METRIC SPACES

R.S.Lalrinzuala

October 15, 2024

Contents

1 Basic Concepts 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.3 Neighbourhood . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.4 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . 3
1.5 Subspace of a Metric Space . . . . . . . . . . . . . . . . . . . . . 6

2 Completeness 7
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Some Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 Complete Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Continuous Function 13
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2.1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2.2 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2.3 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2.4 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.4 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.4.1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.4.2 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.5 Homeomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

i
ii CONTENTS
Chapter 1

Basic Concepts

1.1 Introduction
The system of two numbers has two types of properties. The first type consists
of algebraic, dealing with addition, multiplication, etc. The other type consists
of properties having to do with the notion of distance between two numbers and
with the concept of the limit. The latter properties are called topological or
metric, and the object of the present chapter is to study these properties in a
general space in which the notion of distance is defined.

A metric space is a generalization of the real number system with the abso-
lute value function as the definition of a group which is the generalization of the
algebraic properties of the real number system. We shall generalize the notion
of the absolute value function in R in such a way that it is suitable for the
treatment of convergence sequence in a set and continuous function on the set.

Many of the convergence sequence and continuous function depend only on

the properties of the distance, and not directly on the algebraic nature of the
real ( or complex) number system.

1.2 Definition
A metric (X, d) is a non-empty set x of elements ( which we call points ) together
with a real valued function d defined on X × X such that for all x, y and z in
X:
(i) d(x, y) ≥ 0;
(i) d(x, y) = 0 if and only if x = y;
(i) d(x, y) = d(y, x); (symmetry)
(i) d(x, y) ≤ d(x, z) + d(z, y); (triangle inequality)

1
2 CHAPTER 1. BASIC CONCEPTS

The function d is called a metric.

Example
1. The function d defined by d(x, y) = |x − y| is a metric for the set R of real
number.
With this distance function, R is a metric space defined by (R, d), this
metric d is called the usual metric for R.
The proof that d is actually a metric on R follows from the properties of
the absolute value function on R as:

(i) d(x, y) = |x − y| is a non-negative real number and d(x, y) = 0 if and

only if |x − y| = 0, this implies and implied by x = y;
(i) d(x, y) = |x − y| = | − (y − x)| = |y − x| = d(y, x);
(i) d(x, y) = |x − y| = |x − z + z − y| ≤ |x − z| + |z − y|
i.e. d(x, y) ≤ d(x, z) + d(z, y).

2. (Discrete metric) Let X be and arbitrary non-empty set. For x, y ∈ X,

defined d by (
0, x = y
d(x, y) = (1.1)
1, x ̸= y

Then (X, d) is a metric space. The metric d is called the discrete metric
and the space (X, d) is called discrete metric space and is denoted by Xd .

In the definition of a metric d on a set X it is sometimes convenient to relax

the condition that d(x, y) = 0 only if x = y. When we allow the possibility that
d(x, y) = 0 for some x ̸= y, we call d a pseudometric and (X, p) a pseudometric
space.
Pseudometric space (X, d) is the set X together with a non-negative real valued
function
d:X ×X →R
If d(x, y) = 0 for some x ̸= y, then we call d a pseudometric.

1.3 Neighbourhood
Let (X, d) be a metric space and x ∈ X. A set N ⊂ X is said to be a
neighbourhood(nbd) of x if ∃ an open sphere centred at x and contained in
N.
i.e. if Sr (x) ⊂ N , for some r > 0.
Example:

1. The open interval ]a, b[ is a nbd of each of its points.

2. The set of real numbers is a nbd of each of its points.

1.4. OPEN AND CLOSED SETS 3

Theorem
Let (X, d) be a metric space. A set N ⊂ X is a nbd of a point p ∈ X if and
only if ∃ an open sphere Sr (x) such that p ∈ Sr (x) ⊂ N .
Proof: First we assume that N is a nbd of a point p ∈ X.Then, there exists an
r > 0 such that p ∈ Sr (p) ⊂ N . This proves that there exists an open sphere
Sr (p) containing p and contained in N .
Conversely, assume that p ∈ N and there exists an open sphere Sr (x) such that
p ∈ Sr (x) ⊂ N . We shall prove that there exists an open sphere centred at p
and contained in N . Now,

p ∈ Sr (x) ⇒ d(x, p) < r

Let r1 = r − d(x, p). Then r1 > 0. Let y ∈ Sr1 (p).

