v) = (cos u cos v, cos u sin v, sin u).
A third way of seeing this is to observe that any point x other than the north pole
en+1 lies on a unique great circle and appeal to Item 8g.
Applications:
i. Rn \ {0} is connected, n ≥ 2.
ii. A cylinder {(x, y, z) ∈ R3 : x2 + y 2 = 1} is connected.
iii. An annular region {x ∈ Rn : r < kxk < R} is connected.

9.9 Union, intersection of two connected sets need not be connected. Consider (−1, 0)∪(0, 1),
intersection of the closed upper semi-circle with the closed lower semi-circle.
Complement of a connected set need not be connected. Look at complement of a bounded
interval in R.

9.10 We can give a direct proof of Item 8(n)i. Draw pictures in R2 to understand the proof
below. Let x ∈ Rn be nonzero. Let P = {x ∈ Rn : xn = 1}. Then P is homeomorphic
to Rn−1 and hence is connected (as n > 1). We show that any non-zero x lies on a line
segment which meets P . Hence by Item 8l it will follow that the set of nonzero vectors
in Rn (n > 1) is connected.
Let x ∈ Rn be nonzero. If xj 6= 0 for some j < n, then the line (x1 , . . . , xj , . . . , t) passes
through x, does not contain 0 and it meets P .
If xn is the only nonzero coordinate, the line joining x with (1, 0, . . . , 0, 1) is given by
(1 − t)(0, . . . , 0, xn ) + t(1, 0, . . . , 0, 1). It contains the given point, does not pass through
origin and it meets P .

73
 We now prove that the sphere is connected. The continuous map Rn \ {0} → Rn given
by x 7→ x/ kxk has the sphere as its image.

9.11 A finite metric space is connected iff is a singleton.

9.12 Let X be connected and f : X → R be a continuous non-constant function. Show that

f (X) is uncountable.
Since f is non-constant, there exist points p, q ∈ X such that f (p) = a < b = f (q). Since
f (X) is a connected subset of R, it is an interval.

9.13 Let X be a connected metric space with at least two elements. There X “has at least as
many elements as R.” In particular, X is uncountable.

9.14 What are all the continuous functions from f : R → R that take only rational values?

9.15 Are there continuous functions f : R → R that take irrational values at rational numbers
and rational values at irrational numbers?

9.16 Let f : [a, b] → R be continuous. “Identify” the image f ([a, b]).

9.17 Let f be a one-one continuous function on an interval. Then f is monotone.

9.18 What are all the continuous functions from a connected space to (i) a discrete space, (ii)
a finite Hausdorff space?

9.19 Let f : X → Y be a continuous map from a connected space X onto a finite Hausdorff
space. What can you conclude about Y ?

9.20 Let X and Y be topological spaces and f : X → Y be a map. We say that f is locally
constant if for each x ∈ X, there exists an open set Ux containing x with the property
that f is a constant on Ux .
Show that any locally constant function is continuous.

9.21 Let U ⊂ Rn be a nonempty open set. Let f : U → R be a differentiable function with

derivative 0. Then f is locally constant. (It need NOT be a constant function!)

9.22 Let X be connected and Y be any space. Then any locally constant function f : X → Y
is a constant function on X.
Fix p ∈ X. We show that f (x) = f (p) for x ∈ X. Define a subset A := {x ∈ X : f (x) =
f (p)}. Then p ∈ A. Hence p is non-empty. We shall show that A is both open and closed.
Since X is connected, it will follow that A = X.
Let x ∈ A. Since f is locally constant, there exists an open set U 3 x on which f is a
constant. Hence for any z ∈ U , f (z) = f (x) = f (p). That is, U ⊂ A. Hence A is open.
Let q be a limit point of A. Let Uq 3 q be an open set on which f is a constant. Since
q is a limit point of A, there exists a ∈ A ∩ Uq . Hence f (z) = f (a) for all z ∈ Uq , in
particular, f (q) = f (a) = f (p). Hence q ∈ A, that is, A is closed. Hence A = X.
This is a typical way in which connectedness hypothesis is used. If a result has connect-
edness as hypothesis, define a set which reflects what we want to prove and show that the
set so-defined is non-empty, open and closed. Learn this proof well. For another example,
refer to Item 39.

74
9.23 The converse of the last item is true. So we have the following characterization of con-
nected spaces.
A topological space X is connected iff any locally constant function from X to any space
Y is a constant.
9.24 In Item 21, if we further assume that U is connected, then f is a constant.
This follows from Items 21–22.
9.25 Connected Components. In a topological space X, the relation x ∼ y if there exists
a connected set A with x, y ∈ A is an equivalence relation.
Reflexivity: x ∼ x as x ∈ {x}.
Symmetry: If x, y ∈ C, a connected set, then y, x ∈ C.
Transitivity: Let x, y ∈ C and y, z ∈ D where C and D are connected. Since y ∈ C ∩D,
C ∪ D is connected and x, z ∈ C ∪ D. That is, x ∼ z.

The equivalence classes are called the connected components or components of X. The
following are immediate:
(a) If C is a component, then C is a connected set.
Let C = [x], the equivalence class of x. We claim that any two points y, z ∈ C lie in
a connected set. It follows from Item ????that C is connected. Since y, z ∈ C, we Give Ref!

have x, y ∈ A and y, z ∈ B with A and B connected. As earlier, A ∪ B is a connected

set with y, z ∈ A ∪ B.
(b) Any component C is a maximal connected set in the sense that if A is connected
and C ⊂ A, then C = A.
Let a ∈ A and let C = [x]. Since a, x ∈ A, a connected set, it follows that a ∼ x and
hence a ∈ [x] = C. Thus A ⊂ C.
(c) Any connected component is closed.
If C is a connected component, then C is also connected set with C ⊂ C. Hence
C = C by the ‘maximality.’
(d) If C is a component, x ∈ C and if A is a connected set with x ∈ A, then A ⊂ C.
Easy. x ∈ C ∩ A and hence C ⊂ C ∪ A, a connected set. By maximality, we have
C ∪ A = C, that is, a ⊂ C.
9.26 Examples of components:
(a) The only component of a connected space X is X.
(b) The components of a discrete space are the singleton sets.
(c) The components of Q are the singleton sets. (Note that the topology on Q is not
discrete topology. We gave two proofs of this. One is direct use of subspace topology
and another used existence of non trivial convergent sequences.) Details!

(d) What are the components of R with VIP topology? with outcast topology?
9.27 If f : X → Y is a homeomorphism, then f induces a natural bijective correspondence
between the components of X and those of Y : If C is a component of X, then f (C) is a
component of Y . Application: The pair of intersecting lines is not homeomorphic to R.
(If they are, remove the point of intersection from the pair of lines and its image from R.
Count the components.)

75
9.28 Connectedness can be used to settle questions on homeomorphisms:

(a) The set of irrational numbers in R with subspace topology is not homeomorphic to
R.
(b) A hyperbola cannot be homeomorphic to R.
(c) R cannot be homeomorphic to R2 .
(d) A pair of intersecting lines cannot be homomorphic to a parabola.
(e) The set A of two distinct parallel lines in R2 is not connected. Hence a pair of
intersecting lines cannot be homomorphic to A.

9.29 Path-connected Spaces. A continuous map α : [a, b] → X to a topological space X

is called a path. Since any two intervals are homeomorphic, it is a standard practice to
assume that a = 0 and b = 1. The point p := α(0) is called the initial point and q := α(1)
is called the terminal point of the path α. We also say that p is path connected to q by
the path α.
Note that if p is connected to q by a path to α : [0, 1] → X with α(0) = p and α(1) = q,
then the reverse path α̃ : [0, 1] → X defined by α̃(t) := α(1 − t) is a path from q to p.
Thus if p is connected to q by a path iff q is connected to p by a path. Because of this we
simply say that p and q are path-connected, without specifying which is the initial point
etc.

9.30 It is important not to identify the path α with its image α([0, 1]) in X. (It is called the
trace of α. Mnemonic: the trains could be different but the tracks may be the same.)
The paths α, β : [0, 1] → R2 given by α(t) = (t, 0) and β(t) = (t3 , 0) have the same trace.

9.31 Two point p and q may be connected by more than one path. Think of at least 3 different
paths connecting (−1, 0) to (0, 1) in R2 .

9.32 If x and y are path-connected and y and z are path-connected in a space, then x and z
are path connected.
This is an application of gluing lemma. Assume that α : [0, 1] → X connects x to y and
β : [0, 1] → X connects y to z. Then the map γ : [0, 1] → X defined as
(
α(2t) if 0 ≤ t ≤ 1/2
γ(t) =
β(2t − 1) if 1/2 ≤ t ≤ 1.

Since α(1) = y = β(0), we can apply gluing lemma to conclude that γ is path connecting
x to z.

9.33 We say that a topological space X is path-connected if any two points of X are connected
by a path.

9.34 X is path connected iff there exists p ∈ X such that any point x ∈ X is path connected
to p.

76
9.35 Any path connected space is connected.
Let X be path connected. Let f : X → {±1} be continuous. Let p, q ∈ X and α : [0, 1] →
X be a path joining p to q. Now f ◦α : [0, 1] → {±1} is continuous and hence is a constant.
In particular, f (p) = f (α(0)) = f (α(1)) = f (q), Hence f is a constant.
Or, observe that any two points lie on the trace of a path, which is connected. Hence, by
Item 8g, X is connected.
9.36 The converse is not true. Two examples:
(a) Comb space: Let L := {(x, 0) : 0 ≤ x ≤ 1} and An := {(1/n, y) : 0 ≤ y ≤ 1}, for
n ∈ N. Let P = {(0, 1)}. Then L ∪ (∪n∈N An ) is connected and its closure contains
X := L ∪ (∪n∈N An ) ∪ {P }. Hence X is connected. It is not path connected. If
possible, let γ be path joining P to Q = (1, 0) ∈ X. Choose an open disk B(P, r)
which does not meet the x-axis. Let [0, δ) be such that γ(t) ∈ B(P, r) for t ∈ [0, δ).
Let γ = (γ1 , γ2 ). Then γ1 ([0, δ)) is a connected subset of B(P, r) ∩ X. It follows that
γ1 (t) = 0 for t ∈ [0, δ). Hence γ1 (δ) = 0. We Let t0 := sup{t ∈ [0, 1] : γ1 (t) = 0}.
Then γ1 (t0 ) = 0. We claim that t0 = 1. If not, repeat argument using the continuity
of γ at t0 . We then get there exists s > t0 such that γ1 (s) = 0. Thus we conclude
that γ(t) = P for t ∈ [0, 1].
9.37 The continuous image of a path connected space is path connected.
Let f : X → Y be continuous with X path connected. Let Y = f (X). Given yj = f (xj ) ∈
Y , j = 1, 2. Let γ be a path connecting x1 to x2 . Then f ◦ γ is a path connecting y1 to
y2 .
An application. The proof in Item 10 showed that Rn \ {0}, n ≥ 2, is path connected.
Hence S n , being a continuous image of Rn \ {0} is also path connected.
9.38 The product space of path connected spaces is path connected.
Q
Let x = (xi ), y = (yi ) ∈ i Xi . Let γi be a path connecting xi to yi in the space Xi .
Define γ(t) := (γi (t)). Then γ is a path connecting x to y .
An application. The third proof of the connectedness of S n in Item 8n established its
path connectedness. Hence Rn \ {0} = (0, ∞) × S n−1 is path connected.
9.39 Any open subset of a normed linear space is connected iff it is path connected.
Let A be a connected open subset of a normed linear space X. Fix p ∈ A. It suffices to
show that there is a path connecting p to any q ∈ A. (See Item 34.) Let
E := {x ∈ A : x is path-connected to p}.
From here onwards, the proof is exactly similar to the one in Item 22.
Clearly, p ∈ E and hence E 6= ∅. Let x ∈ E. Then there exists an open ball B(x, r) ⊂ A,
since A is open. Now any z ∈ B(x, r) is connected to x via the line segment t 7→
(1 − t)z + tx. Since x ∈ E, x is path connected to p. Hence by Item 32, z is path-
connected to p and hence B(x, r) ⊂ E. We conclude that E is open.
Let q ∈ A be a limit point of E in A. As earlier, there exists B(q, r) ⊂ A. Since q is
a limit point of E, there exists z ∈ B(q, r) ∩ E. Now, q is path connected by the line
segment (1 − t)q + tz to z which in turn is path connected to p, as z ∈ E. Hence q is path
connected to p, or q ∈ E. Hence E is closed.

77
9.40 Path components are defined in an obvious way. If Cx (resp. Px ) is the component (resp.
path-component) containing x ∈ X, then Px ⊆ Cx .

9.41 Going through the proof in Item 39, we are led to the concept of locally path connected
spaces. First of all a definition.

78
 10 Locally P-Spaces

10.1 Let X be a topological space and x ∈ X. A subset U is called a neighbourhood of x in X

if there exists an open set G such that x ∈ G ⊂ U . Example: [0, 1) is a neighbourhood
of any x ∈ (0, 1) but not of x = 0.

10.2 A set in a topological space is open iff it is a neighbourhood of each of its points.

Locally P spaces

10.3 General Philosophy: Let P be a topological property. We say that a space X is

locally P (or enjoys P locally) if for each x ∈ X and an open set U 3 x, there exists
a neighbourhood N of x where N has the property P and N ⊂ U .

10.4 Let X be a topological space. Then X is said to be locally path connected if for each
x ∈ X and an open set U 3 x, there exists a path connected neighbourhood N of x such
that N ⊂ U .
Now you can similarly define locally connected and locally compact spaces.
Do you see the need for introducing the notion of neighbourhoods? If we replace a
neighbourhood by an open set in the locally P spaces, what will happen if we wanted a
Hausdorff space to be locally compact?

10.5 The proof of Item 9.39 yields the following result: An open set in a locally path connected
space is connected iff it is path-connected.

10.6 An important remark: In general X may have property P but it may not be locally P .
For instance, the complete comb space is connected but not locally connected. (Look
for a connected neighbourhood of the point (0, 1).) Similarly, there exists a compact
space (Item 10.17c) which is not locally compact. (Do NOT get confused with the ‘bad’
definition of Munkres and hence his “note” that any compact space is locally compact!)
Similarly, the space X may be locally P , but X may not enjoy P . For instance, consider
R with discrete topology. Then it is locally connected, locally path-connected and locally
compact. But it is not connected, not path connected and not compact.

10.7 A space X is locally connected iff the components of any open subset (with subspace
topology) are open in X. In particular, the components of X are open. Details!

10.8 The components in a locally path connected space are open.

10.9 Let U be an open subset of a locally path connected space. Then U is connected iff it
is path-connected.

10.10 In a locally path connected space, the components and path components are the same.

10.11 Can we define locally Rn or locally Euclidean spaces?

We say that a space X is locally Euclidean or locally Rn if for each x ∈ X, there exists
a neighbourhood Ux 3 x which is homeomorphic to a neighbourhood in Rn . (Note that
n is fixed.)

79
10.12 Can we define locally Hausdorff spaces? Is it necessarily Hausdorff?
Consider X = R∗ ∪{θ1 , θ2 } where θj are two elements not in R∗ . (We shall think of them
as “two zeros” or “the zero with split personality!”) As a local basis for x ∈ R∗ , we take
{(x − 1/k, x + 1/k) : k ∈ N}. At θj , we take {(−1/k, 0) ∪ {θj } ∪ (0, 1/k) : k ∈ N}. Then
we get a topological space which is locally Euclidean and hence it is locally Hausdorff.
However, it is not Hausdorff.

10.13 Locally Compact Spaces. We say that X is locally compact if for each x ∈ X and an
open set U 3 x, there exists a compact neighbourhood K 3 x such that x ∈ K ⊂ U .
Note that this is seemingly stronger definition than the one found in Munkres but is
equivalent to it. See Theorem 16.

10.14 The following are descendants of Item 8.8.

(a) Let K be a compact subset of a Hausdorff space X and x ∈ / K. Then there exist
disjoint open sets U and V such that x ∈ U and K ⊂ V . (This is Item 9.)
(b) Let A and B be disjoint compact subsets of a Hausdorff space. Then there exist
disjoint open sets U and V such that A ⊂ U and B ⊂ V .
(c) Let X be a compact Hausdorff space. Let A and B be disjoint closed subsets of X.
Then there exist disjoint open sets U and V such that A ⊂ U and B ⊂ V .

10.15 A space X is said to be normal if given two disjoint closed sets A and B, there exist
disjoint open sets U ⊃ A and V ⊃ B.
Last item shows that a compact Hausdorff space is normal.

10.16 Another example of a normal space is any metric space.

To appreciate this, look at A = {(x, y) ∈ R2 : xy = 0}, the set of axes which are
asymptotes of the rectangular hyperbola B := {(x, y) ∈ R2 : xy = 1}.
We now prove the result. If a ∈ A, then A is a not a limit point of B. Hence d(a, B) > 0,
by Item 3.7. There exists ra > 0 such that B(a, ra ) ∩ B = ∅. U := ∪a∈A B(a, ra ). We can
do similarly for B to get a V . Do they intersect? If z ∈ U ∩V , then z ∈ B(a, ra )∩B(b, rb )
for some a ∈ A and b ∈ B. It follows that d(a, b) ≤ ra + rb ≤ 2 max{ra , rb }, no
contradiction! If we replaced rx by rx /2 in forming U and V , we get a contradiction!
Go ahead and work out the details. (Note that we have employed a similar trick in
Theorem 12 and Item 8.22f.)

10.17 Examples of locally compact spaces:

(a) R, Rn are locally compact.

(b) Q is not locally compact.
(c) A compact space need not be locally compact. Example: Consider Q with the
usual topology, adjoin an extra element, say ∞. The neighbourhoods of x ∈ Q are
either the neighbourhoods of x in Q or ∞ added to the standard neighbourhoods.
The neighbourhoods of ∞ are complements in Q of a finite subset of F along with
∞.
(d) An normed linear space is locally compact iff it is finite dimensional. (One way is
easy; the proof of the other is omitted.)

80
 (e) A locally compact metric space need not be complete. A trivial example is (0, 1)!
Theorem 16. . The following are equivalent for a Hausdorff space:
1. X is locally compact.
2. For every x ∈ X and a neighbourhood U of x, there exists an open set V such that
x ∈ V , V is compact and V ⊂ U .
3. Each x ∈ X has a compact neighbourhood.

Proof. (1) =⇒ (2): Let x ∈ X and K be a compact neighbourhood of x. Then there

exists an open set V ⊂ K with x ∈ V . Since X is Hausdorff, K is closed. Hence V ⊂ K.
Hence V being a closed subset of a compact set K, is compact.
(2) =⇒ (3): Take V of (2).
(3) =⇒ (1): Details!

Since locally compact spaces such as Rn arise quite often, whenever we say X is locally
compact, we shall assume that X is Hausdorff also.
10.18 Local compactness is a topological property. In fact, more is true: Let f : X → Y be a
continuous open map of a locally compact space X onto Y . Then Y is locally compact.
Let y ∈ Y and V 3 y be open. Let x ∈ X be such that f (x) = y. Then U := f −1 (V )
is an open set with x ∈ U . Since X is locally compact, there exists an open set W 3 x
with W compact and W ⊂ U . Since f is open f (W ) 3 y is open, since W is compact,
f (W ) is compact. Thus, f (W ) is a compact neighbourhood of y.
10.19 A closed (respectively open) subspace of a locally compact space is locally compact.
10.20 A Hausdorff topological space X is called an n-dimensional topological manifold if for
each p ∈ X, we can find an open set Up 3 p such that Up is homeomorphic to an
open subset of Rn for n fixed. Thus, a manifold is a Hausdorff space which is locally
Euclidean.
Typical examples are (i) open subset of Rn and (ii) S n ⊂ Rn+1 . A non-example is a pair
of intersecting lines in R2 . Modern topology deals mostly with manifolds.
10.21 An interesting example of a topological manifold is Pn (R). Give Ref!

To see this, let Ui := {[(x1 , x2 , , . . . , xn+1 )] : xi 6= 0}, 1 ≤ i ≤ n + 1. Note that Ui

is “well-defined”. It is open in Pn (R). Also, ∪i Ui = Pn (R). The maps ϕi : Ui → Rn
given by [(x1 , . . . , xn )] 7→ (x1 /xi , . . . xi−1 /xi , xi+1 /xi , . . . , xn+1 /xi ) are well-defined and
homeomorphisms.
10.22 The following class of examples arise in Differential Geometry and Differential topology.
Let f : Rn+k → Rk be a continuously differentiable map. Assume that M := f −1 (0) 6=.
We also assume that for each p ∈ M , the derivative Df (p) : Rn+k → Rk is of rank k.
Let us write p = (a, b) ∈ Rn × Rk . Then the implicit function theorem says that for
each p ∈ M , there exist open sets W 3 p and U 3 a and a continuously differentiable
function h : U → Rk such that
U ∩ M = Graph(h) := {(x, h(x)) : x ∈ U }.
It follows from Item 7.34b on Page 51 that M is an n-dimensional manifold.
10.23 Any topological manifold is locally path-connected and locally compact.

81
11 One Point Compactification

11.1 Given X = (0, 1] ⊂ R, by adding just the point 0, we can make it to be compact. Note
that (0, 1] is dense in [0, 1].
Similarly, the subspace topology on the set {1/n : n ∈ N} ⊂ R is discrete. If we add the
point 0 to it, then the resulting space is compact in which the original set is dense.
Can we so something similar to any locally compact, non-compact Hausdorff space X?
That is, can we add a new point, which is denoted by ∞ to X and obtain a compact
Hausdorff space? Let us work backwards. Assume X∞ := X∪{∞} is compact Hausdorff.
We would like to retain open subset of X in tact. So we need to provide a local base at
∞. If U 3 ∞ is an open set , then X∞ \ U is a closed subset of the compact space X∞
and hence is compact. But, it is in fact a subset of X. This suggests a way of defining
a local base at ∞, namely, a subset U 3 ∞ is open if X∞ \ U is a compact subset of X.
11.2 One point compactification. Given a locally compact noncompact Hausdorff space
X, let X∞ := X ∪ {∞} where ∞ ∈
/ X. Let T denote the topology on X. Consider
T∞ := T ∪ {V ⊂ X∞ : X∞ \ V is a compact subset of X.}.
Then Details!

(i) T∞ is a Hausdorff topology on X∞ .

(ii) The subspace topology on X is T .
(iii) (X∞ , T∞ ) is compact.
(iv) X is dense in X∞ .
11.3 Let X be noncompact, locally compact Hausdorff space. Let Y be a compact Hausdorff
space. Assume that there exists q ∈ Y and a homeomorphism f : X → Y \ {q}. Then
the one point compactification X∞ of X is homeomorphic to Y . Details!

Extend f to g : X∞ → Y by setting g(∞) = q. Since g is a bijection of the compact

space onto the Hausdorff space, it suffices to show that g is continuous at ∞. Let
V 3 q = f (∞) be open. Then L := Y \ V is closed in Y and hence compact subset
of Y . Since f is a homeomorphism, K := f −1 (L) is a compact subset of X. If we let
U := {∞} ∪ X \ K, then g(U ) = V . This establishes the continuity of g at ∞.
11.4 Examples:
(a) Consider f : (0, 2π) → S 1 := {z ∈ C : |z| = 1} be defined by f (t) = eit . Then f is
a homeomorphism of (0, 2π) onto S 1 \ {1}. Hence the one point compactification
of (0, 2π) is S 1 . Since homeomorphic (non-compact, locally compact Hausdorff)
spaces have homeomorphic one-point compactifications, it follows that the one-
point compactifications of (a, b) and R are S 1 . More generally, we have the next
item.
(b) We claim that the one point compactification of Rn is (homeomorphic to) S n . Details!

Recall that we have shown (?!) that S n \ {en+1 } is homeomorphic to Rn in

Item 7.56o
(c) Let x : N → X be a sequence in X. Then xn → x∞ iff the function x : N∞ → X
defined by x(n) = xn and x(∞) = x∞ is continuous at ∞.
Application: Use this to give another solution of Item 33b.

82
 (d) Let X be an infinite discrete space. What is its one point compactification? Details!

11.5 Functions vanishing at infinity: Let X be a locally compact Hausdorff space. A contin-
uous function f : X → R is said to vanish at infinity if for any given ε > 0 there exists
a compact set K ⊂ X such that |f (x)| < ε for x ∈ / K. (We can also define continu-
ous function vanishing at ∞ for functions taking values in a normed linear space in an
obvious way.)
A continuous function f : X → R vanishes at infinity iff it extends to a continuous
function f∞ : X∞ → R with f∞ (∞) = 0.

(a) Let f : X → R be given. Its support is by definition the closure of the set {x ∈
X : f (x) 6= 0}, that is,

supp (f ) := {x ∈ X : f (x) 6= 0}.

We say that f has compact support if the support of f is compact. Evidently, any
continuous function with compact support vanishes at infinity.
(b) What are the entire functions f : C → C which vanish at infinity?

11.6 A closely related concept is proper maps between (locally compact Hausdorff) spaces.
See Appendix J.
This concept is so important that any proof of Fundamental theorem of algebra has
to either directly or indirectly use the fact that the any non-constant polynomial with
complex coefficients when considered as a map from C to C is proper.

83
12 Baire Category Theorems

12.1 A subset A ⊂ X of a topological space is said to be nowhere dense in X, if given any

nonempty open set U , we can find a nonempty open subset V ⊂ U such that A ∩ V = ∅.
This definition is equivalent to the standard one found in all text-books: A is nowhere
dense in X iff the interior of the closure of A in X is empty: Int (A) = ∅.

Let A be nowhere dense. We need to show that Int (A) = ∅. Assume the contrary.
Let x ∈ Int (A. Then there exists an open set U such that x ∈ U ⊂ A. Let V be an
open set with x ∈ V . Let W be any open subset such that x ∈ W ⊂ V . Since x ∈ A,
x is a limit point of A and hence W ∩ A 6= ∅. Thus A is not nowhere dense.
To prove the converse, let us assume that Int (A) =. If A is not nowhere dense, then
there exists a nonempty open set U such that for any nonempty open subset V ⊂ U ,
we have V ∩ A 6= ∅. In particular, if x ∈ U , we claim that x ∈ Int (A) and hence
U ⊂ A. If V 3 x is an open set, then x ∈ V ∩ U is a nonempty open subset U and
hence (V ∩ U ) ∩ A 6= ∅. Thus x ∈ A.

12.2 Prototype examples of nowhere dense sets:

(a) Let V be any proper vector subspace of Rn . More generally, any proper vector
subspace of a normed linear space .
(b) The set of zeros of any polynomial map Rn → R.

12.3 Baire Category theorem. We shall give the formulation of Baire category theorem
in a form which will be more useful than the one which uses the notion of category.
Theorem 17. Let (X, d) be a complete metric space.
(1) Let Un be open dense subsets of X, for n ∈ N. Then ∩n Un is dense. (In particular,
∩n Un is non-empty.)
(2) X cannot be a countable union of nowhere dense closed subsets Fn .

We first observe that both the statements are equivalent. For, G is open and dense iff
its complement F := X \ G is closed and nowhere dense. Hence any one of them follows
from the other by taking complements. So, we confine ourselves to proving the first. In
fact, we shall show that ∩n Un is dense in X.
The basic idea is to get into a situation like nested interval theorem. Since we need to
exploit completeness, we need to produce a Cauchy sequence whose limit is likely to be
in the intersection of Un ’s. If we have a nested sequence of open balls, say, (B(xn , rn ))
such that B(xn , rn ) ⊂ B(xn−1 , rn−1 ), we get a sequence (xn ). If we wish to show that
it is Cauchy, the only obvious estimate available (for n > m) is
n
X
d(xn , xm ) ≤ d(xn , xn−1 ) + · · · + d(xm+1 , xm ) ≤ rk .
k=m

Thus we are lead to make the sequence (rn ) of radii as the terms of a convergent series.
The standard way of doing this is to demand 0 < rn < 2−n .
We can also replace (Un ) by a nested sequence (Vn ) of open dense sets. Define V1 := U1 .
Having defined Vn , define Vn+1 = Un+1 ∩ Vn . Clearly, V1 is open dense. Assume that we

84
 have shown Vn is open dense. Let U be any nonempty open set. We need to show that
U ∩ Vn+1 6= ∅. Observe that

U ∩ Vn+1 = (U ∩ Vn ) ∩ Un+1 .

