Table of Contents

Chapter 1 Topological Spaces 1

Section 1.1 Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Section 1.2 Interiors and Closures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Section 1.3 Gδ - and Fσ -Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Section 1.4 The Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Section 1.5 Continuous Functions and Homeomorphisms . . . . . . . . . . . . . . . . . . 7

Section 1.6 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Section 1.7 Subbases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Section 1.8 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Chapter 2 The Separation Axioms 14

Section 2.1 Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Section 2.2 Regular and Completely Regular Spaces . . . . . . . . . . . . . . . . . . . . 15

Section 2.3 Normal Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Section 2.4 Urysohn’s Lemma and Tietze’s Extension Theorem . . . . . . . . . . . . . . 22

Chapter 3 Covering Properties 23

Section 3.1 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Section 3.2 Countably Compact and Sequentially Compact Spaces . . . . . . . . . . . . 25

Section 3.3 Lindelöf Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Section 3.4 Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Section 3.5 The Alexandroff One-Point Compactification . . . . . . . . . . . . . . . . . . 31

Chapter 4 The Countability Axioms 33

Section 4.1 First and Second Countable Spaces . . . . . . . . . . . . . . . . . . . . . . . 33

Section 4.2 Separable Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Chapter 5 Product Spaces 35

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 Section 5.1 Finite Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Section 5.2 Infinite Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Section 5.3 The Stone-Čech Compactification . . . . . . . . . . . . . . . . . . . . . . . . 41

Chapter 6 Connected Spaces 42

Section 6.1 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Section 6.2 Pathwise Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Section 6.3 Locally Pathwise Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . 45

Section 6.4 Locally Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Section 6.5 Totally Disconnected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Chapter 7 The Cantor Set 48

Chapter 8 Ordinals and Cardinals 50

Section 8.1 Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Section 8.2 The Space ω1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Section 8.3 Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Section 8.4 Products of Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Chapter 9 Nets 57

Section 9.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Section 9.2 Nets and Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Section 9.3 Ultranets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Chapter 10 Filters 63

Section 10.1 Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Section 10.2 Filters and Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Section 10.3 Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Section 10.4 Filters and Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Chapter 11 Quotient Spaces 69

Section 11.1 The Quotient Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Section 11.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Chapter 12 Continua 71

Section 12.1 Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

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 Section 12.2 Subcontinua and Irreducible Continua . . . . . . . . . . . . . . . . . . . . . . 72

Section 12.3 Cut and Noncut Points of Continua . . . . . . . . . . . . . . . . . . . . . . . 72

Section 12.4 Metrizable Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Chapter 13 Mysior’s Space 77

Chapter 14 Topological Groups 79

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 Chapter 1: Topological Spaces

Section 1.1: Definitions and Examples

Definition 1: Let X be a set and τ ⊆ P (X). Then τ is a topology on X if the following three

properties are satisfied.

(i) X, ϕ ∈ τ ;

(ii) If U ⊆ τ , then ∪U ∈ τ ;

(iii) If U ⊆ τ is finite, then ∩U ∈ τ .

Note 2: Let X be a set and τ ⊆ P (X). Then the following are equivalent.

(i) If U ⊆ τ is finite, then ∩U ∈ τ ;

(ii) If U, V ∈ τ , then U ∩ V ∈ τ .

Definition 3: A topological space is a pair ⟨X, τ ⟩, where X is a set and τ is a topology on X.

Definition 4: Let ⟨X, τ ⟩ be a topological space.

(i) The elements of τ are called open sets;

(ii) The complements of elements of τ are called closed sets;

(iii) Sets that are open and closed are called clopen sets.

Example 5 (Topology Generated by a Metric): Let ⟨X, d⟩ be a metric space. Then τd = {U ⊆

X | U is open in X} is a topology on X called the topology generated by d.

Example 6 (Trivial Topology): Let X be a set. Then τ = {X, ϕ} is a topology on X called the

trivial topology.

Example 7 (Discrete Topology): Let X be a set. Then τ = P (X) is a topology on X called the

discrete topology.

Example 8 (The Michael Line): Let M = R and τ = {U ∪ A | U is open in R and A ⊆ R \ Q}.

Then (M, τ ) is a topological space.

Proof: Let R denote the reals with the usual topology and M denote the Michael Line. First, since

R and ϕ are open in R, R and ϕ are open in M . Now suppose U = {Uα }α∈Λ is a collection of

open subsets of M . For each α ∈ Λ, Uα = Vα ∩ Aα , where Vα ⊆ R is open and Aα ⊆ R \ Q. Then

S S S S S
∪U = Vα ∪ Aα = ( Vα ) ∪ ( Aα ). Since Vα is open in R and Aα ⊆ R \ Q, ∪U is
α∈Λ α∈Λ α∈Λ α∈Λ α∈Λ
open in M . Finally, suppose that U and V are open in M . Then U = U ′ ∪ A and V = V ′ ∪ B,

where U ′ and V ′ are open in R and A, B ⊆ R \ Q. Then U ∩ V = (U ′ ∪ A) ∩ (V ′ ∪ B) =

(U ′ ∩ V ′ ) ∪ (U ′ ∩ B) ∪ (A ∩ V ′ ) ∪ (A ∩ B) = (U ′ ∩ V ′ ) ∪ [(U ′ ∩ B) ∪ (A ∩ V ′ ) ∪ (A ∩ B)]. Since (U ′ ∩ V ′ )

1
is open in R and [(U ′ ∩B)∪(A∩V ′ )∪(A∩B)] ⊆ R\Q, U ∩V = (U ′ ∩V ′ )∪[(U ′ ∩B)∪(A∩V ′ )∪(A∩B)]

is open in M , as desired. ■

The topological space (M, τ ) above is called the Michael Line.

Example 9 (Co-Finite Topology): Let X be a set and τ = {U ⊆ X | U = ϕ or |X \ U | < ℵ0 }.

Then τ is a topology on X called the co-finite topology.

Proof: First, note that X, ϕ ∈ τ by design. Next, suppose U ⊆ τ . If U ∈ U , then X \ ∪U ⊆ X \U ,

which is finite. Thus, X \ ∪U is finite. Finally, suppose V, W ∈ τ . Then X \ (V ∩ W ) =

(X \ V ) ∪ (X \ W ), which is finite. ■

Example 10 (Co-Countable Topology): Let X be a set and τ = {U ⊆ X | U = ϕ or

|X \ U | ≤ ℵ0 }. Then τ is a topology on X called the co-countable topology.

Proof: First, note that X, ϕ ∈ τ by design. Next, suppose U ⊆ τ . If U ∈ U , then X \∪U ⊆ X \U ,

which is countable. Thus, X \ ∪U is countable. Finally, suppose V, W ∈ τ . Then X \ (V ∩ W ) =

(X \ V ) ∪ (X \ W ), which is countable. ■

Definition 11: A topological space ⟨X, τ ⟩ is metrizable if there is a metric d on X such that

τ = τd .

Definition 12: Let X be a topological space, x ∈ X, and U ⊆ X be open. If x ∈ U , then U is

sometimes called an open neighborhood of X.

Definition 13: Let X be a topological space and {xn }n∈N ⊆ X. Then {xn }n∈N converges to x if

for every open neighborhood U of x, there is an M ∈ N such that {xn }n>M ⊆ U .

Exercises

1. Give an example of a sequence {Bi }i∈N of a topological space X whose union is not closed.

2. Let X be a metrizable topological space and ρ generate the topology on X. For each λ ∈ Λ,

suppose Cλ ⊆ X is closed such that ρ(Cλ1 , Cλ2 ) ≥ ε for all λ1 , λ2 ∈ Λ, where ε > 0. Prove ∪Cλ is

closed.

3. Let X be a set and τ ⊆ P (X). Show that the following are equivalent.

(i) If U ⊆ τ is finite, then ∩U ∈ τ ;

(ii) If U, V ∈ τ , then U ∩ V ∈ τ .

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 Section 1.2: Interiors and Closures
Definition 14: Let X be a topological space and A ⊆ X. The interior of A is the set int(A) =

{x ∈ A | there is an open U ⊆ X such that x ∈ U ⊆ A}.

Note 15: Suppose that X be a topological space, A ⊆ X, and that U ⊆ X is open in X. Then

U ⊆ int(A) is open in X.

Proposition 16: Let X be a topological space and A ⊆ X.

(i) int(A) = ∪{U ⊆ A | U is open in X};

(ii) int(A) is the largest open subset of X contained in A.

Proof: For (i), first note that ∪{U ⊆ X | U is open and U ⊆ A} ⊆ int(A). Now suppose that

x ∈ int(A). Then there is an open set V ⊆ A such that x ∈ V . Then x ∈ V ⊆ ∪{U ⊆ A | U is open

in X}. Therefore, int(A) = ∪{U ⊆ A | U is open in X}.

For (ii), note that by (i), int(A) is open and int(A) ⊆ A. Now suppose that V ⊆ A is open. Then

V ⊆ ∪{U ⊆ A | U is open in X} = int(A). ■

Corollary 17: Let X be a topological space and U ⊆ X. Then U is open in X if and only if

U = int(U ).

Proof: (=⇒) Suppose that U is open in X. Then U is the largest open set contained in U which

means that U = int(U ) by the proposition above.

(⇐=) Suppose that U = int(U ). By the proposition above, U is open in X. ■

Definition 18: Let X be a topological space, x ∈ X, and A ⊆ X. Then x is a cluster (accumu-

lation) point of A if every open neighborhood of x meets A \ {x}.

Definition 19: Let X be a topological space and A ⊆ X. The derived set of A is the set

A′ = {x ∈ A | x is an accumulation point of A}.

Note 20: Let X be a topological space and A ⊆ X.

(i) A′ ⊆ X;

(ii) A′ is not necessarily a subset of A.

 
1
Example 21: Consider R with the usual topology and let A = . Then 0 ∈ A′ but 0 ∈ / A.
n n∈N
Definition 22: Let X be a topological space and A ⊆ X. The closure of A is the set A = cl(A) =

{x ∈ X | every neighborhood of x meets A}.

Note 23: Let X be a topological space and A ⊆ X.

(i) A ⊆ A;

(ii) A′ ⊆ A.

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Proposition 24: Let X be a topological space and A ⊆ X. Then

(i) A = A ∪ A′ ;

(ii) A = ∩{E ⊆ X | A ⊆ E and E is closed in X};

(iii) A is the smallest closed subset of X containing A.

Proof: For (i), since A ⊆ A and A′ ⊆ A, A ∪ A′ ⊆ A. To see that A ⊆ A ∪ A, let x ∈ A \ A and let

U be an open neighborhood of x. Since x ∈ A, U ∩ A ̸= ϕ. Since x ∈

/ A, we have U ∩ (A \ {x}) ̸= ϕ,

and so, x ∈ A′ . Hence, A ⊆ A ∪ A′ . Therefore, A = A ∪ A′ .

For (ii), suppose that x ∈

/ A. Then there is an open U ⊆ X such that x ∈ U and U ∩ A = ϕ. Then

X \ U is closed, A ⊆ X \ U , and x ∈
/ X \ U . Therefore, x ∈
/ ∩{E ⊆ X | A ⊆ E and E is closed in

X}. Now suppose that x ∈

/ ∩{E ⊆ X | A ⊆ E and E is closed in X}. Then there is a closed F ⊆ X

such that A ⊆ F and x ∈

/ F . Since F is closed, X \ F is open. Then, X \ F is an open neighborhood

of x and (X \ F ) ∩ A = ϕ. Therefore, x ∈
/ A.

For (iii), note that by (i), A is closed and A ⊆ A. To see that A is the smallest closed set containing

A, let F be a closed set containing A. Then, by (ii), A = ∩{E ⊆ X | A ⊆ E and E is closed in

X} ⊆ F . ■

Corollary 25: Let X be a topological space and E ⊆ X. Then E is closed in X if and only if

E = E.

Definition 26: Let X be a topological space and D ⊆ X. Then D is dense in X if D = X.

Exercises

Definition: An open subset G in a topological space is regularly open if G = int(G).

Definition: A closed subset E in a topological space is regularly closed if E = int(E).

4. (a) Show that the complement of a regularly open set is regularly closed.

(b) Show that the complement of a regularly closed set is regularly open.

(c) Prove that there are open sets in R that are not regularly open.

(d) If A is a subset of a topological space, show that int(A) is regularly open.

(e) Show that the intersection, but not necessarily the union, of two regularly open sets is regularly

open.

(f ) If A is a subset of a topological space X, let A⊥ = X \ A. Prove that A is regularly open if and

only if (A⊥ )⊥ = A.

Definition: Let X be a topological space and E ⊆ X. The frontier of E is the set FrX (E) =
 
Fr(E) = E ∩ X \ E .

5. Let X be a topological space and E ⊆ X.

4
(a) Prove that E = E ∪ Fr(E);

(b) Prove that int(E) = E \ Fr(E);

(c) Prove that X = int(E) ∪ Fr(E) ∪ int(X \ E).

(d) Prove that any closed subset of R2 is the frontier of some set in R2 .

Definition: Let X be a topological space and A ⊆ X. The boundary of A, denoted ∂A, is defined

as follows:

x ∈ ∂A if and only if for every open set U ⊆ X containing x, U ∩ A ̸= ϕ and U ∩ (X \ A) ̸= ϕ.

6. Let X be a topological space and A ⊆ X.

(a) Prove that ∂A = A ∩ X \ A = A \ int(A).

(b) Prove that A = A ∪ ∂A and int(A) = A \ ∂A.

(c) Prove that X = int(A) ⊔ ∂A ⊔ int(X \ A). (Note: To give a complete solution, you must show

that each set is disjoint and that their union is X.)

Definition: A subset of R2 is radially open if it contains an open line segment in each direction

about each of its points.

7. (Radial Plane) Prove that the collection of radially open sets is a topology for R2 . Compare

this topology with the usual topology on R2 . The plane with this topology is called the radial

plane.

Section 1.3: Gδ - and Fσ -Sets

Definition 27:

(i) A subset of a topological space is a Gδ -set if it is the intersection of a countable collection of

open sets.

(ii) A subset of a topological space is an Fσ -set if it is the union of a countable collection of closed

sets.

Note 28:

(i) The complement of a Gδ -set is an Fσ -set;

(ii) The complement of an Fσ -set is a Gδ -set.

Exercises

8. Prove that the complement of a Gδ -set is an Fσ -set.

9. Prove that the complement of an Fσ -set is a Gδ -set.

10. Show that an Fσ -set can be written as the union of an increasing sequence of closed sets. (Hence,

a Gδ -set can be written as a decreasing intersection.)

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11. Show that a closed set in a metric space is a Gδ -set. [Hence, an open set in a metric space is
1
an Fσ -set.] (Hint: If A is a closed set, then let An = {y | ρ(A, y) < }.)
n
12. Show that Q is an Fσ -set in R.

Definition: A topological space X is perfect if every closed subset of X is a Gδ -set.

13. Prove that every metric space is perfect.

Section 1.4: The Baire Category Theorem

Definition 29: A subset A of a topological space is nowhere dense if int(A) = ϕ.
 
1
Example 30: Consider R with the usual topology and let A = . Then A is nowhere dense
n n∈N
in R.

Definition 31: A topological space is of first category if it is the union of a countable collection

of nowhere dense sets.

Definition 32: A topological space is of second category if it is not of first category.

Theorem 33: A topological space is of second category if and only if the intersection of any

countable collection of dense open sets is nonempty.

Proof: Let X be a topological space.

(=⇒) Suppose that X is of second category and let U = {Un }n∈N be a collection of dense open

sets. Then for each n ∈ N, X \ Un is a closed nowhere dense set. Since X is of second category,
S T
X ̸= X \ Un . Therefore, Un ̸= ϕ by DeMorgan’s Laws.
n∈N n∈N
(⇐=) Suppose that X has the property that the intersection of any countable collection of dense

open sets is nonempty. To see that X is not of first category, let {En }n∈N be a collection of nowhere

dense sets. Since the closure of a nowhere dense set is nowhere dense, we may assume that En

is closed for every n ∈ N. This implies that X \ En is an open dense set for each n ∈ N. Since
T S
X \ En ̸= ϕ, X ̸= En by DeMorgan’s Laws. ■
n∈N n∈N
Definition 34: A topological space is Baire if the intersection of any countable collection of dense

open sets is dense.

Note 35: Every Baire space is of second category.

Definition 36: Let ⟨X, d⟩ be a metric space. The diameter of A is d(A) = sup({d(x, y) | x, y ∈ A}).