Then d(x, p) < r1 . Now, we have

d(y, p) < r1 ⇒ d(y, p) < r − d(x, p)

⇒ d(x, p) + d(y, p) < r
(1.2)
⇒ d(x, y) < r (∵ d(x, y) ≤ d(x, p) + d(y, p))
⇒ y ∈ Sr (x)

Thus, Sr1 (p),Sr (x) ⊂ N . This verified that N is a nbd of p.

1.4 Open and Closed Sets

Let (X, d) be a metric space. A set G ⊂ X is said to be an open set if it is a
nbd of each of its points.
(Equivalently, a set G ⊂ X is said to be an open set if for each x ∈ G, ∃ an
r > 0 such that Sr (x) ⊂ G). Let (X, d) be a metric space and A ⊂ X. The set
A is said to be closed if its complement X − A is open.

Theorem:
The set X and ϕ are open; the intersection of any two open sets is open; and
the union of any collection of open sets is open.

Proof: To show that ϕ is open, we have to show that each points in ϕ is the
centre of an open sphere contained in ϕ. Since there are no points in ϕ, this re-
quirement is vacuously satisfied. If x ∈ X, every open sphere Sr (x) is contained
in X. Hence, X is open.

It is enough if we prove the theorem for two sets, since the argument can be
extended to any finite number of sets by induction. Let G1 and G2 be any two
open subsets of X such that G1 ∩ G2 ̸= ϕ. If x ∈ G1 ∩ G2 , we have to show that
there exists an open sphere Sr (x) contained in G1 ∩ G2 . Since x ∈ G1 and G1
is open, there is an open sphere Sr1 (x) such that Sr1 (x) ⊂ G1 . Similarly there
4 CHAPTER 1. BASIC CONCEPTS

exists an open sphere Sr2 (x) ⊂ G2 . Let r = minr1 , r2 . Then Sr (x) is contained
in G1 and G2 and so Sr (x) ⊂ G1 ∩ G2 . Therefore, G1 ∩ G2 is open subset of X.
S
Let {Gi : i ∈ J} be a family of nonempty open sets. Let H = Gi . We
i∈J
have to show that there is an open sphere Sr (x) contained in H. Let x ∈ H.
Then x ∈ Gi for someSi ∈ J. Since each Gi is open, there exists Sr (x) ⊂ Gi .
Hence, Sr (x) ⊂ Gi ⊂ Gi . So x is in the union implies that there exists Sr (x)
i∈J
contained entirely in H. Therefore, H is an open subset of X.

Theorem:
If A ⊂ B, then A ⊂ B. Also, (A ∪ B) = A ∪ B, and (A ∩ B) ⊂ A ∩ B.

Proof: We have,

A ⊂ B ⇒ A′ ⊂ B ′
∴ A ∪ A′ ⊂ A′ ∪ B ′ i.e., A ⊂ B.

Also,

(A ∪ B) = (A ∪ B) ∪ (A ∪ B)′
= (A ∪ B) ∪ (A′ ∪ B ′ )
= (A ∪ A′ ) ∪ (B ∪ B ′ )
=A∪B

Now, A ∩ B ⊂ A and A ∩ B ⊂ B. So, A ∩ B ⊂ A and A ∩ B ⊂ B

Combining we get A ∩ B ⊂ A ∩ B

Theorem:
The closure E of any set E is closed; that is, E = E.

Proof: The set E is closed if it contains all its points of closure. Let x be
a point of closure of E. Consider a neighborhood Ux of x. There is a point
x′ ∈ E ∩ Ux . Since x′ is a point of closure of E and Ux is a neighborhood of
x′ , there is a point x” ∈ E ∩ Ux . Therefore every neighborhood of x contains a
point of E and hence x ∈ E So the set E is closed. It is clear that if A ⊆ B,
then A ⊆ B and hence E = E.