Since by induction U ∩ Vn is nonempty open, it must have nonempty intersection with

the dense set Un+1 . Thus we have produce a nested sequence (Vn ) of open dense sets
with ∩n Vn = ∩n Un .
We now show that ∩n Vn is dense. Let B(p, r) be an open ball. Since V1 is dense, there
exists x1 ∈ V1 ∩ B(p, r). Since the intersection is open, there exists a positive r1 < 1/2
such that B[x1 , r1 ] ⊂ V1 ∩ B(p, r). Repeating the same argument with B(x1 , r1 ) ∩ V2 ,
we find x2 and 0 < r2 < 2−2 such that

B[x2 , r2 ] ⊂ B(x1 , r1 ) ∩ V2 ⊂ V2 ∩ V1 ∩ B(p, r) = V2 ∩ B(p, r).

By induction (What is the induction hypothesis?) we get sequences (xn ) and (rn ) such Details!

that
B[xn , rn ] ⊂ B(xn−1 , rn−1 ) ∩ Vn ⊂ Vn ∩ B(p, r).
Clearly (xn ) is Cauchy. Since X is complete, (xn ) converges, say, to x ∈ X. Observe
that {xk : k ≥ n} ⊂ B[xn , rn ]. Hence x is a limit point of the closed ball B[xn , rn ] so
that x ∈ B[xn , rn ]. Since this is true for all n, we obtain x ∈ B(p, r) ∩ (∩n Vn ). The
theorem is proved.
12.4 A most often used corollary of Baire’s theorem is the following: If a complete metric
space X can be written as a countable union of closed sets Fn , then at least one Fn will
have a nonempty interior.
12.5 Applications:
(a) Rn cannot written as the union of a countable family of its proper vector subspaces.
In particular, R2 is not the union of a countable family of lines through the origin.
(b) No infinite dimensional complete normed linear space can be countable dimensional.
(Algebraic sense!)
(c) There can exist no metric d on Q such that d induces the usual topology on Q and
(Q, d) is complete.
(d) Let (X, d) be complete and fn : X → R be a sequence of continuous functions. As-
sume that fn → f pointwise on X. Then the set A := {x ∈ X : f is continuous at x}
is dense in X.

Proof. Our proof is a beautiful application of both versions of Baire’s theorem.

Fix ε > 0. Define, for each k ∈ N,

Ek (ε) := {x ∈ X : |fn (x) − fm (x)| ≤ ε, for all m, n ≥ k}.

Then we claim that Ek (ε) is closed for each k.

Reason: Fix m, n ≥ k. Since |fn − fm | is continuous, the set

Ekm,n (ε) := {x ∈ X : |fn (x) − fm (x)| ≤ ε}

is a closed subset of X. Now, since Ek (ε) = ∩m,n≥k Ekm,n (ε), the claim follows.

85
 It is easy to show that X = ∪k Ek (ε).
Reason: Let x0 ∈ X. Since fn (x0 ) → f (x0 ), the sequence (fn (x0 )) is Cauchy.
Hence for the given ε > 0, there exists k0 such that for m, n ≥ k0 , we have
|fm (x0 ) − fn (x0 )| ≤ ε. Hence we conclude that x0 ∈ Ek0 (ε).
Since Xis a complete metric space, at least one of Ek (ε) should have nonempty
interior. Let Uε := ∪k Int (Ek (ε)). Then Uε is a nonempty open subset of X.
Let Un := U1/n . We claim that each Un is dense in X.
Reason: It is enough if we show that every closed ball B := B[x, r] meets Un
non-trivially. (Why?)
Reason: To show a set A is dense in a metric space, it suffices to show
that A∩B(x, r) 6= ∅ for any x ∈ Xand r > 0. Assume that A∩B[z, ρ] 6= ∅
for any z ∈ X and ρ > 0. Then given any B(x, r), we may take z = x
and ρ = r/2. Then ∅ = 6 A ∩ B[x, ρ] ⊂ A ∩ B(x, r).
Observe that the closed set (and hence a complete metric space) B is the union of
a countable family of closed sets: B = ∪n (B ∩ Ek (1/n)). By Baire, at least one of
them has nonempty interior, say, Int (B ∩Ek (1/n)) 6= ∅. Since Int (B ∩Ek (1/n)) ⊂
B ∩ Int Ek (1/n), it follows that B[x, r] ∩ Un 6= ∅ and hence the claim is proved.
Let D := ∩n Un . By Baire, D is dense in X. We claim that every x ∈ D is a point
of continuity of f .
Reason: Fix p ∈ D. Let ε > 0 be given. Choose N  0 such that 1/N < ε.
Since p ∈ D, p ∈ UN and hence there exists k ∈ N such that p ∈ Int (Ek (1/N ).
By continuity of fk at p, there exists an open neighbourhood V of p contained in
Int Ek (1/N ) such that

|fk (x) − fk (p)| < ε, for all x ∈ V. (2)

For x ∈ V , since V ⊂ Ek (1/N ), by the definition of Ek (ε)’s, we have

|fm (x) − fk (x)| ≤ 1/N, for all m ≥ k. (3)

Letting m → ∞ in the above equation, we obtain

|f (x) − fk (x)| ≤ 1/N, for all x ∈ V. (4)

We are now ready for the kill. We claim that |f (x) − f (p)| < 3ε for x ∈ V .

|f (x) − f (p)| ≤ |f (x) − fk (x)| + |fk (x) − fk (p)| + |fk (p) − f (p)|
≤ 1/N + ε + 1/N
< 3ε.

This shows that f is continuous at every point of D.

12.6 An amusing exercise: Let (xn ) be any sequence of real numbers. Show that the set
{x ∈ R : x 6= xn , n ∈ N} is dense in R. Hence conclude that R is uncountable. Details!

12.7 Baire category theorem for locally compact spaces. Let X be a locally compact
Hausdorff space. Let (Un ) be a sequence of open dense sets in X. Then ∩n Un is dense
in X.

86
 Let G be a nonempty open set in X. We need to prove that there exists x ∈ G such that
x ∈ Un for all n. The strategy is to mimic the proof in the case of metric spaces replacing
open balls by the existence of open sets V such that V is compact and x ∈ V ⊂ V ⊂ U
for any given open set U and x ∈ U and then invoking Cantor intersection theorem for
a decreasing sequence of compact sets.
Since G is a nonempty open set and U1 is dense, there exists x1 ∈ G ∩ U1 . Since G ∩ U1
is open, x ∈ G ∩ U1 and X is locally compact hausdorff space, there exists an open set
V1 such that x ∈ V1 , V 1 is compact and V 1 ⊂ G ∩ U1 . Assume, by way of induction,
that we have chosen xi , Vi 3 xi , V i is compact and that xi ∈ Vi ⊂ V i ⊂ Vi−1 ∩ Ui , for
1 ≤ i ≤ n.
Now given a nonempty open set Vn , since Vn ∩ Un+1 is nonempty, there exists xn+1 ∈
Vn ∩Un+1 . Since X is locally compact and hausdorff, there exists an open set Vn+1 3 xn+1
such that V n+1 is compact and xn+1 ∈ Vn+1 ⊂ V n+1 ⊂ Vn ∩ Un+1 . Let Kn := V n . Thus
we have a decreasing sequence (Kn ) of nonempty compact subsets. Hence by Cantor
intersection theorem, there exists x ∈ ∩n Kn . Since x ∈ Kn = V n ⊂ Un , it follows that
x ∈ ∩Un . Also, x ∈ K1 ⊂ U .

12.8 Locally closed sets: A subset A of a topological space is locally closed if for every a ∈ A,
there exists an open set Ua in X such that a ∈ Ua and Ua ∩ A is closed in Ua .

(a) A characterization of locally closed sets: A ⊂ X is locally closed iff there exist an
open set U and a closed set C such that A = U ∩ C.
(b) The characterizations gives us easy examples of locally closed sets: [0, 1) is neither
closed nor open in R but is locally closed in R.

87
13 Completely Regular and Normal Spaces

13.1 Separation axioms. They deal with separating various kinds of disjoint objects by means
of disjoint open sets that contain the given objects. The prominent ones are given below.

(a) Hausdorff spaces: Given two distinct points x 6= y, if we can find open sets U and
V such that x ∈ U , y ∈ V and U ∩ V = ∅.
(b) Regular spaces: Given a point x and a closed set F with x ∈
/ F , there exist open
sets U and V such that x ∈ U and V ⊂ V with U ∩ V = ∅.
(c) Normal spaces: Given two disjoint closed sets A, B, there exist open sets U, V such
that A ⊂ U and B ⊂ V with U ∩ V = ∅.

13.2 There are counterparts of these separation axioms in which we require that the objects
be ‘separated by continuous functions’.

(a) A space X is said to be completely Hausdorff if given two distinct points x 6= y in

X there exists a continuous function f : X → R such that f (x) 6= f (y).
(b) Completely regular spaces: Given two disjoint (nonempty) closed sets, we can find
a continuous function f : X → R such that f = 0 on A and f = 1 on B.
(c) Completely normal spaces: A space X is said to be completely normal if given
two nonempty disjoint closed subsets A, B of X, there exists a continuous function
f : X → R such that f = a on A and f = b on B with a 6= b.
(d) Clearly, a completely Hausdorff (respectivley, completely regular, complletely nor-
mal) space is Hausdorff, (respectively regular, normal).
These ‘complete’ spaces will be useful for analysts since they assure that there is
an ‘abundant’ supply of real valued continuous functions on the given space!

13.3 Some standard examples and facts concerning the above concepts:

(a) Examples of regular spaces.

i. Any metric space is regular.
Let A be a closed subset of a metric space X and x ∈ / A. Let U := X \ A.
Then U is open and x ∈ U . hence there exists r > 0 such that B(x, 3r) ⊂ U .
Then the open sets B(x, r) and X \ B[x, 2r] are open set which separate x and
A.
ii. Any locally compact Hausdorff space is regular. Let A be a closed subset of a
locally compact space X and x ∈/ A. Then x ∈ X \ A and hence there exists
an open set U such that U is compact and x ∈ U ⊂ U ⊂ X \ A. The open sets
U and X \ A separate x and A.
(b) Examples of normal spaces.
i. Any metric space is normal.
We give two proofs of this.
Let A and B disjoint closed subsets of a metric space X. For each a ∈ A,
since a ∈
/ B, a is not a limit point of B. hence there exists ra > 0 such
that B(a, 2ra ) ∩ B = ∅. Similar analysis holds for each b ∈ B. Now consider
U := ∪a∈A B(a, ra ) and V := ∪b∈B B(b, rb ). Then U and V are open sets

88
 containing A and B respectively. If x ∈ U ∩ V . then x ∈ B(a, ra ) ∩ B(b, rb ) for
some a ∈ A and b ∈ B. We observe
d(a, b) ≤ d(a, x) + d(x, b) < ra + rb ≤ 2 max{ra , rb }.
Thus, a ∈ B(b, 2rb ) if rb ≥ ra or b ∈ B(a, 2ra ) if ra ≥ rb . This contradicts our
choice of ra etc. Hence U ∩ V = ∅.
The second is based on Urysohn’s lemma for metric spaces. See Item 13.5.
ii. Any compact Hausdorff space is normal.
We adapt the argument which showed that in a Hausdorff space, compact sets
are closed. Let A and B be disjoint closed subsets of a compact Hausdorff
space X. Fix x ∈ A. For each b ∈ B, there exist open sets Ub 3 a and Vb 3 b
such that Ub ∩ Vb = ∅. Since B is compact, the open cover {Vb : b ∈ B} admits
a finite subcover, say, B ⊂ V := ∪b∈F Vb for a finite subset F ⊂ B. Consider
Ua := ∩b∈F Ub . Then Ua , being a finite intersection of open sets, is open and
a ∈ Ua . Clearly, Ua ∩ V = ∅. Note that this argument shows that a compact
Hausdorff space is regular.
Given a ∈ A, by the last paragraph, there exist open sets Ua 3 a and Va ⊃ B
such that Ua ∩ Va = ∅. Now the open cover {Ua : a ∈ A} of A admits a finite
subcover, say, {Ua : a ∈ G} for a finite subset G ⊂ A. Let U := ∪a∈G Ua and
V := ∩a∈G Va . It is easy to see that U and V separate A and B.
(c) A normal space in which all singleton sets are closed is regular.
13.4 The most important result about normal spaces is the Urysohn’s lemma in Item 13.9.
It says that a space is normal iff it is completely normal.
13.5 We prove Urysohn’s lemma in the case of a metric space. Look at
d(x, A)
f (x) := .
d(x, A) + d(x, B)

Note that f makes sense, as the denominator is nonzero. For, d(x, A) + d(x, B) = 0
implies that each of the non-negative terms is zero. That is, d(x, A) = 0 and d(x, B) = 0.
Hence x is a limit point of the closed sets A and B (Item ???) and hence x ∈ A and
x ∈ B, a contradiction. By Item ??, f is continuous. Clearly, f (x) = 0 iff x ∈ A and
f (x) = 1 iff x ∈ B. (This is stronger than what is required!)
13.6 A key fact needed for Urysohn’s lemma for normal spaces is the following observation.
Lemma 18. A space X is a normal space iff for each closed set F and an open set V
containing A there exists an open set U such that F ⊂ U ⊂ U ⊂ V .

Let X be normal and F , V as above. Then F and X \ V are disjoint closed sets. By
normality of X there exist open sets U and W such that F ⊂ U and X \ V ⊂ W and
U ∩ W = ∅. Since U ⊂ X \ W and X \ W is closed, we see that U ⊂ X \ W ⊂ V . Thus
U is as required.
To see the converse, let A and B disjoint closed subsets of X. Let V1 := X \ B. Then
V1 ⊃ A is an open subset. Hence by hypothesis, there exists U1 such that A ⊂ U1 ⊂
U1 ⊂ V1 . Let V2 = X \ U1 . Then V2 ⊃ B is an open set. Let U2 be an open set such that
B ⊂ U2 ⊂ U 2 ⊂ X \ U1 . We claim that U1 ∩ U2 = ∅. For if x ∈ U1 ∩ U2 , then x ∈ U1
and x ∈ U2 ⊂ X \ U1 and hence x ∈ / U1 . Since U1 ⊂ U 1 , this is a contradiction.

89
13.7 A key step in the proof of Urysohn’s lemma is the construction of a sequence (Un ) of
open sets indexed by dyadic rationals in (0, 1). We bisect (0, 1) successively. At n-th
stage we have 2n subintervals. To proceed to the n + 1-th stage, we insert an open set
(provided by the lemma) indexed by the midpoints of the subintervals
Lemma 19. Let X be a normal space. If A and B are closed subsets of X, for each
dyadic rational r = k2−n ∈ (0, 1), there is an open set Ur with the following properties:
(i) A ⊂ Ur ⊂ X \ B, (ii) U r ⊂ Us for r < s.

Let U := X \ B. Since X is normal, there exist disjoint open sets V and W such that
A ⊂ V and B ⊂ W . Let U1/2 = V . Then, since X \ W is closed, we have
A ⊂ U1/2 ⊂ U 1/2 ⊂ X \ W ⊂ X \ B = U.
Applying the same lemma once again to the open set U1/2 containing A, that is to the
pair (A, U1/2 ) and to the pair (U 1/2 , U ), we get open sets U1/4 and U3/4 such that
A ⊂ U1/4 ⊂ U 1/4 ⊂ U1/2 ⊂ U 1/2 ⊂ U3/4 ⊂ U 3/4 ⊂ U.
Continuing this manner, we construct, for each dyadic rational r ∈ (0, 1), an open set
Ur with the following properties:
(i) U r ⊂ Us , 0 < r < s ≤ 1.
(ii) A ⊂ Ur , 0 < r < 1.
(iii) Ur ⊂ U , 0 < r < 1.
More formally, we proceed as follows. We select Ur for r = k2−n by induction on n.
That is, at n-th stage we have
A⊂U 1 ⊂ · · · ⊂ U k−1 ⊂ U k−1 ⊂U k ⊂ · · · ⊂ U 2nn−1 ⊂ U 2nn−1 ⊂ U. (5)
2n 2 n n 2 2n 2 2

−n n
Assume that we  Ur for r = k2 , 0 < k < 2 , 1 ≤ n ≤ N . To find Ur for
 have chosen
2j+1 1 j j+1 N
r = 2N +1 = 2 2N + 2N ,the midpoint, 0 ≤ j < 2 , observe that U j ⊂ U j+1 . So
2N 2N
once again applying the last lemma to the pair (U j , U j+1 ), we obtain an open set Ur
2N 2N
such that
U j ⊂ Ur ⊂ U r ⊂ U j+1 .
2N 2N

We claim that Ur ’s are as desired. We need only show that r < s implies U r ⊂ Us , the
rest being obvious. Let r = k/2m and s = l/2n . Cast both with the same denominator.
n−m
If m ≤ n, then r = 2 2n k and s = l/2n . Since r < s iff 2n−m k < l, the property (i)
follows from (5). If m > n, similar argument establishes the result.
13.8 We consider the Ur ’s as the level sets of a function to “recover” the function. Let us
explain this with an example.
Let f : X → R be a function on a set X. Fix c ∈ R. The set Xc := {x ∈ X : f (x) < c}
is called the level set of f at the level c.
Let f : X := R2 → R be given by f (x, y) = x2 + y 2 . Then Xc 6= ∅ iff c > 0. If c > 0,
√
then the level set Xc = B(0, c). Thus given the family (Xr2 ) := (B(x, r))r≥0 and given
(a, b) ∈ R2 , can we define a function f : X → R such that f (a, b) = a2 + b2 ? A moment’s
thought tells us to define f (a, b) := inf{c : (a, b) ∈ Xc }.
This example helps us understand the construction in Urysohn’s lemma.

90
 13.9 We are now ready to prove
Theorem 20. Urysohn’s Lemma. A space X is a normal space iff the following
is true: For any two disjoint closed subsets A and B of X there exists a continuous
function f : X → [0, 1] such that f = 0 on A and f = 1 on B.

Let Ur ’s be as in the lemma of the last item. The function f is defined as follows:
(
1, if x lies in no Ur
f (x) :=
inf{r : x ∈ Ur }, otherwise.

The continuity of f falls into three cases:

(i): The continuity of f at x when f (x) = 1. If x ∈/ U r , then f (x) ≥ r.
(ii): The continuity of f at x when f (x) = 0. If x ∈ Ur , then f (x) ≤ r.
(iii): The continuity at other points. If x ∈ Us \ U r , then r ≤ f (x) ≤ s.
We shall attend to (iii). Let 0 < f (x) < 1. Let ε > 0 be given. By the density of
the dyadic rationals (Item 5.2l), we can find two dyadic rational numbers r, s such that
f (x) − ε < r < f (x) < s < f (x) + ε. Consider the open set Us \ U r . If z lies in this
set, then f (z) ≤ s and f (z) ≥ r by the very definition of f . Thus x ∈ Us \ U r and
f (Us \ U r ) ⊂ [r, s] ⊂ (f (x) − ε, f (x) + ε). Therefore f is continuous at x. The other
cases are treated in a similar vein.

13.10 Note that Urysohn’s lemma does not say that f −1 (0) = A and f −1 (1) = B. It simply
says that A ⊂ f −1 (0) etc.
Contrast this with our version of Urysohn’s lemma for metric spaces in Item 13.5. Our
construction says that f (x) = 0 iff x ∈ A etc.

13.11 Let X be a discrete space. Consider A ⊂ X and B := X \ A. What function is

constructed via Uryson?
More generally, if A and B are disjoint subsets of a discre space X, what are all the
possible Urysohn’s functions?

13.12 Let h : [0, 1] → [a, b] be a homeomorphism, say, h(t) := a + (b − a)t. Let f be as in

Urysohn’s lemma. The composition h ◦ f : X → [a, b] is such that f = a on A and f = b
on B.

13.13 We now prove Tietze extension theorem, an important tool for analysts and topologists.
Let X be a normal space. Let A ⊂ X be nonempty closed. Let f : A → [−1, 1] be
continuous. Then there exists an extension F : X → [−1, 1], that is, there exists a
continuous function F : X → [−1, 1] such that F (a) = f (a) for all a ∈ A.

Proof. The subsets

B1 := {x ∈ A : f (x) ≥ 1/3} and C1 := {x ∈ A : f (x) ≤ −1/3}

are closed subsets of A and hence of X. Hence by Urysohn’s lemma, there exits a
continuous function f1 : X → [−1/3, 1/3] such that f = 1/3 on B1 and f = −1/3 on C1 .
Clearly, |f (x) − f1 (x)| ≤ 2/3 for x ∈ A. Hence f − f1 , restricted to A takes values in
[−2/3, 2/3].

91
 We now repeat the proces repalcing f by f − f1 . We divide the interval [−2/3, 2/3] into
three equal parts and define
   
1 2 1 2
B2 := x ∈ A : (f − f1 )(x) ≥ × and C2 := x ∈ A : f (x) ≤ − × .
3 3 3 3

By Urysohn, there exists a continuous function f2 : X → [−2/9, 2/9] such that f2 (x) =
2/9 on B2 and f2 (x) = −2/9 on C2 . It is clear that we have
 2
2
|((f − f1 ) − f2 )(x)| ≤ for all x ∈ A.
3

Continuing this process we have a sequence (fn ) of continuous functions on X such that
(i) fn : X → [− 31 (2/3)n−1 , 31 (2/3)n−1 ] and (ii)
 n
2
|(f − (f1 + f2 − . . . + fn ))(x)| ≤ for all x ∈ A. (6)
3

We letPF (x) := n fn (x) for x ∈ A. Since |fn (x)| ≤ Mn := 31 (2/3)n−1 for x ∈ X, and
P
since n Mn is convergent it follows by Weierstrass M -test that the series is uniformly
convergent to a continuous function F on X. The dsiplayed inequality (6) shows that the
partial sums sn of the series (defining F ) converge to f on A. Thus F is as required.

13.14 We may replace the interval [−1, 1] by any interval [a, b] in the theorem. Justify this.

13.15 We may replace the interval [−1, 1] by R in the theorem.

Let h : R → (−1, 1) be a homeomorphism. Consider the function g := h ◦ f . We may
assume that g takes value in [−1, 1]. By the theorem, there exists a continuous extension,
say, G taking values in [−1, 1] which extends g. Let A1 := {x ∈ X : |G(x)| = 1}. Since
G is an extension of g, if x ∈ A, then |G(x)| = |g(x)| < 1 and hence x ∈ / A1 . Thus
A and A1 are disjoint closed subsets of X. Hence by Urysohn’s lemma, there exists
a continuous function λ such that λ(x) = 1 on A and λ(x) = 0 on A1 . The function
λ(x)G(x) takes values in (−1, 1) and extends g. The function h−1 ◦ (λG) is the desired
extension of f .

13.16 Exercises.

(a) Let X be a normal space and F a closed subset. Assume that f : F → (−R, R) be
a continuous function. Then f can be extended to a continuous function from X
to (−R, R). Hint: You may need Urysohn’s lemma.
(b) Let X be a normal space and F a closed subset. Assume that f : F → R be a
continuous function. Then f can be extended to a continuous function from X to
R. Hint: R is homeomorphic to (−1, 1).
(c) Assuming Tietze extension theorem, prove Urysohn’s lemma.
Consider f : A ∪ B → R where f = 0 on A and 1 on B.
(d) A topological space is normal iff every continuous function from a closed subset to
[0,1] extends to a continuous function from X to [0,1].
By the last item, Urysohn’s lemma is valid for X. Use Item ??.

92
 (e) Let A be a closed subset of a normal space X. Let f : A → S n be continuous. Show
that there exists an open set U ⊃ A (U depends on f ) and an extension g of f to
U.
(f) Show that with the notation of Exer. 198 that f may not extend to all of X. Hint:
What happens (i) if n = 0 and X is connected or (ii) if X := B[0, 1] ⊂ Rn+1 ,
A := S n and f is the identity?

13.17 An example for practice.

Consider X = R with the topology T consisting of sets of the form U \ A where U is
open in the standard topology Td and A ⊂ R is any countable subset.

(a) A subset F is closed in T if F = E ∪ B where E is closed in Td and B is a countable

subset.
(b) If E = U \ A is open in T , show that the closure of E in T is the closure of E in
Td .
(c) Is Q dense in (X, T )?
(d) What are compact subsets in T ?
(e) Show that any open cover of (X, T ) admits a countable subcover.
(f) Show that (X, T ) is not first countable.
(g) Show that any countable subset is closed in T and hence (X, T ) is not separable.
(h) Show that (X, T ) is connected but not path-connected.

93
14 Homotopy

1. Let X be a topological space. A loop in X is a path α : [0, 1] → X with α(0) = α(1).

We say that α s a loop based at α(0).
Recall that if α, β : [0, 1] → X are paths such that α(1) = β(0), then their join α ∗ β is
defined by (
α(2t) for 0 ≤ t ≤ 1/2
α ∗ β(t) :=
β(2t − 1) for 1/2 ≤ t ≤ 1.
Then, α ∗ β is continuous (by gluing lemma) and we say that it is got by concatenation.
Standard Notation in homotopy theory: Let I = [0, 1].

2. Let X, Y be topological spaces. Let f, g : X → Y be continuous maps. We say that they

are homotopic if there exists a continuous map F : X × I → Y such that F (x, 0) = f (x)
and F (x, 1) = g(x) for all x ∈ X. We say that ft (x) := F (x, t) for t ∈ I and x ∈ X.
F
The map F is called a homotopy from f to g and we write f ' g.
If f (a) = g(a) for all a ∈ A ⊂ X and if the homotopy F is such that F (a, t) = f (a) for
all t ∈ I and a ∈ A, we say that f is homotopic to to g relative to A. We denote this
F
by f ' g rel A.
If α and β are paths in X with the same initial and terminal points, then saying that
α is homotopic to β relative to {0, 1} is the same as saying that all the intermediate
paths αt (s) := F (s, t) have the same initial and terminal points, that is, they satisfy
F (0, t) = α(0) and F (1, t) = α(1).

3. Examples:

(a) Let C ⊂ Rn be convex. Let f, g : X → C be continuous maps. Then the map

F (x, t) := (1 − t)f (x) + tg(x) is a homotopy from f to g. If f and g agree on a set
A ⊂ X, then F is a homotopy relative to A.
(b) Let f, g : X → S n be continuous maps such that f (x) 6= −g(x) for x ∈ X. Then
the map
(1 − t)f (x) + tg(x)
F (x, t) :=
k(1 − t)f (x) + tg(x)k
is a homotopy from f to g.
(c) The map f : S 1 := {z ∈ C : |z| = 1} → S 1 defined by f (z) = −z is homotopic to
the identity map g(z) = z.
(d) Let f : X → S n be a continuous map which is not onto. Then it is null-homotopic,
that is, homotopic to a constant map.
(e) Consider X := {p ∈ R2 : 1 ≤ kpk ≤ 2}. Let α be ‘the inner circle’ and β be the
ellipse lying in X and circumscribing α. Assume that they both start and end at
(0, 1). They are homotopic in X. (Note that X is not convex.)

4. The relation of homotopy between the continuous maps from a space X to another
space Y is an equivalence relation.

94
 F G
For, if f ' g and g ' h, then
(
F (x, 2t) 0 ≤ t ≤ 1/2
H(x, t) :=
G(x, 2t − 1) 1/2 ≤ t ≤ 1,

is a homotopy from f to h.

5. The relation of homotopy between the continuous maps from a space X to another
space Y relative to a subset A ⊂ X is an equivalence relation among maps that agree
on A.

6. Homotopy behaves well with respect to composition of maps.

(a) Let f, g : X → Y be homotopic relative to a set A ⊂ X via the homotopy F . Let

h◦F
h : Y → Z be a map. Then h ◦ f ' h ◦ g relative to A.
(b) Let f : X → Y be a map. Assume that g, h : Y → Z are homotopic relative
F
to B ⊂ Y via a homotopy G. Then g ◦ f ' h ◦ f relative to f −1 (B), where
F (x, t) := G(f (x), t).

7. Fix a base point p ∈ X. Let α be a loop at p. The equivalence class hαi of all loops
based at p homotopic to α relative to {0, 1} is called a homotopy class. The collection
of homotopy classes of loops at p is denoted by π1 (X, p).

8. Construction of the fundamental group. We make π1 (X, p) into a group as follows. For
hαi , hβi ∈ π1 (X, p), we let hαi ∗ hβi := hα ∗ βi.

(a) The above multiplication is well-defined.

(
F G H F (2s, t) 0 ≤ s ≤ 1/2
For, α ' and β ' β 0 , then α∗β ' α∗β 0 where H(s, t) :=
G(2s − 1, t) 1/2 ≤ s ≤ 1.
(b) The multiplication is associative.
First of all, we compute

α(4s)
 0 ≤ s ≤ 1/4
((α ∗ β) ∗ γ)(s) = β(4s − 1) 1/4 ≤ s ≤ 1/2

γ(2s − 1) 1/2 ≤ s ≤ 1


α(2s)
 0 ≤ s ≤ 1/2
(α ∗ (β ∗ γ))(s) = β(4s − 2) 1/2 ≤ s ≤ 3/4 .