Theorem 37 (Cantor Intersection Theorem): Let ⟨X, d⟩ be a complete metric space and let

{En }n∈N be a decreasing sequence of closed nonempty subsets of X such that lim d(En ) = 0. Then
n→∞
T
En is a singleton set.
n∈N
T
Proof: First, note that if x, y ∈ En and x ̸= y, then d(En ) ≥ d(x, y) > 0 for all n ∈ N, which
n∈N
T
contradicts the assumption that d(En ) = 0. So, En contains at most one element. It remains
n∈N

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 T
to show that En ̸= ϕ. For each n ∈ N, choose xn ∈ En . Since En+1 ⊆ En for all n ∈ N and
n∈N
lim d(En ) = 0, {xn }n∈N is a Cauchy sequence, and so, since ⟨X, d⟩ is complete, there is an x ∈ X
n→∞
T
such that xn → x. Now, since En+1 ⊆ En for all n ∈ N, x ∈ En = En . Therefore, x ∈ En , as
n∈N
desired. ■

Theorem 38 (Baire Category Theorem): A complete metric space is Baire.

Proof: Let ⟨X, d⟩ be a complete metric space and let {Un }n∈N be a collection of dense open
T
subsets of X. To see that Un is dense, let x ∈ X and let U be an open neighborhood of
n∈N
x. We proceed by induction to define a sequence {xn }n∈N in X and a sequence {rn }n∈N in R.

Let x1 ∈ U ∩ U1 and let r1 ∈ R such that B(x1 , r1 ) ⊆ U ∩ U1 and r1 < 1. For n ≥ 2, let
S T
xn ∈ B(xn−1 , rn−1 ) ∩ Un and let rn ∈ R such that B(xn , rn ) ⊆ B(xn−1 , rn−1 ) ∩ Un and
i≤n i≤n
1 2
rn ≤ . Note that {B(xn , rn )}n∈N is a decreasing sequence of closed sets. Since d(B(xn , rn )) ≤
n n
for all n ∈ N, lim d(B(xn , rn )) = 0. So, by the Cantor Intersection Theorem, there is a unique
n→∞
T T T T
y∈ B(xn , rn ). Since B(xn , rn ) ⊆ Un and B(xn , rn ) ⊆ U , the theorem is proved. ■
n∈N n∈N n∈N n∈N
Corollary 39: A complete metric space is of second category.

Exercises

14. Prove that Z with the restriction of the usual metric on R is complete, even though it is the

union of countably many singleton sets. Why does this not contradict the Baire Category Theorem?

15.

(a) Prove that Q is not a Gδ -set in R with the usual topology.

(b) Conclude that the Michael Line is not perfect.

(c) Prove that Q is not a Gδ -set in the Michael Line.

(d) Conclude that the Michael Line is not metrizable.

Section 1.5: Continuous Functions and Homeomorphisms

Definition 40: Let X and Y be topological spaces and let f : X → Y . Then f is continuous at

x ∈ X if for every open neighborhood V of f (x), there is an open neighborhood U of x such that

f (U ) ⊆ V .

Theorem 41: Let X and Y be topological spaces and let f : X → Y . Then the following are

equivalent.

(i) f is continuous at every x ∈ X;

(ii) f −1 (V ) is open in X for all open V ⊆ Y ;

7
(iii) f −1 (F ) is closed in X for all closed F ⊆ Y .

Proof: ((i) =⇒ (ii)) Let V be open in Y . To see that f −1 (V ) is open in X, let x ∈ f −1 (V ). Since

f is continuous at x and V is an open neighborhood of f (x), there is an open neighborhood U of x

such that f (U ) ⊆ V . Therefore, x ∈ U ⊆ f −1 (V ), as desired.

((ii) =⇒ (iii)) Let F ⊆ Y be closed. Then X \ f −1 (F ) = f −1 (Y \ F ) is open in X by (ii). Since

X \ f −1 (F ) is open in X, f −1 (F ) is closed in X.

((iii) =⇒ (i)) Finally, let x ∈ X. To see that f is continuous at x, let V be an open neighborhood

of f (x). By (iii), X \ f −1 (V ) = f −1 (Y \ V ) is closed in X. Since f (x) ∈ V , x ∈ f −1 (V ).

Therefore, there is an open neighborhood U of x such that x ∈ U ⊆ f −1 (V ), which implies

f (U ) ⊆ f (f −1 (V )) ⊆ V . Therefore, f is continuous at x. ■

Definition 42: Let X and Y be topological spaces and let f : X → Y . Then f is continuous on

X if it satisfies any of the properties above.

Theorem 43: Let X and Y be topological spaces and let f : X → Y be continuous. If {xn }n∈N ⊆ X

is a sequence that converges to x ∈ X, then {f (xn )}n∈N converges to f (x).

Proof: Let V be an open neighborhood of f (x). Since f is continuous, f −1 (V ) is open in X. Since

{xn }n∈N converges to x, there is an M ∈ N such that xn ∈ f −1 (V ) for all n ≥ M . Then, for all

n ≥ M , f (xn ) ∈ V . Therefore, {f (xn )}n∈N converges to f (x). ■

Definition 44: Let X and Y be topological spaces and let f : X → Y . Then f is open if f (U ) is

open in Y for all open U ⊆ X.

Definition 45: Let X and Y be topological spaces and let f : X → Y . Then f is closed if f (E)

is closed in Y for all closed E ⊆ X.

Definition 46: Let X and Y be topological spaces.

(i) A function f : X → Y is a homeomorphism if it is a continuous and open bijection such that

f −1 is continuous.

(ii) If f is not onto, then it is an embedding of X into Y and X is embedded into Y .

Definition 47: A property P is a topological property if every homeomorphic image of a topo-

logical space with property P has property P .

Exercises

16 Prove that the composition of continuous functions is continuous.

17. Prove that metrizability is a topological property.

18. Let X be a topological space. Prove that “X ⊆ R” is not a topological property.

19. Prove that R is homeomorphic to (−1, 1).

8
20. Let X and Y be topological spaces. If X can be embedded into Y and Y can be embedded into

X, prove that X and Y need not be homeomorphic.

21. Let X be a set and A ⊆ X. The characteristic function of A is a function f : A → R defined

1 if x ∈ A;

by f (x) =

0 if x ∈
 / A.
(a) Prove that the characteristic function of A is continuous if and only if A is a clopen subset of

X;

(b) Prove that X has the discrete topology if and only if whenever Y is a topological space and

f : X → Y , then f is continuous.

22. Let X be a topological space such that for every closed A ⊆ X, there is a continuous f : X →

[0, 1] with A = f −1 (0). Prove that X is perfect.

Section 1.6: Bases

Definition 48: Let ⟨X, τ ⟩ be a topological space and B ⊆ τ . Then B is a base (basis) for τ if

for all U ∈ τ , there is a B ′ ⊆ B such that U = ∪B ′ .

Proposition 49: Let ⟨X, τ ⟩ be a topological space and B ⊆ τ . Then B is a base for τ if and only

if for every x ∈ X, and every open neighborhood U of x, there is a B ∈ B such that x ∈ B ⊆ U .

Proof: (=⇒) Let x ∈ X and U be an open neighborhood of x. Then there is a B ⊆ B ′ such that

∪B ′ ⊆ U . So, choose B ∈ B ′ such that x ∈ B. Then, x ∈ B ⊆ ∪B ′ ⊆ U , as desired.

(⇐=) Let U ⊆ X be open. Then for every x ∈ U , there is a Bx ⊆ B such that x ∈ Bx ⊆ U . Then
S
U= Bx , as desired. ■
x∈U
Example 50: The following are bases for the usual topology on R.

(a) {(a, b) | a, b ∈ R with a < b};

(b) {(a, b) | a, b ∈ Q with a < b};

(c) {(a, b) | a, b ∈ R \ Q with a < b};

  
1 1
(d) a − ,a + | a ∈ R and n ∈ N ;
 n n 
1 1
(e) a − ,a + | a ∈ Q and n ∈ N .
n n    
1
Example 51: Let ⟨X, d⟩ be a metric space. Then B x, | x ∈ X and n ∈ N is a base for the
n
topology generated by d.

Theorem 52: Let X be a set and B ⊆ P (X). Then B is a base for a topology on X if and only if

∪B = X and whenever B1 , B2 ∈ B and x ∈ B1 ∩ B2 , there is a B3 ∈ B such that x ∈ B3 ⊆ B1 ∩ B2 .

Proof: (=⇒) Suppose B is a base for a topology τ on X. Since X is open, there is a B ′ ⊆ B such

that X = ∪B ′ ⊆ ∪B ⊆ X. Therefore, X = ∪B. Now suppose that B1 , B2 ∈ B and x ∈ B1 ∩ B2 .

9
Since B1 ∩ B2 is open, there is a U ∈ τ such that x ∈ U ⊆ B1 ∩ B2 . Since B is a base for τ , there

is a B3 ∈ B such that x ∈ B3 ⊆ U ⊆ B1 ∩ B2 , as desired.

(⇐=) Suppose B satisfies ∪B = X and whenever B1 , B2 ∈ B and x ∈ B1 ∩ B2 , there is a B3 ∈ B

such that x ∈ B3 ⊆ B1 ∩ B2 . Let τ = {∪B ′ | B ′ ⊆ B}. Since ∪B = X, X ∈ τ . Also, since

ϕ is the union of the empty collection, ϕ ∈ τ . Now suppose U ⊆ τ . Then, for each U ∈ U ,

there is a BU ⊆ B such that U = ∪BU . Let C = {B ∈ B | B ∈ BU for some U ∈ U }. Then

(∪BU ) = ∪C ∈ τ . Finally, suppose U1 , U2 ∈ τ . Then there are B1 , B2 ∈ B such that

S
∪U =
U ∈U
U1 = ∪B1 and U2 = ∪B2 . For each B ∈ B1 and each C ∈ B2 , there is a BB,C ∈ B such that
S h S i
B ∩ C = ∪BB,C by the second property. Then, U1 ∩ U2 = (∪B1 ) ∩ (∪B2 ) = (B ∩ C) =
B∈B1 C∈B2
S h S i
(∪BB,C ) ∈ τ . ■
B∈B1 C∈B2
Definition 53: Let X be a set and suppose B ⊆ P (X) is a base for a topology on X. Then

τ = {∪B ′ | B ′ ⊆ B} is called the topology generated by B.

Theorem 54: Let S = {[a, b) | a, b ∈ R with a < b}. Then S generates a topology on R stronger

than the usual topology.

S
Proof: First note that R = [−n, n) ⊆ ∪S. Therefore, S generates a topology on R. Let S denote
n∈N
S
R with the topology generated by S. If r, s ∈ R with r < s, then [r, s) = [x, s), which is open
r<x<s
in S. Therefore, the topology on S is stronger than the topology on R. ■

Definition 55: R with the topology generated by the base {[a, b) | a, b ∈ R with a < b} is called

the Sorgenfrey Line.

Note 56: The Sorgenfrey Line is not metrizable.

Theorem 57: Let X and Y be topological spaces and suppose B is a base for Y . Then f is

continuous if and only if f −1 (B) is open in X for all B ⊆ B.

Proof: (=⇒) Suppose that f is continuous and B ∈ B. Since B is open in Y and f is continuous,

f −1 (B) is open in X.

(⇐=) Suppose that f −1 (B) is open in X for all B ∈ B. To see that f is continuous, let U be open

in Y . Then, there is a B ′ ⊆ B such that ∪B ′ = U . Now, f −1 (U ) = f −1 (∪B ′ ) =

S −1
f (B),
B∈B′
which is open in X since f −1 (B) is open in X for all B ∈ B. ■

Definition 58: Let ⟨X, τ ⟩ be a topological space and x ∈ X. Then B ⊆ τ is a local base at x if

x ∈ B for all B ∈ B and whenever x ∈ U ∈ τ , there is a B ∈ B such that x ∈ B ⊆ U .

Example 59: Consider R with the usual topology and let x ∈ R. Then the following are local bases

at x.

(a) {(x − a, x + b) | a, b ∈ R with a, b > 0};

10
   
1 1
(b) x − ,x + |n∈N .
n n

Exercises

23. For each a, d ∈ N, set B(a, d) = {a + nd | n ∈ N} and B = {B(a, d) | a, d ∈ N}.

(a) Show that B is a base for a topology on N.

(b) Let τ denote the topology on N generated by B. Show that {1} is not open in (N, τ ).
T
(c) Set P = {p ∈ N | p is prime}. Show that {1} = B(1, p).
p∈P
(d) Conclude that P is infinite.

24. Prove that the collection of open rectangles is a base for a topology on R2 .

25. For each n ∈ N, let Sn = {n, n + 1, . . .}. Show that {S ⊆ N | Sn ⊆ S} is a base for a topology

on N. Then describe the closure operation in this space.

26. Show that {(a, b) | a, b ∈ R with a < b} ∪ {n}, where n ∈ Z \ {0}, is a base for a topology on R.

Then describe the interior operation in the resulting space.

27. (The Scattered Line) Define a topology on R as follows: A set is open if and only if it is of

the form U ∪ V , where U ⊆ R is open and V ⊆ R \ Q. The resulting space is called the scattered

line and is denoted S.

(a) Use the definition of open set above to show that S is a topological space.

(b) Describe a local base at Q and R \ Q in S. Then describe a base for the topology on S.

28. Let X be a topological space and B be a base for the topology on X. Prove that there is a

dense D ⊆ X such that |D| ≤ |B|.

Section 1.7: Subbases

Definition 60: Let ⟨X, τ ⟩ be a topological space and S ⊆ τ . Then S is a subbase for τ if the

collection {∩S ′ | S ′ ⊆ S and S is finite} is a base for τ .

Example 61: Let S = {(a, ∞) | a ∈ R} ∪ {(−∞, b) | b ∈ R}. Then S is a subbase for the usual

topology on R.

Example 62: Let S = {(a, ∞) | a ∈ R} ∪ {(−∞, b) | b ∈ R}. Then S is a subbase for the

Sorgenfrey Line.

Theorem 63: Let X and Y be topological spaces and suppose S is a subbase for Y . Then

f : X → Y is continuous if and only if f −1 (S) is open in X for all S ∈ S .

Proof: (=⇒) Suppose f is continuous and S ∈ S . Since S is open in Y and f is continuous, f −1 (S)

is open in X.

(⇐=) Suppose f −1 (S) is open in X for all S ∈ S and let B = {∩S ′ | S ′ ⊆ S and S is finite}.

11
By the theorem above, it suffices to show that f −1 (B) is open in X for all B ∈ B. Let U ⊆ S be

finite and note that f −1 (∩U ) =

T −1
f (U ), which is open in X since f −1 (U ) is open in X for all
U ∈U
U ∈U. ■

Exercises

29. Let X be a set. Show that a collection of subsets of X is a subbase for a topology on X.

Then show that the topology generated by the collection of subsets of X is the smallest topology

containing the collection.

Section 1.8: Subspaces

Theorem 64: Let ⟨X, τ ⟩ be a topological space and suppose Y ⊆ X. Then {U ∩ Y | U ∈ τ } is a

topology on Y .

Proof: Let τY = {U ∩ Y | U ∈ τ }. Since ϕ, X ∈ τ , ϕ = ϕ ∩ Y ∈ τY and Y = X ∩ Y ∈ τY .

Now suppose V ⊆ τY . For each V ∈ V , there is a UV ∈ τ such that U = UV ∩ Y . Then

h S  i
UV ∩ Y ∈ τY . Finally, let W ⊆ τY be finite. For each W ∈ W ,
S
∪V = (UV ∩ Y ) =
V ∈V V ∈V h T  i
T
there is an OW ∈ τ such that W = OW ∩Y . Then, ∩W = (OW ∩Y ) = OW ∩Y ∈ τY .
W ∈W W ∈W
Therefore, τY is a topology on Y . ■

Definition 65: Let ⟨X, τ ⟩ be a topological space and suppose Y ⊆ X. Then {U ∩ Y | U ∈ τ } is the

subspace topology on Y inherited from X.

Definition 66: A property P is called hereditary if every subspace of a space with property P

has property P .

Exercises

30. Let X be a topological space and suppose Y ⊆ X. Show that E ⊆ Y is closed in Y if and only

if there is a closed in X set F such that E = Y ∩ F .

31. Let X be a topological space, Y be a subspace of X, and suppose A ⊆ Y . Prove that

clY (A) = clX (A) ∩ Y .

32. Let X and Z be topological spaces and suppose Y is a subspace of X. Prove that if f : X → Z

is continuous, then f ↾Y : Y → Z is continuous.

33. Let X be a linearly ordered set X by a relation ≤. Consider the subbase for a topology on

X consisting of all sets of the form {x ∈ X | x < a} and {x ∈ X | x > a} for every a ∈ X. This

topology is called the order topology on X. An ordered space is a linearly ordered set with

12
its order topology. An interval in a linearly ordered space is any subset which contains all points

between x and y whenever it contains x and y.

(a) If a, b ∈ X with a < b, show that the interval {x ∈ X | a < x < b} is an open set in the order

topology but intervals of the form {x ∈ X | a ≤ x ≤ b} may also be open.

(b) Show that the usual topology on R is the order topology given by the usual order.

13
 Chapter 2: The Separation Axioms

Section 2.1: Hausdorff Spaces

Definition 67: A topological space X is Hausdorff, or T2 , if for each x, y ∈ X with x ̸= y, there

are disjoint open sets U, V ⊆ X such that x ∈ U and y ∈ V .