Theorem:
The sets ϕ and X are closed; the union of any two sets is closed; and the inter-
section of any collection of closed sets is closed.
1.4. OPEN AND CLOSED SETS 5

Proof: Let {Fi : i = 1, 2, . . . , n} be a finite family of closed sets. Then, X − Fi

is an open set for i = 1, 2, . . . , n. Write Gi = X − Fi . Then
n
[ n
[ n
\
Fi = (X − Gi ) = X − Gi (by De M organ′ s law)
i=1 i=1 i=1

n
T
Since the finite intersection of open sets in X is open, then Gi is open.
i=1
n
T n
S
Therefore, X − Gi is closed and hence Fi is closed.
i=1 i=1

T
Let {Fα }α∈A be an arbitrary family of closed sets in X. We claim that Fα
α∈A
is closed. Since Fα is closed for each α ∈ A, X − Fα is open for each α ∈ A.
Write Gα = X − Fα , α ∈ A. Then
\ \ [
Fα = (X − Gα ) = X − Gα (by De M organ′ s law)
α∈A α∈A α∈A
S
Since arbitrary union of open sets in X is open, then is open. Therefore,
S T α∈A
X− Gα is closed and hence Fα is closed.
α∈A α∈A

Theorem:
The complement of an open set is closed; the complement of a closed set is open.

Proof: Suppose F is open. We have to show that its complement F ′ is closed.

For this we shall show that no point x in F can be a limit point of F ′ . Since F
is open, every open sphere with centre x is contained entirely in F and disjoint
from F ′ . So no point of F can be a limit point of F ′ . In other words, F ′ contains
all its limit points. Hence, F ′ is closed.

Now assume that F ′ is closed and show that F is open. If F is empty it is

open. So we suppose that F is non-empty. Let x ∈ F . Since F ′ is closed and x
is not a point of F ′ , x is not a limit point of F ′ . So there exists an open sphere
Sr (x) which is disjoint from F ′ . Hence Sr (x) is an open sphere centred at x and
contained in F . Since x is an arbitrary point of F , F is open subset of X.

Theorem:
A metric space X is separable if and only if there is countable family {Oi } of
open sets such that for any open set O ⊂ X,
[
O= Oi .
Oi ⊂0
6 CHAPTER 1. BASIC CONCEPTS

Proof: If X is separable, let D be a countable dense set. By the ball at x

with radius δ we means the set

Sδ (x) = {y : ρ(x, y) < δ}.

Let {Oi } consist of those balls Sδ (x) for which x is in D and δ is rational. Then
{Oi } is countable collection of open sets. If O is any open set and y ∈ O, then
we want to show that for some Oi we have y ∈ Oi ⊂ O. Since O is open, there
is a ball Sδ (y) such that Sδ (y) ⊂ O. By taking δ even smaller, we may assume
δ is rational. Since y is a point of closure of D, there is a point x ∈ D such that
ρ(x, y) < δ/2. Hence
Sδ/2 (x) ⊂ Sδ (y) ⊂ O.
But Sδ/2 (x) is one of the {Oi }, and the “only if” part of the theorem is proved.
Suppose, on the other hand, we are given the countable collection {Oi }. Let
xi be a point of Oi , and let D be the set of all these points xi . We shall now
see that D is dense. Let x be any point of X and S any spheroid centred at x.
Then we must show that S contains a point of D. But S is an open set, and so
we must have some Oi so x ∈ Oi ⊂ S. Hence xi ∈ S, and we see that x ∈ D.

1.5 Subspace of a Metric Space

Definition
Let (X, d) be a metric space and Y ⊂ X. The mapping dY : Y × Y → R given
by
dY (x, y) = d(x, y), ∀x, y ∈ Y
is a metric on Y. The metric dY is called the relative metric induced on Y by
d. The space (Y, dy ) is called the metric subspace of the metric space (X, d).
Chapter 2

Completeness

2.1 Introduction
One of the main aim to introduce metric spaces is to study convergent sequences
in a context more general than that of classical analysis. First recall the ”Cauchy
principle of convergence” in real numbers which states that a real sequence xn
converges to a limit in R if and only if for given any ϵ > 0, ∃ a positive integer N
such that |xn − xm | < ϵ, for all m, n > N . This leads to a class of metric spaces
called complete metric spaces for which a ”Cauchy principle of convergence”
is true. Complete metric spaces possess enough structure to establish many
important theorems that have wide applications in both topology and analysis.
This chapter deals with the completeness of metric spaces.

2.2 Some Definition

1. Convergent Sequence:
Let (X, d) be a metric spaces. A sequence xn in X is said to be convergent
if there is a point x ∈ X such that for each ϵ > 0, ∃ a positive integer N
such that
d(xn , x) < ϵ, ∀n ≥ N
or equivalently, for each open sphere Sϵ (x), centred on x,∃ a positive
integer N such that
xn ∈ Sϵ (x), ∀n ≥ N
We usually symbolize this by writing

xn → x or lim xn = x
n→∞

and we express it by saying that xn approaches x or that xn converges to

x. The element x is called the limit of the sequence xn .