γ(4s − 3) 3/4 ≤ s ≤ 1


Define f : I → I by setting


2s 0 ≤ s ≤ 1/4
1

f (s) := s + 1/4 ≤ s ≤ 1/2 .

 4
(s + 1)/2 1/2 ≤ s ≤ 1


95
 Since f (0) = 0 and f (1) = 1, we see that f ' 1I , that is, f is homotopic to the
identity map 1I of I relative to {0, 1}. We have

(α ∗ β) ∗ γ = (α ∗ (β ∗ γ)) ◦ f
' (α ∗ (β ∗ γ)) ◦ 1I
= α ∗ (β ∗ γ).

(c) Existence of the identity. Let e = ep denote the constant loop at p: e(t) = p for
0 ≤ t ≤ 1. Then hei serves as the identity for the multiplication. Again, proceeding
as earlier, we have
(
e(2s) 0 ≤ s ≤ 1/2
e ∗ α(s) =
α(2s − 1) 1/2 ≤ s ≤ 1
e∗α = α◦f
(
0 0 ≤ s ≤ 1/2
where f (s) =
2s − 1 1/2 ≤ s ≤ 1.

Thus we have
e ∗ α = α ◦ f ' α ◦ 1I rel I = α.
Similarly, one shows that α ∗ e ' α.
(d) Existence of inverse. The inverse of hαi is α−1 , where α−1 is the reverse path



defined by α−1 (s) := α(1 − s).

F G
i. The inverse s well-defined. If α ' β relative to {0, 1}, then α−1 ' β −1 relative
to {0, 1} where G(s, t) := F (1 − s, t).
ii. We show that α ∗ α−1 = α ◦ f where
(
2s 0 ≤ s ≤ 1/2
f (s) =
2 − 2s 1/2 ≤ s ≤ 1.

Now, f ' g relative to {0, 1} where g(s) = 0 for 0 ≤ s ≤ 1. Hence,

α ∗ α−1 = α ◦ f ' α ◦ g rel {0, 1} = e.

One similarly, shows that α−1 ◦ α ' e.

(e) Explicit homotopies can also be given. (Of what use?)
i. Existence of identity.
• α ∗ e ' α via (  
2t
α s+1 s ≥ 2t − 1
H(s, t) :=
p s ≤ 2t − 1.
• e ∗ α ' α via (
H(s, t) =
p  s ≥ 2t
2t−s
α 2−s s ≤ 2t

96
 ii. Existence of inverse. α ∗ α−1 ' e via

α(2t)
 s ≥ 2t
H(s, t) = α(s) s ≤ 2t and s ≤ 2 − 2t

α(2 − 2t) s ≥ 2 − 2t


iii. Associativity. (α ∗ β) ∗ γ ' α ∗ (β ∗ γ) via


4t
α( s+1 )
 4t − 1 ≤ s
H(s, t) = β(4t − s − 1) 4t − 2 ≤ s ≤ 4t − 1

 4t−2s
γ( 2−s − 1) s ≤ 4t − 2.

I have not verified these, simply copied from a book!

9. Let α, β be two paths such that α(1) = β(0). Then proceeding as in the last item, we
show the following, as the same homotpies work as they take care of the end points!

(a) If α0 ' α relative to {0, 1} and If β 0 ' β relative to {0, 1}, then α ∗ β ' α0 ∗ β 0
relative to {0, 1}.
(b) If α, β, γ are paths such that α ∗ (β ∗ γ) and (α ∗ β) ∗ γ make sense, then

α ∗ (β ∗ γ) ' (α ∗ β) ∗ g relative to {0, 1}.

(c) We have α ◦ α−1 ' eα(0) relative to {0, 1} and α−1 ◦ α ' eα(1) relative to {0, 1}.

10. If X is path connected, then π1 (X, p) is isomorphic to π1 (X, q) for p, q ∈ X. This

isomorphism depends on the choice of path joining p and q.

97
15 Covering Space

1. Let p : E → B be a continuous map. An open subset U ⊂ B is said to be evenly covered

by p if p−1 (U ) is the union ∪i Vi of disjoint open subsets Vi of E such that the restriction
pi of p to Vi is a homeomorphism of Vi onto U .
We say that p is a covering map if (i) p is onto and (ii) each b ∈ B has an open
neighbourhood Ub which is evenly covered by p.
The set p−1 (b) is called the fibre over b.
The sets Vi are called sheets of p−1 (U ).
E is called the total space and B, the base of the covering map p.

2. Properties of a covering map.

(a) Any covering map is open.

(b) Each of the fibres p−1 (b) is discrete.
(c) Each b ∈ B has an open neighbourhood U such that p−1 (U ) is homeomorphic to
p−1 (b) × U .

3. Examples. Details!

(a) The exponential map p : R → S 1 := {z ∈ C : |z| = 1} is a covering.

(b) The quotient map π : S n → Pn (R) is a covering.
(c) Products of covering maps is again a covering map. (precise statement?)
(d) Consider the exponential map exp : C → C∗ . The open set U := C∗ is not evenly
covered by exp.
In fact, an open set U ⊂ C∗ is evenly covered by the exponential map iff there
exists a continuous logarithm L on U , that is, a continuous map L : U → C such
that exp(L(z)) = z for all z ∈ U .
Note however that exp : C → C∗ is a covering map.

4. Let p : E → B a covering map. Let f : X → B be continuous map. Then a map

g : X → E such that p ◦ g = f is called a lift of f . One has the following commutative
diagram. (Figure?)

5. Uniqueness of lifts. Details!

Theorem 21. Let p : E → B be a covering map and X a connected space. Let f : X →

B be a map. If g, h : X → E are lifts of f such that g(x) = h(x) for some x ∈ X, then
g = h.

6. Path lifting lemma. Details!

Theorem 22. Let p : E → B be a covering map. Let c : I → B be a path. Let e0 ∈ E

be such that p(e0 ) = c(0). then there exists a unique path γ : I → E such that γ(0) = e0
and p : γ = c.

7. A Version of homotopy lifting lemma: Details!

98
 Theorem 23. Let p : E → B be a covering map. Let F : I × I → B be a continuous
map. Let e0 ∈ p−1 (F (0, 0)). Then there exists a unique lift G : I × I → E of F such
that G(0, 0) = e0 .

8. Let (E, e) and (B, b) be topological spaces with base points e and b respectively. Let
p : E → B be a covering map. If c is a loop at b and γ is its lift through e, we cannot
conclude that γ is a loop at e but p(γ(1)) = b, that is, γ(1) ∈ p−1 (b). Example: Consider
the spaces (R, 0) and (S 1 , 1). A lift of c(t) = e2πit is γ(t) = t in R.

9. Let c0 and c1 be homotopic loops at b with F as a a homotopy. We thus get a lift

G : I × I → E of F such that G(0, 0) = e and p(G(s, t)) = ct (s), for (s, t) ∈ I × I. Let
γt (s) := G(s, t). Then all these paths start at e and have the same end point γ0 (1).
As a corollary, if hci ∈ π1 (B, b) and γ is a lift of c through e, then

π1 (B, b) → π −1 (b) defined by ϕ : hci 7→ γ(1) (7)

is well-defined.

10. Simply connected space. We say a path-connected topological space X is simply con-
nected if π1 (X, x) is trivial for some (and hence for any) x ∈ X. Examples:

(a) Any convex subset of Rn is simply connected.

(b) The parabola {(x, y) ∈ R2 : y = x2 } is not convex but simply connected.
(c) We shall show below (Item 13) that S n for n ≥ 2 is simply connected.

11. Let p : (E, e) → (B, b) be a covering map. Assume that E is simply connected. Then
the map defined in (7) is a bijection of π1 (B, b) with π −1 (b). Details!

As a corollary (under the above hypothesis), for any q ∈ π −1 (b), if we let γy be a path
joining e to y, then given a loop c at p, we have a unique q ∈ π −1 (b) such that c is
homotopic to p ◦ γy .

12. Applications.

(a) Fundamental group of Pn (R) (n ≥ 2). For n ≥ 2, π1 (Pn (R), [e1 ]) is isomorphic to
Z2 .
(b) Fundamental group of S 1 is isomorphic to Z. The following are the main steps.
i. Given hci ∈ π1 (S 1 , 1), ϕ(hci) ∈ Z. The integer hci is called the index of c.
ii. The map hci 7→ ϕ(hci) is a group homomorphism of π1 (S 1 , 1) to Z.

13. Let X be a space, U, V be simply connected open subsets of X such that (i) X = U ∪ V
and (ii) U ∩ V is path connected. Then X is simply connected.
Application. S n is simply connected for n ≥ 2.

14. Applications of the index of loops in S 1 . Details!

(a) No retraction theorem. There is no continuous map f : B 2 → S 1 such that f (z) = z

for z ∈ S 1 .
(b) Brouwer fixed point theorem. Any continuous map of B 2 to itself has a fixed
point.

99
 (c) Borsuk-Ulam theorem. Let f : S 2 → R2 be a continuous map. Then there exist
antipodal points ±v ∈ S 2 such that f (v) = f (−v).
This has a physical interpretation.
(d) Ham-Sandwich theorem. Let A, B, C be bounded connected open subsets of R3 .
Then there exists a plane in R3 that divides each of the sets into two subsets of
equal volume.
Proof of this relied on some intuitively obvious facts on volumes.
(e) Fundamental theorem of algebra.

For proofs, you may refer to my relevant articles in Expository Articles.

Give exact references in the Appendix. Give Ref!

To add as appendices:

1. Finite sets

2. Cardinality

3. Subspace Topology

4. Quotient Topology

5. Generating Topologies

6. Tykonoff’s theorem

7. Compact Spaces

8. Connected Spaces

9. Existence of Continuous Functions

10. Proper maps

11. Covering spaces

12. Topological groups

13. Discrete Subgroups of Rn .

14. Characterization of Compact Metric Spaces

15. Vector fields on spheres

16. Rm is not homeomorphic to Rn for n 6= m via dimension theory.

17. Maps into punctured plane

18. Winding Numbers

19. No Retraction of R2 onto S 1 .

100
A Concepts Summary

• About sets: Open, closed, dense, nowhere dense subsets. Bounded and totally bounded
subsets in metric space.

• Examples: Metric spaces, R and Rn with the Euclidean topologies, discrete and indis-
crete spaces, R with the lower limit topology, R2 with order topology, co-finite topology,
co-countable topology, VIP topology, outcast topology.

• About Points: interior point, boundary point, limit point, cluster point and isolated
point

• About maps: continuity, homeomorphism, open maps, closed maps and proper maps.
Universal mapping properties of the constructions.

• About Spaces: compact, connected, path-connected spaces; their local versions: locally
compact, locally connected and locally path-connected spaces.

• Bases: base for the topology and base for a topology; countability axioms and separa-
bility.

• Separation Axioms: Hausdorff spaces, regular spaces and normal spaces; their counter-
parts where separation is by means of continuous functions.

• Constructions: Subspace, product and quotient spaces

• Some of the major/most useful results:

1. Product of compact/connected spaces, Heine-Bore theorem

2. Urysohn’s lemma
3. A bijective continuous map of a compact space onto a Hausdorff space is a home-
omorphism.
4. Baire Category theorem
5. A closed subset of a compact space is compact and a a compact subset of a Haus-
dorff space is closed.
6. Characterization of compact metric spaces
7. Any continuous map from a compact space to a metric space is bounded, such a
map to R has a maximum and minimum
8. A map from Y to the product space is continuous iff each of its component maps
is continuous.
9. Cantor intersection theorem.

101
B Finite Sets

For n ∈ N, let In := {1, 2, . . . , n} be the initial segment.

Definition 24. A set A is said to be finite if either A = ∅ or there exists a bijection f : A →

{1, 2, . . . , n} for some n ∈ N.

Lemma 25. If m < n, there is no one-to-one map of In into Im .

Proof. Let m = 1 and n > 1. No map f : In → I1 = {1} can be 1-1. For, f (1) = f (n) = 1
and n 6= 1. Thus the result is true for m = 1.
Let P (m) be the statements: Given n > m, no map f : In → Im can be 1-1.
Thus we have seen P(1) is true. Assume the result P (m). Consider m + 1. Let n > m + 1.
Let f : In → Im+1 be 1-1. There are two possibilities for f (n).
Case 1: Let f (n) = m + 1. Then consider the map g : In−1 → Im given by g(j) = f (j).
Then g is 1-1 and hence Im is not true.
Case 2: Let f (n) = r < m+1. Then there is at most one 1 ≤ k < n such that f (k) = m+1.
We define g : In−1 → Im by setting g(j) = f (j) for j 6= k and g(k) = r = f (n). Then g is 1-1
and hence P (m) is not true.
Thus we conclude that such an f cannot exist. In other words, P (m + 1) is also true. By
the principle of induction, we conclude that P (m) is true for all m and hence the lemma is
proved.

Lemma 26. If m < n, then there is no onto map f : Im → In .

Proof. Let f : Im → In be onto where m < n. We define g : In → Im as follows: Let r ∈ In .

Let i := min f −1 (r). We set g(r) = i. Then g is 1-1: If g(r) = g(s), then there exists k ∈ Im
such that f (k) = r, s, i.e., f is not a function!

Corollary 27. If f : Im → In is a bijection, then m = n.

Definition 28. A finite set A is said to have n elements, if there is a bijection f : A → In .

Note that in view of the corollary this is well-defined. For any finite set A, we let |A| denote
the number of elements in A.

Lemma 29. Let f : A → In be 1-1. Then A is finite, and we have |A| ≤ n.

Proof. Let r1 = min{f (a) : a ∈ A} ⊂ N, r2 = min{f (A) \ {r1 }}. Note that r1 ≥ 1 and r2 > r1
so that r2 ≥ 2. We proceed by induction to construct r1 < r2 < · · · < rk where rk ≥ k. This
process will stop at some stage in the sense that f (A) \ {rj : 1 ≤ j ≤ k} = ∅ for some k ≤ n.
For, otherwise, if k > n, then rk ≥ k > n. This contradicts the fact that rk ∈ In . We now
construct a bijection g : Ik → A as follows: g(i) = a where f (a) = ri . One easily checks that
g is a bijection.

Corollary 30. If A is finite and B ⊂ A, then B is finite and |B| ≤ |A|.

102
Proof. Let f : A → In be a bijection. Then the composition of the inclusion B ,→ A followed
by f is a 1-1 map of B into In . By Lemma 29, the result follows.

Lemma 31. Let f : In → A be onto. Then A is finite and |A| ≤ n.

Proof. Define g : A → In by setting g(a) = min f −1 (a). Then g is 1-1 and the result follows
from the last corollary.

Proposition 32 (Pigeonhole Principle). Let m, n ∈ N be such that m < n. If f : In → Im is

a map, then there exists i, j ∈ In such that i 6= j and f (i) = f (j).

Ex. 33. Let A and B be finite sets with A ∩ B = ∅. Show that A ∪ B is finite. What is the
number of elements in A ∪ B?

Ex. 34. Let X be a finite set. Let f : X → X be map. Show that the following are equivalent:
(a) f is a bijection.
(b) f is one-one.
(c) f is onto.

Ex. 35. Let A and B be finite sets and f : A → B be a map. Prove the following:
(a) If f is one-one, than |A| ≤ |B|.
(b) If f is onto, than |A| ≥ |B|.
(c) If f : A → B and g : B → A are one-one, then |A| = |B| and f , and g are bijections.

103
C Cardinality and Countability

The primitive idea of “counting” a set is to set up a bijection with a known set. The words
‘calculus’ and ‘calculation’ have their origin with such a correspondence with a pile of stones!
Definition 36. We say that two sets A and B have the same cardinality if there is a bijection
from one onto the other. (Intuitively, this means that A and B “have the same number of
elements.” Because of this we may even say that A and B are equinumerous.) Note that
“having the same cardinality” is “an equivalence relation.”
Example 37. (i) N and 2N, the set of even positive integers have the same cardinality.
(ii) Any two closed intervals [a, b] and [c, d] have the same cardinality.
(iii) Any two open intervals (a, b) and (c, d) have the same cardinality.
x
(iv) (−1, 1) and R have the same cardinality. Hint: Consider the map x 7→ . Or,
1 − x2
π π
observe that tan : (− 2 , 2 ) → R is a bijection.
(v) Z and N have the same cardinality.
(vi) R and (0, ∞) have the same cardinality.
(vii) (0,1) and (1, ∞) have the same cardinality.
Lemma 38 (Knaster). Let F : P (X) → P (X) be a map. Assume that it is increasing in
the sense that if A ⊆ B, then F (A) ⊆ F (B). Then F has a fixed point, that is, there exists
S ⊂ X such that F (S) = S.
Hint: Consider the set C := {C ⊆ X : C ⊆ F (C)}. Let S be the union of all members of
C. Then F (S) = S.

The next theorem is very useful. See Example 40.

Theorem 39 (Schroeder-Bernstein). Let A and B be sets. Assume that f : A → B and
g : B → A be one-one. Then there exists a bijection h : A → B.
Hint: Consider F : P (A) → P (A) given by F (C) := A \ g(B \ f (C)). Apply the last
lemma.
Example 40. (i) N × N and N have the same cardinality. Hint: The map f : N × N → N
given by f (m, n) := 2m 3n is one-one. For an explicit bijection, see Example 45.
(ii) There exists a bijection between the intervals [a, b] and (c, d). Hint: The interval [a, b]
and a closed subinterval of (c, d) have the same cardinality by Example 37.
(iii) The sets A := [0, 1) and B := A × A have the same cardinality. Hint: Use non-recurring
decimal expansion to get a one-map of B into A. For example, consider g(0.x1 x2 . . . , 0.y1 y2 . . .) :=
0.x1 y1 x2 y2 . . ..

For any n ∈ N, let In denote the subset {k : 1 ≤ k ≤ n} of N.

Definition 41. A set A is said to be finite if either A = ∅ or there is a bijection f : A → In
for some n ∈ N.
A set which is not finite is said to be infinite.
Theorem 42. Let A be a finite set. Let f : A → Im and g : A → In be bijections. Then
m = n.

104
Definition 43. If A is finite with f : A → In is a bijection, then n is unique by the last
theorem. (Note that f need not be unique.) We say that A has n elements. If A is empty,
we say that A has zero elements.

Definition 44. A set A is said to be countable if either A is finite or if there exists a bijection
f : A → N . A set of the latter type is said to be countably infinite.

Example 45. (i) Z+ , Z are countably infinite.

(ii) Any infinite subset of N is countably infinite.
(m + n − 1)(m + n − 2)
(iii) N×N is countably infinite. Hint: Consider the map f (m, n) := +
2
n. How did one arrive at this map? What is the inverse of this map? The inverse is given by
m 7→ ( n(n−1)
2 − m + 1, m − (n−1)(n−2)
2 ) where (n−2)(n−1)
2 < m ≤ n(n−1)
2 .

Proposition 46. Let A be a set. The following are equivalent.

(i) A is countable.
(ii) There is a one-one map of A into N.
(iii) There is an onto map from N onto A.

Corollary 47. (i) A subset of a countable set is countable.

(ii) Let I be a countable set and let Ai be countable for each i ∈ I. Then A := ∪i∈I Ai is
countable, that is, a countable union of countable set s is countable.
(iii) A finite product of countable sets is countable.

Example 48. The set of positive rational numbers is countable.

Example 49. A complex number is said to be an algebraic number if it is a root of a

polynomial with integer coefficients. The set of algebraic numbers is countable. Hint: Show
that the set of polynomials with integer coefficient is countable.

Lemma 50. The set of functions from N to {0, 1} is not countable.

Theorem 51 (Cantor). Let X be any set and P (X), the power set of X. There is no onto
function from P (X) onto X.

Example 52. R is uncountable. Hint: Enough to show that [0, 1] is uncountable. Use Nested
interval theorem.

A complex number is transcendental if it is not algebraic.

Corollary 53. The set of transcendental numbers is uncountable.

Remark 54. Why is the last result historically important?

Theorem 55. The following are equivalent for a set X.

(i) The set X is infinite.
(ii) There exists an infinitely countable subset S of X.
(iii) There exists a proper subset Y of X such that X and Y have the same cardinality.

Reference:
J.R. Munkres, Topology, especially Sections 1.6, 1.7 and 1.9

105
D Subspace Topology

Let Y ⊂ X of a topological space (X, T ). We say that a set V ⊂ Y is open in Y if there exists
an open set U in X such that V = U ∩ Y . Let TY denote the set of all subsets V ⊂ Y which
are open in Y . Then TY is a topology on Y . It is called the subspace topology. Given below
are some examples-cum-exercises which will help you master this concept. We concentrate on
“basic” open sets in Y , that is, those sets whose arbitrary unions will produce all elements
of TY . In the following any
qP Rn is endowed with the standard topology coming from the
n 2 2
Euclidean metric (x, y) := i=1 (xi − yi ).

Ex. 56. Let (X, d) be a metric space. If we restrict d to Y × Y we get a metric on Y .

Observe that BY (y, r) := B(y, r) ∩ Y , where BY (y, r) := {z ∈ Y : d(z, y) < r}. The collection
{BY (y, r) : y ∈ Y, r > 0} is a family of basic open sets for TY .
Ex. 57. Let Y := [0, ∞) ⊂ R. Then the sets of the form [0, x) with x > 0 are open in Y . In
fact, the basic open sets are [0, r), r > 0 and sets of the form (a, b), 0 < a < b.
Ex. 58. Let Y := {(x, 0) : x ∈ R} ⊂ R2 . Then the sets of the form (a, b) × {0} are basic
open sets.
Ex. 59. Let S := {(x, y) ∈ R2 : x2 + y 2 = 1} ⊂ R2 be the unit circle in R2 . The basic open
sets in S are open arcs of the circle.
Ex. 60. Consider two circles in R2 which ‘touch’ (or, which are tangential) at the origin.
Then the basic open sets around the origin are two arcs (through the origin) of the two circles.
Ex. 61. Consider Y := {(x, y) : xy = 0} ⊂ R2 be the two axes. Then the basic open sets
near (0,0) are crosses (of two line segments along the x and y-axes.) At other points, just
intervals around them.
Ex. 62. This is a generalization of Ex. 58. It requires the knowledge of product topology.
Let X and Y be topological spaces. We consider the product topology on X × Y . Fix
y0 ∈ Y . Let S := X × {y0 }. Then the basic open sets of S are of the form U × {y0 } where U
is an arbitrary open set in X.
Ex. 63. Let Y ⊂ X be open in X. Then Z ⊂ Y is open in Y iff Z is open in X.
The result is not true if Y is not open in X.
Ex. 64. Let Y ⊂ X be closed in X. Then Z ⊂ Y is closed in Y iff Z is closed in X.
The result is not true if Y is not closed in X
Ex. 65. Let A := {1/n : n ∈ N} ∪ {0}. Then the basic open sets are the singletons {1/n}
for n ∈ N and {1/n : n ≥ n0 } ∪ {0}. The latter are basic opens sets near 0 in A.
Ex. 66. Z ⊂ R has discrete topology as the subspace topology.
Ex. 67. The basic open sets in Q with the subspace topology from R are of the form
(a, b)Q := {x ∈ Q : a < x < b} for a, b ∈ R.
Is the collection {(a, b)Q : a, b ∈ Q} a family of basic open sets in Q?

106
Ex. 68. Let A ⊂ X. If the subspace topology on A is the discrete topology on A, then
every a ∈ A is an isolated point in X, that is, there exists an open set Ua 3 a in X such that
Ua ∩ A = {a}.

Ex. 69. Let Y := {(x, y) ∈ R2 : x ≥ 0, y ≥ 0} be the first quadrant in R2 . Let A := {(x, y) ∈

Y : 0 ≤ x < 1, 0 ≤ y < 1}. Is A open in Y ?

Ex. 70. Let Y ⊂ X. Let f be the restriction of the identity of X to Y . Then f : (Y, TY ) →
(X, T ) is continuous.
We can say more. If T 0 is a topology on Y such that f : (Y, T 0 ) → (X, T ) is continuous,
then TY ⊂ T 0 . Thus, the subspace topology is the smallest topology w.r.t. which the natural
inclusion map f is continuous.

Ex. 71. Let X and Y be topological spaces. Let f : X → Y be a (not necessarily continuous)
map. Let G(f ) := {(x, f (x)) : x ∈ X} ⊂ X × Y be the graph of f . Let X × Y be equipped
with the product topology. Let G(f ) ⊂ X × Y be given with the subspace topology. What
are the basic open sets of G(f )?

107
E Generating Topologies — A Unified View of
Subspace, Product and Quotient Topologies

Many constructions of new topologies in point set topology arise out of the following type of
questions.
(i) Let X be a topological space and Y a nonempty set. Assume that a map f : X → Y is
given. Can one find a topology TY such that the function f : X → (Y, TY ) is continuous.
(ii) The situation is reversed now. Assume that X is a nonempty set and Y a topological
space. We are given a map g : X → Y . We wish to endow X with a topology TX so that
g : (X, TX ) → Y becomes continuous.
Both the questions have trivial answers. In case (i), we may take the topology on Y to
be the indiscrete topology TY := {∅, Y }. In case (ii), we can endow X with the discrete
topology. If we look at the questions a little more closely, it transpires that we need to ask
for ‘the largest topology’ on Y in Case (i) while in Case (ii), we should go for ‘the smallest
topology’ on X so that the respective maps (f or g) become continuous.
These questions and their generalizations occur very naturally in various contexts in topol-
ogy. In abstract setup, we have ‘very natural’ maps from a set into another set and one of them
is topological space. We are looking for an ‘optimal’ topology on the set with no topology.
Let us look at two such natural instances.
Example 72. Let X be a topological space. Let ∼ be an equivalence relation on X. Let
Y := X/ ∼ be the set of equivalence classes of ∼ in X. We then have a natural map, namely,
the quotient map π : x 7→ [x], that is, each x ∈ X is mapped to its equivalence class.
Example 73. If X is a subset of a topological space Y , we have the natural map of inclusion
of X into Y , namely, g is the restriction of the identity map Y to X. Here we are looking for
the ‘smallest topology’ on X so that g is continuous.

We now return to answer the general questions posed above.

Case (i). If TY is a topology on Y so that f is continuous, then for any V ∈ TY , the
set f −1 (V ) must be open in X. This suggests us the following definition. We say a subset
V ⊂ Y is open if and only if f −1 (V ) is open in X. If we set

TY := {V ⊂ Y : f −1 (V ) is open in X},

then it is easily verified that TY is a topology on Y such that f : X → (Y, TY ) is continuous.

It is also the largest topology with this property. For, if TY0 is another such topology and if
W ∈ TY0 , then f −1 (W ) must be open in X. Hence, W ∈ TY .
Case (ii). Reasoning as in Case (i), we arrive at the following.

TX := {U ⊂ X : there exists an open set V ⊂ Y such that U = f −1 (V )}.

It is easy to see that TX is a topology on X such that g becomes continuous. It is the smallest
topology with this property. For, if TX0 is another topology with this property, then for any
open set V ⊂ Y , the set f −1 (V ) must be in TX0 . Thus, any arbitrary subset of TX belongs to
TX0 .

108
 Let us apply these constructions to the specific examples cited above. In Example 1,
according to our construction, a set U ⊂ X is open iff there exists an open set V ⊂ Y such
that U = g −1 (V ). Since g is IdY restricted to X, we see that g −1 (V ) = V ∩ X. Thus the
open sets in X are all of the open V ∩ X, as V varies over all open sets in Y . This topology
is known as the subspace topology of X!
In Example 2, if π : X → X/ ∼ is the quotient map, then V ⊂ Y is open iff π −1 (V ) is
open in X. This topology is known as the quotient topology on the quotient set X/ ∼.
Now that we have created these new objects, how do we work with them? The answer is
provided by the so-called universal mapping properties. Before stating them precisely, let us
see what kind of working knowledge we need about these topologies. If X and Y are given
as in Case (i), and we define the topology on Y as above, the natural questions would be:
if Z is a topological space and if we are given a map h either from Z to Y or from Y to Z,
how do we know the map h is continuous. The Universal mapping property gives a definitive
answer to precisely one of the question by transferring the onus of proving the continuity of
h to that of a ‘natural composite’ map. Note that if h : Y → Z is a map, then the composite
h ◦ f : X → Z makes sense. Universal mapping property of the topology on Y says that if
h : Y → Z is a map, then h is continuous iff the natural composite h◦f : X → Z is continuous.
In Case (ii), the natural composite map is from Z → X followed by the given map g from
X to Y . Universal mapping property of the topology on X says that if h : Z → X is a map,
then h is continuous iff the natural composite g ◦ h : Z → Y is continuous. These are easily
proved and we prove these and more as Theorem after generalizing the Cases (i) and (ii).
Note that we keep mum on the continuity h : Z → Y in Case (i) and that of h : Z → X!
Thus we have dealt with both the Cases rather easily and satisfactorily. We can generalize
these questions further and at least one of them is of immense interest.
Case I. Let {Xi : i ∈ I} be a family of topological spaces and Y a nonempty set. Assume
that we are given maps fi : Xi → Y for each i ∈ I. We are interested in finding a topology
on Y so that each of the maps fi is continuous. As pointed out earlier, we need to look for
the largest topology on Y with this property. Again arguing as in Case (i), we arrive at the
following collection of sets which must be declared to be open sets in Y :

O := {V ⊂ Y : fi−1 (V ) is open for some i ∈ I}.