Proposition 68: Metric spaces are Hausdorff.

d(x, y)
Proof: Let ⟨X, d⟩ be a metric space and suppose x, y ∈ X with x ̸= y. Let r = . To see
2
that B(x, r) ∩ B(y, r) = ϕ, suppose by means of contradiction that there is a z ∈ B(x, r) ∩ B(y, r).
d(x, y) d(x, y)
Then d(x, y) ≤ d(x, z) + d(y, z) < r + r = + = d(x, y), a contradiction. Therefore,
2 2
B(x, r) ∩ B(y, r) = ϕ, and so, X is Hausdorff. ■

Definition 69: A topological space is T1 if for each x, y ∈ X with x ̸= y, there are open X, Y ⊆ X

such that x ∈ U , y ∈ V , x ∈
/ V , and y ∈
/ U.

Note 70: Every T2 -space is T1 .

Example 71:

(a) Set X = N with the co-finite topology. Then X is T1 but not T2 .

(b) Set X = R with the co-countable topology. Then X is T1 but not T2 .

Theorem 72: A topological space is T1 if and only if singleton sets are closed.

Proof: Let X be a topological space.

(=⇒) Suppose X is T1 and let x ∈ X. To see that X \ {x} is open, let y ∈ X \ {x}. Since X

is T1 , there is an open neighborhood U of y such that x ∈

/ U . Therefore, U ⊆ X \ {x}, and so,

y ∈ U ⊆ X \ {x}, which implies that X \ {x} is open.

(⇐=) Suppose that singletons are closed in X and let v, w ∈ X with v ̸= w. Since {v} is closed,

there is an open neighborhood W of w such that w ∈ W ⊆ X \ {v}. Likewise, there is an open

neighborhood V of v such that v ∈ V ⊆ X \ {w}. So, we have v ∈ V , w ∈ W , v ∈

/ W , and w ∈
/ V.

Therefore, the space X is T1 . ■

Definition 73: A topological space is T0 if for every x, y ∈ X with x ̸= y, there is an open U ⊆ X

that contains exactly one of x and y.

Example 74: Let X = {1, 2, 3} and τ = {ϕ, {1}, {1, 2}, {1, 2, 3}}. Then X is T0 but not T1 .

Note 75: Every T1 -space is T0 .

Example 76: Let X be an infinite set and τ be the co-finite topology on X. Then ⟨X, τ ⟩ is T1 but

not T2 . ■

14
Note 77: The properties T2 , T1 , T0 are hereditary properties.

Exercises

34. Let X and Y be topological spaces and suppose Y is Hausdorff. If f : X → Y is a continuous

injection, show that X is Hausdorff.

35. Let X and Y be topological spaces and suppose Y is Hausdorff, D is dense in X, and f : X → Y

and g : X → Y are continuous. If f (x) = g(x) for all x ∈ D, show that f (x) = g(x) for all x ∈ X.

36. If C(R, R) = {f : R → R | f is continuous}, show that |C(R, R)| = |R|.

37. For any n ∈ N, show that there is a set A ⊆ R such that A, A1 , A2 , . . . , An−1 are nonempty and

An = ϕ, where An = (An−1 )′ .

38. Let X be a topological space. Define a relation ⪯ on X by x ⪯ y if and only if x ∈ {y}. Note

that the relation ⪯ is reflexive since x ∈ {x} ⊆ {x}.

(a) Prove that the relation ⪯ is transitive.

(b) Prove that X is a T0 space if and only if the relation ⪯ is antisymmetric.

Definition: A topological space X is a Tsc -space if for each x, y ∈ X, x ⪯ y if and only if y ⪯ x,

where ⪯ is the relation defined in the exercise above.

39. Prove that a topological space is T1 if and only if it is T0 and Tsc .

Definition: Let X be a topological space and A ⊆ X. The sequential closure of A is the set

[A]seq = {x ∈ X | there is a sequence {xn }∞

n=1 ⊆ A such that xn → x}.

40. Let X be a topological space and A ⊆ X.

(a) Show that [A]seq ⊆ A.

(b) Give an example to show that [A]seq need not be equal to A.

Section 2.2: Regular and Completely Regular Spaces

Proposition 78: Let X be a topological space and suppose A ⊆ X and U ⊆ X is open. Then

U ∩ A = ϕ if and only if U ∩ A = ϕ.

Proof: (=⇒) We prove the contrapositive. Suppose x ∈ U ∩ A. Since x ∈ A and U is an open

neighborhood of x, there is an a ∈ U ∩ A. Therefore, U ∩ A ̸= ϕ.

(⇐=) Clear since A ⊆ A. ■

Definition 79: A topological space X is regular if whenever E ⊆ X is closed and x ∈

/ E, there

are disjoint open U, V ⊆ X such that x ∈ U and E ⊆ V .

Theorem 80: A topological space X is regular if and only if whenever U ⊆ X with x ∈ U , there

is an open V ⊆ X such that x ∈ V ⊆ V ⊆ U .

15
Proof: (=⇒) Suppose that X is regular, U ⊆ X is open and x ∈ U . Since X is regular, there are

disjoint open sets V, W ⊆ X such that x ∈ V and X \ U ⊆ W . Since W ∩ V = ϕ and W is open,

W ∩ V = ϕ by the proposition above. Since W ⊇ X \ U , V ∩ (X \ U ) = ϕ, and so, x ∈ V ⊆ V ⊆ U .

(⇐=) Suppose that for each x ∈ X and each open neighborhood U of x, there is an open V ⊆ X

such that x ∈ V ⊆ V ⊆ U . Let y ∈ X and suppose E is closed in X with y ∈

/ E. Then there is an

open O ⊆ X such that x ∈ O ⊆ O ⊆ X \ E. Since O ⊆ X \ E, X \ O ⊇ E. So we have that x ∈ O,

E ⊆ X \ O, and O ∩ X \ O = ϕ. Therefore, X is regular. ■

   
1
Example 81: Let τR denote the usual topology on R. Also, let X = R and τ = τR ∪ U \ | U ∈ τR .
n n∈N
Then ⟨X, τ ⟩ is T2 but not regular.

Proof: Note that ⟨X, τ ⟩ is T2 since τR ⊆ τ . Also, note that ⟨X, τ ⟩ is not regular since 0 and
 
1
cannot be separated by disjoint sets. ■
n n∈N
Example 82: Let X be a nonempty set and τ = {X, ϕ}. Then X is regular but not Hausdorff.

Definition 83: A regular T1 space is called a T3 -space.

Note 84: Every T3 space is T2 .

Proposition 85: A regular T0 space is T2 .

Proof: Let X be a regular T0 space. To see that X is T2 , let x, y ∈ X. Then there is an open

U ⊆ X such that U contains exactly one of x and y. Without loss of generality, suppose x ∈ U and

y∈
/ U . Since X is regular, there are open V, W ⊆ X such that V ∩ W = ϕ, x ∈ V , and X \ U ⊆ W .

Note that y ∈ X \ U ⊆ W . Therefore, X is Hausdorff. ■

Definition 86: A topological space X is completely regular if whenever E ⊆ X is closed and

nonempty and x ∈
/ E, there is a continuous f : X → [0, 1] such that f (x) = 0 and f (E) ⊆ {1}.

Proposition 87: A completely regular space is regular.

Proof: Let X be a completely regular space. To see that X is regular, let E ⊆ X be closed and

x ∈ X \ E. Then there is a continuous f : X → [0, 1] such that f (x) = 0 and f (E) ⊆ {1}. Let
   
−1 1 −1 1
U =f 0, and V = f , 1 . Then U and V are disjoint open subsets of X such
2 2
that x ∈ U and E ⊆ V . Hence, X is regular. ■

Definition 88: A completely regular T1 space is called a Tychonoff (T3 12 ) space.

Note 89: The properties T0 , T1 , T2 , T3 , and T3 21 are hereditary properties.

Definition 90: A topological space is zero-dimensional if it has a base of clopen sets.

Note 91: A topological space X is zero-dimensional if and only if for each x ∈ X and each closed

A ⊆ X with x ∈
/ A, there is a clopen set U ⊆ X such that x ∈ U and U ∩ A = ϕ.

Proposition 92: A zero-dimensional space is completely regular.

Proof: Let X be a topological space and B is a base for the topology on X consisting of clopen

16
sets. To see that X is completely regular, suppose that E ⊆ X is closed and x ∈ X \ E. Since X \ E

is open, there is a B ∈ B such that x ∈ B ⊆ X \ E. Define f : X → [0, 1] by f (x) = χX\B = 1 − χB .

Clearly, f (x) = 0 and f (E) ⊆ {1} by the way f is defined. To see that f is continuous, suppose

that a, b ∈ R with 0 < a < b < 1. Then f −1 ([0, a)) = B, f −1 ((b, 1]) = X \ B, and f −1 ((a, b)) = ϕ,

all of which are open in X. Therefore, X is completely regular. ■

Proposition 93: A countable completely regular space is zero-dimensional.

Proof: Let X be a countable completely regular space. To see that X is zero-dimensional, let

B = {A ⊆ X | A is a clopen set}. To see that B is a base, suppose that x ∈ X and U ⊆ X is open

with x ∈ U . Since X is completely regular, there is a continuous f : X → [0, 1] such that f (x) = 0

and f (X \ U ) ⊆ {1}. Since X is countable and [0, 1] is uncountable, f is not a surjection. Then

f −1 ([0, a)) = f −1 ([0, a]) is clopen. Since f (x) = 0 and f (X \ U ) ⊆ {1}, f −1 ([0, a)) − f −1 ([0, a]) ⊆ U .

So, B is a base, and so, X is zero-dimensional. ■

Exercises

41. Prove that Tychonoff is a hereditary property.

42. Let X be a topological space and suppose A ⊆ X is infinite. Show that there is a sequence

{Un }∞
n=1 ⊆ X such that Uk is open for all k ∈ N, Un ∩ Um = ϕ if n ̸= m, and Un ∩ A ̸= ϕ for all

n ∈ N. (Hint: Use induction.)

43. Show that every subspace of a T3 -space is T3 .

44. Show that continuous images need not preserve regularity, even if the domains and ranges are

T1 . (Hint: Consider R with the co-finite topology.)

45. Let X be a set and τ = {ϕ} ∪ {A ⊆ Y | Y ⊆ X}. Note that τ is a topology on X. Under what

conditions is the space ⟨X, τ ⟩ regular? Under what conditions is it completely regular?

Definition: A zero-set in a topological space X is a set of the form f −1 (0) for some continuous

f : X → R.

Definition: A cozero-set is the complement of a zero-set.

46. Let X be a topological space. If f : X → R is continuous, show that the sets {x ∈ X | f (x) ≥ a}

and {x ∈ X | f (x) ≤ a} are zero-sets for each a ∈ R.

47. Show that the following are equivalent for any topological space X.

(i) X is completely regular.

(ii) The zero-set neighborhoods of each point in X form a local base.

(iii) Every closed set in X is an intersection of zero-sets.

48. Prove that a zero-dimensional space is Tychonoff.

17
Definition: A topological space is semiregular if its regularly open sets form a base for the topol-

ogy.

49.

(a) Prove that every regular space is semiregular. Is the converse true?

(b) Prove that a semiregular T1 space need not be Hausdorff.

(c) Prove that every topological space can be embedded in a semiregular space.

Definition: A topological space is Urysohn if whenever x, y ∈ X with x ̸= y, there are neighbor-

hoods U of x and V of y such that U ∩ V = ϕ.

50.

(a) Prove that every regular T1 space is Urysohn and every Urysohn space is Hausdorff.

(b) Prove that not every Urysohn space is semiregular.

(c) Prove that there is a semiregular Hausdorff space which is not Urysohn.

Definition: A topological space X is completely (functionally) Hausdorff if whenever x, y ∈ X

with x ̸= y, there is a continuous f : X → [0, 1] such that f (x) = 0 and f (y) = 1.

51. Prove that every completely Hausdorff space is Hausdorff.

Section 2.3: Normal Spaces

Proposition 94: Let X be a topological space and Y be an open subspace of X. Then U ⊆ Y is

open in Y if and only if U is open in X.

Proof: (=⇒) Suppose U ⊆ Y is open. Then there is an open UX ⊆ X such that UX ∩ Y = U .

Since Y is open, U = UX ∩ Y is open in X.

(⇐=) Suppose U ⊆ Y is open in X. Then U = U ∩ Y is open in Y . ■

Proposition 95: Let X be a topological space and Y be a closed subspace of X. Then E ⊆ Y is

closed in Y if and only if E is closed in X.

Proof: (=⇒) Suppose E ⊆ Y is closed in Y . Then there is a closed EX ⊆ X such that EX ∩ Y = E.

Since Y is closed, E = EX ∩ Y is closed in X.

(⇐=) Suppose E ⊆ Y is closed in X. Then E = E ∩ Y is closed in Y . ■

Definition 96: A topological space is normal if whenever E, F ⊆ X are closed and disjoint, there

are disjoint open U, V ⊆ X such that E ⊆ U and F ⊆ V .

Theorem 97: Metric spaces are normal.

Proof: Let ⟨X, d⟩ be a metric space and suppose E, F ⊆ X are closed and disjoint. For each
   
dx
x ∈ E, let dx = d(x, F ) and for each x ∈ F , let dx = d(x, E). Let U = ∪ B x, | x ∈ E and
    2
dx
V = ∪ B x, | x ∈ F . Clearly, E ⊆ U and F ⊆ V . It remains to show that U ∩ V = ϕ.
2

18
  
ds
Attempting a contradiction, suppose w ∈ U ∩ V . Then there is an s ∈ E such that w ∈ B s,
  2
dt
and there is a t ∈ F such that w ∈ B t, . Without loss of generality, suppose ds ≤ dt . Then
2
ds dt dt dt
we have d(t, E) ≤ d(s, t) ≤ d(s, w) + d(w, t) < + ≤ + = dt = d(t, E), a contradiction.
2 2 2 2
Therefore, U ∩ V = ϕ, as desired. ■

Theorem 98: A topological space X is normal if and only if whenever E ⊆ X is closed and U ⊆ X

is an open set containing E, there is an open V ⊆ X such that E ⊆ V ⊆ V ⊆ U .

Proof: (⇐=) Let E ⊆ X be closed and suppose U ⊆ X is an open set containing E. Then E and

X \U are closed subsets of X. Since X is normal, there are disjoint open V, W ⊆ X such that E ⊆ V

and (X \ U ) ⊆ W . Since V ∩ W = ϕ and W is open, V ∩ W = ϕ. So V ∩ (X \ U ) ⊆ V ∩ W = ϕ.

Therefore, E ⊆ V ⊆ V ⊆ U , as desired.

(⇐=) Let E, F ⊆ X be disjoint and closed. Since E ⊆ X \ F and X \ F is open, there is an open

U ⊆ X such that E ⊆ U ⊆ U ⊆ (X \ F ). Since U ⊆ (X \ F ), F ⊆ (X \ U ). Therefore, E ⊆ U ,

F ⊆ (X \ U ), and U ∩ (X \ U ) = ϕ, as desired. ■

Proposition 99: A closed subspace of a normal space is normal.

Proof: Let X be a normal space and suppose Y ⊆ X is closed. To see that Y is normal, let

E, F ⊆ Y be closed in Y and disjoint. Since Y is closed in X, E and F are closed in X. Since X is

normal, there are disjoint open U, V ⊆ X such that E ⊆ U and F ⊆ V . Then, U ∩ Y, V ∩ Y ⊆ Y

are disjoint and open such that E ⊆ U ∩ Y and F ⊆ V ∩ Y . ■

Lemma 100 (Jones’ Lemma): Let X be a normal space, D ⊆ X be dense, and S ⊆ X be a

closed and discrete subspace. Then |P (S)| ≤ |P (D)|.

Proof: Since S is closed and discrete, every subset of S is closed in X. So, for each A ⊆ S, there

are disjoint open UA , VA ⊆ X such that A ⊆ UA and S \ A ⊆ VA . Define f : P (S) → P (D) by

f (A) = UA ∩ D. To see that f is one-to-one, let B, C ⊆ S such that B ̸= C. Without loss of

generality, suppose C \ B ̸= ϕ and let c ∈ C \ B ⊆ S \ B ⊆ VB . Since c ∈ C ⊆ UC , c ∈ VB ∩ VC .

Hence, VB ∩UC ̸= ϕ. Since D is dense, there is a d ∈

/ UB ⊇ UB ∩D = f (B). Therefore, f (B) ̸= f (C),

and so, f is one-to-one. ■

Example 101 (The Moore-Niemytzki Plane): Let Γ = R × [0, ∞) and S = R × {0}. Also, let

d denote the usual metric on R × R. For each x ∈ Γ \ S, let Bx = {Bd (x, r) | r < d(x, S)}. Now

suppose x = (s, 0) ∈ S. Then Bx = {Bd ((s, r), r) ∪ {x} | r > 0}. Let B =
S
Bx .
x∈Γ
Claim: The collection B is a base for a topology on Γ.