7
8 CHAPTER 2. COMPLETENESS

2. Bounded Sequences:
In a metric space, a sequence is said to be bounded, if the range of the
sequence forms a bounded set.

3. Cauchy Sequences:
A sequence xn in a metric space (X, d) is said to be a Cauchy sequence if
for each ϵ > 0, ∃ a positive integer N such that

d(xm , xn ) < ϵ, ∀m, n ≥ N

Remark: In a metric space every convergent sequence is a Cauchy se-

quence. But the converse need not be true.

2.3 Complete Spaces

Definition
A metric space (X, d) is said to be complete if every Cauchy sequence in X is
convergent.

Examples
1. The usual metric spaces Ru are complete.

Proof: Let {xn } be a c Cauchy sequence of Ru , that is, each term of

the sequence belongs to Ru . We can choose ϵ > 0 as ϵ = 21q , for this
∃ nq ∈ N such that

1 1
m, n ≥ nq ⇒ d(xm , xn ) < ⇒ |xm −xn | < ∀ m, n ≥ n0 and xm , xn ∈ X.
2(q+1) 2(q+1)
1
Then we can have |xn − xnq | < 2(q+1)

If Iq is the closed interval such that nq ∈ Iq , where Iq = [xnq − 2−q , xnq +

2−q ] and since |xnq − xn(q+1) | < 2(q+1)
1

We get, I(q+1) ⊂ Iq

2 1
The length of Iq = 2q = 2(q−1)
→ 0 if q → ∞ that means |Iq | → 0
as k → ∞

Let I1 , I2 , . . . , In , . . . be a collection of closed intervals such that I1 ⊇

I2 ⊇ . . . ⊇ In ⊇ . . . then the intersection of this collection is non-empty
and equal to a singleton set.

Thus, the intersection of all Iq consists of exactly one point of Ru , say

s.
2.3. COMPLETE SPACES 9

So s ∈ Iq ∀ q ∈ N such that
1
|s − xnq | <
2(q+1)
Hence for all n ≥ nq , we have
1 1 1
|xn −s| = |xn −xnq −s+xnq | ≤ |xn −xnq |+|s−xnq | < + =
2(n+1) 2(n+1) 2q
1
⇒ |xn − s| =
2q
From the above, it follows that the Cauchy sequence {xn } converges to a
point s in Ru

Thus, every Cauchy sequence in R converge to a point in Ru .

Therefore, Ru is complete.
2. The discrete metric space Xd is complete.

Proof: Let {xn } be a Cauchy sequence in metric space Xd . Then we

can choose ϵ = 21 > 0 accordingly there exist n0 depending on ϵ such that

1
For n, m ≥ n0 ⇒ d(xm , xn ) < 2

⇒ d(xm , xn ) = 0 ∀ n, m ≥ n0 (since d is a discrete metric)

⇒ xn = x(n+1) = x(n+2) = . . . = x (say)

1
Therefore, d(xn , x) = 0 < 2 ∀ n ≥ n0

That means the Cauchy sequence {xn } converges to a point x in X.

Thus, every Cauchy sequence in Xd converges which makes Xd a com-

plete metric space.
3. The Euclidean space Rn is a complete metric space.

Proof: Let {xn } be a Cauchy sequence in Rn , where

(m) (m)
xm = (α1 , α2 , . . . , αn(m)

Then, for each ϵ > 0, ∃ a positive integer N such that

n
!1/2
(m) (p)
X
d(xm , xn ) = (αi − αi )2 < ϵ, ∀ m, p ≥ N
i=1
10 CHAPTER 2. COMPLETENESS

On squaring both sides, we get

n
(m) (p)
X
(αi − αi )2 < ϵ2
i=1
(m) (p)
⇒(αi − αi )2 < ϵ2
(m) (p)
⇒|αi − αi | < ϵ, ∀ m, p ≥ N, (i = 1, 2, . . . , n)

(m)
This shows that for each fixed i(1 ≥ i ≥ n), the sequence {αi }m is a
Cauchy sequence in the usual metric space Ru . Since Ru is complete, it
(m)
converges in Ru . Let αi → αi as m → ∞. Using these n limits, we
define x = (α1 , α2 , . . . , αn ). Clearly, x ∈ Rn . Letting p → ∞ in (1), we
obtain
d(xm , x) ≥ ϵ, ∀ m ≥ N

xm → x in Rn

Hence Rn is a complete metric space.