Unfortunately this time, this collection need not be a topology on Y ! Before we see how to
resolve this difficulty, let us consider
Case II. Let now X be a set and Yi be topological spaces for i ∈ I. Let fi : X → Yi be
given. We look for the smallest topology on X so that the maps fi are continuous. Proceeding
as earlier, we are led to conclude that any topology on X which makes fi continuous must
contain the collection

O := {U ⊂ X : fi−1 (Vi ) for some open set Vi ⊂ Yi , where i ∈ I}.

Once again there is no guarantee that this collection is a topology on X.

Thus, in both the cases, we are faced with the following type of problem. We have a set
X and we have a collection O of subsets of X and we are looking for the smallest topology
T on X which will contain O. In theory, there is a smallest topology containing O, namely,

109
the intersection of all topologies which contain O. (Note that P (X), the power set of X is a
topology on X which contains O and hence we are not taking the intersection over an empty
collection!) What we would like to have is a practical way of dealing with this topology. Here
is a neat description of the topology.
Definition 74. Given a collection O ⊂ P (X), we say a subset U ⊂ X is open if given x ∈ U ,
there exists a finite collection G1 , . . . , Gn of members of O such that

x ∈ G1 ∩ · · · ∩ Gn ⊂ U.

It is an easy exercise to show that the collection T of all open sets (according to this definition)
is a topology on X which contains O. It is also clear that it is the smallest topology that
contains O. This topology is said to be the topology generated by O.
Remark 75. It is not hard to motivate this definition. We shall relegate the motivation to
the end of this article. It is more important to see this construction of the topology in a
concrete context.
Q
Example 76. Let Xi be topological spaces, i ∈ I. Let X := i∈I Xi be the Cartesian product
of the sets Xi . We then have natural maps π : X → Xi , the projections on the i-th factor:
πi (x) = xi . (Recall that

X := {x : I → ∪i∈I Xi such that x(i) ∈ Xi }.

For x ∈ X, xi stands for x(i).) Then any set in the collection O of Case IIQis of the form
πi−1 (Ui ) where Ui is an open subset of Xi and i ∈ I. Note that π −1 (Ui ) := j∈I Gj where
Gj = Xj for j 6= i and Gi = Ui . Thus a subset U ⊂ X is open iff for each x ∈ U , we can find
i1 , . . . , in ∈ I and open sets Uik ⊂ Xik for 1 ≤ k ≤ n such that
Y
x∈ Gj ⊂ U, where Gj = Xj for j 6= ik , and Gik = Uik , 1 ≤ k ≤ n.
j

(Note that j Gj = ∩k πi−1

Q
k
(Uik ).) This topology on X is known as the product topology. To
be able to work with this, we must have a universal mapping property of this topology. What
is it? The only natural composite maps are of the form Y → X followed by X → Xi . Thus,
we may predict that f : Y → X is continuous iff πi ◦ f : Y → Xi are continuous for all i ∈ I.
These adumbrations are formulated precisely in the next theorem and explained in detail in
the couple of remarks that follow the theorem. The topologies constructed above in Cases
(i)-(ii) and Case (II) are referred to as the topologies generated by the respective maps whose
continuity was sought after.

Theorem 77 (Universal Mapping Property).

(1.) Let f : X → Y be a map from a set X to a topological space Y . Let X be given the
topology generated by f . Then a function h : Z → X is continuous iff f ◦ h : Z → Y is
continuous.
(2.) Let g : X → Y be a map from a topological space X to a set Y . Let Y be endowed with
the topology generated by g. Then a map h : Y → Z is continuous iff the map h ◦ g : X → Z
is continuous.

110
(3.) Let πi : X → Xi be maps from the set X to topological spaces Xi for i ∈ I. Let X
be given the topology generated by πi ’s. Then a map h : Y → X is continuous iff the maps
πi ◦ h : Z → Xi are continuous.

Proof. Let us prove (1) as a sample, as the proofs are all similar and easy. To prove the
nontrivial part, let us assume that the map f ◦ h is continuous. Let U ⊂ X be open. We need
to show that h−1 (U ) is open in Z. By the very definition of the topology on X, there exists
an open set V ⊂ Y such that U = f −1 (V ). Now

h−1 (U ) = h−1 (f −1 (V )) = (f ◦ h)−1 (V ),

which is open by the continuity of f ◦ h.

To prove (2), let W be open in Z. We need to show that h−1 (W ) is open in Y . By the
continuity of h ◦ g, the set
(h ◦ g)−1 (W ) = g −1 (h−1 (W ))
is open in X. By the definition of topology on Y , the subset h−1 (W ) is open.
To prove (3), we observe that it is enough to show that h−1 (U ) is open for

U ∈ O := {U : πi−1 (Vi ) for an open Vi ⊂ Xi for some i ∈ I}.

(Prove this. Or, see Remark 82.) If U = πi−1 (V ), we have h−1 (U ) := h−1 πi−1 (U ) =

(πi ◦ h)−1 (U ) is open, by the continuity of πi ◦ h.

Remark 78. The most important thing to observe in the theorem is that the problem of
establishing continuity of a map either from or to a newly constructed space is reduced to
showing the continuity of a ‘natural composite map’ between the ‘known spaces’. Go back to
the statement and understand this remark. Also, go through the next remark.

Remark 79. Let us explicate the theorem in the concrete situations.

If f : X → Y is the inclusion map (that is, the restriction of the identity of Y to X) of
a subset X into a topological space Y , then (1) of the theorem says that a map h : Z → X
is continuous iff we think of h as a map form the space Z to Y (taking values only in X) is
continuous.
Let X be a topological space, ∼ be an equivalence relation on X and Y = X/ ∼ be the
quotient set with the quotient map π : X → Y . Then a map h : Y → Z form the quotient
space Y to a space Z is continuous iff the map h ◦ π : X → Z is continuous.
Q
Let X := i∈I Xi is the Cartesian product of topological spaces with the product topology
as in Example 76. Then a map h : Z → X can be written as h(z) = (hi (z)) where the
coordinate maps hi (z) := πi ◦ h(z). Thus h : Z → X is continuous iff the coordinate maps hi
are continuous.

To complete the story we should say something about the way B was got out of O. We
start with the following definition.

Definition 80. Let (X, T ) be a topological space. Then a collection B of open sets is called
a base for the topology if it satisfies the following two conditions:

111
 (i) for each U ∈ T and x ∈ U , there exists B ∈ B such that x ∈ B ⊂ U .
(ii) Given B1 , B2 ∈ B and x ∈ B1 ∩ B2 , then there exists B ∈ B such that x ∈ B ⊂ B1 ∩ B2 .
A typical and motivating example: the family of open sets {B(x, r) : x ∈ X, r > 0} is a
base for the metric topology on X

This definition leads to the following question. Given a set X and a collection B ⊂ P (X)
of subsets of X when is B a base for some topology on X? Since condition (i) of the definition
is any way will be used to declare open sets, the decisive condition is (ii). We state this as a
proposition.

Proposition 81. Let X be a nonempty set and B be a collection of subsets of X. Then there
exists a topology T on X for which B is a base if B satisfies the following condition:

For all B1 , B2 ∈ B and x ∈ B1 ∩B2 , there exists B ∈ B such that x ∈ B ⊂ B1 ∩B2 .

Proof. Declare U ⊂ X to be open if for each x ∈ U , there exists B ∈ B such that x ∈ B ⊂ U .

It is easy to check that the collection of open sets is a topology T on X and that B is a base
for T .

If a family O of subsets of a set X is given and if we are looking for the smallest topology
on X which contains O, we need only find a family B which could be base for a topology on
X. The family
B := {B1 ∩ B2 ∩ · · · ∩ Bn : Bi ∈ O, n ∈ N}
is a base for some topology on X. Go back to Definition 74 and try to see how we side-stepped
the intermediate construction of B and defined the topology straight away starting from O.

Remark 82. If h : Z → X is any map from a topological space Z to X, to show that h

is continuous, it suffices to verify that h−1 (B) is open for B ∈ O. If B ∈ B is of the form
B = B1 ∩ · · · ∩ Bn with Bi ∈ O then h−1 (B) = ∩i h−1 (Bi ) is the intersection of a finite
number of open sets and hence is open in Z. Now any open set U in X is an arbitrary union
of members from B. For, U = ∪x∈U Bx where for each x ∈ U , Bx ∈ B is chosen so that
x ∈ Bx ⊂ U . Clearly, h−1 (U ) = ∪x h−1 (Bx ) is open in Z.

Remark 83. The universal mapping property is the most important to deal with the newly
constructed topologies. For instance, in the case of quotient space Y = X/ ∼, giving a map
h : Y → Z is the same as giving a map h̃ : X → Z which is constant on the equivalence classes.
Hence the continuity of h is the same as that of h̃. As a concrete example, let X = R and
x ∼ y iff x − y ∈ Z. Let S := {z ∈ C : |z| = 1} be with the subspace topology. We have a
natural map h̃ : X → S given by h̃(t) := e2πit . Then h̃ gives rise to map h : Y → S which is
a bijection from well-known properties of the exponential map. Also, h is continuous since h̃
is so. Since any continuous map from a compact space to a Hausdorff space is a closed map,
h is a homeomorphism. Thus Y is the circle in C! For more such applications, see my article
on Quotient spaces.

112
F Quotient Topology

This article is devoted to the mathematical formulation of gluing geometric objects to get
new geometric objects. For example, one may form a circle from a closed line segment by
bending it around and gluing the ends together. Or, one can form a cylinder from a rectangle
by bending the rectangle around and gluing two opposite sides together. If we further bend
the cylinder around and glue the two circular rims together we get a torus or a cycle tube. In
this article, we concentrate on some of the very basic results of the theory which will enable
the reader to deal with quotient spaces with confidence. The theory is full of pathologies and
often text-books and teachers tend to frighten the beginner with the macabre rather than
emphasizing the positive aspects and initiating him into a working knowledge of quotient
spaces. This article attempts to make it easy for a student to learn quotient spaces.
Let X be a set and ∼ be an equivalence relation on X. Let X/∼ be the quotient set or the
set of equivalence classes of ∼. Let π : X → X/∼ be the quotient map defined by π(x) = [x],
the equivalence class of x. If we further assume that X is a topological space, we then want
to introduce a topology on the quotient set so that the quotient map π is continuous. Note
that the indiscrete topology on X/∼ will be one such. However we would like to have the
largest possible topology on X/∼ with this property. If τ is such a topology and V is open
in X/∼, then π −1 (V ) must be open in X. This suggests the following

Definition 84. With the notation as above, we define τ to be the set of V ⊂ X/∼ such that
π −1 (V ) is open in X. It is easy to check that τ is indeed a topology on the quotient set. The
space (X/∼, τ ) is called the quotient space of X relative to the equivalence ∼.

We record the following fact which is an immediate consequence of the definition of the
quotient topology.

Proposition 85. Let X be a topological space and ∼ an equivalence relation on X. Then

the quotient topology on X/ ∼ is the largest topology for which the natural quotient map
π : X → X/∼ is continuous.

The next theorem, though easy, is quite often used to check the continuity of maps from
quotient spaces to others.

Theorem 86 ( Universal Mapping Property). Let π : X → X/∼ be a quotient map. A map

f : X/∼→ Y is continuous iff f ◦ π is continuous.

Proof. If f is continuous, then certainly so is f ◦ π. To prove the converse, let V be an open

set of Y . Then (f ◦ π)−1 (V ) = π −1 (f −1 (V )) is an open subset of X. By the definition of
quotient topology, f −1 (V ) is open set in X/∼. Hence f is continuous.

The next theorem tells us how to generate quotient spaces.

Theorem 87. Let f : X → Y be continuous. Let ∼ be the equivalence relation on X defined

by x1 ∼ x2 iff f (x1 ) = f (x2 ). Then there exists a continuous function g : X/∼→ Y such that
f = g ◦ π.

113
Proof. Define g([x]) = f (x) for any x ∈ [x] ∈ X/∼. Then g is well-defined and g ◦ π = f .
Thm. 86 assures the continuity of g.

The following result allows us to identify quotient spaces with other concrete spaces.

Theorem 88. Let X and Y be compact. Assume further that Y is hausdorff. Let f : X → Y
be a surjective continuous map. Define the equivalence relation ∼ by declaring x1 ∼ x2 iff
f (x1 ) = f (x2 ). Then X/∼ is homeomorphic to Y .

Proof. X/∼ is compact, being the image of X under the continuous map π. Let g : X/∼→ Y
be the continuous function defined by g([x]) = f (x). Then g is continuous, 1-1 and onto.
Hence by Exer. 89 g is a homeomorphism.

Ex. 89. Let f : X → Y be a 1-1 continuous map from a compact space to a Hausdorff space.
Then f is a homeomorphism of X onto f (X).

We give four simple applications to illustrate the power of Thm. 88.

Example 90. Consider the quotient space obtained from [0, 1] got by identifying the end
points 0 and 1. That is, Y is the space X/∼ where X := [0, 1] and the equivalence classes are
{t} for 0 < t < 1 and {0, 1}. Consider the map f : [0, 1] → S 1 given by f (t) := e2πit . Y and
S 1 are homeomorphic by Thm. 88.

Example 91. We show that the quotient space got by identifying two of the opposite sides of
a rectangle is homeomorphic to a cylinder. Let X := {(u, v) ∈ R2 : −π ≤ u ≤ π, −1 ≤ v ≤ 1}.
Let Y := {(x, y, z) ∈ R3 : x2 + y 2 = 1, |z| ≤ 1}. On X we define the equivalence relation by
setting (u, v) ∼ (u0 , v 0 ) iff u = ±π = u0 and v = v 0 if u = ±π = u0 and otherwise u = u0 and
v = v 0 . Consider the map f : X → Y given by f (u, v) := (cos u, sin u, v). Then the level sets
of f are precisely the equivalence classes. f induces a homeomorphism via Thm. 88.

Example 92. Let X := [0, 1]×[0, 1] and f : X → S 1 ×S 1 be defined by f (s, t) := (e2πis , e2πit ).
The level sets of f are the singletons inside the square, the pairs of opposite points on the
interiors of the bounding intervals and the set of four vertices of X. Thus we see that the
quotient space of a square obtained from the equivalence is a torus.

For any space X and a subset A of X, the space X/A stands for the quotient space of X
with respect to the equivalence: x1 ∼ x2 iff x1 = x2 or x1 , x2 ∈ A. Thus X/A is the space
obtained from X by collapsing A to a single point.

Example 93. Consider the map f : B[0, 1] ⊂ Rn → S n ⊂ Rn+1 defined by setting

f (tx) := (cos π(1 − t), x1 sin π(1 − t), . . . , xn sin π(1 − t)),

where x ∈ S n−1 and 0 ≤ t ≤ 1. Then f (S n−1 ) = e0 = (1, 0, . . . , 0) ∈ S n . f induces a

homeomorphism of B[0, 1]/S n−1 with S n . See also Exer. 95 below.

Ex. 94. Let F be a closed subset of a compact Hausdorff space X. Prove that the quotient
space obtained from X by identifying F to a single point is homeomorphic to the one-point
compactification of X \ F .

114
Ex. 95. Let B be the closed unit ball in Rn . Prove that the quotient space obtained from B
by identifying its boundary S n−1 to a point is homeomorphic to the n-sphere.
Ex. 96. Let X denote the union of circles (in R2 ) centred at (0, 1/n) and of radius 1/n
with the subspace topology of R2 . Let Y denote the quotient space R/Z obtained from R by
collapsing all of Z to a single point. Show that X and Y are not homeomorphic.

Thm. 88 is a special case of the next theorem which can be used the same way as the
former was used in the examples above. Thm. 86, Thm. 88 and Thm. 97 are thus the most
useful results in practice.
Theorem 97. Let f : X → Y be an open (or closed) continuous surjective map. Then Y
is homeomorphic to the quotient space of X obtained by identifying each level set of f to a
point.

Proof. Argue as in Thm. 88.

The following exercises introduce some of the important quotient spaces. They will help
the reader understand the concept of quotient spaces well.
Ex. 98 (Real Projective Spaces). Let Pn (R) be the n-dimensional projective space over R.
It is the quotient space obtained from S n with respect to the equivalence relation x ∼ y iff
x = ±y. Prove the following:
(a) Pn (R) is a compact Hausdorff space.
(b) The projection π : S n → Pn (R) is a local homeomorphism, that is, each x ∈ S n has an
open neighbourhood which is mapped homeomorphically onto an open neighbourhood of π(x)
by π.
(c) P1 is homeomorphic to S 1 .
(d) Pn (R) is homeomorphic to the quotient space obtained from the closed unit ball B in Rn
by identifying the antipodal points of its boundary S n−1 .
(e) On X := Rn+1 \ {0} introduce the equivalence relation x ∼ y iff there is a nonzero α ∈ R
such that x = αy. Show that X/∼ is homeomorphic to Pn (R). (Thm. 97 may be of use here.)
Ex. 99 (Complex Projective Spaces). Think of X := S 2n+1 as a subset of Cn+1 :
X 2
S 2n+1 = {z ∈ Cn+1 : |zi | = 1}.
i

Define an equivalence relation on X by declaring that z ∼ w iff there exists a λ ∈ S 1 ⊂ C

such that z = λw. The resulting quotient space, denoted by Pn (C) is called the n-dimensional
complex projective space. Prove the following:
(a) Pn (C) is a compact Hausdorff space.
(b) P1 (C) is homeomorphic to S 2 .
(c) Let π denote the projection π : X → Pn (C). Show that π −1 (x) is homeomorphic to S 1
for all x ∈ Pn (C). Show that each x ∈ Pn (C) has a neighbourhood U such that π −1 (U ) is
homeomorphic to U × S 1 .
(d) Let Y := Cn+1 \ {0}. Let ∼ be the equivalence relation defined by x ∼ y iff there exists
a nonzero λ ∈ C such that x = λy. Then the quotient space Y /∼ is homeomorphic to Pn (C).
Thus Pn (C) can be regarded as the set of one-dimensional subspaces of Cn+1 .

115
 Quotient spaces are full of pathologies. It is necessary to recognize which of the topological
properties of X are inherited by X/∼ and which are not. The following exercises deal with
this concern.
Ex. 100. Let X be a topological space and X/∼ be its quotient with respect to an equivalence
relation. Prove the following:
(a) If X is compact, so is X/∼.
(b) If X is connected, so is X/∼.
(c) If X is path connected, so is X/∼.
Ex. 101. Pathologies
(a) Let ∼ be an equivalence relation on a space X. Prove that X/∼ is a T1 -space iff each
equivalence class is closed. Give an example of a T1 space X and a quotient space of X which
is not T1 .
(b) Define an equivalence relation on X := [0, 1] × [0, 1] by setting (s, t) ∼ (s0 , t0 ) iff t = t0 > 0.
Describe the quotient space and show that it is not Hausdorff.

One of the most important ways of defining equivalence relations is by means of group
actions. So it should not be a surprise that many quotient spaces arise as the quotients of
group actions on spaces. Below we indicate some of these instances.
Definition 102. We say a group G acts on a space X if there is a map ϕ : G × X → X with
the following properties (Below we write g · x for ϕ(g, x)):
(i) e · x = x for the identity e ∈ G and x ∈ X.
(ii) For g, h ∈ G and x ∈ X, we have (gh) · x = g · (h · x).
Note that these conditions mean that for each g ∈ G the map ϕg : x 7→ g · x is a
homeomorphism of X onto itself. Thus we have a group homomorphism of G into the group
of homeomorphisms of X.
We say X is a G-space if an action of G on X is given.

On any G-space, we have a natural equivalence: x ∼ y iff there exists a g ∈ G such that
y = g · x. The equivalence classes are called the orbits of G, for, [x] ≡ {g · x : g ∈ G}. The
corresponding quotient space X/∼ is denoted by X/G.
Ex. 103. Let X = R and G = Z. Let G act on X by n · x = x + n. Then the quotient space
R/Z is homeomorphic to S 1 .
Ex. 104. Let X = R2 and G = Z. G acts on X by n · (x, y) = (x + n, y). Show that the
resulting quotient space is the infinite cylinder {(x, y, z) ∈ R3 : x2 + y 2 = 1}.
Ex. 105. Let X = S n and G = Z2 , the multiplicative group of two elements. If −1 is the
nontrivial element of G, then define −1 · x := −x. Then X/G is Pn (C).
In a similar way, if we let G := R∗ , the multiplicative group of nonzero reals act on
X := Rn+1 \ {0} via scalar multiplication, then X/G is the n-dimensional real projective
space.
Ex. 106. Let X := S 2n+1 ⊂ Cn+1 . Let S 1 ⊂ C act on X by

eit · (z1 , . . . , zn+1 ) := (eit z1 , . . . , eit zn+1 ).

116
The quotient S 2n+1 /S 1 is homeomorphic to Pn (C).
Let Y := Cn+1 \ {0}. Let G := C∗ , the multiplicative group of complex numbers act on
Y via scalar multiplication. The quotient Y /G is Pn (C).

Ex. 107. Let X = R2 and G = Z2 . The action is (m, n) · (x, y) = (x + m, y + n). The
quotient R2 /Z2 is homeomorphic to the torus in R3 got by revolving around the z-axis a circle
of unit radius centered at (2, 0, 0) and of radius 1 in the (x, z)-plane. Hint: Consider the map
(u, v) 7→ ((2 + cos 2πu) cos 2πv, (2 + cos 2πu) sin 2πv, sin 2πu).

Ex. 108 (Möbius Strip). On the unit square X we define the equivalence relation as follows:

(x, y) ∼ (x0 , y 0 ) ⇐⇒ (x, y) = (x0 , y 0 ) or {x, x0 } = {0, 1} and y = 1 − y 0 .

Thus two points of opposite vertical sides are identified cross-wise. The quotient space is
known as the Möbius strip.
Let Y := {(x, y) ∈ R2 : −1/2 ≤ y ≤ 1/2}. Let Z act on Y by m·(x, y) := (m+x, (−1)m y).
Show that the quotient space Y /Z is homeomorphic to the Möbius strip.

Ex. 109 (Klein Bottle). Let X be the unit square. Define an equivalence relation on X
whose nontrivial relations are given by

(0, y) ∼ (1, y) and (x, 0) ∼ (1 − x, 1).

The quotient space is called the Klein’s bottle.

Let Y = R2 . Let ϕ, ψ : R2 → R2 be given by

ϕ(x, y) := (x + 1, y), ψ(x, y) := (1 − x, y + 1).

Thus ϕ is a translation parallel to the x-axis and ψ is a glide reflection along the line x = 1/2.
Show that ϕ and ψ are homeomorphisms of R2 , ψ ◦ ϕ = ϕ−1 ◦ ψ so that G := {ϕm ψ 2n ψ  :
m, n ∈ Z,  ∈ {0, 1}} is a group of homeomorphisms of R2 . Show that Y /G is homeomorphic
to the Klein bottle.

Ex. 110. Can you identify the quotient spaces X/G?

(a) Let S 1 ⊂ C act on S 2 via rotations about the z-axis.
(b) Let Zn act on S 2 via rotations by an angle which is a multiple of 2π/n.
(c) Let O(n), the orthogonal group, act on Rn via the usual linear action.

Ex. 111 (Lens Spaces). Consider S 3 ⊂ C2 . Let p and q be relatively prime integers. Let
a generator g ∈ Zp act on S 3 by g · (z1 , z2 ) := (e2πi/p z1 , e2πqi/p z2 ). The quotient space is
denoted by L(p, q) and called a Lens space. Show that L(2, 1) is homeomorphic to ¶3 (R). If
p divides q − q 0 , then L(p, q) is homeomorphic to L(p, q 0 ).
If L(p, q) and L(p0 , q 0 ) are homeomorphic then p = p0 . Hint: What is the fundamental
group of X/G if X is simply connected and G acts properly discontinuosly? (G acts properly
discontinuously on X if for any x ∈ X there exists a neighbourhood U of x such that g·U ∩U =
∅ for every g 6= e in G.)

Ex. 112. Let X := Rn \ {0}. Fix any real number α ∈ / {0, ±1}. Let G := Z act on X by
m · x := αm x. Show that G acts properly discontinuously. Identify the quotient X/G.

117
Hausdorff Quotient Spaces

Definition 113. Let ∼ be an open equivalence on a space X. For any set A ⊂ X, we let
[A] stand for the set of all elements x ∈ X which are equivalent to some element of A. The
equivalence is called open if [A] is open whenever A is open in X.

Ex. 114. An equivalence relation ∼ on a topological space is open iff the quotient map
π : X → X/∼ is open. Hint: Observe that [A] = π −1 (π(A)).

Proposition 115. Let ∼ be an equivalence on a topological space X. Then R := {(x, y) :

x ∼ y} is closed in X × X iff the quotient space X/∼ is hausdorff.

Proof. Assume that X/∼ is hausdorff and that (x, y) ∈ / R. Then there exist disjoint neigh-
bourhoods U of π(x) and V of π(y). We denote by Ũ and Ṽ the open sets π −1 (U ) and
π −1 (V ), which contain x and y respectively. If the open set Ũ × Ṽ containing (x, y) intersects
R, then it must contain a point (x0 , y 0 ) for which x0 ∼ y 0 , so that π(x0 ) = π(y 0 ), contrary to
the assumption that U ∩ V = ∅. This contradiction shows that Ũ × Ṽ does not intersect R.
Hence R is closed.
Conversely, suppose that R is closed. Given any distinct pair of points π(x) and π(y) in
X/∼, there is an open set of the form Ũ × Ṽ containing (x, y) and having no points in R. It
follows that U := π(Ũ ) and V := π(Ṽ ) are disjoint. Exer. 114 and hypothesis imply that U
and V are open. Thus X/∼ is hausdorff.

Example 116. We show a typical application of the last result by proving that Pn (R) is
hausdorff. Let X = Rn+1 \ {0}. Let the equivalence be as in (e) of Exer. 98. We first show
that π : X → Pn (R)is open. If α ∈ R is nonzero, the map ϕα : X → X given be ϕα (x) = αx
is a homeomorphism. If U ⊂ X is open, then [U ] = ∪ϕα (U ) where the union is taken over
all nonzero reals. Since each ϕα (U ) is open, their union [U ] is also open. By Exer. 114, π is
open.
n
We now Pshow that P (R)2 is hausdorff. Consider the function f : X × X → R is given by
f (x, y) := i6=j (xi yj − xj yi ) . Then f is continuous and vanishes iff y = αx for some nonzero
real α, that is, iff x ∼ y. Thus R = {(x, y) : x ∼ y} = f −1 (0) is a closed subset of X × X. By
Prop. 115, Pn (R) is hausdorff.

Collapsing and Attaching

We now discuss two construction which are very important Algebraic Topology and which
arise as quotient spaces.

Example 117. Let A be a nonempty subset of a topological space. Define ∼ to be the

equivalence relation on X that identifies all points of A with each other:

x ∼ y ⇐⇒ (x = y) or (x ∈ A and y ∈ A).

The points of the quotient set X/ ∼ are the singletons {x} for x ∈
/ A and the distinguished
point A. The quotient space is most often denoted by X/A. One says that it is obtained by
collapsing A to a single point.