Proof (Claim): Clearly, ∪B = Γ. Now suppose B1 , B2 ∈ B and y ∈ B1 ∩ B2 . If B1 , B2 ∈ Bx

for some x ∈ S, then B1 ∩ B2 is either B1 or B2 . Otherwise, B1 ∩ B2 is open in R2 with the usual

19
topology. Then there is an r > 0 such that B(y, r) ⊆ B1 ∩ B2 . Note that B(y, r) ∈ By ⊆ B.

Therefore, B is a base for a topology on Γ. ■Claim

Let τ be the topology on Γ generated by B. Then (Γ, τ ) is a topological space called the Moore-

Niemytzki Plane. ■

Note 102: The Moore-Niemytzki Plane is Hausdorff since the topology on the Moore-Niemytzki

Plane is stronger than the subspace topology inherited from R.

Proposition 103: The Moore-Niemytzki Plane is not normal.

Proof: Let Γ denote the Moore-Niemytzki Plane, S = R × {0}, and D = (Q × Q) \ S. Then D is

dense in Γ and S is closed and discrete. Since |Q×Q| = |N|, |P (Q×Q)| = P (N) = |R| = |S| < |P (S)|.

Therefore, Γ is not normal by Jones’ Lemma. ■

Theorem 104: The Moore-Niemytzki Plane is Tychonoff.

Proof: Let Γ denote the Moore-Niemytzki Plane, S = R × {0}, and d denote the usual metric on

R2 . For each x = (s, 0) ∈ S and each r > 0, let C(x, r) = {x} ∪ {(v, w) ∈ R2 | d[(v, w), (s, r)] < r}.

In other words, C(x, r) is the union of {x} and the interior of the circle centered at (s, r) of radius

r. To see that Γ is Tychonoff, suppose E is closed in Γ and x ∈ Γ \ E. If x ∈

/ S, then there is

 r > 0 such that Bd (x, r) ⊆ X \ E and Bd (x, r) ∩ S = ϕ. Define f : Γ → [0, 1] by f (y) =

an

1
 if y ∈
/ Bd (x, r);

 d(x, y) otherwise.


r
To see that f is continuous, note that for any t ∈ (0, 1), f −1 ([0, t)) = Bd (x, rt) and f −1 ((t, 1]) =

Γ \ Bd (x, rt), and if (a, b) ⊆ (0, 1), then f −1 ((a, b)) = B(x, rb) \ B(x, ra). Now suppose x ∈ S. Then

x = (s, 0) for some s ∈ R. Since x ∈  is an r > 0 such that C(x, r) ⊆ Γ \ E. For each
/ E, there




 0 if y = x;


y = (u, v) ∈ Γ, define f : Γ → [0, 1] by f (y) = 1 if y ∈
/ C(x, r);


2 2
 (s − u) + v



 otherwise.
2rv
It remains to show that f is continuous.

Claim: If y = (u, v) ∈ C(x, r) \ {x}, then f (y) = t if and only if y is on the circle centered at (s, rt)

with radius rt.

Proof (Claim): Consider the following.

f (y) = t
(s − u)2 + v 2
⇐⇒ =t
2rv
⇐⇒ (s − u)2 + v 2 = 2rtv

⇐⇒ (s − u)2 + v 2 − 2rtv = 0

20
⇐⇒ (s − u)2 + v 2 − 2rtv + (rt)2 = (rt)2

⇐⇒ (s − u)2 + (rt − v)2 = (rt)2 .

Thus, (u, v) lies on the circle centered at (s, rt) with radius rt. ■Claim

Now, to see that f is continuous note that for any t ∈ (0, 1), f −1 ([0, t) = C(x, rt) and f −1 ((t, 1]) =

Γ \ C(x, rt), and if (a, b) ⊆ (0, 1), then f −1 ((a, b)) = C(x, rb) \ C(x, ra). ■

Definition 105: A normal T1 space is called a T4 space.

Example 106 (Sierpinski Space): Let X = {0, 1} and τ = {ϕ, {0}, {0, 1}}. Then X is normal

and T0 but not T1 .

Definition 107: A topological space X is completely normal if whenever A, B ⊆ X such that

A ∩ B = A ∩ B = ϕ, there are disjoint open U, V ⊆ X such that A ⊆ U and B ⊆ V .

Theorem 108: For any topological space X, the following are equivalent.

(i) X is completely normal;

(ii) Every subspace of X is normal;

(iii) Every open subspace of X is normal.

Proof: ((i) =⇒ (ii)) Let Y be a subspace of X and suppose that E, F ⊆ Y are closed and

disjoint. Then there are EX , FX ⊆ X such that E = EX ∩ X and F = FX ∩ X. Since E ⊆ Y ,

E ∩ FX = E ∩ FX ∩ Y , and so, E ∩ FX = E ∩ FX = E ∩ FX ∩ Y = E ∩ F = ϕ. Similarly, F ∩ EX = ϕ.

Since X is completely normal, there are disjoint open U, V ⊆ X such that E ⊆ U and F ⊆ V .

Then E = E ∩ Y ⊆ U ∩ Y and F = F ∩ Y ⊆ V ∩ Y . Since U, V ⊆ X are disjoint and open in X,

U ∩ Y, V ∩ Y ⊆ Y are disjoint and open in Y . Therefore, Y is normal.

((ii) =⇒ (iii)) Clear.

X X X X
((iii) =⇒ (i)) Suppose that A, B ⊆ X such that A ∩B = A∩B = ϕ. Let Y = (X\A )∪(X\B ).

Then Y is an open subspace of X such that A ⊆ Y and B ⊆ Y . By DeMorgan’s Laws, Y =

X X X X X X X X
(X \ A ) ∪ (X \ B ) = X \ (A ∩ B ). So A ∩B ∩ Y = ϕ. Since A ∩ Y and B ∩ Y are

disjoint and closed subsets of Y and Y is normal, there are disjoint and open in Y U, V ⊆ Y such
X X
that A ∩ Y ⊆ U and B ∩ Y ⊆ V . Since Y is open in X, note that U and V are open in X. So,
X X
A = A ∩ Y ⊆ A ∩ Y ⊆ U and B = B ∩ Y ⊆ B ∩ Y ⊆ V . Therefore, X is completely normal. ■

Exercises

52. Why can’t the method used to show every subspace of a regular space is regular be used to

construct a proof that every subspace of a normal space is normal?

53. Is it true that every metric space is completely normal?

21
 Section 2.4: Urysohn’s Lemma and Tietze’s Extension

Theorem
Proposition 109: Let X be a topological space and A, B ⊆ X. Then there is a continuous

f : X → [0, 1] such that f (A) ⊆ {0} and f (B) ⊆ {1} if and only if for any a, b ∈ R with a < b, there

is a continuous g : X → [a, b] such that f (A) ⊆ {a} and f (B) ⊆ {b}.

Proof: (=⇒) Suppose there is a continuous f : X → [0, 1] such that f (A) ⊆ {0} and f (B) ⊆ {1}.

Define h : [0, 1] → [a, b] by h(x) = a + x(b − a). Then h ◦ f : X → [a, b] is continuous.

(⇐=) Suppose that for any a, b ∈ R with a < b, there is a continuous g : X → [a, b] such that

f (A) ⊆ {a} and f (B) ⊆ {b}. Then it is clear that f : X → [0, 1] where f (A) ⊆ {0} and f (B) ⊆ {1}

is continuous. ■

Lemma 110 (Urysohn’s Lemma): A topological space X is normal if and only if whenever E, F

are disjoint closed subsets of X, there is a continuous f : X → [0, 1] such that f (E) ⊆ {0} and

f (F ) ⊆ {1}.

Proof: (=⇒) Suppose X is normal and that E and F are disjoint closed subsets of X. Let U1 ⊆ X

be open such that E ⊆ U1 and U1 ∩ F = ϕ and let U0 ⊆ X be open such that E ⊆ U0 ⊆ U0 ⊆ U1 .

Enumerate [0, 1] ∩ Q by {qn }n∈ω with q0 = 0 and q1 = 1. For n ∈ ω with n ≥ 2, let pn =

sup({qk | 0 ≤ k < n and qk < qn }) and rn = inf({qk | 0 ≤ k < n and qn < qk }). Note that

0 = q0 ≤ pn < qn < rn ≤q1 = 1. Let Uqn ⊆ X be open such that Upn ⊆ Uqn ⊆ Uqn ⊆ Urn . Define

inf({qn | n ∈ ω and x ∈ Uqn }) if x ∈ U1

f : X → [0, 1] by f (x) =

1
 otherwise.
Clearly, f (E) ⊆ {0} and f (F ) ⊆ {1}. So, it remains to show that f is continuous. To see that f is

continuous on X, let z ∈ X. If f (z) = 0, let b ∈ (0, 1] and choose s ∈ (0, 1) ∩ Q such that s < b and

f (x) ≤ s for each x ∈ Us . Then, f (Us ) ⊆ [0, s]. Since f (z) = 0 < s, z ∈ Us . So, f (z) ∈ f (Us ) ⊆

[0, s] ⊆ [0, b). If f (z) = 1, let a ∈ [0, 1) and choose r ∈ (0, 1) ∩ Q such that r > a and f (x) ≥ r

for each x ∈ X \ Ur . Since f (z) = 1 > r, z ∈

/ Ur . Therefore, f (z) ∈ f (X \ Ur ) ⊆ [r, 1] ⊆ (a, 1].

Finally, if f (z) ∈ (0, 1), let a, b ∈ (0, 1) with f (z) ∈ (a, b) and choose r, s ∈ (0, 1) ∩ Q such that

a < r < f (z) < s < b and f (x) ∈ [r, s] for each x ∈ Us \ Ur . Thus, f (z) ∈ f (Us \ Ur ) ⊆ [r, s] ⊆ (a, b),

as desired.

(⇐=) Let E, F ⊆ X be closed and disjoint. Since E and F are closed and disjoint, there is a
 
−1 1
continuous f : X → [0, 1] such that f (E) ⊆ {0} and f (F ) ⊆ {1}. Let U = f 0, and
  2
1
V = f −1 , 1 . Clearly, U, V ⊆ X are open and disjoint so that E ⊆ U and F ⊆ V . ■
2

22
Note 111: The following is an equivalent way of stating Urysohn’s Lemma.

Urysohn’s Lemma: A topological space X is normal if and only if whenever E, F are disjoint

closed subsets of X and a, b ∈ R with a < b, there is a continuous f : X → [a, b] such that

f (E) ⊆ {a} and f (F ) ⊆ {b}.

Corollary 112: Every T4 space is Tychonoff.

Theorem 113 (Tietze’s Extension Theorem): A T1 space X is normal if and only if whenever

E is a closed subspace of X and f : E → [0, 1] is continuous, there is a continuous F : X → [0, 1]

such that F ↾E = f .

Corollary 114 (Jones’ Lemma): Let X be a normal space, D ⊆ X be dense, and S ⊆ X be a

closed and discrete subspace. Then |P (S)| ≤ |P (D)|.

Proof: Exercise. ■

Exercises

Definition: A topological space X is perfectly normal if for each pair of disjoint closed A, B ⊆ X,

there is a continuous f : X → [0, 1] such that A = f −1 (0) and B = f −1 (1).

54. Prove that every metric space is perfectly normal.

55. Prove that a T1 -space is perfectly normal if and only if it is T4 and every closed subset is a

Gδ -set.

56. Show that the Michael Line is normal.

57. Prove that every ordered space is T4 .

58. Use Tietze’s Extension Theorem to prove Jones’ Lemma.

Chapter 3: Covering Properties

Section 3.1: Compact Spaces

Definition 115: Let X be a topological space and A ⊆ X. Then a collection U of open subsets of

X is an open cover of A if A ⊆ ∪U .

Definition 116: Let X be a topological space and A ⊆ X. Then A is compact if every open cover

of A has a finite subcover.

Example 117: The closed interval [0, 1] is a compact subset of R.

23
Example 118: The open interval (0, 1) is not a compact subset of R.

Note that U = {(0, r) | r ∈ (0, 1)} has no finite subcover.

Theorem 119: Let X and Y be topological spaces and suppose f : X → Y is continuous. If A is

compact, then f (A) is compact.

Proof: Let V be an open cover of f (A) and define U = {f −1 (V ) | V ∈ V }. Since f is continuous,

U is an open cover of A. Let U ′ be a finite subcover of U . For each U ∈ U ′ , choose VU ∈ V such

that U = f −1 (VU ). Let V ′ = {VU | U ∈ U ′ }. Clearly, V ′ is finite. So it remains to show that V ′

covers f (A). To see this, let y ∈ f (A). Then there is an a ∈ A such that f (a) = y. Also, there is a

W ∈ U ′ such that a ∈ W . Then y = f (a) ∈ f (f −1 (W )) ⊆ VW ∈ V ′ . Therefore, V ′ covers f (A).

Definition 120: A collection of sets F has the finite intersection property if for every finite

F ′ ⊆ F , ∩F ̸= ϕ.

Theorem 121: A topological space is compact if and only if the intersection of any collection of

closed sets with the finite intersection property is nonempty.

Proof: (=⇒) Let X be a compact topological space and suppose F is a collection of closed subsets

of X with the finite intersection property. Attempting a contradiction, suppose ∩F = ϕ. By

DeMorgan’s Laws, {X \ F | F ∈ F } covers X. Since X is compact, there are {Fi }ni=1 ⊆ F such
n n
Fi = ϕ. This is a contradiction since F has the finite
S T
that X ⊆ Fi . By DeMorgan’s Laws,
i=1 i=1
intersection property.

(⇐=) Suppose X is a topological space such that every collection of closed subsets with the finite

intersection property has a nonempty intersection. To see that X is compact, let U be an open

cover of X. Since U covers X, ∩{X \ U | U ∈ U } = ϕ. So, {X \ U | U ∈ U } does not have

n
the finite intersection property. Therefore, there are {Ui }ni=1 ⊆ U such that
T
(X \ Ui ) = ϕ. By
i=1
n
S
DeMorgan’s Laws, X = Ui . Therefore, X is compact. ■
i=1
Proposition 122: Closed subsets of compact spaces are compact.

Proof: Exercise. ■

Lemma 123: Compact T2 spaces are regular.

Proof: Let X be a compact T2 space and suppose E ⊆ X is closed in X and x ∈ X \ E. For each

y ∈ X \ E, let Uy , Vy ⊆ X be disjoint and open such that x ∈ Uy and y ∈ Vy . Since E is a closed

n
subset of a compact space, E is compact. So, there are {yi }ni=1 ⊆ E such that E ⊆
S
Vyi . Let
i=1
n
T n
S
U = Uyi and V = Vyi . Then, x ∈ U , E ⊆ V , and U ∩ V = ϕ by design. Therefore, X is
i=1 i=1
regular. ■

24
Theorem 124: Compact T2 spaces are normal.

Proof: Let X be a compact T2 space. To see that X is normal, let E, F ⊆ X be disjoint and closed.

By the lemma above, X is regular. So, for each x ∈ E, there are disjoint open Ux , Vx ⊆ E such

that x ∈ Ux and F ⊆ Yx . Since X is compact and E is closed, E is compact. Therefore, there are
n n n
{xi }ni=1 ⊆ E such that E ⊆
S S T
Uxi . Let U = Uxi and V = Vxi . Then, E ⊆ U , F ⊆ V , and
i=1 i=1 i=1
U ∩ V = ϕ by design. Therefore, X is normal. ■

Exercises

59. Show that if X is a T2 -space and every subspace of X is compact, then X is discrete.

60. Suppose that X is a T2 -space and {Kn }n∈ω is a decreasing sequence of nonempty compact
T
subsets of X. Show that Kn ̸= ϕ.
n∈ω
Note: The exercise above proves the nested interval theorem from Analysis, which says that
∞
T
[an , bn ] ̸= ϕ whenever {[an , bn ]}n∈ω is a nested sequence of closed bounded intervals.
n=0
61. Suppose ⟨X, τ ⟩ is a compact Hausdorff space.

(a) If σ is a topology on X with τ ⊊ σ, prove that ⟨X, σ⟩ is not compact.

(b) If σ is a topology on X with σ ⊊ τ , prove that ⟨X, σ⟩ is not Hausdorff.

62. Prove that every compact subset of a compact T1 -space has an accumulation point.

63. Prove that closed subsets of compact spaces are compact.

64. Prove that compact subsets of Hausdorff spaces are closed.

Section 3.2: Countably Compact and Sequentially

Compact Spaces
Definition 125: Let X be a topological space and A ⊆ X. Then A is countably compact if

every countable open cover of A has a finite subcover.

Note 126: Compact spaces are countably compact.

Theorem 127: A topological space is countably compact if and only if the intersection of any

countable collection of closed sets with the finite intersection property is nonempty.