Theorem
Let (X, d) be a complete metric space and (Y, dY ) be a subspace of (X, d). Then,
Y is complete if and only if Y is closed.

Proof: Suppose that Y is complete. To prove that Y is closed, let x be a

limit point of Y . Then, every open sphere centred on x contains points (other
than x) of Y . In particular, the open sphere S1/n (x), where n is a positive
integer, contains a point xn of Y , other than x. Thus {xn } is a sequence in Y
such that xn → x in X since d(xn , x) < n1 . Since {xn } is a Cauchy sequence in
X and hence in Y . But Y being complete, x ∈ Y . Hence Y is closed.

Conversely, Assume that Y is closed. Let {xn } be a Cauchy sequence in Y .

Then, it is also Cauchy sequence in X since Y ⊂ X. But X being complete,
{xn } converges to a point x ∈ X. We claim that x ∈ Y . Now, there arise two
cases:

Case (i) If {xn } has only finitely many distinct points, then xn = x, for in-
finitely many values of n. Since {xn } is in Y , it follows that x ∈ Y .

Case (ii) If {xn } consists of infinitely many distinct points, then the limit of the
sequence is also the limit point of the range of the sequence {xn }. Therefore x
is also the limit point of Y since {xn } ⊂ Y . But Y being closed, x ∈ Y . Hence
Y is complete.
2.3. COMPLETE SPACES 11

Theorem
Let (X, d) be a complete metric space and let {Fn } be a decreasing sequence
of non-empty closed subset of X such that d(Fn ) → 0. Then, the intersection
∞
T
Fn contains exactly one point.
n=1

Proof: Construct a sequence {xn } in X by selecting a point xn ∈ Fn for each

n. Since the sets Fn are nested, xn ∈ Fm , ∀n ≥ m. We now prove that {xn } is
a Cauchy sequence.

Let ϵ > 0 be given. Since d(Fn ) → 0, ∃ a positive integer N such that

d(Fn ) < ϵ

Note that xn , xm ∈ FN , ∀n, m ≥ N and as such, we have

d(xn , xm ) ≤ d(FN )
< ϵ, ∀n, m ≥ N

Thus {xn } is a Cauchy sequence. Since (X, d) is a complete, ∃x ∈ X such that

xn → x.
∞
T
Now, we claim that x ∈ Fn .. Let n be fixed. Then, the subsequence
n=1
{xn , xn+1 , · · · } of xn is contained in Fn and still converges to x. But Fn being a
closed subspace of a complete metric space (X, d), it is complete and so x ∈ Fn .
∞
T ∞
T
This is true for each n ∈ N. Hence x ∈ Fn . This verifies that Fn is
n=1 n=1
non-empty.
∞
T
Finally, to establish that x is the only point in the intersection Fn , let
n=1
∞
T
y∈ Fn . Then x and y both are in Fn , for each n. Therefore
n=1

0 ≤ d(x, y) ≤ d(Fn ) → 0 as n → ∞

=⇒
d(x, y) = 0

=⇒
x=y

This completes the prove of the theorem.

12 CHAPTER 2. COMPLETENESS
Chapter 3

Continuous Function

3.1 Introduction
A number of characterisation of continuous functions are given in terms of open
sets, closed sets etc. The concept of uniform continuity and homeomorphism
are discussed.

3.2 Definition
Let (X, d) and (Y, ρ) be two metric spaces. A function f : X → Y is said to be
continuous at a point x0 ∈ X if for each ϵ > 0, ∃ a δ > 0 such that

d(x, x0 ) < δ ⇒ ρ(f (x), f (x0 )) < ϵ, x ∈ X

that is
x ∈ Sδ (x0 ) ⇒ f (x) ∈ Sϵ (f (x0 ))
which means the same thing as

f (Sδ (x0 )) ⊂ Sϵ (f (x0 ))

The function f is said to be continuous on X or simply continuous if it is

continuous at each point of X.

3.2.1 Theorem
Let (X, d) and (Y, ρ) be metric spaces and f : X → Y be a function. Then, f is
continuous at a point x0 ∈ X if and only if f (xn ) → f (x0 ), for every sequence
{xn } ⊂ X with xn → x0 .