118
 Let π : X → X/A be the quotient map. Then π maps the open subspace x \ A of X onto
the complement (X \ A) \ {A} of the special point A of X/A. Thus the complement of A in
X/A is open, as its inverse image under π is X \ A.
In fact, π induces a homeomorphism X \ A ∼
= (X/A) \ {A}. Note that the continuous map
π is one-one on X \ A. If U ⊂ (X \ A) is open in X \ A, then π −1 (π(U )) = U is open in X.
Therefore, π(U ) is open in X/A. We thus see that X/A contains a point whose complement
is the same as X \ A topologically.
We now consider a specific example. Let X = B[0, 1] ⊂ R2 , the closed unit disc in the
plane and A = S 1 , its boundary. Can you imagine what the quotient space look like? Imagine
a circular piece of rubber with a drawstring along its boundary. When the string is drawn
tight, a kind of spherical bag results. Therefore, we should expect X/A ∼
= S 2 , the unit sphere
3
in R .
How do we prove this rigorously? Let p be the north pole of S 2 . Geometrically thinking,
can we find a map f : X → S 2 that sends each point of S 1 to p and maps X \ S 2 injectively
onto S 2 \ {p}? The induced map on X/A would be as required. Look at the closed unit disk
D := {(x, y, z) ∈ R2 : x2 + y 2 ≤ π, z = −1} in the plane tangent to S 2 at the south pole −p.
We wrap it S 2 by wrapping each radial line segment in this disk onto a meridian of S 2 . If you
still remember cylindrical and spherical coordinates, what we plan to do is to send a point
P ∈ D with cylindrical coordinates (r, θ, −1) to a point on S 2 whose spherical coordinates
are (1, π − r, θ). Thus f will be the composite of two maps: one is the homeomorphism
(x, y) 7→ (πx, πy, −1) of X onto D and the second is as described earlier. Since the resulting
map is from a compact space to a Hausdorff space, it is closed. It is clearly continuous.

Example 118. Let X and Y be topological spaces and A ⊂ X be nonempty and closed.
Let f : A → Y continuous. Imagine joining X and Y together by gluing each point a ∈ A to
f (a) ∈ Y . The resulting space should be a topological space Z which contains (homeomorphic)
copies of X \ A and Y in which each y ∈ f (A) represents an identification of all a ∈ f −1 (y)
with y. Our aim is to construct such a space.
Before doing this, we need the notion of sum of two topological spaces. Consider two
topological spaces X and Y . Assume that as sets they are disjoint. Consider the union
Z = X ∪ Y . We wish to endow a topology on Z in an obvious manner: call W ⊂ Z open iff
W ∩ X and W ∩ Y are open in X and Y respectively. Note that this is the same as saying any
open set W ⊂ Z is of the form W = U ∪ V with U and V being open in X and Y respectively.
One easily shows that this defines a topology on Z in such a way that both X and Y are
open subspaces in Z. The space Z is called the topological sum of X and Y
If X and Y are not disjoint, we resort to a standard trick. In stead of X and Y , we
consider X1 := X × {1} and Y := Y × {2}. The open sets of X1 are U × {1}. Similarly, we
define open sets in Y1 . Clearly, X1 ∼
= X and Y1 ∼
= Y . Let Z be the topological sum of X1
and Y1 . The space Z is denoted as X1 + Y1 . Via the maps x 7→ (x, 1) and y 7→ (y, 2), we see
that X and Y are open subsets of Z.
Let us now return to our original notation. Let us first assume that X and Y are disjoint.
Then the space X + Y contains both X and Y as open, closed subspaces. We define an

119
equivalence relation ∼ on X + Y as follows:

u∼v ⇐⇒ (u = v) or (u ∈ A & v = f (u)) or (v ∈ A & u = f (v))

or (u ∈ A & v ∈ A & f (u) = f (v)).

We let Z = (X + Y )/ ∼. The standard notation for Z is X ∪f Y . We say that X ∪f Y is

obtained by attaching X to Y by f . The map f is called the attaching map.
We now show that X ∪f Y has the desired properties. Let π : X + Y → X ∪f Y be the
quotient map. Clearly, π(a) = π(f (a)) for any a ∈ A. Also, we have

X ∪f Y = π(X) ∪ π(Y ) = π(X \ A) ∪ π(A) ∪ π(Y ),

as well as
π(X) ∩ π(Y ) = π(A) = π(f (A)).
We make the following claims: (i) π(Y ) is closed in X ∪f Y and π maps Y homeomorphically
onto π(Y ), (ii) π(X \A) is open in X ∪f Y and π maps X \A homeomorphically onto π(X \A).
Proof of Claim (i): The set π(Y ) is closed since π −1 (π(Y )) = A ∪ Y is closed in X + Y .
The restriction of π to Y is continuous and injective. If F is closed in Y , then π(F ) is closed
in π(Y ) since π −1 (π(F )) = A ∪ F is closed in X + Y . This proves (i). The proof of (ii) is
similar.
We now take up a specific example. Let X = [0, 1] and A = {0, 1}. Let Y = B[0, 1], the
closed unit disk in R2 . Let f : A → Y be such that f (0) = f (1) = 0 ∈ Y . Can you imagine
the space X ∪f Y ? It looks line a circle touching a unit disk tangentially at the origin of the
disk. More precisely, we show that X ∪f Y is homeomorphic to the subspace

Z = {(x, y, z) ∈ R3 : z = 0, x2 + y 2 ≤ 1} ∪ {(x, y, z) ∈ R3 : x = 0, y 2 + (z − 1)2 = 1}.

See Figure??? To complete the proof, consider the map g : X + Y → Z given by

g(x) = (0, sin 2πx, 1 − cos 2πx), for x ∈ X,

g(y) = (y1 , y2 , 0), for y ∈ Y.

The general case is done considering X1 and Y1 as earlier and form their sum. Let
h : X → X1 +Y1 and k : Y → X1 +Y1 be the imbeddings defined earlier. Define an equivalence
relation on X1 + Y1 using these imbeddings and so on. The details are left to the reader.

120
G Tychonoff ’s Theorem
Q
Theorem 119 (Tychonoff). The product of compact spaces is compact. That is, if X = Xα
where each Xα is compact, X is compact.

Proof. Let F0 be a family of closed sets in X with the finite intersection property (f.i.p). We
shall show that there is a point common to all the sets F ∈ F0 .
We apply Zorn’s lemma to get a family F ⊆ F0 of (not necessarily closed) sets in X
with finite intersection property: Two families F and G are related iff F ⊆ G. Now let C
be any totally ordered chain of families with finite intersection property. That is, if there
exists F, G ∈ C, then either F ⊆ G or G ⊆ F. This chain has an upper bound, viz.,
H = ∪F ∈C F, where H has the finite intersection property. To see that H has the finite
intersection property, let A1 , . . . , An ∈ H. Then there exists Fj ∈ C such that Aj ∈ Fj ∈ C.
Since C is totally ordered, and F1 , . . . , Fn are finite in number, there exists k with 1 ≤ k ≤ n
such that Fj ⊆ Fk for all k. But then A1 , . . . , An ∈ Fk and Fk has the finite intersection
property. Hence A1 ∩ · · · ∩ An 6= ∅.
Hence by Zorn’s lemma, there exists a maximal family F ∈ C, with F ⊇ F0 . Let F α
denote {E α := Pα (E), E ∈ F}. Then Fα ⊆ P (Xα ) has the finite intersection property,
(here Pα : X → Xα is the canonical projection map). For otherwise, E1α ∩ · · · ∩ Enα = ∅ will
imply E1 ∩ · · · ∩ En = ∅, where Pα (Ei ) = Eiα . Hence, F α = {E α } has finite intersection
property.
SinceQXα is compact, there exists xα ∈ ∩E α where the intersection is over all E α ∈ F α .
Let x ∈ Xα be such that x(α) := xα .

Claim: x ∈ ∩F ∈F F .

Since F ⊇ F0 , the claim completes the proof of the theorem.

Proof of the Claim:

Let U be an open set in X. By definition of product topology, there exists α1 , . . . , αn and

open sets Uαi ⊆ Xαi , 1 ≤ i ≤ n such that x ∈ ∩ni=1 Pα−1 i
(Uαi ) ⊆ U with x ∈ ∩ni=1 Pα−1 i
(Uαi ).
α
This implies xαi ∈ Uαi for all i. By hypothesis on xα ’s, xαi ∈ F αi for all Fαi ∈ F . Thati

is, Uαi ∩ F αi 6= ∅, for all Fαi ∈ F αi . Hence Pα−1

i
(Uαi ) has a non-empty intersection with
every F ∈ F. Therefore Pα−1 i
(Uαi ) ∈ F (otherwise F ∪ {Pα−1i
(Uαi )} ⊃ F and the former has
finite intersection property, contradicting the maximality of F). This being true for all i,
and F has finite intersection property, it follows that ∩ni=1 Pα−1
i
(Uαi ) ∈ F. Since F has the
finite intersection property, this basic open set and hence U intersects each member of F
non-trivially. Since U was an arbitrary open neighborhood of x, this means that x ∈ F , for
all F ∈ F. Hence the claim.

121
H Compact Spaces

H.1 Heine-Borel Theorem

Definition 120. A family {Uα : α ∈ I} of open sets in a topological space X is said to be

an open cover of a subset K ⊂ X if K ⊂ ∪α Uα .
We say that a subset K ⊆ X of topological space X is compact if for any given open cover
{Uα : α ∈ I}, we can find a finite subset F ⊂ I such that K ⊂ ∪α∈F Uα . We then say the
given open cover admits a finite subcover.
Lemma 121. Any compact subset of a hausdorff space is closed.

Proof. Let K be a compact subset of a hausdorff space X. Let z ∈ X be a limit point of

K in X. If z ∈ / K, then by Hausdorff property of X, for each x ∈ K, we can find open sets
Ux 3 x and Vx 3 z such that Ux ∩ Vx = ∅. The collection {Ux : x ∈ K} is an open covering
of the compact set K and hence it admits a finite subcover, say, Ui := Uxi , 1 ≤ i ≤ n. If we
let V := ∩ni=1 Vxi , then V is an open set containing z. Also, we observe that

V ∩ K ⊆ V ∩ (U1 ∪ · · · ∪ Un ) = ∪ni=1 (V ∩ Ui ) ⊆ ∪ni=1 (Vxi ∩ Uxi ) = ∅.

This contradicts the fact that z is a limit point of K.

Definition 122. We say that a subset A of a metric space (X, d) is said to be bounded if
there exists x0 ∈ X and R > 0 such that A ⊂ B(x0 , R).
It is easy to see that this is equivalent to requiring that for any given x ∈ X, there exists
R = R(x) such that A ⊂ B(x, R).
Remark 123. Our definition of bounded subsets uses only the primitive notion of a metric
space and is intuitive. By definition the empty set is bounded. The standard definition runs
as follows. Given a nonempty set A ⊂ X, we define its diameter

diam (A) := sup{d(x, y) : x, y ∈ A}, as an extended real number, possibly + ∞.

We say that A is bounded if A is empty or if A is nonempty and diam (A) < ∞. We leave
the equivalence of both the definitions as an easy exercise to the reader.
Lemma 124. Any compact subset of a metric space (X, d) is bounded.

Proof. Fix x ∈ X. Observe that K ⊂ ∪n∈N B(x, n). By compactness of K and by the fact
that B(x, m) ⊂ B(x, n) for m ≤ n, it follows that A ⊂ B(x, N ) for some N .

Theorem 125. Any closed and bounded interval [a, b] ⊂ R is compact.

Proof. Given an open cover U of [a, b], let

E := {x ∈ [a, b] | [a, x] is covered by finitely many elements of U}.

We note that E 6= ∅, since a ∈ E: For, [a, a] = {a} and since U is an open cover there exists
U ∈ U such that a ∈ U . Hence [a, a] is covered by the single element U . In fact, we can say

122
more. Since a ∈ [a, b], there exists U in the open cover and δ > 0 such that (a − δ, a + δ) ⊂ U .
Hence [a, a + δ/2] ⊂ U . In other words, a + 2δ ∈ E.
E is bounded by b. Hence the supremum of E, say β exists.
We claim that β ∈ E and that β = b. The claim proves the result. Suppose the claim is
false.
Now β ∈ [a, b] since [a, b] is closed. There exists V ∈ U such that β ∈ V . Hence there
exists ε > 0 such that (β − ε, β + ε) ⊆ V , as V is open. Assume that β 6= b. Then we may
assume that ε is so small that (β − ε, β + ε) ⊆ [a, b]. Since β = sup E, β − ε is not an upper
bound of E. Thus, there exists x ∈ E, such that β − ε < x ≤ β. Since x ∈ E, there exists
finitely many Ui ∈ U, 1 ≤ i ≤ n such that [a, x] ⊆ ∪ni=1 Ui . But then [a, β + 2ε ] ⊆ ∪ni=1 Ui ∪ V .
Hence β + 2ε ∈ E, a contradiction since β ≥ x, for all x ∈ E. Hence β = b.

Ex. 126. Let a, b, c, d ∈ R be such that b−a = d−c. Let S := [a, b]×[c, d] be the square in R2 .
The vertices of S are (a, c), (b, c), (b, d) and (a, d). We call the point (a, c) as the bottom left
vertex of S. The pair of midpoints of its opposite sides are given by ([a + b]/2, c), ([a + b]/2, d)
and (a, [c+d]/2), (b, [c+d]/2]). By joining the midpoints of opposite sides, we get four smaller
squares. Observe that if (a1 , c1 ) is the bottom left vertex of any of these squares, we have
a ≤ a1 and c ≤ c1 .

Theorem 127. A subset of R2 is compact iff it is closed and bounded.

Proof. Any compact subset of a metric space is closed and bounded in view of first two
lemmas.
Let K be a closed and bounded set in R2 . Then there exists R > 0 such that K ⊂ S :=
[−R, R] × [−R, R]. Since a closed subset of a compact set is compact, it suffices to show that
S is compact.
Suppose that S is not compact. Then there is an open cover {Ui : i ∈ I} of which there
is no finite subcover of S. Let us divide the square S into four smaller squares by joining the
pairs of midpoints of opposite sides. (See Exercise 126 above.) One of these square will not
have a finite subcover from the given cover. For, otherwise, all these four squares will have
finite subcovers so that S itself will admit a finite subcover. Choose one such smaller square
and call it S1 . Note that the length of its sides is R and that if (a1 , c1 ) is the bottom left
vertex of S1 , then a1 ≥ a0 = −R and c1 ≥ c0 = −R. We repeat the argument by subdividing
S1 into four squares and choosing one of the smaller squares which does not admit a finite
subcover of {Ui }. Call this smaller square as S2 . Note that the length of its sides is R/2 and
that if (a2 , c2 ) is the bottom left vertex of S2 , then a1 ≤ a2 and c1 ≤ c2 .
Proceeding recursively, we have a sequence of squares Sn such that Sn dose not admit a
finite subcover and the length of sides of Sn is 2−n+1 R and its bottom left vertex (an , cn ) is
such that an−1 ≤ an and cn−1 ≤ cn . Thus we have two sequences of real numbers (an ) and
(cn ). They are bounded and monotone. Hence there exist real numbers a and c such that
an → a and cn → c. It follows that (an , cn ) → (a, c) ∈ R2 . Since S is closed, we infer that
(a, c) ∈ S. Hence there is Ui0 in the open cover such that (a, c) ∈ Ui0 . Since Ui0 is open there
exists an r > 0 such that B((a, c), r) ⊂ Ui0 .
√
Choose n ∈ N so that (1) diam Sn = 2−n+1 2R < r/2 and (2) d((a, c), (an , cn )) < r/2.

123
We then have, for any (x, y) ∈ Sn ,
√
d((a, c), (x, y)) ≤ d((a, c), (an , cn )) + d((an , cn ), (x, y)) < r/2 + 2−n+1 2R < r.

Thus Sn ⊂ B((a, c), r) ⊂ Ui0 . But then {Ui0 } is a finite subcover for Sn , contradicting our
choice of Sk ’s. Therefore, our assumption that S is not compact is not tenable.

Theorem 128. A subset of Rn is compact iff it is closed and bounded.

Proof. One can adapt the proof of Thm. 127 to prove the theorem including the case when
n = 1. We leave the details to the reader.

H.2 Characterization of Compact Metric Spaces

Ex. 129. Let A be a compact subset of (X, d). Let ε > 0 be given. Show that there exist
finitely many points xk ∈ X, 1 ≤ k ≤ n such that A ⊂ ∪nk=1 B(xk , ε). Hint: Consider the
open cover {B(x, ε) : x ∈ X} of X.

This exercise motivates the following definition which should be thought of as the back-
door entry of compactness!

Definition 130. A subset A ⊂ X of a metric space is said to be totally bounded if for every
ε > 0 there exists a finite number of points in X, say, xj , 1 ≤ j ≤ n, such that A ⊂ ∪k B(xk , ε).
(The number n may depend up on ε.

Ex. 131. Show that any totally bounded subset is bounded. Is the converse true? Hint:
Consider an infinite set with discrete metric.

Ex. 132. Show that in R with the usual metric a set is bounded iff it is totally bounded.
Extend this result to Rn .

Theorem 133. For a metric space (X, d), the following are equivalent:
(1) X is compact: every open cover has a finite subcover.
(2) X is complete and totally bounded.
(3) Every infinite set has a cluster point.
(4) Every sequence has a convergent subsequence.

Proof. (1) =⇒ (2): Let (X, d) be compact. Given ε > 0, {B(x, ε) | x ∈ X } is an open
cover of X. Let {B(xi , ε) | 1 ≤ i ≤ n } be a finite subcover. Hence X is totally bounded.
Now let (xn ) be a Cauchy sequence in X.
1
 every k ∈ N1
 Then for there exits nk such that
d(xn , xnk ) < k for all n > nk . Let Uk := x ∈ X  d(x, xnk ) > k . Then Uk is open: If
y ∈ Uk and δ := d(xnk , y) − k1 then B(y, δ) ⊆ Uk . Now xn ∈ / Uk for n > nk . Hence no finite
subcover of Uk ’s cover X: For, if they did, say, X = ∪m i=1 i we take n > max{n1 , . . . , nm }.
U
Then xn ∈ / Uk for any k with 1 ≤ k ≤ m. This implies that {Uk } cannot cover X. Thus there
exists x ∈ X \ ∪∞ 1
k=1 Uk . But then d(x, xnk ) < k . Hence xnk → x. Since (xn ) is Cauchy we see
that xn also converges to limk xnk . Thus X is complete. We have thus shown (1) implies (2).

124
 (2) =⇒ (3): Let E be an infinite subset of X. Let Fn be a finite subset of X such
that X = ∪x∈Fn B(x, n1 ). Then for n = 1 there exists x1 ∈ F1 such that E ∩ B(x1 , 1) is
infinite. Inductively choose xn ∈ Fn such that E ∩ (∩nk=1 B(xk , k1 )) is infinite. Since there
1
is a ∈ E ∩ B(xm , m ) ∩ B(xn , n1 ) we see that d(xm , xn ) ≤ d(xm , a) + d(a, xn ) < m
1
+ n1 < m
2

if m < n. Thus (xn ) is Cauchy. Since X is complete xn converges to some x ∈ X. Also

d(x, xn ) < n2 for all n. Thus B(x, 3/n) includes B(xn , n1 ) which includes infinitely many
elements of E. Thus x is a cluster point of E. Hence (3) is proved.
(3) =⇒ (4): If (xn ) is a sequence in X we let {xn | n ∈ N } be its image. If this set is
finite then (4) trivially follows. So assume that {xn | n ∈ N } is infinite. Let x be a cluster
point of this set. Then there exist elements xnk such that d(x, xnk ) < k1 for all k. Thus
xnk → x and (4) is thereby proved.
(4) =⇒ (1): Let {Uα } be an open cover of X. For x ∈ X, let

rx := sup {r ∈ R | B(x, r) ⊆ for some Uα } .

We claim that ε := inf {rx | x ∈ X } > 0. If not there is a sequence (xn ) such that rxn → 0.
But (xn ) has a convergent subsequence, say, xnk → x. Now x ∈ Uα for some α and hence
there is an r > 0 such that B(x, r) ⊂ Uα . For k large enough d(x, xnk ) < 2r so that rxnk > 2r
for all sufficiently large k – a contradiction. Hence the claim is proved.
Let ε := inf {rx | x ∈ X }. Choose any x1 ∈ X. Inductively choose xn such that xn ∈ /
∪n−1
k=1 B(xi , ε/2).We cannot do this for all n. For otherwise, (xn ) will not have a convergent
subsequence since d(xn , xm ) > 2ε for all m 6= n. Hence X = ∪N ε
k=1 B(xk , 2 ) for some N . But
ε N
then for each k there is an αk such that B(xk , 2 ) ⊂ Uαk . Hence X = ∪k=1 Uαk . Thus {Uα }
has a finite subcover or X is compact.

Ex. 134. Prove Thm. 127 using the fourth characterization (in Thm. 133) of compact metric
spaces.

A slight reordering of the equivalences is established in the following result.

Theorem 135. For a metric space (X, d), the following are equivalent:
(1) X is compact.
(2) Every infinite set has an accumulation point.
(3) Every sequence has a convergent subsequence.
(4) X is complete and totally bounded.

Proof. (1) =⇒ (2) Let E ⊆ X be an infinite set. Assume that E has no accumulation point.
Then for any ε > 0, and x ∈ X, the ball B(x, ε) has only finitely many points of E, i.e.,
B(x, ε) ∩ E is a finite set for every x ∈ E. Since {B(x, ε) | x ∈ X} is an open covering of X
and X is compact there exists a finitePnumber of x, say, x1 , . . . , xn such that X ⊆ ∪ni=1 B(xi , ε).
But then E = E ∩ X, so that |E| ≤ |E ∩ B(xi , ε)|, a finite number - a contradiction.
(2) =⇒ (3) Let n 7→ xn be a sequence in X. If there exists an x ∈ X such that for an
infinite number of n’s we have x = xn , (3) follows. For we take n1 to be the first n such that
xn = x. Let n2 > n1 be the first n such that xn = x and so on. Then k 7→ nk 7→ xnk is a
subsequence of the original sequence and it is convergent, as it is a constant sequence.

125
 So, we now assume the set E := {xn | n ∈ N} is infinite. If it were finite, we are through
by the last paragraph. By (2) there exists an x ∈ X which is an accumulation point of E. But
this means that for every k ∈ N, there exists a nk such that xnk ∈ B(x, k1 ). k 7→ nk 7→ xnk is
thus a subsequence which converges to x. Thus (3) is proved.
(3) =⇒ (4) Let n 7→ xn be a Cauchy sequence in X. By (3), there exists a convergent
subsequence k 7→ nk 7→ xnk with limn→∞ xnk = x. We claim that limn→∞ xn = x. For, given
ε > 0, we choose N > 0 such that d(xn , xm ) < 2ε for n, m ≥ N . Also, we choose M > 0 such
that d(xn , x) < 2ε for n ≥ M . Let n0 = max{N, M }. Then for all m ≥ n0 , we have

d(x, xn ) ≤ d(x, xnk ) + d(xnk , xm )

 
≤ +
2 2
= ε

for any k such that nk ≥ n0 . Hence the claim.

We need to prove that X is totally bounded. If X is not totally bounded, then there
exists ε > 0 such that no finite number of ε-balls cover X. Let x1 ∈ X be arbitrary. Since
B(x1 , ε) 6= X, there exists x2 ∈ X \ B(x1 , ε). Assume that we have chosen x1 , . . . , xk ∈ X
j−1
such that xj ∈/ ∪i=1 B(xi , ε). Then there exists xk+1 ∈ X \ ∪ki=1 B(xi , ε). Thus k 7→ xk is a
sequence in X. It cannot have any convergent subsequence as d(xi , xj ) ≥ ε for all i 6= j. This
contradicts our hypothesis that X satisfies (3).
(4) =⇒ (1) Suppose X is not compact. Then there exists an open cover {Uα } of X which
admits no finite subcover. Since X is totally bounded, there exists a finite number of y’s such
that X = B(y1 , 1) ∪ B(y2 , 1) ∪ · · · ∪ B(yn , 1). Since X is not covered by finitely many Uα ’s
it follows that there exists yj such that B(yj , 1) is not covered by finitely many of Uα ’s. Let
x1 := yj . Now consider X1 := B(x1 , 1). Then (X1 , d) is totally bounded. Hence there exists
finitely many y ∈ X1 such that x1 = ∪nj=1 [B(yj , Y2 ) ∩ X1 ]. Note that
1 1 1
BX1 (y, ) = {z ∈ X1 | d(z, y) < } = B(y1 , ) ∩ X1 .
2 2 2
As above, there exists y2 ∈ X1 such that BX1 (yj , 21 ) is not covered by finitely many {X1 ∩Uα }.
Call it X2 . We proceed by induction to choose xk such that BXk−1 (xk , k1 ) is not covered by
1
finitely many {Xk ∩ Uα }α . Now clearly {xk } is a Cauchy sequence in X: d(xm , xn ) ≤ m if
m ≤ n. Since X is complete xn → x for some x ∈ X. Let x ∈ Uα for some α. Then, all
xn ∈ Uα for n ≥ N (α). In particular, B(xn , n1 ) ⊆ Uα for n sufficiently large. This contradicts
1
our choice of xn , viz., B(xn , m ) is not a subset of finitely many Uα ’s.

Theorem 136. [a, b] is compact.

Proof. Given an open cover U of [a, b], let

E := {x ∈ [a, b] | [a, x] is covered by finitely many elements of U}.

We note that E 6= ∅, since a ∈ E: For, [a, a] = {a} and since U is an open cover there exists
U ∈ U such that a ∈ U . Hence [a, a] is covered by the single element U .
E is bounded, obviously by b. Hence the supremum of E, say β exists. If β = b, we are
through. Suppose not.

126
 Now β ∈ [a, b] and hence there exists V ∈ U such that β ∈ V . Hence there exists ε > 0
such that (β − ε, β + ε) ⊆ V , as V is open and (β − ε, β + ε) ⊆ [a, b]. Since β = sup E,
β − ε is not an upper bound of E. Thus, there exists x ∈ E, such that β − ε < x ≤ β. Since
x ∈ E, there exists finitely many Ui ∈ U, 1 ≤ i ≤ n such that [a, x] ⊆ ∪ni=1 Ui . But then
[a, β + 2ε ] ⊆ ∪ni=1 Ui ∪ V . Hence β + 2ε ∈ E, a contradiction since β ≥ x, for all x ∈ E. Hence
β = b.

H.3 Tychonoff ’s Theorem

Let X be a set. We say that a collection A of (nonempty) subsets of X has finite intersection
property (f.i.p., in short) if every finite family A1 , . . . , An of elements in A has a nonempty
intersection.
Ex. 137. A topological space is compact iff every family of closed sets with f.i.p. has
nonempty intersection. Hint: Start with an open cover U which does not admit a finite
subcover. Look at {X \ U : U ∈ U}.

Let us briefly review the product topology.

Q Given a family {Xα : α ∈ I} of topological
spaces, the topology on the product set α∈I Xα is the weakest (or the smallest topology)
which makes the canonical projection maps Pα : X → Xα continuous. Hence for any open
set Uα ⊂ Xα , the set Pα−1 (Uα ) must be declared open in X. Finite intersections of such sets
form a basis for the product topology, that is, U ⊂ X is open iff for each x ∈ U , there exists
a finite subset F ⊂ I such that x ∈ ∩α∈F Pα−1 (Uα ) for some open sets Uα ⊂ Xα .
To bring out the main ideas of the proof clearly, let us list the following two exercises
which are set-theoretic in nature. Their solutions can be gleaned from the proof of the
theorem below.
Ex. 138. Let A be a family of subsets of a set X with f.i.p. Then there exists a ‘maximal’
family B containing A with f.i.p., that is, a family B of subsets such that (i) A ⊂ B, (ii) B
has f.i.p. and (iii) if C is any family with f.i.p. such that A ⊂ C , then C ⊂ B. Hint: Partially
order the set of all collections with properties (i) and (ii) by inclusion. Apply Zorn’s lemma.
Ex. 139. Let B be a family of subsets of a set X which is maximal with respect to finite
intersection property. Then (i) if A ⊂ X has nonempty intersection with each member of B,
then A ∈ B and (ii) the intersection of any finite number of elements of B again lies in B.
Theorem 140 (Tychonoff). The product
Q of compact spaces is compact. That is, if Xα is
compact for each α ∈ I and if X := α∈I Xα is endowed with the product topology, then X
is compact.