Proof: (=⇒) Let X be a countably compact topological space and suppose F is a collection

of closed subsets of X with the finite intersection property. Attempting a contradiction, suppose

∩F = ϕ. By DeMorgan’s Laws, {X \F | F ∈ F } covers X. Since X is countably compact, there are

n n
{Fi }ni=1 ⊆ F such that X ⊆
S T
(X \ Fi ). By DeMorgan’s Laws, Fi = ϕ, which is a contradiction
i=1 i=1
since F has the finite intersection property.

(⇐=) Suppose X is a topological space such that every countable collection of closed subsets with

25
the finite intersection property has a nonempty intersection. To see that X is countably compact, let

U be a countable open cover of X. Since U covers X, ∩{X \ U | U ∈ U } = ϕ. By the hypothesis,

{X \ U | U ∈ U } does not have the finite intersection property. Therefore, there are {Ui }ni=1 ⊆ U
n
T Sn
such that (X \ Ui ) = ϕ. By DeMorgan’s Laws, X = Ui . Therefore, X is countably compact.
i=1 i=1
■

Corollary 128: Closed subsets of countably compact spaces are countably compact.

Proof: Let X be a countably compact space, E ⊆ X be closed, and F be a countable collection of

closed subsets of E with the finite intersection property. Since E is closed, F is closed in X for all

F ∈ F . Since X is countably compact, ∩F ̸= ϕ. Therefore, E is countably compact. ■

Lemma 129: Let X be a T1 space and suppose A ⊆ X. Then x ∈ A′ if and only if for every open

neighborhood U of x, U ∩ A is infinite.

Proof: (=⇒) Suppose x ∈ A′ , U is an open neighborhood of x, and F ⊆ X is finite with x ∈

/ F.

Since X is T1 , U \ F is an open neighborhood of x. Since x ∈ A′ , (U \ F ) ∩ (A \ {x}) ̸= ϕ. Since F

was chosen arbitrarily, U ∩ A is infinite.

(⇐=) Clear. ■

Theorem 130: A T1 space is countably compact if and only if every countably infinite subset has

an accumulation point.

Proof: (=⇒) We prove the contrapositive. Suppose X is a T1 space and A ⊆ X is a countably

infinite subset that has no accumulation point. Then A′ = ϕ, which means A = A ∪ A′ = A. Hence,

A is closed. For each x ∈ A, let Ux ⊆ X be an open neighborhood of x such that Ux ∩ A = {x}.

Then {Ux | x ∈ A} ∪ (X \ A) is a countable open cover of X that has no finite subcover.

(⇐=) We prove the contrapositive. Suppose U = {Un }n∈ω is a countable open cover of X with
n−1
S
no finite subcover. Without loss of generality, suppose Un ̸= ϕ for all n ∈ ω and Un ⊈ Ui for
i=0
n−1
S
all n ∈ N. Let x0 ∈ U0 . For all n ∈ N, choose xn ∈ Un \ Ui . To see that {xn }n∈ω has no
i=0
accumulation points, let x ∈ X. Since {Un }n∈ω covers X, there is a k ∈ ω such that x ∈ Uk . By the

construction of {xn }n∈ω , Uk ∩ {xn }n∈ω ⊆ {xk }k≤n . Since X is T1 , x is not an accumulation point of

{xn }n∈ω by the lemma above. Since x was chosen arbitrarily, {xn }n∈ω has no accumulation points.

Example 131: Let X = N and τ denote the discrete topology on N.

(i) ⟨X, τ ⟩ is not T1 ;

(ii) ⟨X, τ ⟩ is T0 ;

(iii) ⟨X, τ ⟩ is not countably compact;

26
(iv) Every countably infinite subset of X has an accumulation point.

Definition 132: A topological space is sequentially compact if every sequence has a convergent

subsequence.

Theorem 133: Sequentially compact spaces are countably compact.

Proof: Let X be sequentially compact. To see that X is countably compact, let U = {Un }n∈ω

be an open cover of X. Attempting a contradiction, suppose U has no finite subcover. Without

n−1
S
loss of generality, suppose Un ̸= ϕ for all n ∈ ω and Un ⊈ Ui . Since X is sequentially com-
i=0
pact, {xn }n∈ω has a convergent subsequence {xnk }k∈ω . Let x = lim xnk and let m ∈ ω such that
k→∞
x ∈ Um . Since x = lim xnk , Um ∩ {xnk }k∈ω is infinite. This is a contradiction by the way {xn }n∈ω
k→∞
was constructed. Therefore, U has a finite subcover. ■

Note 134: There are T4 spaces that are compact but not sequentially compact.

Example 135: The space βN is T4 and compact but not sequentially compact.

Note 136: There are T4 spaces that are sequentially compact but not compact.

Example 137: The space ω1 is T4 and sequentially compact but not compact.

Exercises

65. Show that a metric space is compact if and only if it is countably compact.

66. Prove that a compact metric space is bounded.

Section 3.3: Lindelöf Spaces

Definition 138: Let X be a topological space and A ⊆ X. Then A is Lindelöf if every open cover

of A has a countable subcover.

Definition 139: A collection F of sets has the countable intersection property if for every

countable F ′ ⊆ F , ∩F ′ ̸= ϕ.

Theorem 140: A topological space is Lindelöf if and only if the intersection of any collection of

closed sets with the countable intersection property is nonempty.

Proof: (=⇒) Let X be a Lindelöf space and suppose F is a collection of closed subsets of X with

the countable intersection property. Attempting a contradiction, suppose ∩F = ϕ. By DeMorgan’s

Laws, {X \ F | F ∈ F } covers X. Since X is Lindelöf, there is a countable collection {Fn }n∈ω ⊆ F

Fn = ϕ, which is a contradiction since F has

S T
such that X ⊆ (X \ F ). By DeMorgan’s Laws,
n∈ω n∈ω
the countable intersection property.

(⇐=) Suppose X is a topological space such that every collection of closed subsets of X with

27
the countable intersection property has a nonempty intersection. To see that X is Lindelöf, let

U be an open cover of X. Since U covers X, ∩{X \ U | U ∈ U } = ϕ. This implies that

{X \ U | U ∈ U } does not have the countable intersection property. Therefore, there is a countable

collection {Un }n∈ω } ⊆ U such that

T S
(X \Un ) = ϕ. By DeMorgan’s Laws, X = Un . Therefore,
n∈ω n∈ω
X is Lindelöf, as desired. ■

Corollary 141: A closed subset of a Lindelöf space is Lindelöf.

Proof: Let X be a Lindelöf space and suppose E ⊆ X is closed. Let F be a collection of closed

subsets of E with the countable intersection property. Since E is closed in X, F is closed in X for

all F ∈ F . Since X is Lindelöf, ∩F ̸= ϕ. Therefore, E is Lindelöf. ■

Theorem 142: Regular Lindelöf spaces are normal.

Proof: Let X be a regular Lindelöf space and suppose E, F ⊆ X are closed and disjoint. For

each x ∈ E, there is an open neighborhood Ux of x such that Ux ∩ F = ϕ. Likewise, for each

y ∈ F , there is an open neighborhood Vy of y such that Vy ∩ E = ϕ. Since E is Lindelöf there is a

S
countable {xn }n∈ω ⊆ E such that E ⊆ Uxn . Likewise, there is a countable {yn }n∈ω ⊆ F such
n∈ω
S n
S n
S S
that F ⊆ Vyn . For each n ∈ ω, let Wn = Uxn \ Vyi and On = Vyn \ Uxi . Set W = Wn
n∈ω i=0 i=0 n∈ω
S
and O = On .
n∈ω
Claim 1: E ⊆ W and F ⊆ O.
S S S
Proof (Claim 1): Since E ∩ Vyn = ϕ for all n ∈ ω, E ⊆ ( Uxn ) \ (Vyn ) = [Uxn \ ( Vyn )] ⊆
n∈ω
S S S S S n∈ω n∈ω
Wn = W . Likewise, F ⊆ ( Vyn ) \ (Uxn ) = [Vyn \ ( Uxn )] ⊆ On = O. ■Claim 1
n∈ω n∈ω n∈ω n∈ω n∈ω
Claim 2: W ∩ O = ϕ.

Proof (Claim 2): Let j, k ∈ ω. To see that Wj ∩ Ok = ϕ, suppose without loss of generality that
k
S
j < k. Then Ok = Vyk \ Uxi ⊆ Vyk \ Uxj . Since Wj ⊆ Uxj , Wj ∩ Uk = ϕ. ■Claim 2
i=0
By the claims above, X is normal. ■

Definition 143: Let X be a topological space and A ⊆ X. Then x ∈ X is a condensation point

of A if for every open neighborhood U of x, U ∩ A is uncountable.

Exercises

67. Show that every uncountable subset of a Lindelöf space has a condensation point.

68. Show that the Michael Line is not Lindelöf.

69. Show that the Sorgenfrey Line is Lindelöf.

70. Prove that a regular hereditarily Lindelöf space is perfect.

71. Prove that every closed and discrete subset of a Lindelöf space is countable.

72. Prove that the following are equivalent for any Lindelöf space X.

28
(a) Every subspace of X if Lindelöf;

(b) Every open subspace of X is Lindelöf.

73. Prove that every open subset of a T3 , hereditarily Lindelöf space is an Fσ -set.

74. Prove that every T3 , hereditarily Lindelöf space is perfectly normal.

75. Let τ1 and τ2 be topologies on a set X with τ1 ⊆ τ2 . If the space ⟨X, τ2 ⟩ is Lindelöf, prove that

the space ⟨X, τ1 ⟩ is Lindelöf.

Section 3.4: Locally Compact Spaces

Definition 144: A topological space X is locally compact if every point in X has a local base

consisting of compact sets.

Theorem 145: A Hausdorff space X is locally compact if and only if every point in X has a

compact neighborhood.

Proof:

(=⇒) Clear.

(⇐=) Let x ∈ X and suppose that K is a compact neighborhood of x. Let U be a neighborhood of

x and V = int(K ∩ U ). Then V is an open neighborhood of x and clX (V ) is compact and Hausdorff,

and hence regular. Since V is a neighborhood of x in clX (V ), there is a neighborhood W of x

in clX (V ) such that clclX (V ) (W ) ⊆ V . This implies that W is open in V , and hence in X, and

clclX (V ) (W ) is closed in clX (V ) and hence is compact. This means that clclX (V ) (W ) is a compact

neighborhood of x in U which is contained in U . Therefore, x has a base of compact neighborhood

in X, and so, X is locally compact. ■

Corollary 146: Every compact Hausdorff space is locally compact.

Example 147:

(a) The space R is locally compact;

(b) The space Q is locally compact;

(c) The space R \ Q is not locally compact.

Theorem 148:

(i) The intersection of an open and a closed set in a locally compact Hausdorff space is locally

compact;

(ii) A locally compact subset of a locally compact Hausdorff space is the intersection of an open

and a closed set.

Proof: For (i), let X be a locally compact Hausdorff space. If U ⊆ X is open and u ∈ U , then

there is a compact neighborhood K of u in X contained in U . This implies that K is a compact

29
neighborhood of u in U , and so, U is locally compact. If F ⊆ X is closed in X and v ∈ F , then there

is a compact neighborhood K of v in X and K ∩ F is a compact neighborhood of v in F , and so, F

is locally compact. Clearly, the intersection of two locally compact subsets of X is locally compact,

and so, U ∩ (K ∩ F ) is locally compact.

For (ii), suppose that X is a Hausdorff space and that Y ⊆ X is locally compact. It suffices to

show that Y is open in clX Y . Let y ∈ Y . Since Y is locally compact, there is a neighborhood U

of y in Y such that clY U is compact. In other words, U = Y ∩ V , where V ⊆ X is open. Then

clX (Y ∩ V ) ∩ X = clX (U ) ∩ Y = clY (U ) and clY (U ) is compact. Thus, clX (Y ∩ V ) ∩ X is closed in

X. Since Y ∩ V ⊆ clX (Y ∩ V ) ∩ X, clX (Y ∩ V ) ⊆ clX (Y ∩ V ) ∩ X. This implies that clX (Y ∩ V ) ⊆ Y ,

and hence, clX (Y ) ∩ V ⊆ Y . Thus, clX (Y ) ∩ V is a neighborhood of x in clX (Y ) which is contained

in Y . Hence, Y is open in clX (Y ). ■

Corollary 149: A dense subset of a compact Hausdorff space is locally compact if and only if it is

open.

Theorem 150: Let X and Y be topological spaces and f : X → Y . If f is continuous and open

and X is locally compact, then Y is locally compact.

Proof: Let y ∈ Y , V ⊆ Y be an open neighborhood of y, and x ∈ f −1 (y). Since f is continuous

and locally compact, there is a compact neighborhood K of x such that f (K) ⊆ V . This implies

that x ∈ intX (K), and so, y ∈ f (intX (K)) ⊆ f (K). Since f is open, f (intX (K)) is open, and so,

f (K) is a compact neighborhood of y contained in V . ■

Exercises

76. Prove that Q is not locally compact.

77. Prove that the Sorgenfrey Line is not locally compact.

78. Prove that the Moore-Niemytzki Plane is not locally compact.

79. Let X be a set, A ⊆ X, and τ = {ϕ} ∪ {B ⊆ X | A ⊆ B}. Is the space ⟨X, τ ⟩ locally compact?

80 (Slotted Plane). If x ∈ R2 , then define local neighborhoods of x to be the sets {x} ∪ A, where

A ⊆ R2 is a disk about x with finitely many straight lines through x removed.

(a) Verify that this gives a topology on R2 .

(b) Compare this topology with the usual topology on R2 .

(c) Can we replace the word “finite” with the word “countable?”

This space is called the slotted plane and is sometimes denoted A.

(d) Prove that the slotted plane is not locally compact.

81. Is the radial plane locally compact?

30
82.

(a) Prove that the closed continuous image of a locally compact space need not be locally compact.

(b) Prove that the closed continuous image of a locally compact space is locally compact if and only

if the preimage of each point is compact.

83. Prove that any closed subspace of a locally compact space is locally compact.

84. Prove that a Hausdorff space is locally compact if and only if there is a base B such that B is

compact for each B ∈ B.

85. Prove that any open subspace of a locally compact Hausdorff space is locally compact.

86. Prove that every locally compact Hausdorff space is Baire.

Section 3.5: The Alexandroff One-Point Compactification

Theorem 151 (The Alexandroff One-Point Compactification): Let X be a locally compact

Hausdorff space. Then there is a topological space X ∗ containing X such that the following are

satisfied.

(i) The space X ∗ is a compact Hausdorff space;

(ii) The set X ∗ \ X is a singleton set;

(iii) If the space X is not compact, then X is dense in the space X ∗ .

/ X and set X ∗ = X∪{∞}. Define τ = {U ⊆ X ∗ | U is an open subset of X or X ∗ \

Proof: Choose a point ∞ ∈

U is a compact subset of X}.

Claim 1: The collection τ is a topology of X.

Proof (Claim 1): Since X ∗ \ X ∗ = ϕ and ϕ ⊆ X is open, ϕ ∈ τ . Suppose that U, V ∈ τ . If one of

U and V is a subset of X, then U ∩ V = U ∩ V ∩ X = (U ∩ X) ∩ (V ∩ X). Since X is T2 , U ∩ X and

V ∩ X are open in X. If neither is a subset of X, then X ∗ \ (U ∩ V ) = (X ∗ \ U ) ∪ (X ∗ \ V ), which

is the union of two compact subsets of X, and thus is compact. In either case, U ∩ V ∈ τ . Finally,
S T
suppose that {Uα | α ∈ Λ} ⊆ τ . If ∞ ∈ Uβ for some β ∈ Λ, then X \ Uα = (X \ Uα ) =
" # α∈Λ α∈Λ

(X ∗ \ Uβ ) ∩
T S
(X \ Uα ) . This implies that X \ Uα is a closed subset of the compact space
α∈Λ\{β} α∈Λ
X ∗ \ Uβ , and hence, it is compact. If ∞ ∈
S
/ Uα for every α ∈ Λ, then Uα is an open subset of X.
α∈Λ
Uα ∈ τ . Therefore, τ is a topology on X ∗ .
S
In either case, ■Claim 1
α∈Λ
Clearly, X ∗ \ X is a singleton set and if X is not compact, then X is dense in X ∗ . It remains to

show that X ∗ is compact and Hausdorff.

Claim 2: The space X ∗ is Hausdorff.

Proof (Claim 2): Let x, y ∈ X ∗ . If {x, y} ⊆ X, then since X is T2 , there are disjoint and open

subsets U, V ⊆ X such that x ∈ U and y ∈ V . Since U and V are open in X, they are open in X ∗ .