Proof: We first assume that f is continuous at x0 . Let ϵ > 0 be given and

{xn } be sequence in X such that xn → x0 . Then, ∃ a δ > 0 such that

13
14 CHAPTER 3. CONTINUOUS FUNCTION

f (Sδ (x0 )) ⊂ Sϵ (f (x0 )). Also, since xn → x0 , ∃ a positive integer N such that
xn ∈ Sδ (x0 ), ∀n ≥ N . Hence

f (xn ) ∈ Sϵ (f (x0 )), ∀n ≥ N

This proves that f (xn ) → f (x0 ).

Conversely, let f (xn ) → f (x0 ), for every sequence {xn } with xn → x0 . Let,
if possible, f be not continuous at x0 . Then,∃ an ϵ > 0 such that there is no
open sphere centred on x0 whose f -image is contained in Sϵ (f (x0 )). Consider
the sequence of open spheres

S1 (x0 ), S1/2 (x0 ), · · · , S1/n (x0 ), · · ·

Let xn ∈ S1/n (x0 ) be such that f (xn ) ∈

/ Sϵ (f (x0 )). Thus, we get a sequence
{xn } ⊂ X with xn → x0 such that f (xn ) ↛ f (x0 ). This is contradiction to
hypothesis. Hence f is continuous a x0 .

3.2.2 Theorem
Let (X, d) and (Y, ρ) be metric spaces and f : X → Y be a function. Then, f
is continuous if and only if f −1 (G) is open in X whenever G is open in Y .

Proof: Let f be continuous and G be open in Y. We shall prove that f −1 (G)

is open in X. Take some x ∈ f −1 (G). Then f (x) ∈ G. Since G is open and
f (x) ∈ G, it follows that Sϵ (f (x)) ⊂ G, for some ϵ > 0. But f being continuous,
∃ a δ > 0 such that
f (Sδ (x)) ⊂ Sϵ (f (x)) ⊂ G
⇒ Sδ (x) ⊂ f −1 (G)
−1
Hence, f (G) is open.
Conversely, assume that f −1 (G) is open in X whenever G is open in Y .
Let x ∈ X be arbitrary and ϵ > 0 be given. Then, f (x) ∈ Y and Sϵ (f (x))
is an open set. Therefore, by assumption, f −1 (Sϵ (f (x))) is an open set and
x ∈ f −1 (Sϵ (f (x))). Consequently, ∃ a δ > 0 such that Sδ (x) ⊂ f −1 (Sϵ (f (x))),
i.e., f (Sδ (x)) ⊂ Sϵ (f (x)). This verifies that f is continuous at x. Hence f is
continuous.

3.2.3 Theorem
Let (X, d) and (Y, ρ) be a metric space and f : X → Y be a function. Then, f
is continuous if and only if f −1 (F ) is closed in X whenever F is closed in Y .

Proof: Let f be continuous and F be closed in Y . Then, Y − F is open in

Y and therefore f −1 (Y − F ) is open in X, Since f −1 (F ) is the complement of
f −1 (Y − F ) in X and f −1 (Y − F ) is open, it follows that f −1 (F ) is closed.
Conversely, suppose that f −1 (F ) is closed in X whenever F is closed in Y .
We shall prove that f is continuous, Let G be and open subset of Y . Then,
Y − G is closed in Y and by the hypothesis f −1 (Y − G) is closed in X. Using
3.3. UNIFORM CONTINUITY 15

the argument as in the first part, we note that f −1 (G) is open in X. Hence f
is continuous.

3.2.4 Theorem
Let (X, d), (Y, ρ), (Z, σ) be three metric spaces. Suppose f : X → Y and
g : Y → Z be continuous functions. Then g ◦ f , the composite of f and g,
is continuous.

Proof: We know that g ◦ f : X → Z. Let G be an open set in Z. Then

g −1 (G) is open in Y (∵ g is continuous)
⇒ f −1 (g −1 (G)) is open in X (∵ f is continuous)
⇒ (f −1 ◦ g −1 )(G) is open in X
⇒ (g ◦ f )−1 (G) is open in X (∵ f −1 ◦ g −1 = (g ◦ f )−1 )
⇒ g ◦ f is continuous .

3.3 Uniform Continuity

3.3.1 Definition
Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y is said to be
uniformly continuous if for given ϵ > 0, ∃δ > 0 such that

d(x, x′ ) < δ, x, x′ ∈ X ⇒ ρ(f (x), f (x′ )) < ϵ

Remark Uniform continuity implies continuity. But the converse need not be
true.