Proof. We plan to use Ex. 137. Let F0 be a family of closed sets in X with the finite
intersection property (f.i.p.). It suffices to show that there is a point common to all the
sets F ∈ F0 . We use Ex. 138 to get maximal family F ⊇ F0 . The details are in the next
paragraph.
Consider the class of all families F of (not necessarily closed) subsets such that F0 ⊂ F
which have f.i.p. For two families F and G in this class we say that F ≤ G iff F ⊆ G. Now let

127
C be any totally ordered chain in this class, that is, if F, G ∈ C, then either F ⊆ G or G ⊆ F.
This chain has an upper bound, viz., H = ∪F ∈C F. We need only show that H has f.i.p. Let
A1 , . . . , An ∈ H. Then there exists Fj ∈ C such that Aj ∈ Fj ∈ C. Since C is totally ordered,
and F1 , . . . , Fn are finite in number, there exists k with 1 ≤ k ≤ n such that Fj ⊆ Fk for all
j. Hence A1 , . . . , An ∈ Fk . Since Fk has f.i.p., A1 ∩ · · · ∩ An 6= ∅. Thus H has f.i.p. and hence
is an upper bound for the chain. Therefore, by Zorn’s lemma, there exists a maximal family
F ∈ C, with F ⊇ F0 .
Let F α denote {E α := Pα (E), E ∈ F }, where Pα : X → Xα is the canonical projection
map. Then Fα ⊆ P (Xα ), the power set of Xα , has f.i.p. For otherwise, E1α ∩ · · · ∩ Enα = ∅
will imply E1 ∩ · · · ∩ En = ∅, where Pα (Ei ) = Eiα . Hence, F α = {E α } has finite intersection
property.
SinceQXα is compact, there exists xα ∈ ∩E α where the intersection is over all E α ∈ F α .
Let x ∈ Xα be such that x(α) := xα . We claim that x ∈ ∩F ∈F F . Since F ⊇ F0 and since
every element of F0 is closed, the claim completes the proof of the theorem.
We now prove the claim. Let U be an open set in X. By definition of product topology,
there exists α1 , . . . , αn and open sets Uαi ⊆ Xαi , 1 ≤ i ≤ n such that x ∈ ∩ni=1 Pα−1
i
(Uαi ) ⊆ U .
This implies xαi ∈ Uαi for all i. By hypothesis on xα ’s, xαi is in the closure of Fαi for all
Fαi ∈ F αi . Select yαi ∈ Uαi ∩Fαi and a y ∈ F such that Pαi (y) = yαi . Then y ∈ Pα−1 i
(Uαi )∩F .
Thus Pα−1i
(U αi ) has a non-empty intersection with every F ∈ F. Therefore P −1 (U ) ∈ F by
αi αi
Ex. 139.
(Reason: Otherwise F ∪ {Pα−1i
(Uαi )} ⊃ F and the former has finite intersection property,
contradicting the maximality of F.)
Using Ex. 139 again, we infer that ∩ni=1 Pα−1
i
(Uαi ) ∈ F.
(Reason: Since F, Pα−1 −1 (U ) 6= ∅. Thus ∩ P −1 (U )


i
(U α i ) ∈ F, we see that F ∩ ∩ i P α i α i i αi αi
meets every element of F. Thus F ∪ {∩i Pα−1 i
(U α i )} has f.i.p. Since F is maximal with respect
to f.i.p. it follows that ∩i Pα−1
i
(Uαi ) ∈ F.)
Since F has f.i.p., this basic open set and hence U intersects each member of F non-
trivially. Since U was an arbitrary open neighborhood of x, this means that x ∈ F , for all
F ∈ F. Hence the claim.

Ex. 141. Prove Thm.128 using Tychonoff’s theorem.

Acknowledgement: I thank Vikram Aithal and Rohit Gupta whose discussions with me on
this subsection helped me improve its readability.

H.4 Continuous Functions on Compact Spaces

Theorem 142. Let f : X → C be a continuous function from a compact (metric) space to C.

Then
(1) f is bounded, i.e., there exists a constant A > 0 such that |f (x)| ≤ A for all x ∈ X.
(2) If we further assume that f (x) > 0 for all x ∈ X, then there exists B > 0 such that
f (x) ≥ B for all x ∈ X.

128
(3) If f is real-valued, then there exists p ∈ X and q ∈ X such that f (p) ≥ f (x) for all x ∈ X
and f (q) ≤ f (x) for all x ∈ X.

Proof. To prove (1), consider the open sets Un := {x ∈ X : |f (x)| < n} for n ∈ N. Then
Un ⊂ Un+1 and X = ∪Un . By compactness, we conclude that X = UN for some N . (1)
follows.
To prove (2), consider Vn := {x ∈ X : f (x) > 1/n} and argue as in (1).
We now prove (3). By (1), there exits α ∈ R such that |f (x)| ≤ α for all x ∈ X. Let
M := sup{f (x) : x ∈ X}. By (1), M exists. If there is no p ∈ X such that f (p) = M , then
the sets
1
Un := {x ∈ X; f (x) < M − }
n
form an open cover of X. As in (1), we conclude that X = UN for some N . But this leads to
the contradiction: sup{f (x) : x ∈ X} ≤ M − N1 !
Note that (2) can be deduced from (3).
The following is an easy but a most useful result.

Theorem 143. Let f : X → Y be a bijective continuous map from a compact space X to a

hausdorff space Y . Then f is a homeomorphism.

Proof. It is enough to show that f is a closed map, that is, it maps closed sets of X to closed
sets of Y . Let K be a closed subset of X. Then K is compact being a closed subset of a
compact space. Since f is continuous, f (K) is compact. Since f (K) is a compact subset of
a hausdorff space Y , we deduce that f (K) is closed in Y . Thus, any closed subset of X is
mapped to a closed subset of Y .

H.5 Characterization of Compact Metrizable Spaces

via Metrics and Continuous Functions

A topological space X is said to be metrizable if there exists a metric on X such that the
given topology coincides with the topology defined by the metric.

Theorem 144. For a metrizable topological space X, the following properties are equivalent:
1. X is compact.
2. Every metric on X inducing the given topology is bounded.
3. Every continuous (real valued) function on X is bounded.

Proof. We now prove the theorem according to the pattern:

(1) =⇒ (2) =⇒ (3) =⇒ (1).

(1) =⇒ (2): If X is compact, so is X × X. Any metric d on X inducing the given

topology on X is a continuous function on X × X, whence bounded.

129
 (2) =⇒ (3): Let f be a continuous function on X. We then push points of X apart
at distances bounded from below by f using the following standard technique. Consider the
graph Z of f :

Z := {(x, f (x)) : x ∈ X} ⊆ X × R.

The map i : X ,→ Z given by x 7→ (x, f (x)) is then a homeomorphism of X onto Z, its

inverse being given by the restriction of the first projection p : X × R −→ X to Z. The space
X × R with the product topology is metrizable; e.g. one may take the metric δ defined as

δ((x, s), (y, t)) := d(x, y) + |t − s|.

Pulling this metric back to X using the map i therefore equips X with a metric d0 inducing
the topology given by d, which therefore by assumption is bounded, by a constant B > 0,
say. Now by construction

d0 (x, y) = d(x, y) + |f (y) − f (x)| ,

so d0 being bounded by B implies

|f (y)| ≤ |f (x)| + B,

for all x, y ∈ X. If we fix x ∈ X, the inequality above shows that f is bounded.

(3) =⇒ (1): We show that on any noncompact metrizable space X there exists a
continuous unbounded function. Let d be any metric on X inducing the given topology and
let X 0 be the completion of X with respect to d. We distinguish the cases X 0 being compact
and being not so.

Case (i): X 0 compact. Since X is assumed to be noncompact, X 6= X 0 whence X 0 \ X is

not empty. Let x∞ be a point in X 0 \ X. Since X is dense in X 0 , the function f defined by
f (x) := 1/d(x, x∞ ) is then a continuous function on X which is not bounded.

Case (ii): X 0 noncompact. If f is a continuous unbounded function on X 0 , its restriction

to X is a continuous unbounded function on X. So we may assume X itself is complete.
According to the standard characterization of compactness, X cannot be totally bounded
since it is assumed to be noncompact. So there is a real number ε > 0 such that X cannot
be covered by finitely many closed ε-balls. Let x1 be any point in X and put r1 := ε.
Then the closed ball B[x1 , r1 ] does not cover X. So there is x2 in X \ B[x1 , r1 ]. The latter
complement being open there is r2 with r1 ≥ r2 > 0 such that B[x2 , r2 ] ⊆ X \ B[x1 , r1 ]. The
balls B[x1 , r1 ] and B[x2 , r2 ] together do not cover X, so there are x3 and r3 with B[x3 , r3 ] ⊆
X \ B[x1 , r1 ] ∪ B[x2 , r2 ] and r1 ≥ r2 ≥ r3 > 0. Continuing this way we obtain sequences
x1 , x2 , . . . and r1 ≥ r2 ≥ · · · > 0. They have the property that the balls B[xk , rk ] are
mutually disjoint. Now we define f : X → R as follows:

∞
X d(x, X \ B(xk , rk ))
f (x) := k· .
d(x, xk ) + d(x, X \ B(xk , rk ))
k=1

130
(Visualize f in the case of X = R and xk = k and rk = 2−k , say.) If the k-th term of the
sum that defines f (z) is nonzero, it means that z ∈ B[xk , rk ] and hence all other terms of
the series that defines f (z) are zero, since the balls B[xk , rk ] are mutually disjoint. Hence the
series is convergent and f (x) makes sense for any x ∈ X. We thus get a well-defined function
f on X. Since f (xk ) = k, f is not bounded. It is easily seen to be continuous on X. For,
if x ∈ U := X \ ∪k∈N B[xk , rk ], then f (x) = 0 and since U is open (why?), f is zero in an
open set containing x. If x ∈ B[xk , rk ], then f is just the k-th term of the series, which is
continuous. This finishes the proof.

Remark 145. The case (ii) of (3) =⇒ (1) can also be seen as follows. If X is not totally
bounded, there is some ε > 0 such that no finite collection of balls of radius ε covers X. So
we can pick x1 in X, x2 ∈ X \ B(x1 , ε), x3 ∈ X \ (B(x1 , ε) ∪ B(x2 , ε)), and so on. Each
d(xi , xj ) ≥ ε for i 6= j. Hence there are no nonconstant Cauchy sequences among the xi . So,
the set {xi } is closed in X and also discrete. If we now define f (xi ) = i, then f is continuous
function on the discrete set. We can extend this by the Tietze theorem to f : X −→ R. The
function f is clearly an unbounded continuous function.

131
I Connected and Path Connected spaces

The aim of this article is to introduce the readers to an easier way of working with connect-
edness concept. If the reader’s background does not include general (abstract) topological
spaces, he may assume that the spaces are metric spaces.

I.1 Connectedness

Definition 146. A topological space X is said to be connected, if the only subsets of X which
are both open and closed are the empty set ∅ and X. In other words, a topological space is
connected whenever a subset A is both open and closed in X, then either A = ∅ or A = X.
A subset A of a topological space X is said to be connected if A is a connected space
when considered as a topological space with the induced (or subspace) topology. In the
case of metric space (X, d), this amounts to saying that (A, δ) is connected, where δ is the
restriction of the metric d on X to A.

Therefore, if a topological space X is not connected, there will be a proper non-empty

sub set A of X which is both open and closed in X. If A is a proper non-empty sub set of
X and both open and closed, then B = Ac , its complement is also a proper non-empty sub
set of X which is both open and closed in X. In other words a topological space X is not
connected iff there exist two disjoint proper non-empty sub sets A and B such that A and B
are both open and closed in X and X = A ∪ B. In such case we also say that the pair (A, B)
is a disconnection of X.
Example 147. Let X be a set such that | X |≥ 2 with discrete topology (or discrete metric).
Then X is not connected.
Example 148. The subset {±1} ⊂ R with the subspace topology is not connected. (Why?)
In the sequel, we consider {±1} as a subset of R.

Now we prove a single most important theorem in connectedness which supplies us an

abundance of examples of connected and non-connected spaces.
Theorem 149. A topological space X is connected iff every continuous function f : X →
{±1} is a constant function.

Proof. Let X be a connected space and f : X → {±1} a continuous function. We want to

show that f is a constant function. If f is non-constant, then it is on-to. Let A = f −1 (1)
and B = f −1 (−1). Then A and B are disjoint non-empty subsets of X such that A and
B are both open and closed subsets of X and X = A ∪ B.(Why?). This is a contradiction.
Therefore f is constant.
Conversely, let us assume that X is not connected. Therefore there exist two disjoint
proper non-empty subsets A and B in X such that A and B are both open and closed in X
and X = A ∪ B. Now we define a map f : X → {±1} as
(
1 if x ∈ A
f (x) =
−1 if x ∈ B

132
Then f : X → {±1} is a continuous non-constant function. (Why?). This completes the
proof.

We shall now use this theorem to get examples of connected spaces.

Example 150. A set J ⊆ R is connected iff J is an interval.

Proof. Let J be a connected sub set of R. Let us assume that J is not an interval. This
means that there exist points a < b in J and c ∈ R such that a < c < b but c ∈
/ J. Now we
define a map f : J → {±1} as (
1 if x < c
f (x) =
−1 if x > c
Now we claim that f is a continuous function. We need only to check that f −1 (1) and f −1 (−1)
are open in J. By our definition f −1 (1) = J ∩ (−∞, c) and f −1 (−1) = J ∩ (c, ∞) which are
open, proper and nonempty subsets in J. (Why?) This is a contradiction to the fact that J
is connected. Therefore for every pair of points a and b in J such that a < b, all the points c
such that a < c < b are also in J. This means that J is an interval in R.
Conversely, let us assume that J is an interval in R. Let f : J → {±1} be a continuous
function. We need to show that f is constant. If not, then there exist a, b ∈ J such that
f (a) = 1 and f (b) = −1. Since a, b ∈ J and J is an interval, [a, b] ⊂ J. Hence, by applying
the intermediate value theorem to the restriction f to [a, b], there exists c ∈ (a, b) ⊂ J such
that f (c) = 0. This is a contradiction, since the codomain is {±1}.

Example 151. Let M(2, R) denote the set of all 2 × 2 matrices of real numbers. and
GL(2, R) := {A ∈ M(2, R) : det(A) 6= 0} is not connected.
 
a b
Proof. Here we identify M (2, R) with R4 via the map 7→ (a, b, c, d) ∈ R4 . Let
c d
f : GL(2, R) → R be defined by f (A) := det(A). Complete the proof.

Example 152. O(2, R) := {A ∈ GL(2, R) : AAt = Id} is not connected.

Proof. The equation AAt = Id shows that det(A) = ±1 for every A ∈ O(2, R). This suggests
us that we define the map f : O(2, R) → {±1} by f (A) := det(A). Complete the proof.

Proposition 153. Let X be a topological space. Let A and B be two connected subsets of X
such that A ∩ B 6= ∅. Then A ∪ B is connected.

Proof. Let f : A ∪ B → {±1} be a continuous function. We have to show that f is constant.

Let c ∈ A ∩ B. Since A is connected, the function f |A : A → {±1} is constant so that
f (a) = f (c) for all a ∈ A. Similarly, f (b) = f (c) = 1 for all b ∈ B. Thus f (x) = f (c) for all
x ∈ A ∪ B. i.e., f is a constant.

Proposition 154. Let A be a connected subset of a space X. Let A ⊂ B ⊂ A. Then B is

connected.

133
Proof. Let f : B → {±1} be a continuous function. Without loss of generality, let us assume
that f = 1 on A. Let x ∈ B. Since {f (x)} is open in {±1}, the set U := f −1 (f (x)) is an
open containing x. Hence, there exists a point a ∈ A ∩ U . Since a, x ∈ U and f = f (x) on U ,
it follows that f (x) = f (a) = 1. Thus f = 1 on B.

Proposition 155. Let {Ai : i ∈ I} be a collection of connected subsets of a space X with the
property that for all i, j ∈ I we have Ai ∩ Aj 6= ∅. Then A := ∪i Ai is connected.

Proof. Fix Ai . Let f : ∪ Aj → {±1} be continuous. Since Ai is connected, f is a constant on

it, say, f = 1 on Ai . Let x ∈ A. Then x ∈ Aj for some j. Let y ∈ Ai ∩ Aj . Then f (x) = f (y)
since Aj is connected and x, y ∈ Aj . Since y ∈ Ai , we have f (y) = 1. Hence for all x ∈ A, we
conclude f (x) = 1. Hence A is connected.

Proposition 156. Let X be a connected topological space and g : X → Y be a continuous

map. Then g(X) is connected.

Proof. We will show that any continuous map f : g(X) → {±1} is constant.
Let f : g(X) → {±1} be a continuous map. Then the map f ◦ g : X → {±1} is continuous.
(Why?). Since X is connected, it follows that g ◦ f is constant. Hence f is constant. For,
otherwise, there exist y1 , y2 ∈ g(X) such that f (y1 ) 6= f (y2 ). Since yj ∈ g(X), this implies
the existence of xj ∈ X such that g ◦ f (x1 ) 6= g ◦ f (x2 ). In particular, g ◦ f is not a constant.
Hence we are forced to conclude that f is constant. Thus g(X) is connected.

Corollary 157. In the above proposition, if the map g is onto then Y is connected.
Ex. 158. Show that the set GL(2, R) is not connected.
Ex. 159. Show that the circle {(x, y) ∈ R2 : x2 + y 2 = 1} is connected.
Ex. 160. Show that the set SO(2, R) := {A ∈ O(2, R) : det A = 1} is connected. Hint:
Write down all elements of SO(2, R) explicitly.
Proposition 161. Let X and Y be connected spaces. Then the product space X × Y is
connected.

Proof. Let f : X × Y → {±1} be a continuous map. Let (x0 , y0 ) ∈ X × Y be fixed. Let (x, y)
be an arbitrary point in X × Y . If we show that f ((x, y)) = f ((x0 , y0 )), we are through.
To prove the above claim, let us first observe that for every point y ∈ Y , the map
iy : X → X × Y defined by iy (x) := (x, y) is continuous; similarly the map ix : Y → X × Y
defined by ix (y) := (x, y) is continuous for every point x in X. Therefore for every point y
in Y , the subset X × {y} := {(x, y) : x ∈ X} is a connected subset of X × Y ; similarly, the
subset {x} × Y := {(x, y) : y ∈ Y } is a connected subset of X × Y for every point x in X.
Now the point (x, y0 ) lies in both sets X × {y0 } and {x} × Y . The restrictions of f to
either of these sets are continuous and hence constants. We see that f (x0 , y0 ) = f (x, y0 )) for
all x ∈ X and similarly, f (x, y) = f (x, y0 ) for all y ∈ Y . In particular, f (x, y) = f (x, y0 ) =
f (x0 , y0 ). (See Figure ??.)

The following is a typical way in which connectedness hypothesis is used.

134
Theorem 162. Let X be connected. Let f : X → R be a locally constant function, i.e., for
each x ∈ X, there exists an open set Ux containing x with the property that f is a constant
on Ux . Then f is a constant on X.

Proof. First of all note that any locally constant function is necessarily continuous.
Fix x0 ∈ X. We show that f (x) = f (x0 ) for all x ∈ X. Consider the set E := {x ∈
X | f (x) = f (x0 )}. As x0 ∈ E, we see that E is nonempty. Since E = f −1 (f (x0 )), E is the
inverse image of a closed set under the continuous map f and hence is closed.
If x ∈ E, since f is locally constant, there exists an open set Ux with x ∈ Ux and f
is constant on Ux . Thus for each y ∈ Ux , we have f (y) = f (x). Since x ∈ E, we have
f (x) = f (x0 ). Hence it follows that f (x) = f (x0 ) for all x ∈ Ux . In other words, Ux ⊂ E.
Hence E is open. Thus E is nonempty, open and closed subset of the connected space X.
Hence we must have E = X.

As an immediate corollary we have

Theorem 163. Let U be an open connected subset of Rn and f : U → R be a differentiable

function such that Df (p) = 0 for all p ∈ U . Then f is a constant function.

Proof. To prove this theorem we will use the following fact which follows from mean value
theorem. Let U be an open convex subset of Rn and f : U → R be a differentiable function
such that Df (p) = 0 for all p ∈ U . Then f is constant on U .
Now let f be as in the theorem. Then for each x ∈ U , since U is open, there exists an
open ball B(x, rx ) ⊂ U . It is easy to see that any ball in Rn is convex. Thus an application
of the calculus result quoted above shows us that f locally constant.

I.2 Path Connected spaces

Definition 164. 1. Let X be a topological space. A continuous map γ : [0, 1] → X is

called a path in X. If γ(0) = x and γ(1) = y, then γ is also called a path joining the
points x and y or simply a path from x to y.

2. A topological space X is said to be path connected if for all points x and y in X, there
exists a path γ : [0, 1] → X such that γ(0) = x and γ(1) = y.

I.3 Examples & Exercises

Example 165. The space Rn is path connected. Any two points can be joined by a line
segment: γ(t) := x + t(y − x), for 0 ≤ t ≤ 1. We call this path γ a linear path.

Example 166. For every r > 0, the circle Cr := {(x, y) ∈ R2 : x2 + y 2 = r2 } is path

connected.(Why?)

Example 167. The set {(x, y) ∈ R2 : x ≥ 0 & x2 − y 2 = 1} is path connected. Draw the
picture and see that it is the “right” hand of the hyperbola x2 − y 2 = 1. Similarly the left

135
hand of a hyperbola is also path connected. However the hyperbola is not path connected.
(Why?)

Example 168. The parabola {(x, y) ∈ R2 : y 2 = x} is path connected.

Example 169. The union of the two parabolas {(x, y) ∈ R2 : y 2 = x} and {(x, y) ∈ R2 : y =
x2 } is path connected.

Example 170. The union of the parabolas {(x, y) ∈ R2 : y 2 = x} and {(x, y) ∈ R2 : y 2 = −x}
is path connected.

Example 171. The set S 2 := {(x1 , x2 , x3 ) ∈ R3 : x21 + x22 + x23 = 1} is path connected. Let X
X+t(Y −X)
and Y be two points in S 2 . Then define γ : [0, 1] → S 2 by γ(t) := k X+t(Y −X) k . Then check
that this gives us a path from X to Y . (Does it?).

Proposition 172. Let X be a topological space. Let γ1 : [0, 1] → X and γ2 : [0, 1] → X be two
paths such that γ1 (1) = γ2 (0). Then there exists a path γ3 : [0, 1] → X such that γ3 (0) = γ1 (0)
and γ3 (1) = γ2 (1).

Proof. Define the map γ3 : [0, 1] → X such that

(
γ1 (2t) if t ≤ 21
γ3 (t) : = 1
γ2 (2t − 1) if t ≥ 2

Now we leave it as an exercise to verify that γ3 is a path in X meeting our requirements.

(Draw pictures and see geometrically).

Proposition 173. Let X be path connected. Then X is connected.

Proof. Let f : X → {±1} be a continuous function. We need to show that f is constant.

Let x 6= y be two points in X. Since X is path connected, there exists a continuous
map γ : [0, 1] → X such that γ(0) = x and γ(1) = y. Now, the map f ◦ γ : [0, 1] → {±1}
is continuous. Since [0, 1] is connected, the map f ◦ γ is constant. Therefore f is constant.
(Why?). This proves that X is connected.

The converse is not always true. However, in the case of open subsets of Rn , the converse
is also true and we prove this in

Theorem 174. Let U be an open connected subset of Rn . Then U is path connected.

Proof. Let x0 be a point in U and let

E := {x ∈ U : there exists a path γ such that γ(0) = x & γ(1) = x0 }.

We will show that the set E is non-empty, both open and closed in U . Then since U is
connected, it will follow that E = U and this will prove the theorem. (Why?)
First we note that the set E is non-empty. The map γ : [0, 1] → X defined by γ(t) = x0
for all t is a path in X. Therefore x0 is in E. Let x be a point in E. Since U is open there

136
exists r > 0 such that B(x, r) ⊆ U . Let y be a point in B(x, r). Since B(x, r) is convex, there
exists a linear path, say, γ1 , joining the points y and x. Since x is in E there exists a path
γ2 from x to the point x0 . From Proposition 172, it follows that there exists a path γ3 from
y to x0 . This means that B(x, r) ⊆ E. Hence E is open.
We will now show that E is also closed in U . Let x ∈ U be a limit point of E. Therefore
there exists a sequence xn of points in E such that the sequence xn converge to the point x.
Since U is open there exists an r > 0 such that the open ball B(x, r) ⊆ U . Since the sequence
xn converges to the point x, there exists N in N such that the points xn ∈ B(x, r) for all
n ≥ N . Let γ1 be the linear path from x to the point xN and γ2 be a path from xN to x0 .
From Proposition 172, there exists a path γ3 from x to x0 . This means that the point x is in
E. Hence E is closed and therefore E = U .

137
J Proper Maps

Definition 175. A map f : X → Y is said to be proper if for every compact subset L ⊂ Y ,

the inverse image f −1 (L) is a compact subset of X.
Example 176. Any continuous map from a compact space to any hausdorff space Y is
proper.
Example 177. Let p be a nonconstant polynomial with complex coefficients. A most im-
portant and typical example of a proper map is the function z 7→ p(z). Recall the standard
estimate: There exists R > 0 such that
n
|an | n X
|p(z)| ≥ |z| , for |z| ≥ R, where p(z) = ak z k .
2
k=0

Let K be a compact subset of C. Since p is continuous, p−1 (K) is closed. If p−1 (K) is not
compact, we conclude that p−1 (K) is not bounded. (Why? Heine-Borel theorem!) Hence
there exists a sequence zn ∈ p−1 (K) such that |zn | → ∞, but p(zn ) ∈ K for all n. By the
estimate quoted above, p(zn ) → ∞. But since p(zn ) ∈ K and K is compact, {p(zn ) : n ∈ N}
is bounded. This contradiction shows that p is proper.
Ex. 178. The exponential map exp : R → R or exp : C → C is not proper.
Lemma 179. Let f : X → Y be a closed map. Assume that f −1 (y) is compact for each
y ∈ Y . Then f is proper.

Proof. Let L ⊂ Y be a compact subset. Let {Ui : i ∈ I} is an open cover of K := f −1 (L).

For each y ∈ L, by hypothesis, f −1 (y) is compact. Hence, there exists a finite set Jy ⊂ I such
that {Ui : i ∈ Jy } is a finite subcover of f −1 (y). Let Uy := ∪i∈Jy Ui . Then Uy is open and so
Ay := X \ Uy is closed in X. Since f is closed, the set Vy := Y \ f (Cy ) is open in Y . Note
that f −1 (Vy ) ⊂ Uy . Since y ∈ Vy , the collection {Vy : y ∈ L} is an open cover of the compact
set L. Hence there exists a finite number of points yj , 1 ≤ j ≤ n such that L ⊂ V1 ∪ · · · ∪ Vn
where Vj := Vyj . But then

f −1 (L) ⊂ f −1 (V1 ) ∪ · · · ∪ f −1 (Vn )

⊂ U1 ∪ · · · ∪ Un
= ∪{Ui : i ∈ Jyi , 1 ≤ i ≤ n},

a finite subcover.

Lemma 180. Let X be compact. Then for any topological space Y , the projection πY : X ×
Y → Y is closed.

Proof. Let L ⊂ X × Y be closed. We have to show that πY (L) is closed in Y . We show

that its complement is open in Y . Let y ∈ Y but y ∈ / πY (L). Note that this means that
(x, y) ∈ L for any x ∈ X. What we plan to do is something similar to the preliminary step,
the so-called tube lemma, in the proof of compactness of X × Y : There exists an open set V
such that y ∈ V and (x, y 0 ) ∈
/ L for any x ∈ X and y 0 ∈ V . From this it follows that such a
V ⊂ Y \ πY (L).

138
 Since L is closed and (x, y) ∈
/ L, we can find a basic open set Ux × Vx such that (x, y) ∈
Ux × Vx ⊂ (X × Y ) \ L. By the compactness of X, we can find x1 , . . . , xn ∈ X such that
Ui := Uxi , 1 ≤ i ≤ n, cover X. Let V := V1 ∩ · · · ∩ Vn , where, as is our standard practice
Vi := Vxi , 1 ≤ i ≤ n. Note that V is an open set containing y. We have

(X × Y ) ∩ L = [(U1 ∪ · · · ∪ Un ) × (V1 ∩ · · · ∩ Vn )] = ∅.

Proposition 181. If X is compact, then πY : X × Y → Y is proper.

Proof. Immediate consequence of the last two lemmas.

Theorem 182. If X and Y are compact, then X × Y is compact.

Proof. By the last proposition, the projection πY is proper and hence X × Y = πY−1 (Y ) is
compact.

The next theorem is the philosophical reason for the introduction of proper maps. Loosely
speaking, a continuous map is proper iff it maps points near to infinity to points near to
infinity. Compare and contrast the non-constant polynomial maps and the exponential maps.
We have a characterization of proper maps between locally compact hausdorff spaces in
terms of their one-point compactifications.
Given a locally compact noncompact hausdorff space X, let X∞ := X ∪ {∞} where
∞∈
/ X. Let T denote the topology on X. Consider

T∞ := T ∪ {V ⊂ X∞ : X∞ \ V is compact}.

Then
(i) T∞ is a hausdorff topology on X∞ .
(ii) The subspace topology on X is T .
(iii) (X∞ , T∞ ) is compact.
(iv) X is dense in X∞ .

Theorem 183. Let X and Y be locally compact hausdorff spaces. Then a continuous map
f : X → Y is proper iff it extends to a continuous map of X∞ to Y∞ with f (∞X ) = ∞Y .