31
If one of x and y is ∞, then x = ∞. Since X is locally compact, there is a compact neighborhood K

of y in X. Set U = X ∗ \ K and V = intX (K). Then U and V are disjoint open sets in X ∗ , ∞ ∈ U ,

and y ∈ V . Therefore, X ∗ is Hausdorff. ■Claim 2

Claim 3: The space X ∗ is compact.

Proof (Claim 3): Suppose that U is an open cover of X ∗ . Choose a U0 ∈ U such that ∞ ∈ U0 .

Since X ∗ \ U0 is a compact subset of X, we may choose a finite V ⊆ U such that X ⊆ ∪V . Thus,

{U0 } ∪ V is the desired subcover, and so, X ∗ is compact. ■Claim 3

Therefore, X ∗ is compact and Hausdorff. ■

Example 152:

(a) The space R∗ is homeomorphic to the circle S1 ;

(b) The one-point compactification of Rn is the surface of the unit ball in Rn+1 .

(c) The space C∗ is homeomorphic to the sphere S2 .

Corollary 153: Any locally compact Hausdorff space which is not compact can be embedded into

an open dense subset of a compact Hausdorff space.

Proposition 154: If X is a locally compact Hausdorff space and X ∗ is the Alexandroff one-point

compactification of X, then X is an open subset of X ∗ .

Proof: Exercise. ■

Corollary 155: Every locally compact Hausdorff space is completely regular.

Proof: Let X be a locally compact Hausdorff space. Then the space X ∗ is compact and Hausdorff,

and hence normal. By Urysohn’s Lemma, X ∗ is completely regular. Recall that every subspace of

a completely regular space is completely regular. Hence, X is completely regular. ■

Exercises

87. Find the one-point compactification of the space ω1 .

88. If X is a topological space which is not compact, prove that X is dense in X ∗ .

89. Prove proposition 154.

Chapter 4: The Countability Axioms

32
 Section 4.1: First and Second Countable Spaces
Definition 156: A topological space is second countable if it has a countable base.

Definition 157: A topological space is first countable if every point has a countable local base.

Note 158: Second countable spaces are first countable.

Theorem 159: Every second countable space is Lindelöf.

Proof: Let X be a second countable space. Then X has a countable base B. To see that X is

Lindelöf, let U be an open cover of X. For each x ∈ X, let Ux ∈ U such that x ∈ Ux and let Bx ∈ B

such that x ∈ Bx ⊆ Ux . Set V = {B ∈ B | B = Bx for some x ∈ X}. Then ∪V =

S
Bx = X.
x∈X
Since B is countable, V is countable. For each V ∈ V , let UV ∈ U such that V ⊆ UV . Since V

covers X, {UV }V ∈V covers X. Therefore, X is Lindelöf. ■

Corollary 160: A second countable T3 space is normal.

Example 161: Let X = [0, ∞) and τ = {ϕ, X} ∪ P [(0, ∞)] ∪ {[0, ∞) \ A | A ⊆ [0, ∞) and |A| < ℵ0 }.

Then X is Hausdorff and Lindelöf, but not first or second countable.

Proof: We will show that X is not first countable by showing there is no countable local base at

0. Let {Bn }n∈ω be a collection of open neighborhoods of 0. For each n ∈ ω, let An = [0, ∞) \ Bn .
S
Note that An is countable. Then, A = An is countable. Let x ∈ X \ A. Then x ∈ Bn for all
n∈ω
n ∈ ω. Therefore, Bn ⊈ [0, ∞) \ (A ∪ {x}) for all n ∈ ω, and so, {Bn }n∈ω is not a local base at 0.

Thus, X is not first countable, and hence, not second countable. ■

Exercises

90.

(a) Show that the continuous image of a first countable space is not necessarily first countable.

(b) Show that the continuous open image of a first countable space is first countable.

91. Prove that a compact Hausdorff space in which every singleton is a Gδ -set is first countable.

92. Show that every base for the open sets of a second countable space has a countable basic

subcollection.

93. Prove that every second countable space is hereditarily Lindelöf.

94. Prove that any topological space X can be embedded as a dense subset of a Lindelöf space.

95. Let X be a hereditarily Lindelöf space and E ⊆ X. Set E ∗ = {x ∈ X | x ∈

/ E ′ }. Prove that E ′

is countable.

33
 Section 4.2: Separable Spaces
Definition 162: A topological space is separable if it contains a countable dense subset.

Theorem 163: Second countable spaces are separable.

Proof: Let X be a second countable space. Then there is a countable base B = {Bn }n∈ω . For

each n ∈ ω, let dn ∈ Bn . It remains to show that {dn }n∈ω is dense in X. Let x ∈ X and suppose U

is an open neighborhood of x. Then there is a k ∈ ω such that x ∈ Bk ⊆ U . Hence, dk ∈ Bk ⊆ U ,

and so, {dn }n∈ω is dense in X. ■

Example 164: Let X be an uncountable set and τ be the co-countable topology. Then ⟨X, τ ⟩ is

Lindelöf but not separable.

Example 165: Since Q × Q is dense, the Moore-Niemytzki Plane is separable. However, since the

Moore-Niemytzki Plane is not normal, it not Lindelöf.

Theorem 166: The following are equivalent for any metric space ⟨X, d⟩.

(i) ⟨X, d⟩ is second countable;

(ii) ⟨X, d⟩ is Lindelöf;

(iii) ⟨X, d⟩ is separable.

Proof: Exercise.

Exercises

96. Prove that every collection of disjoint open sets in a separable space must be countable.

97. Show that every increasing chain of real numbers that is well-ordered by the usual order is

countable.

98. Prove theorem 154.

99. If possible, give an example of a separable first countable space that is not second countable.

Definition: A topological space is a ccc-space if every pairwise disjoint collection of open subsets

is countable.

100. Prove that every separable space is a ccc-space.

101 (Souslin Hypothesis). Do you believe that every linearly ordered ccc-space is separable?

Chapter 5: Product Spaces

34
 Section 5.1: Finite Products
Theorem 167: Let (X, τX ) and (Y, τY ) be topological spaces. Then {U × V | U ∈ τX and V ∈ τY }

is a base for a topology on X × Y .

Proof: Let B = {U × V | U ∈ τX and V ∈ τY }. Since X × Y ∈ B, ∪B = X × Y . Now

suppose B1 , B2 ∈ B and (x, y) ∈ B1 ∩ B2 . Then there are U1 , U2 ∈ τX and V1 , V2 ∈ τY such that

B1 = U1 × V1 and B2 = U2 × V2 . Since (x, y) ∈ B1 ∩ B2 , x ∈ U1 ∩ U2 and y ∈ V1 ∩ V2 . Then

(x, y) ∈ (U1 ∩ U2 ) × (V1 ∩ V2 ) ⊆ B1 ∩ B2 . ■

Definition 168: Let (X, τX ) and (Y, τY ) be topological spaces. Then the topology on X × Y

generated by {U × V | U ∈ τX and V ∈ τY } is called the product topology on X × Y .

Example 169: Examples of products.

Theorem 170: The product of two separable spaces is separable.

Proof: Exercise. ■

Theorem 171: The product of two second countable spaces is second countable.

Proof: Exercise. ■

Example 172: Let S denote the Sorgenfrey Line. Recall that S is Lindelöf but S × S is not Lindelöf.

Proof: Note that S × S is not normal by Jones’ Lemma. Hence, S × S is not Lindelöf. ■

Lemma 173: A topological space is compact if and only if every open cover of basic open sets has

a finite subcover.

Proof: (⇐=) Let X be a topological space and suppose that every open cover of basic open sets

has a finite subcover. To see that X is compact, suppose U is a cover for X and B is a base for

the topology on X. Then for each x ∈ X, let Ux ∈ U and Bx ⊆ B with x ∈ Bx ⊆ Ux . Note that

Bx has a finite subcover {Bxi }ni=1 . Hence, since Bx ⊆ Ux and {Bxi }ni=1 covers Bx , {Bxi }ni=1 covers

Ux for all x ∈ X, as desired.

(=⇒) Clear. ■

Lemma 174 (Tube Lemma): Let X and Y be topological spaces such that Y is compact and

suppose U is an open cover of X × Y . Then for each x ∈ X, there is an open neighborhood W of

x and a finite V ⊆ U such that W × Y ⊆ ∪V .

Proof: Exercise. ■

Theorem 175: The product of two compact spaces is compact.

Proof: Let X and Y be compact spaces and suppose U is an open cover of X × Y . By the

Tube Lemma, for each x ∈ X, there is an open neighborhood Vx of x and a finite Ux ⊆ U such

that Vx × Y ⊆ Ux . Since X is compact, there are {xi }ni=1 ⊆ X such that X ⊆

S
Vxi . Now let
i≤n

35
W = Uxi and note that W is a finite subcollection of U . It remains to show that W covers
S
i≤n
X × Y . To do this, suppose (a, b) ∈ X × Y . Then, there is a 1 ≤ k ≤ n such that a ∈ Vxk . Now,

(a, b) ∈ Vxk × Y ⊆ Uxk ⊆ ∪W . Therefore, W covers X × Y . ■

Theorem 176: The product of a Lindelöf and a compact space is Lindelöf.

Proof: Exercise. ■

Exercises

102. Suppose X and Y are topological spaces, A ⊆ X, and B ⊆ Y .

(a) Show that A × B = A × B.

(b) Show that int(A × B) = int(A) × int(B).

103. Prove the Tube Lemma.

104. Prove theorem 170.

105. Prove theorem 171.

106. Prove theorem 176. (Hint: If the proof can be modified to show that the product of two

Lindelöf spaces is Lindelöf, there is a flaw.)

107. If X and Y are locally compact spaces, prove that X × Y is locally compact.

108. Prove that a topological space X is Hausdorff if and only if the diagonal ∆ = {(x, x) | x ∈ X}

is a closed subset of X × X.

Section 5.2: Infinite Products

Q
Definition 177: Let Λ be an index set and suppose Xα is a set for all α ∈ Λ. Then Xα = {f :
α∈Λ
S
Λ→ Xα | f (α) ∈ Xα for all α ∈ Λ}.
α∈Λ
Definition 178: Let Λ be an index set and suppose Xα is a set for all α ∈ Λ. Then for each β ∈ Λ,
Q
define the projection map πβ : Xα → Xβ by πβ (x) = x(β).
α∈Λ
Notation 179: πβ (x) = x(β) is usually denoted xβ .

Definition 180: Let Λ be an index set and suppose (Xα , τα ) is a topological space for all α ∈ Λ.

Xα formed by taking {πα−1 (U ) | α ∈ Λ and U ∈ τα } as a subbase is called

Q
Then the topology on
α∈Λ
Q
the product topology on Xα .
α∈Λ
Note 181:
Q Q
(i) The basic open sets of the product space Xα have the form Uα where each Uα is open
α∈Λ α∈Λ
in Xα and Uα = Xα for all but finitely many α ∈ Λ.
Q Q
(ii) The subbasic open sets of the product space Xα have the form Uα where each Uα is
α∈Λ α∈Λ
open in Xα and Uα = Xα for all but one α ∈ Λ.

36
 Q Q
Example 182: For n ∈ N, let Rn = R. Consider Rn . Then (0, 1)n is not open.
n∈N n∈N
Note 183: The projection maps are continuous by the way subbasic open sets are defined.

Theorem 184: The projection maps are open.

Q
Proof: Let (Xα , τα ) be a topological space for all α ∈ Λ and suppose U = Uα is a basic open
  α∈Λ
Q Q
set in Xα and β ∈ Λ. Then πβ Uα = Uβ , which is open in Xβ . Since images of basic open
α∈Λ α∈Λ
sets are open and functions preserve union, the projection maps are open. ■

Lemma 185 (Alexander’s Subbase Lemma): Let ⟨X, τ ⟩ be a topological space and suppose S

is a subbase for τ . Then X is compact if and only if every cover of X consisting of elements of S

has a finite subcover.

Proof: (=⇒) Clear.

(⇐=) Suppose that every cover of X consisting of elements of S has a finite subcover. Toward a

contradiction, suppose that X is not compact. Then X has an open cover with no finite subcover.

Claim 1: The collection of open covers of X with no finite subcover has a maximal element with

respect to subcollection inclusion.

Proof (Claim 1): Exercise. ■Claim 1

Let U be an open cover of X with no finite subcover that is not properly contained in another

open cover with no finite subcover. Since every open cover consisting of elements of S has a finite

subcover, U ∩ S does not cover X. Let x ∈ X \ ∪(U ∩ S ), U ∈ U with x ∈ U and S1 , . . . , Sn ∈ S

n
Si ⊆ U . Since x ∈ X \ ∪(U ∩ S ), Si ∈
/ U for every 1 ≤ i ≤ n. By the maximality
T
such that x ∈
i=1
of U , U ∪ {Si } has a finite subcover for each 1 ≤ i ≤ n. For each 1 ≤ i ≤ n, let Ui be a finite

subcollection of U such that Ui ∪ {Si } covers X. In other words, X ⊆ (∪Ui ) ∪ Si .

n
Ui ∪ {U } covers X.
S
Claim 2:
i=1
Proof (Claim 2): Let y ∈ X. Recall that X ⊆ (∪Ui ) ∪ Si for every 1 ≤ i ≤ n. So, either y ∈ ∪Ui
n n
Si ⊆ U . In any case, y ∈ ∪[( Ui ) ∪ {U }].
T S
for some i or y ∈ Si for all i, in which case y ∈
i=1 i=1
■Claim 2

By claim 2, U has a finite subcover, a contradiction. ■

Theorem 186 (Tychonoff ’s Theorem): Let {Xα }α∈Λ be an indexed collection of topological
Q
spaces. Then Xα is compact if and only if Xα is compact for all α ∈ Λ.
α∈Λ
Q
Proof: (=⇒) Suppose that the product space Xα is compact and let β ∈ Λ. Since πβ is con-
 Q  α∈Λ

tinuous, πβ = Xα = Xβ is compact.
α∈Λ
Xα is compact, let U be a cover of
Q
(⇐=) Suppose Xα is compact for all α ∈ Λ. To see that
α∈Λ
Xα consisting of subbasic open sets. For each α ∈ Λ, set Uα = {U ∈ τα | π −1 (U ) ∈ U }.
Q
α∈Λ

37
Claim: There is an α ∈ Λ such that Uα covers Xα .

Proof (Claim): Toward a contradiction, suppose that for all α ∈ Λ, there is an xα ∈ Xα \ ∪Uα .

Then for all β ∈ Λ and all U ∈ Uβ , / πβ−1 (U ). This is a contradiction since {πα−1 (U ) | U ∈ Uα
Q
xα ∈
α∈Λ
and α ∈ Λ} = U covers
Q
Xα . ■
α∈Λ
Choose λ ∈ Λ such that Uλ covers Xλ . Since Xλ is compact, there are {Ui }ni=1 ⊆ Uλ such that Xλ =
n Sn  n
Xα = πλ−1 (Xλ ) = πλ−1 πλ−1 (Ui ), which means πλ−1 (U1 ), πλ−1 (U2 ), . . . πλ−1 (Un )
S Q S 
Ui . Then Ui =
i=1 α∈Λ i=1 i=1
is a finite subcover of U . ■

Theorem 187: Suppose that Λ is an index set and that Xα is nonempty for every α ∈ Λ. Then
Y
the product space Xα is locally compact if and only if each Xα is locally compact and all but
α∈Λ
finitely many Xα are compact.

Proof:
Y
(=⇒) Suppose that the product space Xα is locally compact. Since the projection maps are
α∈Λ
open and continuous, each Xα is locally compact. To see that all but finitely Xα are compact, let
Y
x∈ Xα and W be a compact neighborhood of x. Then W contains a basic neighborhood of the
α∈Λ
n
πα−1
T
form i
(Uαi ). Hence, if α ̸= αi for all i, 1 ≤ i ≤ n, then πα (W ) = Xα . This implies that all
i=1
but finitely many Xα are compact.
Y Y
(⇐=) Let x ∈ Xα and U be a basic neighborhood of x. Then U = Uα1 × . . . × Uαn × Xα ,
α∈Λ α∈Λ
where the set S = {α1 , . . . , αn } is expanded to include all α such that Xα is not compact. It

suffices to find a compact neighborhood contained in U . For each αi , 1 ≤ i ≤ n, there is a

compact neighborhood Kαi of xαi such that Kαi ⊆ Uαi . Since Xα is compact for each α ∈ / S,
Y Y
K = Kα1 × . . . × Kαn × {Xα | α ∈
/ S} is a compact neighborhood of x and K ⊆ U . Thus, Xα
α∈Λ α∈Λ
is locally compact. ■

Exercises

109.
Q Q
(a) Show that if Hα ⊆ Xα for all α ∈ Λ, then Hα = Hα . [In particular, any product of
α∈Λ α∈Λ
closed sets will again be a closed set.]

(b) Prove that in the product space Rω , int[(0, 1)ω ] = ϕ. Hence, the corresponding result for interiors

fails.

(c) Show that (b) can be extended to infinite products and (a) can be extended to finite products.