3.4 Theorem
Let (X, d) be a metric space and A ⊂ X. Then, the function f : X → R given
by
f (x) = d(x, A), x ∈ X
is uniformly continuous.

Proof: By triangle inequality

d(x, a) ≤ d(x, y) + d(y, a), ∀a ∈ A, x, y ∈ X

On taking infimum, we get

inf d(x, a) ≤ d(x, y) + inf d(y, a) (∵ d(x, y) is independent of a)

a∈A a∈A
⇒ d(x, A) ≤ d(x, y) + d(y, A)
⇒ d(x, A) − d(y, A) ≤ d(x, y)
16 CHAPTER 3. CONTINUOUS FUNCTION

This is true for all x, y ∈ X. Therefore, on interchanging x and y, we get

d(y, A) − d(x, A) ≤ d(x, y)

Thus
|d(x, A) − d(y, A)| ≤ d(x, y)
Therefore, for a given ϵ > 0, choosing a δ such that 0 < δ ≤ ϵ, we have

|f (x) − f (y)| = |d(x, A) − d(y, A)| ≤ d(x, y) < δ ≤ ϵ

i.e. |f (x) − f (y)| < ϵ, whenever, d(x, y) < δ

Hence f is uniformly continuous on X.

3.4.1 Theorem
Composition of two uniformly continuous function is a uniformly continuous
function.

Proof: Let f : A → B and g : B → C be two uniformly continuous func-

tions. We have to show that the composition of these functions, g ◦ f , is also
uniformly continuous.
Assume f and g are uniformly continuous functions. This means that for
any ϵ1 > 0, there exists a δ1 > 0 such that for every x, y ∈ A, if |x − y| < δ1 ,
then |f (x) − f (y)| < ϵ1 . Similarly, for any ϵ2 > 0, there exists a δ2 > 0 such
that for every u, v ∈ B, if |u − v| < δ2 , then |g(u) − g(v)| < ϵ2 .
Choose ϵ2 to be equal to ϵ > 0, where ϵ is the desired precision for the
composition function g ◦ f . Since g is uniformly continuous, there exists a
δ2 > 0 such that for every u, v ∈ B, if |u − v| < δ2 , then |g(u) − g(v)| < ϵ2 .
Choose ϵ1 to be equal to δ2 from the previous step. Since f is uniformly
continuous, there exists a δ1 > 0 such that for every x, y ∈ A, if |x − y| < δ1 ,
then |f (x) − f (y)| < ϵ1 .
Now, let’s show that g ◦ f is uniformly continuous. Let x, y ∈ A such that
|x − y| < δ1 . Since f is uniformly continuous, we have |f (x) − f (y)| < ϵ1 = δ2 .
Now let u = f (x) and v = f (y). Hence, |u − v| < δ2 . Since g is uniformly
continuous, we have |g(u) − g(v)| < ϵ2 = ϵ which implies |g(f (x)) − g(f (y))| < ϵ.
Therefore, g ◦ f is uniformly continuous. This completes the proof that the
composition of two uniformly continuous functions is also uniformly continuous
function.

3.4.2 Theorem
Let (X, d) and (Y, ρ) be metric spaces and f : X → Y be a uniformly continuous
function. If {xn } is a Cauchy sequence in X, then {f (xn )} is a Cauchy sequence
in Y .
3.5. HOMEOMORPHISM 17

Proof: Since f is uniformly continuous, for a given ϵ > 0, ∃ a δ > 0 such

that

d(x, x′ ) < δ, x, x′ ∈ X ⇒ ρ(f (x), f (x′ )) < ϵ

In particular, we have

d(xn , xm ) < δ ⇒ ρ(f (xn ), f (xm )) < ϵ (1)

But {xn } being a Cauchy sequence in X, given δ > 0, ∃ a positive integer N

such that

d(xn , xm ) < δ, ∀n, m ≥ N (2)

Therefore, from (1) and (2) it follows that

ρ(f (xn ), f (xm )) < ϵ, ∀n, m ≥ N

Hence {f (xn )} is a Cauchy sequence in Y .

3.5 Homeomorphism
3.5.1 Definition
Let (X, d) and (Y, ρ) be two metric spaces. A function f : X → Y is said to be
a homeomorphism if

(i) f is bijective.
(ii) f is continuous.
(iii) f −1 is continuous.

If a homeomorphism from X to Y exists, we say that the spaces X and Y are

homeomorphic.