Proof. Let f be proper. Extend f as above. Then we need to check its continuity. Let V
be open in Y . The f −1 (V ) is an open subset of X and hence of X∞ . If V 3 ∞Y , then
L := Y∞ \ V is a compact subset of Y and hence f −1 (L) is a compact subset of X, since f is
proper. Since X is hausdorff, f −1 (L) is closed. Hence X \ f −1 (L) is open. But it is nothing
but f −1 (V ).
Let f , the extension as in the statement, be continuous. Then f −1 (Y ) = X, since
f (∞X ) = ∞Y . If L ⊂ Y is compact, then L is closed in Y and hence in Y∞ . So f −1 (L) is
closed in X∞ . Since X∞ is compact, f −1 (L) is compact. It is clearly a subset of X. Hence f
is proper.

139
Proposition 184. Let f : X → Y be a proper map (i) either between two locally compact
hausdorff spaces or (ii) between two metric spaces. Then f is closed.

Proof. Assume Case (i). Let g denote the extension of f to X∞ . If F is closed in Y , then
F∞ := F ∪ {∞X } is closed in X∞ and hence is compact. Hence g(F∞ ) is compact in Y∞
and hence is closed, since Y∞ is hausdorff. But then f (F ) = g(F∞ ) ∩ Y is closed in Y . This
proves the result in the first case.
We can also prove this directly without recourse to the one-point compactifcations as
follows. Let C be closed in X. Let q ∈ Y be a limit point of f (C). Let V be an open set
such that q ∈ V and L := V is compact. (This is possible since Y is locally compact and
hausdorff.) Consider K := f −1 (L). Then K is closed, since f is proper. As K ∩ C is compact,
we have f (K ∩ C) = L ∩ f (C) (verify!) is compact and hence closed since Y is hausdorff.
Since q ∈ f (C), and V is an open neighbourhood of q, we see that

q ∈ L ∩ f (C) = L ∩ f (C) = f (K ∩ C) ⊂ f (C).

This shows that any limit point q of f (C) lies in f (C) and hence f (C) is closed.
Assume that X and Y are metric spaces. Let C ⊂ X be closed. Let w be a limit point of
f (C). Then there exists a sequence wn ∈ f (C) such that wn → w. Since wn ∈ f (C), there
exists zn ∈ C such that wn = f (zn ). Now the subset L := {wn : n ∈ N} ∪ {w} is a compact
subset of Y . Since f is proper, its inverse image K := f −1 (L) is compact. By our choice,
(zn ) is a sequence in the compact set K and hence has a convergent subsequence, say, (znk )
converging to z ∈ K. Since C is closed, we conclude that z ∈ C. By continuity of f at z, we
see that f (znk ) → f (z). Since f (zn ) → w, it follows that f (z) = w. Hence we have shown
that w ∈ f (C), that is, f (C) is closed.

140
K Existence of Continuous Functions

If x 6= y are two distinct points of a space X, is there a continuous function f : X → R

such that f (x) 6= f (y)? In general, this may not be true. There may not exist continuous
functions on the given space other than the constants. For each pair of distinct points, if there
is an f ∈ C(X, R) with f (x) 6= f (y), we say that the family C(X, R) separates points. This
is the reason for defining the completely regular and normal spaces which ensures plenty of
continuous functions. One kind of spaces for which existence of an abundance of continuous
real valued functions is assured is the class of metric spaces. We shall look at them first.

K.1 Case of Metric Spaces

The crucial fact here is the simple observation: If (X, d) is a metric space and x ∈ X, then
the function fx (y) := d(x, y) is continuous on X. For, by triangle inequality we have
|fx (y) − fx (z)| = |d(x, y) − d(x, z)| ≤ d(y, z).
Thus {fx : x ∈ X} is a separating family of continuous functions on X. More generally, we
have
Lemma 185. Let A be any nonempty subset of a metric space (X, d). Define d(x, A) ≡
dA (x) := inf{d(x, a) : a ∈ A}. Then |dA (x) − dA (y)| ≤ d(x, y) and hence dA is uniformly
continuous on X.

Proof. Observe from the triangle inequality d(x, a) ≤ d(x, y) + d(y, a), we obtain
inf d(x, a) ≤ inf (d(x, y) + d(y, a))
a∈A a∈A
= d(x, y) + inf d(y, a),
a∈A

so that dA (x) ≤ d(x, y) + dA (y). Thus, da (x) − dA (y) ≤ d(x, y). Interchanging x and y yields
the result.
Ex. 186. dA (x) = 0 iff x is a limit point of A. Hence if A is a closed set then d(x, A) = 0 iff
x ∈ A.
Lemma 187 (Urysohn’s Lemma for Metric Spaces). Let A and B be nonempty disjoint closed
subsets of a metric space X. Then there exists an f ∈ C(X, R) such that 0 ≤ f (x) ≤ 1 for
x ∈ X and f = 0 on A and f = 1 on B.

Proof. Note that for any x ∈ X, d(x, A) + d(x, B) 6= 0. For, if it were so, then d(x, A) = 0 =
d(x, B). Since A and B are closed x ∈ A and x ∈ B by the last exercise. This contradicts
our hypothesis that A ∩ B = ∅.
d(x,A)
The function f (x) := d(x,A)+d(x,B) meets our requirements.

Theorem 188 (Tietze extension theorem for metric spaces). Let Y be a closed subspace of
a metric space (X, d). Let f : Y → R be a bounded continuous function. Then there exists a
continuous function g : X → R such that g(y) = f (y) for all y ∈ Y and
inf{g(x) : x ∈ X} = inf{f (y) : y ∈ Y }, sup{g(x) : x ∈ X} = sup{f (y) : y ∈ Y }.

141
Proof. Assume that f ≥ 0. Consider the function Mx (r) := sup{f (y) : y ∈ Y ∩ B(x, r)}.
Then, for each x ∈ X, Mx is real valued, bounded and monotonic increasing in r. Hence it is
Riemann integrable as a function of r over any finite interval. Let δ(x) := d(x, Y ). Note that
δ(x) > 0 iff x ∈
/ Y . We define g by g(x) = f (x) if x ∈ Y and if x ∈
/ Y,
Z 2δ(x)
1
g(x) := Mx (r) dr.
δ(x) δ(x)

We claim that g is continuous at any y ∈ Y . If x ∈

/ Y , then

min f ≤ g(x) ≤ max f, where δ := d(x, y).

Y ∩B(y,3δ) Y ∩B(y,3δ)

Since 3δ → 0 as x → y (with x ∈ / Y ), it follows that, for any ε > 0, |g(x) − f (y)| < ε if x ∈
/Y
and δ(x) < δ0 for δ0 sufficiently small. On the other hand, |g(x) − f (y)| = |f (x) − f (y)| < ε
if x ∈ Y and d(x, y) < δ1 by continuity of f on Y . Thus g is continuous at y.
Consider next the continuity of g at z ∈ / Y . Let x be any point in X with d(x, z) <
d(z, Y )/3. Let α := d(x, z). Then 2α < d(x, Y ). For, otherwise, d(x, Y ) ≤ 2α so that
d(z, Y ) ≤ d(z, x) + d(x, Y ) < 3α. Hence α > d(z, X)/3, contradicting our assumption on x.
Since |δ(x) − δ(z)| ≤ d(x, z) = α (by Lemma 185), Mz (r) ≥ Mx (r−α) as B(x, r−α) ⊂ B(z, r)
and Mz (r) ≥ 0, we have
Z 2δ(x) Z 2δ(z)
1 1
g(x) − g(z) = Mx (r) dr − Mz (r) dr
δ(x) δ(x) δ(z) δ(z)
Z 2δ(x) Z 2δ(x)−2α
1 1
≤ Mx (r) dr − Mx (r − α) dr
δ(x) δ(x) δ(x) + α δ(x)+α
Z 2δ(x)−3α Z 2δ(x)
1 1
= Mx (r) dr + Mx (r) dr
δ(x) δ(x) δ(x) 2δ(x)−3α
Z 2δ(x)−3α
1
− Mx (s) ds, using a change of variable
δ(x) + α δ(x)
Z 2δ(x)−3α Z 2δ(x)
α 1
= Mx (r) dr + Mx (r) dr
δ(x)[δ(x) + α] δ(x) δ(x) 2δ(x)−3α
4M α
≤ ,
δ(x)
where M = supY f . A similar inequality holds with x and z interchanged. Hence g(x) → g(z)
as x → z. One easily checks that g is as desired.
To treat the general case, let m := inf Y f . Consider F := f − m. Apply the first case to
F to get a continuous extension G. Then g := G + c is as required.

We shall use Weierstrass approximation theorem to give a proof of Tietze theorem for Rn .

Proof. (of Tietze theorem for Rn .) Let us prove the result when the closed set is compact.
So, we assume that f : K → R is a continuous function on a compact subset of Rn . By

142
Weierstrass approximation theorem, for each k ∈ Z+ , there exists a polynomial pk such
P that
− p(x)| < 2−k−2 for all x ∈ K. We let q0 = p0 and qk := pk − pk−1 . Then pk = ki=1 qi
|f (x)P
and qk converges uniformly to f on K.
Let M := sup{|f (x)| : x ∈ K}. Then |p0 (x)| ≤ 2−2 + M for x ∈ K. Also, |qk (x)| < 2−k
for k ≥ 1 and x ∈ K. We let

h0 := max{−2−2 − M, min{q0 , 2−2 + M }},

hk := max{−2−k , min{qk , 2−k }}, for k ≥ 1.
n −k n
P for x ∈ K, hk is continuous non R and |hk (x)| ≤ 2 for x ∈ R and
Then hk (x) = qk (x)
for all k. Hence hk converges uniformly on R to a continuous function h. Then h is
continuous and h(x) = f (x) for x ∈ K.
We now extend to result if the subset K is any arbitrary closed subset. If K is bounded
the result follows from the previous paragraph. So, we assume that K is not bounded. Let
k ∈ N be such that B[0, k] ∩ K is nonempty. Let fk be the restriction of f to this nonempty
compact set. Then there exists a continuous function hk on Rn which extends fk . Define
(
hk (x), if x ∈ B[0, k]
gk (x) :=
f (x), if x ∈ K ∩ B[0, k + 1].

Then gk is continuous on the compact set B[0, k] ∪ (K ∩ B[0, k + 1]). There is an extension
hk+1 on Rn . Let (
hk+1 (x), if x ∈ B[0, k + 1]
gk+1 (x) :=
f (x), if x ∈ K ∩ B[0, k + 2].
Continuing in this way, we obtain a sequence (gm ) whose domains are increasing to Rn . Define
g(x) := gm (x) if x ∈ B[0, m]. Then g is an extension of f .

K.2 Normal Spaces

Lemma 189. A space X is a normal space iff for each closed set F and an open set V
containing F there exists an open set U such that F ⊂ U ⊂ U ⊂ V .

Proof. Let X be normal and F , V as above. Then F and X \ V are disjoint closed sets.
By normality of X there exist open sets U and W such that F ⊂ U and X \ V ⊂ W and
U ∩ W = ∅. Since U ⊂ X \ W and X \ W is closed, we see that U ⊂ X \ W ⊂ V . Thus U is
as required. The converse is left as an exercise.

Ex. 190. Recall that a dyadic rational is a rational number of the form p/2n , where p and
n are integers. Show that the set of dyadic rationals is dense in R.

Lemma 191. urys2 Let X be a normal space. If A and B are closed subsets of X, for each
dyadic rational r = k2−n ∈ (0, 1], there is an open set Ur with the following properties: (i)
A ⊂ Ur ⊂ X \ B, (ii) U r ⊂ Us for r < s.

143
Proof. Let U1 := X \ B. By the last lemma, there exist disjoint open sets V and W such that
A ⊂ V and B ⊂ W . Let U1/2 = V . Then, since X \ W is closed, we have

A ⊂ U1/2 ⊂ U 1/2 ⊂ X \ W ⊂ X \ B = U1 .

Applying the same lemma once again to the open set U1/2 containing A and to the open set
U1 containing U 1/2 , we get open sets U1/4 and U3/4 such that

A ⊂ U1/4 ⊂ U 1/4 ⊂ U1/2 ⊂ U 1/2 ⊂ AU3/4 ⊂ U 3/4 ⊂ V.

Continuing this manner, we construct, for each dyadic rational r ∈ (0, 1), an open set Ur with
the following properties:
(i) U r ⊂ Us , 0 < r < s ≤ 1.
(ii) A ⊂ Ur , 0 < r ≤ 1. (iii) Ur ⊂ U1 , 0 < r ≤ 1.
More formally, we proceed as follows. We select Ur for r = k2−n by induction on n.
Assume that we have chosen Ur for r = k2−n , 0 < k < 2n , 1 ≤ n ≤ N − 1. To find Ur for
r = (2j + 1)2−N , 0 ≤ j < 2N −1 , observe that U j21−N and X \ U(j+1)21−N are disjoint closed
sets. So once again appealing to the last lemma, we can choose an open set Ur such that

U j21−N ⊂ Ur ⊂ U r ⊂ U(j+1)21−N .

These Ur ’s are as desired.

Theorem 192. Urysohn’s Lemma. A space X is a normal space iff the following is
true: For any two disjoint closed subsets A and B of X there exists a continuous function
f : X → [0, 1] such that f = 0 on A and f = 1 on B.

Proof. Let Ur ’s be as in the last lemma. We define the function f so that the sets ∂Ur are
the level sets of f for the value r. We achieve this by defining
(
0, x ∈ Ur for all r
f (x) =
sup{r : x ∈
/ Ur }, otherwise.

Clearly, 0 ≤ f ≤ 1, f = 0 on A and f = 1 on B. We need only establish the continuity of f .

Let x ∈ X be such that 0 < f (x) < 1. Let ε > 0. Choose dyadic rationals r and s
in (0, 1) such that f (x) − ε < r < f (x) < s < f (x) + ε. Then x ∈ / Ut for dyadic rationals
t ∈ (r, f (x)). By (i), x ∈ / U r . On the other hand x ∈ Us . Hence W = Us \ U r is an open
neighbourhood of x. If y ∈ W , then from the definition of f we see that r ≤ f (y) ≤ s. In
particular, |f (y) − f (x)| < ε for y ∈ W . Thus f is continuous at x. The cases when f (x) = 0
or 1 are easier and left to the reader.

Lemma 193. Let X and Y be Banach spaces. Let T : X → Y be a bounded linear map.
Assume that for y0 ∈ Y there exist constants M and r ∈ (0, 1) such that there exists x ∈ X
such that kxk ≤ M ky0 k and ky0 − T xk ≤ r ky k. Then there exists z ∈ X such that T z = y0
with kz k ≤ M/(1 − r).

144
Proof. Let y ∈ Y be given. We may assume without loss of generality that ky k = 1. Given
y ∈ Y let z1 = x as given in the lemma. For y0 = y − T z1 , we can find a z2 ∈ X such
that kz2 k ≤ M ky − T z1 k ≤ M r and ky − T z1 − T z2 k ≤ r ky − T z1 k ≤ r2 . Proceeding
Pn by
kz k ≤ n−1 and (ii) ky −
induction, we get
P∞ a sequence (z n ) in X such that (i) n mr i=1 T zi ≤
k
n
r . The series n=1 zn converges to an element z ∈ X. We have T z = y0 .

Theorem 194 (Tietze Extension Theorem). Let X be a normal space and Y a closed subset
of X. Let f ∈ Y := Cb (Y, R). Then there exists a g ∈ X := Cb (X, R) such that g(y) = f (y)
for all y ∈ Y and sup{g(x) : x ∈ X} = sup{f (y) : y ∈ Y }.

Proof. Let T : X → Y denote the restriction map g 7→ g|Y . We show that T satisfies the
hypothesis of the previous lemma. Without loss of generality, assume that |f (y)| ≤ 1 for all
y ∈ Y . Let A := f −1 ([−1, −1/3]) and B := f −1 ([1/3, 1]). Then A and B are closed in Y and
hence in X. By Urysohn’s lemma, there exists a g ∈ X such that |g(x)| ≤ 1/3 for x ∈ X
and g = −1/3 on A and g = 1/3 on B. One easily checks that kT g − f kX ≤ 1/3. If we take
M = 1/3 and r = 2/3, then T satisfies the previous lemma. Note that the assertion about
the equality of the norms is also obtained.

Ex. 195. Let X be a normal space and F a closed subset. Assume that f : F → (−R, R) be
a continuous function. Then f can be extended to a continuous function from X to (−R, R).
Hint: You may need Urysohn’s lemma.

Ex. 196. Let X be a normal space and F a closed subset. Assume that f : F → R be a
continuous function. Then f can be extended to a continuous function from X to R. Hint:
R is homeomorphic to (−1, 1).

Ex. 197. Assuming Tietze extension theorem, prove Urysohn’s lemma.

Ex. 198. Let A be a closed subset of a normal space X. Let f : A → S n be continuous.

Show that there exists an open set U ⊃ A (U depends on f ) and an extension g of f to U .

Ex. 199. Show that with the notation of Exer. 198 that f may not extend to all of X. Hint:
What happens (i) if n = 0 and X is connected or (ii) if X := B[0, 1] ⊂ Rn+1 , A := S n and f
is the identity?

145
L Topological Groups — via Problems

Definition 200. A topological group is a triple (G, τ, ·) such that (G, τ ) is a topological space
and (G, ·) is a group. Both these structures are inter-related in the sense that the group
operations are continuous, that is,
(i) the group multiplication G × G → G given by (x, y) 7→ xy and
(ii) inversion map G → G given by x 7→ x−1 are continuous.

Example 201. (Rn , +) is a topological group with the usual topology.

Example 202. Let GL(n, K) denote the group of all n × n invertible matrices with entries
2
in K = R or K = C. Then GL(n, K) ⊂ M (n, K) ' Kn is open. GL(n, K) is a topological
group.

Example 203. The group T := {z ∈ C : |z| = 1} is a topological group with the subspace
topology.

Example 204. Let G be any group. If G is endowed with the discrete topology, then G is a
topological group.

Ex. 205. If we endow an abstract group with the indiscrete topology, does it become a
topological group?

Ex. 206. Show that the triple (G, τ, ·) is a topological group iff the map (x, y) 7→ xy −1 is
continuous.

Ex. 207. If Gi , 1 ≤ i ≤ n, are topological groups, then so is their product with the product
topology.

Ex. 208. If H is a subgroup of a topological group, then H is a topological group with the
subspace topology.

Ex. 209. How do you define a topological subgroup of a topological group?

Ex. 210. If H is a subgroup of G, then the closure H is also a subgroup.

Ex. 211. The following subgroups (of the respective groups) are topological groups with the
subspace topology.
(a) Let SL(n, K) denote the subgroup of GL(n, K) with determinant 1.
(b) Let O(n, R) denote the subgroup of GL(n, R) of all orthogonal matrices.
(c) Let U (n) denote the subgroup of all unitary matrices in GL(n, C).
(d) Let SO(n, R) and SU (n, R) denote respectively the subgroups consisting of elements of
O(n, R) and U (n) whose determinant is one.
(e) Let GL+ (n, R) denote the subgroup of all elements with positive determinant.
Show that O(n, R), SO(n, R), U (n) and SU (n) are compact.

Ex. 212. The left translations La : x 7→ ax are homeomorphisms. So are the right transla-
tions Ra .

Ex. 213. If U is an open set, so is U −1 := {x−1 : x ∈ U }. If A is an arbitrary subset of G,

then AU and U A are open.

146
Ex. 214. Let U denote the set of all neighborhoods of e ∈ G. Show that the topology on G
is completely determined by the knowledge of U.

Ex. 215. With the notation of the last exercise, prove that U has the following properties:
(i) e ∈ U for all U ∈ U .
(ii) If U1 , U2 ∈ U, then there exists U ∈ U such that U ⊂ U1 ∩ U2 . Hint: Use the continuity
of the group multiplication at (e, e).
(iii) If U ∈ U, there exists V ∈ U such that V V −1 ⊂ U .
(iv) If U ∈ U and a ∈ U , then there exists V ∈ U such that aV ⊂ U .
(v) If U ∈ U and a ∈ G, there exists V ∈ U such that aV a−1 ⊂ U .

Ex. 216. Let G be a group. Let U be a collection of subsets with the properties enumerated
in the previous exercise. Show that there exists a topology on G such that G becomes a
topological group with this topology and the neighbourhood basis at e ∈ G is precisely U.

Ex. 217. Let G be a topological group. Let U be the neighbourhood base at e. Show that
G is Hausdorff iff ∩U ∈U = {e}.

Ex. 218. If H is a normal subgroup of G, then the closure H is also a normal subgroup.

Ex. 219. If H is an open subgroup of G, then H is closed. Hint: Consider the coset
decomposition of G with respect to H. That is, observe that H = G \ ∪x∈H
/ xH.

Ex. 220. If H is a closed subgroup of finite index in a topological group, then H is open.

Ex. 221. Let G be a topological group and E ⊂ G. Show that

E = ∩U ∈U U E = ∩U ∈U EU.

Ex. 222. How will you define the uniform continuity of f : G → C on any topological group?
(G need not be metrizable.)

Ex. 223. Let f : G → C be a continuous function with compact support. Show that f is
uniformly continuous.

Ex. 224. Let G and H be topological groups. The a group homomorphism f : G → H is

continuous iff it is continuous at e.

Ex. 225. If G is a topological group and H is a subgroup, then the coset space G/H is
endowed with the quotient topology. The quotient map π : G → G/H is an open continuous
map.

Ex. 226. With the notation of the previous exercise, if we further assume that H is normal
in G, then the quotient group G/H becomes a topological group with the quotient topology.

Ex. 227. Show that the quotient group G/H is Hausdorff iff H is closed in G. Is it still true
if H is only a subgroup rather than a normal subgroup?

Ex. 228. When is G/H discrete?

Ex. 229. Let G be a connected topological group. Let U be a symmetric neighbourhood of

e, that is, U = U −1 . Then G = ∪∞ n
n=1 U . Hint: Observe that the union is an open subgroup.

147
Ex. 230. With the notation of the last exercise, assume that G is also compact. Can you
sharpen the result in this case?

Ex. 231. Let H be a subgroup of a topological group G. If G/H and H are connected then
G is connected.

Ex. 232. Let G be a topological group. Let G0 denote the connected component of G
containing e. Show that G0 is a closed normal subgroup.

Ex. 233. Show that GL+ (n, R) is connected and that it is the connected component of
GL(n, R) containing the identity. Hint:By induction.
 For, n > 1, consider the subgroup H
1 v
consisiting fo elements of the form g = where v ∈ Rn−1 and h ∈ GL+ (n − 1, R).
0 h
Ex. 234. Show that GL(n, C) is connected.

Ex. 235. How will you define the action of a topological group on a topological space X?

Definition 236. We say that a topological group G acts on a topological space X if the group
G acts on X in the algebraic sense and if the group action G × X → X given by (g, x) 7→ gx
is continuous. We then say X is a G-space.

Ex. 237. Examples of such actions. SL(2, R) acts on the upper half plane via fractional linear
transformations. O(n, R) acts on the unit sphere S n−1 . The group of affine transformations
fA,v : x 7→ Ax + v on Kn where A is a nonsingular linear map and v ∈ Kn is fixed. The group
law is the composition of maps. This group acts on Kn .

Ex. 238. Let G the a topological group and H a closed subgroup. Let X := G/H be the
quotient space. Then G acts on X via (g, xH) 7→ gxH. This action is transitive.

Ex. 239. When do we say two G-spaces X and Y are G-isomorphic?

Ex. 240 (Baire’s Theorem). Let X be a locally compact Hausdorff space. Assume that
X = ∪∞n=1 Fn where Fn is closed for each n. Show that at least one Fn is open in X. Hint:
Go through the proof in the case of metric spaces.

Ex. 241. Let X be a locally compact Hausdorff space. Let a locally compact Hausdorff
group with a countable basis. Assume that G acts on X transitively. Let x0 ∈ X be fixed.
Let H be the isotropy of x0 , that is, H := {g ∈ G : gx0 = x0 }. Then X is “isomorphic” to
the quotient space G/H as G-spaces.

Ex. 242. What is the isotropy at i when SL(2, R) acts on the upper half plane? Same
question when O(n) acts on S n−1 ⊂ Rn .

Ex. 243. Show that O(n, R)/O(n − 1, R) is G-isomorphic to S n−1 . How does O(n − 1, R) sit
in O(n, R)?

Definition 244. A subgroup Γ ⊂ G is called a discrete subgroup if it is a subgroup of G

which is both closed and discrete as a subset of G.

Ex. 245. Let Γ be a discrete subgroup of Rn . Then there exist v1 , . . . , vd ∈ Γ such that
(i) v1 , . . . , vd are linearly independent over R.
(ii) Every element of Γ is uniquely written as an integral linear combination of vj ’s.

148
 (iii) d is unique though vj ’s need not be. P
Hint: Consider the L1 -norm on Rn : kxk1 := ni=1 |xi |. Show that inf{kγ k : γ ∈ Γ, γ 6= 0} is
positive and attained, say, v1 ∈ Γ. If Γ 6= Zv1 , extend it to a basis {u1 = v1 , . . . , un } of Rn .
Show that
X n
X
inf{ |xj | : where γ ∈ Γ \ {0}, γ = xi ui }
j6=1 i=1

is attained at some v2 ∈ Γ.

Ex. 246. If f : G → H is a continuous homomorphism into a locally compact Hausdorff

group H, then f is necessarily open.

Ex. 247. Let G be a connected group and H a discrete normal subgroup of G. Then H is
contained in the center of G.

149
M Discrete subgroups of Rn

Proposition 248. Let Γ be a discrete subgroup of Rn . Then there exists a basis u1 , u2 , . . . , un

of Rn such that

Γ = {x ∈ Rn : x is of the form x = n1 u1 + · · · + nr ur , nj ∈ Z},

for some r ≤ n.

Proof. The proof is an inductive construction. Let W be a vector subspace of Rn such that
Γ ∩ W = Zw1 + · · · + Zwk for some basis w1 , . . . , wk of W . Such W ’s exist, for instance
W = {0}! Suppose that there exists u ∈ Γ that does not lie in W . Consider the set BW of
points
a1 w1 + · · · + ak wk + bu, 0 ≤ ai ≤ 1, 0 ≤ b ≤ 1. (8)
This set is bounded in Rn . Since Γ is discrete, this set BW can contain only finitely many
points of Γ. Hence there exists a point v ∈ BW ∩ Γ such that the coefficient b of u in v will be
the least positive coefficient, say β. If a1 w1 + · · · + ak wk + bu lies in Γ with ai , b ∈ Z, then b is
a multiple of β. For, otherwise, by division algorithm, we write b = mβ + r where 0 < r < β.
Hence the element

a1 w1 + · · · + ak wk + ru = (a1 w1 + · · · + ak wk + bu) − mβu ∈ Γ.

Since wj ∈ Γ, by subtracting suitable multiples of wj , we can assume that 0 ≤ aj ≤ 1. In

other words, a1 w1 + · · · + ak wk + ru ∈ BW . This contradicts our choice of µ. Thus we have
established that

Γ ∩ (W + Ru) = (Γ ∩ W ) + Zv = Zw1 + · · · + Zwk + Zu.

Note that the set {w1 , . . . , wk , u} is a basis of W + Ru. If there exists u0 ∈ Γ, we can proceed
as above. This process has to stop in a finite number of steps.

150
N Non-contractibility of the circle and
Brouwer Fixed Point Theorem

The aim of this article is to classify the homotopy classes of maps from a circle to the punctured
plane
We prove that the circle S 1 := {z ∈ C : |z| = 1} is not contractible and derive its
consequences. We start with a lemma from complex analysis which says that it is possible to
assign the argument of a complex number in a continuous fashion if we restrict ourselves to
C minus {z ∈ C : Re z ≤ 0}, or the complex plane minus any closed half line starting from
the origin.

Lemma 249. There exists a continuous map

α : X := C \ {z ∈ C : z ∈ R and ≤ 0} → (−π, π)

such that z = |z|eiα(z) for all z ∈ X.

Proof. Let us define the following open half-planes whose union is X: H1 := {z ∈ C : Re z >
0}, H2 := {z ∈ C : Im z > 0} and H3 := {z ∈ C : Im z < 0}. We define αi on Hi which glue
together to give the required map.
Let z ∈ H1 . Then Re z = |z| cos θ for some θ ∈ [−π, π] and hence cos θ > 0. This
means that θ ∈ (−π/2, π/2). sin is increasing on (−π/2, π/2) so that we have the continuous
inverse sin−1 : (−1, 1) → (−π/2, π/2). We define α1 (z) := sin−1 ( Im z
|z| ). We can similarly
define α2 : H2 → (0, π) and α3 : H3 → (−π, 0) by

Re z
α2 (z) = cos−1 ( )
|z|
Re z
α3 (z) = cos−1 ( ).
|z|

One easily sees that they agree upon their common domains. Thus we get the required
function α.