110. In this exercise, you will show that Tychonoff’s Theorem implies the Axiom of Choice by

proving the intermediate claims.

38
Proof: Suppose that Tychonoff’s Theorem is true and let A = {Aα | α ∈ I}, where I is an index

set, be a nonempty collection of nonempty sets.

Y
Claim 1: There is a choice function on A if and only if the product Aα ̸= ϕ.
α∈I
Y
By claim 1, it suffices to prove that the product Aα ̸= ϕ. Let x ∈
/ Aα for every α ∈ I. For each
α∈I
α ∈ I, define τα = {U ∈ P (Aα ∪ {x}) | U = ϕ, U = {x}, or the set (Aα ∪ {x}) \ U is finite}.

Claim 2: The collection τα is a topology on the set Aα ∪ {x}.

Claim 3: The pair ⟨Aα ∪ {x}, τα ⟩ is a compact topological space for every α ∈ I.
Y
By Tychonoff’s Theorem, the space X = (Aα ∪ {x}) is compact with its product topology. Set
α∈I
C = {πα−1 (Aα ) | α ∈ I} ⊆ P (X).

Claim 4: Every element of the collection C is a closed subset of X.

We now show that the collection C has the finite intersection property. Let C1 , . . . , Cn ∈ C . Then

for each i, there is an αi ∈ I such that Ci = πα−1

i
(Aαi ). Since the projection maps are onto, we

obtain that Ci ̸= ϕ for every i. Choose a bi ∈ Aαi for each i = 1, . . . , n. Define an element f ∈ X by



b

i if α = βi for some i = 1, . . . , n;
f (α) =

x
 otherwise.

n
T
Claim 5: The function f ∈ Ci .
i=1
Claim 6: The collection C ̸= ϕ.
Y
By claim 6, the set Aα ̸= ϕ, and the proof is complete. ■
α∈I
111. In this exercise, you will show that the Well-Ordering Principle implies Tychonoff’s Theorem

by proving the intermediate claims.

Proof: Let (X, τ1 ) and (Y, τ2 ) be topological spaces such that X is compact and let B be the usual

base on X × Y as in theorem 167. Let A ⊆ B be a collection of basic open sets such that no finite

subcollection of A covers X × Y .

Claim 1: There is an x ∈ X such that no finite subcollection of A covers {x} × Y .

Let I ̸= ϕ be an index set and suppose that X = {(Xα , τα ) | α ∈ I} is a collection of compact

Y
topological spaces. Set X = Xα . We must show that X is compact in the product topology. Let
α∈Λ
B denote the base for the product topology. By the Well-Ordering Principle, we may put without

loss of generality a well-ordering ≤ on I such that (I, ≤) has a largest element. We proceed by

induction on the order ≤. Let β ∈ I and for each γ < β, let pγ ∈ Xγ . For each α < β, define

Yα ⊆ X by Yα = {x ∈ X | πγ (x) = pγ for every γ ≤ α}. Note that the sets Yα are nested since if

39
 T
α1 < α2 , then Yα2 ⊆ Yα1 . Let Zβ = Yα .
α<β
Claim 2: If A ⊆ B is a finite collection of basic open sets that covers Zβ , then A covers Yα for

some α < β.

We now verify that X is compact by proving the contrapositive. Let U ⊆ B be a collection of basic

open sets such that no finite subcollection of U covers X.

Claim 3: For each α ∈ I, it is possible to choose points pα ∈ Xα such that for each β ∈ I, the set

Yβ cannot be covered by a finite subcollection of U . (Hint: Use transfinite induction on α.)

By the Well-Ordering Principle, the set I has a maximal element.

Claim 4: If β = max I, then the element f ∈ X defined by f (α) = pα is an element of X that is

not covered by any finite subcollection from U .

By claim 4, the collection U is not a cover, and so, the space X is not compact. ■

Note: It is known that Tychonoff’s Theorem is equivalent to the Axiom of Choice.

40
 Section 5.3: The Stone-Čech Compactification
Lemma 188: Let Λ be an index set, and suppose X is a topological space and Xα is a topological
Q
space for all α ∈ Λ. Also, suppose f : X → Xα . Then f is continuous if and only if (πα ◦ f ) :
α∈Λ
X → Xα is continuous for all α ∈ Λ.

Proof:

(=⇒) Recall that the composition of continuous functions is continuous.

(⇐=) Note that for all α ∈ Λ and all U ∈ τα , f −1 (πα (U )) = (πα ◦ f )(U ) is open in X. So, the

preimage of any subbasic open set is open. Thus, since preimages preserve intersections and unions,

f is continuous. ■

Example 189 (The Stone-Čech Compactification): Let X be a Tychonoff space and set

F = {f : X → R | f is continuous and bounded}. For each f ∈ F , let If ⊆ R be a closed bounded

Q Q
interval with f (X) ⊆ If . Define e : X → If by e(x) = f (x).
f ∈F f ∈F
Claim 1: The function e is one-to-one.

Proof (Claim 1): Since X is Tychonoff, F separates points from closed sets. Also, since F

separates points from closed sets and singletons are closed, F separates points. Therefore, the

function e is one-to-one. ■

Claim 2: The function e is continuous.

Proof (Claim 2): Note that for all f ∈ F , πf ◦ e = f by the way e is defined. By the lemma

above, e is continuous. ■

Claim 3: The collection {f −1 (V ) | f ∈ F and V ⊆ If is open} is a base for the topology on X.

Proof (Claim 3): Suppose that x ∈ X and U ⊆ X is open with x ∈ U . Since X is Tychonoff,

there is a continuous f : X → [0, 1] such that f (x) = 0 and f (X \ U ) ⊆ {1}. Then (−1, 21 ) ∩ If is

open in If , and so, x ∈ f −1 ((−1, 12 ) ∩ If ) ⊆ U . ■

Claim 4: The function e : X → e(X) is open.

Proof (Claim 4): Suppose that U ⊆ X is open and x ∈ U . By claim 3, there is an f ∈ F and an

open V ⊆ If such that x ∈ f −1 (V ) ⊆ U . Now note the following.

x ∈ f −1 (V ) ⊆ U

⇐⇒ e(x) ∈ e[f −1 (V )] ⊆ e(U )

⇐⇒ e(x) ∈ e[(πf ◦ e)−1 (V )] ⊆ e(U )

⇐⇒ e(x) ∈ e(e−1 [f −1 (V )]) ⊆ e(U )

⇐⇒ e(x) ∈ πf−1 (V ) ∩ e(X) ⊆ e(U ).

Therefore, e(U ) is open in e(X). ■

41
So, e : X → e(X) is a homeomorphism. Set βX = e(X). Note that βX is compact since it is a

closed subset of a compact space. Also, by design, note that e(X) is dense in βX. The space βX is

called the Stone-Čech Compactification of X.

Example 190: The space βN is compact but not sequentially compact.

Exercises

112. Prove that the space βR is separable.

Chapter 6: Connected Spaces

Section 6.1: Connected Spaces

Definition 191: A topological space X is connected if whenever U, V are nonempty open sets

such that X = U ∪ V , then U ∩ V ̸= ϕ. A topological space is disconnected if it is not connected.

Proposition 192: The following are equivalent for any topological space X.

(i) X is disconnected;

(ii) There are disjoint, nonempty, and open U, V ⊆ X such that U ∪ V = X;

(iii) X has a nontrivial proper clopen subset.

Corollary 193: A topological space is connected if and only if the only clopen sets are the set itself

and the empty set.

Example 194:

(a) The Sorgenfrey Line is disconnected.

(b) The set (−∞, 1) ∪ (1, ∞) is disconnected. (It is the union of two connected spaces, however.)

Definition 195: A subset I ⊆ R is an interval if whenever x, y ∈ I and x < z < y, then z ∈ I.

Theorem 196: A subspace of R with the usual topology is connected if and only if it is an interval.

Proof: (=⇒) We prove the contrapositive. Suppose that X is a subspace of R in the usual topology

which is not an interval. Then there are x, y, z ∈ R such that x < y < z, x ∈ X, z ∈ X, and y ∈
/ X.

This implies that {X ∩ (−∞, y), X ∩ (y, ∞)} is a disconnection of X.

(⇐=) Suppose that X is a subspace of R which is an interval. Attempting a contradiction, suppose

that {H, K} is a disconnection of X. Since H and K are nonempty, choose an x ∈ H and a z ∈ K,

and, without loss of generality, suppose that x < z. Set y = sup({t < z | t ∈ H}). Since X is an

42
interval, y ∈ [x, z] ⊆ X, which means that y ∈ X. By the definition of supremum, for every ε > 0,

H ∩ (y − ε, y] ̸= ϕ. Hence, y ∈ H = H. Also, for every ε > 0, [y, y + ε) ∩ K ̸= ϕ, which means that

y ∈ K = K. Thus, y ∈ H ∩ K, a contradiction. Hence, K is connected. ■

Corollary 197:

(i) The set R with the usual topology is connected.

(ii) Subsets of connected spaces need not be connected.

Proof: Exercise. ■

Theorem 198: The continuous image of a connected space is connected.

Proof: We will prove the contrapositive. Suppose that X and Y are topological spaces, f : X → Y

is continuous, and Y is disconnected. Then there are disjoint, nonempty, and open sets U, V such

that U ∪ V = Y . Now, X = f −1 (Y ) = f −1 (U ∪ V ) = f −1 (U ) ∪ f −1 (V ). Since f is continuous,

f −1 (U ) and f −1 (V ) are open. Also, f −1 (U ) ∩ f −1 (V ) = f −1 (U ∩ V ) = f −1 (ϕ) = ϕ. So, X is

disconnected. ■

Corollary 199: Suppose that f : R → R is continuous and a < b. Then f ([a, b]) is either a closed

bounded interval or a singleton.

Proof: Exercise. ■

Corollary 200 (Intermediate Value Theorem): Suppose that a, b ∈ R with a < b, and f :

[a, b] → R is continuous. Then f attains every value between f (a) and f (b).

Proof: Let m be between f (a) and f (b). By the corollary and the theorem above, f ([a, b]) is an

interval, and so, m ∈ f ([a, b]). ■

Proposition 201: Let X be a countable Tychonoff space with |X| ≥ 2. Then,

(i) X is disconnected;

(ii) X is zero-dimensional.

Proof: Exercise. ■

Exercises

113. Show that the Sorgenfrey Line is disconnected.

114. Show that any infinite set with the co-finite topology is connected.

115. Show that no countable subset of R is connected.

Definition: Let X be a topological space and x ∈ X. A component in X is a maximal connected

set containing x and is denoted Cx .

116.

(a) Prove that the components of a topological space are closed sets.

43
(b) If X is a topological space is a T1 zero-dimensional space and x ∈ X, prove that Cx = {x}.

117. Prove that a connected, completely regular, T1 -space is uncountable.

118. Let Y = {0, 1} be endowed the discrete topology. Prove that a topological space X is

disconnected if and only if there is a continuous surjection f : X → Y .

119. Suppose that X is a Tychonoff space. Show that βX is connected if and only if X is connected.

(Hint: One direction is clear.)

120. Prove that any continuous f : [0, 1] → [0, 1] has a fixed point.

121. Prove that a polynomial of odd degree has at least one real root.

122. Prove that a connected and separable metric space has either 1 point or |R| points.

123. Prove corollary 162.

124. Prove corollary 164.

125. Prove proposition 166.

Definition: Let X be a topological space and x, y ∈ X. A simple chain connecting x and y is a

sequence {Ui }ni=1 ⊆ X of open sets such that U1 = {x}, Un = {y}, and for each i, j, 1 ≤ i, j ≤ n,

Ui ∩ Uj ̸= ϕ if and only if |i − j| ≤ 1.

126. If X is a connected space and U is an open cover of X, prove that any two points of X can

be connected by a simple chain consisting of elements of U .

Section 6.2: Pathwise Connected Spaces

Definition 202: A topological space X is pathwise connected if for any x, y ∈ X, there is a

continuous f : [0, 1] → X, called a path, such that f (0) = x and f (1) = y.

Definition 203: A topological space X is arcwise connected if for any x, y ∈ X, there is a

homeomorphism f : [0, 1] → X, called an arc, such that f (0) = x and f (1) = y.

Theorem 204: Every pathwise connected space is connected.

Proof: Let X be a pathwise connected space and x, y ∈ X. To see that X is connected, suppose

not. Then there are nonempty, disjoint, and open H, K ⊆ X such that X = H ∪ K, x ∈ H, and

y ∈ K. Let f : [0, 1] → X be a path from x to y. Then f −1 (H) and f −1 (K) disconnect [0, 1], which

is a contradiction since [0, 1] is connected. ■

  
1
Example 205 (Topologist’s Sine Curve): Set X = x, sin | x > 0 ∪ ({0} × [−1, 1]) and
x
endow X the subspace topology from R2 . Then X is connected but not pathwise connected.

Proof: Exercise. ■

Exercises

127. Prove that the continuous image of a pathwise connected space is pathwise connected.

44
Definition: Let X be a topological space and x, y ∈ X. Define a relation ∼ on X by x ∼ y if and

only if there is a path joining x to y. The path components of X are the equivalence classes [x]∼ .

128. Let X be a topological space and x, y ∈ X. Prove that the path component of x is pathwise

connected and contained in the component of x.

129. Verify that the topologist’s sine curve is connected but not pathwise connected.

130. Let E ⊆ R2 be countable. Prove that R2 \ E is pathwise connected.

Section 6.3: Locally Pathwise Connected Spaces

Definition 206: A topological space is locally pathwise connected if each point has a local base

consisting of pathwise connected subsets.

Definition 207: Suppose that X is a topological space and f : [0, 1] → X and g : [0, 1] → X are

paths such that f (1) = g(0). The sum h = f ∗ g : [0, 1] → X is defined by

  
1
f (2t) if t ∈ 0, ;


2
h(t) =  

g(2t − 1) 1
 if t ∈ ,1 .
2

Theorem 208: A connected, locally pathwise connected space is pathwise connected.

Proof: Let X be a connected, locally pathwise connected space, a ∈ X, and H = {x ∈ X |

there is a path joining x to a}. Since a ∈ H, H ̸= ϕ.

Claim 1: The set H is open.

Proof (Claim 1): Suppose that b ∈ H and let U be a pathwise connected neighborhood of b.

Then there is a z ∈ U such that there is a path joining z to b. Add the path to b to the path to a

to obtain a path from z to a. Hence, b ∈ H. ■Claim 1

Claim 2: The set H is closed.

Proof (Claim 2): Suppose that b ∈ H and let U be a pathwise connected neighborhood of b.

Then there is a z ∈ U ∩ H and there is a path joining b to z and a path joining z to a. Hence, by

the addition of paths, b ∈ H. ■Claim 2

Since H is a nonempty clopen set and H is connected, H = X, and so, X is pathwise connected.

Corollary 209: Every open connected subset of Rn is pathwise connected.

Exercises

131. Prove that a topological space is locally pathwise connected if and only if every path component

of each open set is open.

45
132. If a topological space X is locally pathwise connected, prove that the path components of X

are clopen sets.

Section 6.4: Locally Connected Spaces

Definition 210: A topological space X is locally connected if every point of X has a local base

consisting of connected sets.

Example 211:

(a) Set X = (0, 1)∪(2, 3) with the subspace topology from R. Then the space X is locally connected

but not connected.

 
1
(b) Set A = {0} ∪ | n ∈ N and Y = ([0, 1] × A) ∪ ({0} × [0, 1]). Then the space Y is connected
n
and pathwise connected but not locally connected.

Note 212: Locally pathwise connected spaces are locally connected.

Theorem 213: A topological space X is locally connected if and only if for any U ⊆ X and any

x ∈ U , the component of the point x in U is an open set.

Proof:

(=⇒) Suppose that X is a locally connected space, U ⊆ X is open, C ⊇ U is a component of U in X,

and x ∈ C. Since X is locally connected, there is an open connected V ⊆ U such that x ∈ V ⊆ U .

Hence V ⊆ C, and so, C is open.

(⇐=) Suppose that each component of each open set in X is open and x ∈ X. If U is an open

neighborhood of x, then the component of U containing x is an open connected neighborhood of x

contained in U . Hence, X is locally connected. ■

Corollary 214: The components of a locally connected space are clopen sets.

Corollary 215: A compact locally connected space has a finite number of components.

Theorem 216:

(i) The continuous open image of a locally connected space is locally connected.

(ii) The continuous closed image of a locally connected space is locally connected.

Theorem 217: Suppose that Λ is an index set and that Xα is nonempty for every α ∈ Λ. Then
Y
the product space Xα is locally connected if and only if each factor space is locally connected
α∈Λ
and all but finitely many factor spaces are connected.

Exercises

133. Prove that the Sorgenfrey Line is not locally connected.

134. Prove that the topologist’s sine curve is not locally connected.

46
Definition: A topological space X has property S if every open cover of X can be refined to a

cover consisting of finitely many connected sets.