Definition 250. Let f and g be continuous functions from a space X to Y . Then f and g
are homotopic iff there is a continuous function H : I × X → Y such that H(0, x) = f (x) and
H(1, x) = g(x) for all x ∈ X. H is called a homotopy from f to g. Thus a homotopy enables
one to pass continuously from one map to another.

Lemma 251. Assume that f : S 1 → S 1 is homotopic to a constant map. Then there is a

continuous function ϕ : S 1 → R such that f (x) = eiϕ(x) for all x ∈ S 1 .

Proof. Let H : I × S 1 → S 1 be a homotopy with H(0, x) = c and H(1, x) = f (x) for x ∈ S 1 .

Since H is uniformly continuous, for ε = 2, there is a δ > 0 such that

|H(s, x) − H(t, x)| < 2, for |s − t| < δ, x ∈ S1.

151
Let 0 = t0 < t1 · · · < tn = 1 be a partition of I such that |ti − ti+1 | < δ for 0 ≤ i ≤ n − 1. Note
that H(0, x) = c = eiψ(x) for some constant map ψ : S 1 → R. We show that H(t1 , x) = eiϕ1 (x)
for some ϕ1 .
Since |H(t1 , x)−H(0, x)| < 2, we see that H(t1 , x) 6= −H(0, x) and hence that H(t 1 ,x)
H(0,x) 6= −1
for x ∈ S 1 . We define a continuous function α : S 1 → R by setting α(x) to be the argument
of x taking values in (−π, π). (This is possible by Lemma 263.) Thus H(t 1 ,x)
H(0,x) = e
iα(x) and

consequently
H(t1 , x) = eiα(x) H(0, x) = ei(ψ(x)+α(x)) = eiϕ1 (x) ,
where ϕ1 (x) = ψ(x) + α(x). Continuing this way proves the lemma.

Definition 252. A space is said to be contractible if there is a homotopy between the identity
map and a constant map.

Ex. 253. Any convex subset of Rn is contractible.

Theorem 254. The circle S 1 is not contractible.

Proof. If it were, then by Lemma 251 there is a function ϕ : S 1 → R such that Id(x) ≡ x =
eiϕ(x) for all x ∈ S 1 . Hence ϕ is 1-1 and in particular ϕ(x) 6= ϕ(−x). Define g : S 1 → {±1}
by
ϕ(x) − ϕ(−x)
g(x) := .
|ϕ(x) − ϕ(−x)|
Then g maps S 1 continuously onto {±1}. This contradicts the connectedness of S 1 .

Definition 255. A subset A of a space X is a retract of X if there is a continuous function

r : X → A such that r(a) = a for all a ∈ A. r is called a retraction of X onto A.

Corollary 256. There is no retraction of R2 onto S 1 .

Proof. Let r : R2 → S 1 be retraction. Let p = (0, 0). Define a homotopy H : I × S 1 → R2

by H(t, x) = tp + (1 − t)x. Then r ◦ H : I × S 1 → S 1 is a contraction — contradicting
Thm. 279.

Corollary 257 (Brouwer Fixed Point Theorem). Let f : B[0, 1] → B[0, 1] be a continuous
map. Then f has a fixed point, i.e., there is an x ∈ B[0, 1] such that f (x) = x.

Proof. If there is no point x such that f (x) = x, then the two distinct points f (x) and x
determine a line joining f (x) and x. We let g(x) be the point on the boundary at which
the line starting from f (x) and going to x meets S 1 . Then g is a retraction of B[0, 1] onto
S 1 —a contradiction to Corollary
q 281. In analytical terms, we have g(x) = x + tv, where
x−f (x)
v= k x−f (x) k and t = − hx, vi + 1 − kxk2 + (hx, vi)2 .

Corollary 258 (Generalised Brouwer Fixed Point Theorem). Let f : B[0, 1] → R2 be contin-
uous such that f (S 1 ) ⊂ B[0, 1]. Then f has a fixed point.

152
Proof. Define r : R2 \ {(0, 0)} → S 1 by r(x) = x/|x|. If f (x) 6= x for all x ∈ B(0, 1) then S 1
can be contracted via the homotopy
(
r(x − 2tf (x)), 0 ≤ t ≤ 1/2,
H(t, x) =
r((2 − 2t)x − f ((2 − 2t)x)), 1/2 ≤ t ≤ 1.

This contradicts Thm. 279.

153
O Maps into Punctured Plane

There is a lot of overlap between this and the last sections. Need to edit this carefully to
avoid meaningless repetition. Details!

CARE!
The aim of this article is to classify the homotopy classes of maps from a circle to the
punctured plane C∗ := C \ {0}. Such a classification can be obtained from the knowledge
of the fundamental group π1 (S 1 ) of the circle. Our approach will be more analytic and will
yield an alternative proof of the isomorphism π1 (S 1 ) ' Z.

Definition 259. Let f and g be continuous functions from a space X to Y . Then f and g
are homotopic iff there is a continuous function H : I × X → Y such that H(0, x) = f (x) and
H(1, x) = g(x) for all x ∈ X. H is called a homotopy from f to g. Thus a homotopy enables
one to pass continuously from one map to another.

Let X be a topological space. We consider maps from X into C∗ . Such functions form a
group under pointwise multiplication.

Definition 260. A map f : X → C∗ is an exponential if f = exp(g) = eg for some continuous

map g : X → C.

Ex. 261. The exponential maps form a subgroup of the group of maps from X to C∗ .

Theorem 262. It is impossible to make a continuous choice θ(z) ∈ arg (z) on C∗ . That is,
there is no continuous map θ : C∗ → R such that z = |z| exp(θ(z)) for z ∈ C∗ .

Proof. Assuming such a θ exists, consider f : [0, 2π] → R by setting f (t) := [θ(eit )+θ(e−it )]/2π.
Then f is a real valued continuous function on [0, 2π]. Then 2πf (t) is a choice of arg (eit e−it )
and hence of arg (1). Thus it is integer valued continuous function on the interval [0, 2π]. By
intermediate value theorem, it is a constant. In particular, f (0) = f (π). This implies that
[θ(1) + θ(1)]/2π = [θ(−1) + θ(−1)]/2π. Or, θ(1) = θ(−1), which is impossible as the arg (1)
and arg (−1) are disjoint.

However, the following lemma says that it is possible to assign the argument of a complex
number in a continuous fashion if we restrict ourselves to C minus {z ∈ C : Re z ≤ 0}, or the
complex plane minus any closed half line starting from the origin.

Lemma 263. There exists a continuous map

α : X := C \ {z ∈ C : z ∈ R & z ≤ 0} → (−π, π)

such that z = |z|eiα(z) for all z ∈ X.

Proof. Let us define the following open half-planes whose union is X: H1 := {z ∈ C : Re z >
0}, H2 := {z ∈ C : Im z > 0} and H3 := {z ∈ C : Im z < 0}. We define αi on Hi which glue
together to give the required map.
Let z ∈ H1 . Then Re z = |z| cos θ for some θ ∈ [−π, π] and hence cos θ > 0. This
means that θ ∈ (−π/2, π/2). sin is increasing on (−π/2, π/2) so that we have the continuous

154
inverse sin−1 : (−1, 1) → (−π/2, π/2). We define α1 (z) := sin−1 ( Im z
|z| ). We can similarly
define α2 : H2 → (0, π) and α3 : H3 → (−π, 0) by
Re z
α2 (z) = cos−1 ( )
|z|
Re z
α3 (z) = cos−1 ( ).
|z|
One easily sees that they agree upon their common domains. Thus we get the required
function α.

Every continuous function f from X to positive reals is an exponential. In this case f = eg

where g := log f . More generally we have
Lemma 264. Suppose f : X → C∗ is a map that omits the negative real axis (that is, f (X) ∩
(−∞, 0] = ∅). Then f is an exponential.

Proof. We use the previous lemma. Recall that the principal logarithm Log is defined on the
given open subset of C by Log z = |z|eiθ , where θ ∈ (−π, π). Thus Log depends continuously
on z. If we set g(z) := Log f , then we have f = eg .

The following result is related to Rouche’s theorem in Complex Analysis.

Theorem 265. Let f and g be functions from X to C satisfying

|f (x) − g(x)| < |f (x)| + |g(x)|, x ∈ X. (9)

Then f /g and g/f are exponential. In particular f is an exponential iff g is.

Proof. Observe that the strict inequality in Eq. 9 implies that neither f nor g can vanish on
X. Dividing Eq. 9 by f (x) we obtain

|1 − g(x)/f (x)| < 1 + |g(x)/f (x)|, x ∈ X.

It follows that g/f cannot assume negative real values, for, then the RHS will equal the LHS.
Hence by Lemma 264, g/f is an exponential. As Eq. 9 is symmetric in f and g this means
that f /g is also an exponential. The last statement is a consequence of the fact that the
product of exponentials is an exponential.

Theorem 266. Let X be a compact metric space and f, g : X → C∗ . Then f and g are
homotopic iff f /g is an exponential.

Proof. Suppose that f /g is an exponential, say, f /g = eh . Then F (t, x) := g(x)eth(x) defines

a homotopy from f to g.
Conversely, suppose that f and g are homotopic. Let F be a homotopy from f to g.
Since [0, 1] × X is compact, the continuous positive function |F | attains its minimum. The
minimum m := inf{|F (t, x| : t ∈ [0, 1], x ∈ X} is positive. F is uniformly continuous on the
compact metric space I × X. Thus, for ε := m there exists a δ > 0 such that

|s − t| < δ =⇒ |F (s, x) − F (t, x)| < m, ∀x ∈ X. (10)

155
We now choose an integer N > 1/δ and consider the maps fj : X → C∗ , defined by fj (x) :=
F (j/N, x). Now f0 = f and fN = g. We see from Eq. 10 that

|fj (x) − fj−1 (x)| < m ≤ |fj (x)|, x ∈ X, 1 ≤ j ≤ N.

By Thm. 265, each fj /fj−1 is an exponential. As f /g = (f0 /f1 )(f1 /f2 ) · · · (fN −1 /fN ), we see
that f /g is an exponential.

Corollary 267. Let X be a compact metric space and f : X → C∗ . Then f is an exponential

iff f is homotopic to a constant map.

Definition 268. A space is said to be contractible if there is a homotopy between the identity
map and a constant map.

Ex. 269. Any convex subset of Rn is contractible.

Corollary 270. Let X be a compact contractible metric space. Then every map f from X
to C∗ is an exponential.

Proof. Let F : [0, 1] × X → X be the homotopy of the identity map of X and a constant
map x0 . Then f ◦ F is a homotopy of f to the constant map f (x0 ). By Cor. 267, f is an
exponential.

Now we restrict our attention to maps of S 1 to C∗ . We wish to assign to any such map
an index that corresponds to the number of times the functions wraps around the origin.

Definition 271. Let f : S 1 → C∗ be a map. Consider the map θ 7→ f (eiθ ) of [0, 2π] into C∗ .
Since the interval is contractible, by Corollary 270,

f (eiθ ) = eg(θ) for some g : [0, 2π] → C∗ . (11)

Let g1 : [0, 2π] → C∗ be another map which satisfies Eq. 11. Then eg(θ)−g1 (θ) = 1. Hence
g(θ) − g1 (θ) must assume values from the discrete set 2πiZ. Since g − g1 is continuous, it
follows that g − g1 is a constant. Thus the number g(2π) − g(0) is independent of the choice
of g satisfying Eq. 11. Consequently the number

ind (f ) := [g(2π) − g(0)]/2πi

is well defined. This integer is called the index of the map f .

Ex. 272. Let fn (z) = z n for n ∈ Z. Then ind (f ) = n.

Theorem 273. The index function, defined on the maps from S 1 to C∗ has the following
properties:
(i) ind (f g) = ind (f ) + ind (g).
(ii) ind (f ) = 0 iff f is an exponential.
(iii) ind (f /|f |) = ind (f ).
(iv) If f : S 1 → S 1 is a map such that f (1) = 1, then ind (f ) coincides with that of the loop
α defined by α(s) = f (e2πis ), 0 ≤ s ≤ 1.

156
Proof. (i) is easy and left to the reader.
iθ
Suppose f (eiθ ) = eh(e ) . If we set g(t) = h(eit ) then g satisfies Eq. 11. Since g(2π) = g(0),
ind (f ) = 0. Conversely, assume that ind (f ) = 0. Write f (eit ) = eig(t) for 0 ≤ t ≤ 2π. Then
g(0) = g(2π) so that the function h : S 1 → C defined by setting h(eit ) = g(t), 0 ≤ t ≤ 2π, is
well defined and continuous. Since f = eh f is an exponential. This proves (ii).
Since |f | is an exponential, ind (|f |) = 0 by (ii). By (i), ind (f ) = ind (f /|f |) + ind (|f |).
(iii) follows.
Let f and α be as in (iv). Choose h : [0, 1] → R such that h(0) and α(s) = e2πih(s) , for
0 ≤ s ≤ 1. Thus h is a lift of α and hence index ind (α) = h(1). Define g : [0, 2π] → C by
g(t) := 2πih(t/2π). Then g satisfies Eq. 11 so that

ind (f ) = [g(2π) − g(0)]/2πi = h(1) = ind (α).

This proves (iv).

Theorem 274. Let f, g : S 1 → C∗ be maps. Then the following are equivalent:

(i) f is homotopic to g.
(ii) ind (f ) = ind (g).
(iii) f /g is an exponential.

Proof. The equivalence of (i ) and (iii) is a special case of Thm. 266.

If f /g is an exponential, then by (ii) of Thm. 273, ind (f /g) = 0. Write f = g·(f /g). By (i)
of Thm. 273, we obtain ind (f ) = ind (f /g)+ind (g) = ind (g). Conversely, if ind (f ) = ind (g),
then ind (f /g) = 0. So, by (ii) of Thm. 273, f /g is an exponential.

Corollary 275. Each map f : S 1 → C∗ is homotopic to precisely one of the maps fm : z 7→ z n

where n = ind (f ).

Corollary 276. We have π1 (S 1 , 1) ≡ Z.

Proof. Since the maps z n are not homotopic, the loops αn : [0, 1] → S 1 defined by αn (t) =
e2πint cannot be homotopic with end points fixed. On the other hand, let α : [0, 1] → S 1
be an arbitrary loop based at 1. Define f : S 1 → S 1 by f (e2πis ) = α(s). Let n := ind (f ).
2πis
By Thm. 274, f /αn is an exponential, say, f (e2πis )/e2πins = eh(e ) for 0 ≤ s ≤ 1. Then
2πis
F (t, s) := eth(e ) e2πins for 0 ≤ s, t ≤ 1 is a homotopy from αn and the loop α with end
points fixed. Thus the correspondence ϕ : [αn ] 7→ n is a bijection between π1 (S 1 , 1) and Z.
One easily checks that the product path αm αn corresponds to a map from S 1 to itself of index
m + n, so that αm αn is homotopic to am+n . Thus ϕ is a group homomorphism.

Theorem 277 (Fundamental Theorem of Algebra). A polynomial p(z) := z n + an−1 z n−1 +

· · · + a1 z + a0 of degree n ≥ 1 and with complex coefficients has a zero in C.

Proof. Choose R so large that

an−1 n−1 a1 a0
| w + · · · + n−1 + n | < 1, |w| ≤ 1.
R R R

157
This can be done, if, for instance, we take R > |an−1 | + · · · + |a0 | + 1. Define a map
g : B[0, 1] → C by setting

p(Rw) an−1 n−1 a0

g(w) := n
= wn + w + ··· + n, |w| ≤ 1.
R R R
The estimate above shows that |g(w) − wn | < 1 for w = 1. By Thm. 265, the restriction of
wn /g to the unit circle is an exponential. Now wn is not an exponential since its index is
n ≥ 1 (Theorem 273). Hence g = wn · (g/wn ) is not an exponential. Corollary 270 shows that
g must have a zero on B[0, 1], and hence p has a zero in C.

We now apply some of our earlier results to arrive at some standard theorems of the
topology of the plane.

Corollary 278. Assume that f : S 1 → S 1 is homotopic to a constant map. Then there is a

continuous function ϕ : S 1 → R such that f (x) = eiϕ(x) for all x ∈ S 1 .

Proof. A special case of Corollary 267.

Theorem 279. The circle S 1 is not contractible.

Proof. If it were, then by Corollary 278 there is a function ϕ : S 1 → R such that Id(x) ≡
x = eiϕ(x) for all x ∈ S 1 . Thus, there is a continuous argument on S 1 and hence on C∗ ,
contradicting Theorem 262.
Or, more directly, such a ϕ is 1-1 and in particular ϕ(x) 6= ϕ(−x). Define g : S 1 → {±1}
by
ϕ(x) − ϕ(−x)
g(x) := .
|ϕ(x) − ϕ(−x)|
Then g maps S 1 continuously onto {±1}. This contradicts the connectedness of S 1 .

Definition 280. A subset A of a space X is a retract of X if there is a continuous function

r : X → A such that r(a) = a for all a ∈ A. r is called a retraction of X onto A.

Corollary 281. There is no retraction of R2 onto S 1 .

Proof. Let r : R2 → S 1 be retraction. Let p = (0, 0). Define a homotopy H : I × S 1 → R2

by H(t, x) = tp + (1 − t)x. Then r ◦ H : I × S 1 → S 1 is a contraction — contradicting
Thm. 279.

Corollary 282 (Brouwer Fixed Point Theorem). Let f : B[0, 1] → B[0, 1] be a continuous
map. Then f has a fixed point, i.e., there is an x ∈ B[0, 1] such that f (x) = x.

Proof. If there is no point x such that f (x) = x, then the two distinct points f (x) and x
determine a line joining f (x) and x. We let g(x) be the point on the boundary at which
the line starting from f (x) and going to x meets S 1 . Then g is a retraction of B[0, 1] onto
S 1 —a contradiction to Corollary
q 281. In analytical terms, we have g(x) = x + tv, where
x−f (x)
v= k x−f (x) k and t = − hx, vi + 1 − kxk2 + (hx, vi)2 .

158
 We end this article with some exercises.

Ex. 283. Let X be compact and f : X → C∗ be an exponential. Show that there exists
ε > 0 such that any map g : X → C∗ which satisfies |f (x) − g(x)| < ε is an exponential.

Ex. 284. Let X be locally compact hausdorff space. Show that two maps f and g from X to
C∗ are homotopic iff f /g is an exponential. Hint: Consider first the case when X is compact.

Ex. 285. Let X be a locally compact contractible metric space. Show that any map f : X →
C∗ is an exponential.

Ex. 286. Let f : S 1 → C∗ be given. Show that there exists a ε > 0 such that any map
g : S 1 → C∗ with |f (z) − g(z)| < ε for z ∈ S 1 has the same index as f .

Ex. 287. Assume that f, g : S 1 → S 1 be maps such that f and g do not assume antipodal
values at any point of S 1 . Show that ind (f ) = ind (g).

Ex. 288. Show that any map from S n (n ≥ 2) to C∗ is an exponential.

Ex. 289. Show that any map from Pn (R) (n ≥ 2) to C∗ is an exponential. (Note that Pn is
not simply connected. Can you explain what is happening here?)

Ex. 290. Classify the maps from the figure eight to C∗ .

159
P Rm is not homeomorphic to Rn if m 6= n

The aim of this article is to prove the theorem of the title. As a rule no first course in topology
proves this result. Even if they raise the question of homeomorphism between Rm and Rn
they refer to Brouwer’s theorem on the invariance of domain which is proved as an application
of Homology Theory. We wish to make the following elementary proof more widely known.
It could be taught in any first course on Topology. There is nothing original in the following
proof except the organization of the material available in the literature.

Outline of the Proof

Definition of a simplex, standard simplex; barycentric subdivision.

Dimension of a compact metric space. dim sn ≤ n where sn is the standard n-simplex in Rn .
Sperner’s lemma — restricted version (applicable only to triangulation arising from barycen-
tric subdivision).
Nagata’s Lemma (restricted version): Let Dk (s) be the k-th barycentric subdivision of an
n-simplex s. Let {Ui }n0 be an open cover of s such that Ui ⊆ s \ Fi where Fi is the face
opposite to the vertex ei . Then there exists an r and an n-simplex in Dr (s) which intersects
each of Ui .
dim sn = n.

Definition 291. A topological space X is said to have dim X ≤ n if given any open cover A
of X there exists an open cover B with the following properties:
For each B ∈ B there is an A ∈ A such that B ⊂ A.
There exists an element of X which lies in n + 1 members of B and no element of X
lies in more than n + 1 members of B.
We say that X is of (covering) dimension n if n is the least integer m such that dim X ≤ m.
If no such n exists then we write dim X = ∞.

Ex. 292. Homeomorphic spaces have the same dimension.

Ex. 293. Let X be a compact metric space. We say that dim X ≤ n if for every ε > 0 there
is a finite open cover A of X by sets of diameter < ε such that some point of X lies in n + 1
members of A and no point of X lies in more than n + 1 members of A. Hint: Use Lebesgue
covering lemma.

Ex. 294. Let X be a compact metric space of dimension n. Let K be a closed subset of X.
Then dim K ≤ n.

Definition 295. A set of vectors {v0 , v1 , . . . , vk } in Rn is said to be geometrically (or affinely)

independent if the set of vectors {v1 − v0 , v2 − v0 , . . . , vk − v0 } is linearly independent. A
singleton set is geometrically dependent by definition.

Example 296. A set {v0 , v1 } is geometrically independent iff they are not multiples of
each other. A set {v0 , v1 , v2 } is geometrically independent iff they are not collinear. A set
{v0 , v1 , v2 , v3 } is geometrically independent iff they are not coplanar.

160
Ex. 297. A set of vectors {v0 , v1 , . . . , vk } in Rn is
P said to be geometrically (or affinely)
independent iff for any set of real numbers λi with ki=0 λi = 0 and ki=0 λi vi = 0 we have
P
λi = 0 for 0 ≤ i ≤ k.

Definition 298. Let {v0 , v1 , . . . , vk } in Rn be geometrically independent. The (open) simplex

sk is the set
Xk X
sk := {x ∈ Rn : x = λi vi , λi > 0 and λi = 1}.
i=0 i

We refer to k as the algebraic dimension of s. s is also called a k-simplex. We denote the

simplex sk by (v0 v1 . . . vk ). vi are called the vertices of sk . The simplex σi spanned by
{v0 , . . . , vˆi , . . . , vk } is called the i-th face opposite to the vertex vi . (vˆi means vi is omitted.)
More generally, one defines a r-face of the simplex (v0 . . . vk ) as the r-simplex (vi1 . . . vir ) for
0 ≤ i1 < i2 · · · ir ≤ k. If σ is an r-face of s we write σ ≺ s.

Example 299. Any one simplex with vertices v0 and v1 is the open line segment joining v0
and v1 . Any two simplex spanned by three noncollinear vectors is the open interior of the
triangle of which they are vertices. Any three simplex spanned by four coplanar vectors is
the open tetrahedron. The faces are respectively the endpoints of the line segment, sides of
the triangle and the faces of the tetrahedron.

The proof of the theorem of the title depends on the following

Theorem 300. Let T n be the closure of any n-simplex in Rn . Then dim T n = n.

Assuming the theorem let us complete the proof of the main result. Let f : Rn → Rm be
a homeomorphism. Let us assume that, if possible, that m 6= n. Let m < n without loss of
generality. Let T n be as above. Then f (T n ) is a compact subset of Rm . Hence there exists
an m-simplex, say, s such that f (T n ) ⊂ s ⊂ s. Since dim s ≤ m and dim T n = n, we infer
that n = dim f (T n ) ≤ m < n in view of Exercises 292 and 294. This contradiction shows
that m = n.
We break the proof of Theorem 300 into two statements: dim T n ≤ n and dim T n ≥ n.
In the next section we establish the first and the second in the last section.

P.1 Barycentric Subdivision

Definition 301. A complex K is a collection of simplices {s} with the following property: If
s ∈ K and σ ≺ s then σ ∈ K. The set |K| = ∪s∈K s is called the geometric realization of the
complex K and K is called a triangulation of |K|.

Definition 302. Let s := (v0 . . . vk ) be any simplex. Then the barycenter of s is the vector
(v0 + · · · + vk )/(k + 1). We denote this vector by b(s).

Example 303. The barycenter of any 0-simplex is itself. The barycenter of (v0 v1 ) is the
midpoint of the line segment, that of (v0 v1 v2 ) is the centroid of the triangle and that of
(v0 . . . v3 ) is the centre of gravity of the tetrahedron.

161
Definition 304. Let a complex K be given. The first barycentric subdivision of K is the
complex D1 (K) whose vertices are b(s) where s runs through all the faces of all simplices
of K. The other simplices in the new complex are of the form (b(s1 )b(s2 ) . . . b(sr )) where
s1 ≺ s2 · · · ≺ sr . We leave it to the reader that D1 K is indeed a complex. Recursively we
define Dr (K) := D1 (Dr−1 (K)), called the r-th barycentric subdivision of K.

The following simple exercise will be repeatedly used in the sequel.

Ex. 305. Let µ(K) := max{diam s : s ∈ K} be mesh of the complex. Let s be any simplex.
Show that µ(s) is the length of the longest side, i.e.,1-dimensional face and µ(Dr (s)) → 0 as
r → ∞.

Theorem 306. Let s be an n-simplex. Then dim s ≤ n.

Proof. Given ε > 0 let us choose r sufficiently large so that the mesh of Dr (s) is less than
ε/2. Then the closed simplices in Dr−1 (s) is an ε-covering such that each vertex vi ∈ s lies in
∩ni=0 star(vi ). This does the job. (Details are to be given.)

P.2 Sperner’s Lemma and its Corollaries

We shall give a very special version of Sperner’s lemma which will be sufficient for our purpose.
For more general versions, see references.

Theorem 307. Let s = (v0 . . . vn ) be a simplex. Let Dr (s) be the r-th barycentric subdivision
of s. Let a map f : V (Dr (s)) → V (s) be given such that f (v) = vi where vi is a vertex of the
carrier of v. (Such maps will be called Sperner maps.) Then there exists a simplex σ ∈ Dr (s)
such that f (σ) = {v0 , . . . , vn }.

Proof. We shall prove this for r = 1 by induction on n. For n = 1 this is clear. Let us
assume that the result is true for n − 1. Assume without loss of generality that f (z) = vn
where z = b(s) ∈ D1 (s). Then f induces a Sperner map on the n − 1-simplex (v0 . . . vn−1 ).
By induction hypothesis, we there is an n − 1-simplex, say, σ n−1 = (b0 . . . bn−1 ) such that
f (V (σ)) = {v0 , . . . , vn−1 }. Clearly the simplex τ = (b0 . . . bn−1 vn ) is of the required type.
Thus the result is true for D1 (s) for any n-simplex s.
To complete the proof we now use induction on r and the previous paragraph. Let a
Sperner map f : V (Dr (s)) → V (s) be given. Then it induces a Sperner map on Dr−1 (s) by
restriction. By induction there exists a σ ∈ Dr−1 (s) such that f (σ) = {0, 1, . . . , n}. By the
first part of the proof there exists a simplex τ ∈ D1 (σ) such that τ is completely labeled. But
then τ ∈ Dr (s) and meets our requirement.

P.3 dim T n = n

Theorem 308. Let s := (v0 . . . vn ) be an n-simplex and T n := s. Let {Ui : 0 ≤ i ≤ n} be an

open cover of T such that Ui does not intersect the i-th face. Then there a sufficiently large
positive integer r such that there is an n-simplex σ in Dr s such that σ ∩ Ui 6= ∅.

162
Proof. Let ε be the Lebesgue number of the covering {Ui }. We choose r so that the mesh of
Dr (s) is less than ε/2. We define a Sperner map f : f (v) = i if v ∈ Ui and vi is a vertex of the
carrier of v. This is possible by our assumption on the cover. Sperner lemma gives a simplex
of the required kind.

Theorem 309. Let the notation be as above. Then dim sn ≥ n.

Proof. To prove this we need to exhibit an open cover A such that any B is in Definition 291
will be of order greater than or equal to n + 1. Let us take Ai = T n \ Fi , the complement of
the i-th face. Let B be any open cover such that B ≤ A. After doing a little jugglery we may
assume that B has n + 1 members. By the last result the order of B is n + 1.

1. Alexandrov, Combinatorial Topology,vol. 1.

2. Munkres, Topology, A First Course.
3. Nagata, Dimension Theory.

163