135. Prove that every topological space with property S is locally connected.

136. Prove that a compact Hausdorff space is locally connected if and only if it has property S.

137. Prove that there is a locally connected Hausdorff space which does not have property S.

138. Prove that the continuous Hausdorff image of a compact locally connected space is compact

and locally connected.

Section 6.5: Totally Disconnected Spaces

Definition 218: A topological space X is totally disconnected if the nonempty connected subsets

of X are singleton sets.

Theorem 219:

(i) Every product of totally disconnected spaces is totally disconnected;

(ii) Every subspace of a totally disconnected space is totally disconnected;

(iii) Every zero-dimensional T1 -space is totally disconnected.

Lemma 220: A compact Hausdorff space X is totally disconnected if and only if whenever x, y ∈ X,

there is a clopen set A ⊆ X such that x ∈ A and y ∈

/ A.

Proof: Exercise. ■

Theorem 221: A locally compact Hausdorff space is zero-dimensional if and only if it is totally

disconnected.

Proof:

(=⇒) Clear.

(⇐=) Let X be a locally compact, totally disconnected Hausdorff space. To see that X is zero-

dimensional, let A ⊆ X be closed, x ∈ X \ A, and U ⊆ X be an open neighborhood of x such that

U is compact and U ∩ A = ϕ. For each p ∈ Fr(U ), let Vp ⊆ U be a clopen set such that x ∈ Vp

and p ∈/ Vp . Then the sets X \ Vp form an open cover of Fr(U ), and so, there is a finite subcover
n
T
corresponding to points p1 , . . . , pn . Set V = Vpi . Then V ⊆ U is a clopen set such that y ∈ V
i=1
and V ∩ Fr(U ) = ϕ. Hence, V ⊆ U and hence is a clopen set in X which contains x and V ∩ A = ϕ.

Hence, X is zero-dimensional by the note above. ■

Theorem 222: Every totally disconnected, compact metric space is homeomorphic to a subset of

a countable product of discrete spaces.

Exercises

139. Prove that the space R \ Q is zero-dimensional.

47
140. Prove that the Sorgenfrey Line is zero-dimensional.

141. Prove that the spaces βN and βQ are totally disconnected.

142. Prove that a compact Hausdorff space X is totally disconnected if and only if for each x, y ∈ X,

there is a clopen U ⊆ X such that x ∈ U and y ∈

/ U . (Hint: First show that the quasi-components

of X are the components of X.)

143. Prove lemma 203.

Chapter 7: The Cantor Set

Definition 223: Let X be a topological space and A ⊆ X. Then A is perfect in X if A = A′ .

Note 224: A topological space is perfect if and only if it is dense in itself.

Example 225: The Cantor set C is perfect.

Lemma 226: If U is a nonempty open subset of a totally disconnected, perfect T2 -space and n ∈ N,
n
S
then U = Ui where Ui is a nonempty disjoint open set for all i, 1 ≤ i ≤ n.
i=1
Proof: Note that it suffices to prove the theorem for n = 2. If U ⊆ X is nonempty and open, then

U is not a singleton set since X is perfect. Let p, q ∈ U . By a previous lemma, there is a clopen

V ⊆ X such that p ∈ V and q ∈

/ V . Set U1 = U ∩ V and U2 = U \ V . ■

Theorem 227: Any two totally disconnected, perfect, and compact metric spaces are homeomor-

phic.

Proof: Let X and Y be totally disconnected, perfect, and compact metric spaces. Let {Un }∞
n=1 ⊆ X

and {V }∞ n=1 ⊆ Y be sequences of finite covers of X and Y , respectively, by disjoint open sets so
1
that the sets of the nth cover have diameter < n . By using the lemma above, we suppose without
2
loss of generality that for each n ∈ N, |Un | = |Vn |. If U1 = {U11 , . . . , U1n } and V = {V11 , . . . , V1n },

then each U1j and V1j is a union of elements of U2 and V2 , respectively. By the lemma above, we

suppose without loss of generality that U1j and V1j are the union of the same number of elements

of U2 and V2 , respectively, such that U2k ⊆ U1j if and only if V2k ⊆ V1j . Continue this process to
f1 g1
match the covers of Un and Vn for all n ∈ N. Let X0 ←− X1 ← · · · and Y0 ←− Y1 ← · · · be the

derived sequence of the covers Un and Vn , respectively. Define ϕn : Xn → Yn by ϕn (Unj ) = Vnj .

Then ϕn is a homeomorphism from Xn to Yn , and it is readily verified that the map ϕ : X∞ → Y∞

is a homeomorphism. Hence, X∞ is homeomorphic to X and Y∞ is homeomorphic to Y . ■

48
Corollary 228:

(i) The Cantor set C is the only totally disconnected, perfect, and compact metric space, up to

homeomorphism.

(ii) The Cantor set C is homeomorphic to the product space 2ℵ0 .

(iii) The Cantor set C is homeomorphic to the product space Cℵ0 .

Theorem 229: Every compact metric space is a continuous image of the Cantor set.

Proof: Let X be a compact metric space. Let {Un }∞ n=1 ⊆ X be a sequence of finite covers of X
1
by the closures of open sets such that the sets of Un have diameter < n and Un < Un−1 for each
2
n ∈ N. In other words, Un = {Un1 , . . . , Unkn }. For each U1i ∈ U1 , define Vi1 = {(u, i) | u ∈ U1i } so

that V1 = V11 ∪ . . . ∪ Vk1 is the disjoint union of the sets U1i . Then each U2j ∈ U2 is contained in
k2
some U1k ∈ U1 . Define V2ij = {(u, i, j) | u ∈ U2j } whenever U2j ⊆ U1i and let V2 =
S S
V2ij .
j=1 U2j ⊆U1i
Note that each U2j occurs in the disjoint union once for each for each U1i so that U2j ⊆ U1i . Define

f2 : V2 → V1 . by f2 ((u, i, j)) = (u, i). Then f2 is continuous on each V2ij and hence on V2 . Also,

there is a ϕ2 : V2 → X defined by ϕ2 (u, i, j) = u. Continue this process to obtain a pair of inverse

sequences and a mapping between them. Note that each ϕn satisfy the composition condition nec-

essary to be a map of inverse sequences. The result is a map ϕ : V∞ → X of the inverse limit spaces,

which is continuous and onto since X and each Vn are compact Hausdorff spaces and each ϕn is

continuous and onto. Note that each Vn is a compact metric space for each n ∈ N. Let dn be the

metric on Vn induced by the metrics on Unj . Also, if (x1 , . . .) ∈ V∞ , then ϕ(x1 ) = ϕ2 (x2 ) = . . . and

if zx = ϕ(x1 ) = ϕ2 (x2 ) = . . ., then for any (y1 , . . .) ∈ V∞ , dn (xn , yn ) ≥ d(zx , zy ). Note that V∞ is a

compact metric space.

Claim: The space V∞ is totally disconnected.

Proof (Claim): Suppose that x = (x0 , . . .), y = (y0 , . . .) ∈ V∞ with x ̸= y. Then xn ̸= yn for

some n ∈ ω. This implies that ϕn (xn ) ̸= ϕ(yn ). Suppose that ϕn (xn ) = zx and ϕn (yn ) = zy .

If dm is the metric on Vm , then it is clear that dm (xm , ym ) ≥ d(zx , zy ). Since the diameters of

the sets Vm1 , . . . , Vmkn approach zero as m → ∞, xm and ym belong to different sets in Vm . In

other words, xm ∈ Vm1 and ym ∈

/ Vm1 . However, Vm1 is a clopen set in Vm and hence the set

{(z0 , z1 , . . .) ∈ V∞ | zm ∈ Vm1 } is a clopen neighborhood of x in V∞ which does not contain y. Thus,

V∞ is totally disconnected. ■Claim

Note that the space V∞ may not be perfect. However, if C denotes the Cantor set, then V∞ × C is

a perfect, totally disconnected, compact metric space such that the space V∞ , and hence the space

X, is a continuous image. ■

49
 Exercises

Notation: In these exercises, the symbol C will denote the Cantor set.

144. Prove that the set C is nowhere dense in the interval [0, 1].

145. Prove that every totally disconnected, compact metric space is homeomorphic to a subset of

the set C.

146. Prove that every perfect set in a complete metric space contains a compact perfect set.

147. Prove that a countable compact Hausdorff space is not perfect.

148. Let X be a topological space and A ⊆ X. If A contains no isolated points, prove that A is

perfect in X.

149. Prove that every open subset of the set C can be expressed as a union of clopen subsets.

150. Prove that every open subset of the set C is homeomorphic to either C or C \ {0}.

151. Prove that every compact metric space is a continuous image of the space βN.

152. Is the space βN a continuous image of the Cantor set?

Definition: A topological space is scattered if it contains no nonempty dense-in-itself subset.

153. Prove that every discrete space is scattered.

154. Prove that every topological space can be expressed as a union of two disjoint sets: one perfect

and the other scattered.

Definition: In the above, the perfect set is called the perfect kernel of the topological space.

155. Prove that every closed subset of the set C is a retract of C.

Chapter 8: Ordinals and Cardinals

Section 8.1: Ordinals

Definition 230: A set α is an ordinal if the following are satisfied.

(i) If whenever x ∈ y ∈ α, then x ∈ α.

(ii) If whenever x ∈ y ∈ z ∈ α, then x ∈ z.

Note 231: The following are equivalent for any set α.

(i) α is an ordinal.

(ii) The relation ∈ is transitive and whenever x ∈ y ∈ z ∈ α, then x ∈ z.

(iii) The relation ∈ is transitive and α is strictly well-ordered by inclusion.

50
Notation 232: The class of ordinals is denoted Ord.

Example 233: ϕ ∈ Ord.

Proof: It is vacuously true that the properties of an ordinal are satisfied. So, ϕ is an ordinal. ■

Proposition 234: If α ∈ Ord, then α ∪ {α} ∈ Ord.

Proof: First, suppose that x ∈ y ∈ α ∪ {α}. If y ∈ α, then x ∈ α ⊆ α ∪ {α}. If y ∈

/ α, then y ∈ {α},

which implies that y = α. Then x ∈ y = α ⊆ α ∪ {α}. Now suppose that x ∈ y ∈ z ∈ α ∪ {α}. To

see that x ∈ z, note that if z ∈ α, then x ∈ z since α ∈ Ord. Also, if z ∈

/ α, then z ∈ {α}, which

implies that z = α. So, x ∈ y ∈ z = α, which means that x ∈ α = z. Therefore, α ∪ {α} ∈ Ord. ■

Proposition 235: If A ⊆ Ord, then ∪A ∈ Ord.

Proof: First, suppose that x ∈ y ∈ ∪A. Then there is an α ∈ A such that y ∈ α. Since α ∈ Ord

and x ∈ y ∈ α, we obtain that x ∈ α ⊆ ∪A. So, x ∈ ∪A. Now suppose that x ∈ y ∈ z ∈ ∪A. To

see that x ∈ z, note that z ∈ α for some α ∈ A. Since x ∈ y ∈ z ∈ α and α ∈ Ord, we obtain that

x ∈ z. Therefore, ∪A ∈ Ord. ■

Proposition 236: If A ⊆ Ord and A ̸= ϕ, then ∩A ∈ Ord.

Proof: First, suppose that x ∈ y ∈ ∩A. To see that x ∈ ∩A, note that y ∈ α for every α ∈ A,

which implies that x ∈ α for every α ∈ A, and so, x ∈ ∩A. Now suppose that x ∈ y ∈ z ∈ ∩A. To

see that x ∈ z, note that z ∈ α for every α ∈ A. Since z ∈ α for every α ∈ A and α ∈ Ord, we

obtain that x ∈ z. Therefore, ∩A ∈ Ord. ■

Theorem 237: Ord is well-ordered by ≤, where α < β if α ∈ β.

Corollary 238: If α ∈ Ord, then α = {β ∈ Ord | β < α}.

Proposition 239: If α, β ∈ Ord, then α ∈ β if and only if α ⊆ β and α ̸= β.

Proof: (=⇒) Suppose that α ∈ β. To see that α ⊆ β, suppose that γ ∈ α. Since γ ∈ α ∈ β ∈ Ord,

we obtain that γ ∈ β, and so, α ⊆ β. Also, since α ∈ β, we obtain that α ̸= β.

(⇐=) Suppose that α ⊆ β and α ̸= β. To see that α ∈ β, note that since Ord is well-ordered, we

obtain that α ∈ β or β ∈ α. Since α ⊆ β, we obtain that β ∈

/ α, and so, α ∈ β, as desired. ■

Definition 240: For any α ∈ Ord, α + 1 = α ∪ {α}.

Proposition 241: Suppose that α, β ∈ Ord and α ∈ β. Then α + 1 ∈ β or α + 1 = β.

Proof: If α + 1 ∈
/ β, then β ⊆ α + 1 = α ∪ {α}. Since α ∈ β, β ⊈ α. Therefore, β = α + 1. ■

Example 242: Set 0 = ϕ. Then,

1 = 0 + 1 = ϕ ∪ {ϕ} = {0}

2 = 1 + 1 = 1 ∪ {1} = {0, 1}

3 = 2 + 1 = 2 ∪ {2} = {0, 1, 2}
..
.

51
n = n − 1 ∪ {n − 1} = {0, 1, . . . , n − 1}

Set ω = {0, 1, 2, . . .}.

Notation 243: ω0 = ω.

52
Example 244:
..
ω + 1 = ω ∪ {ω} .

ω + 2 = (ω + 1) + 1 (ω · 2) + ω = ω · 3
.. ..
. .

ω+ω =ω·2 ω·4

..
(ω · 2) + 1 .

(ω · 2) + 2 ω·ω
..
..

Section 8.2: The Space ω1

Definition 245: The first uncountable ordinal is ω1 .

Definition 246: Let α ∈ Ord. If α = β + 1 for some β ∈ Ord, then α is a successor ordinal.

Otherwise, α is a limit ordinal.

Note 247: For any α ∈ Ord, the set (α, α + 1] = {α + 1}. (i.e., All successor ordinals are isolated.)

Definition 248: Let τ be the topology on ω1 formed by taking the set B = {(α, β] | α, β ∈ ω1 and

α < β} as a base.

Theorem 249: Every countable subset of ω1 is bounded.

S
Proof: Suppose that A ⊆ ω1 is countable. Set B = [0, α + 1] and note that B is countable since
α∈A
it is a countable union of countable sets. So, there is a γ ∈ ω1 \ B. If β ≤ α ∈ A, then β ∈ B. Since

γ∈
/ B, we obtain that α < γ for every α ∈ A. ■

Corollary 250: Every countable subset of ω1 has a supremum.

Proof: Suppose that A ⊆ ω1 is countable. Then the set {x ∈ A | x is an upper bound of A}, which

exists by the previous theorem, has a least element since ω1 is well-ordered. ■

Proposition 251: There are no strictly decreasing sequences in ω1 .

Proof: Toward a contradiction, suppose that {xn }n∈N ⊆ ω1 is strictly decreasing. Since ω1 is well

ordered, the sequence {xn }n∈N has a least element xk . Since the sequence is strictly decreasing,

xk+1 < xk . This is a contradiction since xk is the least element of {xn }n∈N . ■

Proposition 252: The supremum of a strictly increasing sequence in ω1 is the limit of the sequence.

Proof: Suppose that {xn }n∈N ⊆ ω1 is strictly increasing. Set γ = sup({xn }n∈N ), which exists since

ω1 is well ordered. Since {xn }n∈N is strictly increasing, γ ∈

/ {xn }n∈N . Also, if γ = β + 1 for some

β ∈ ω1 , then β is an upper bound of the sequence {xn }n∈N . So, γ is a limit ordinal. For any α ∈ ω1

with α < γ, note that (α, γ] ∩ {xn }n∈N ̸= ϕ. Since the sequence {xn }n∈N is increasing, it converges

to γ. ■

Corollary 253: The space ω1 is countably compact.

53
Proof: Suppose that A ⊆ ω1 is countable and infinite. Let x0 = inf(A) For n > 0, let xn =

inf(A \ {xi }i<n . Then {xn }n∈ω is a strictly increasing sequence. By the previous proposition,

the sequence {xn }n∈ω converges to a limit ordinal γ, which is therefore an accumulation point of

{xn }n∈ω , and hence, an accumulation point of A. ■

Corollary 254: The space ω1 is not separable.

Proof: Let A ⊆ ω1 be countable and set γ = sup(A). Since γ = sup(A), we obtain that (γ, γ + 1] ∩

A = ϕ. Therefore, A is not dense, and so, ω1 is not separable. ■

Corollary 255: The space ω1 is not Lindelöf.

Proof: Set U = {[0, α] | α ∈ ω1 }. Suppose that {Un }n∈N ⊆ U . For each n ∈ N, there is an αn ∈ ω1

Un . So, U has no countable

S
such that Un = [0, αn ]. Set γ = sup(αn ) and note that γ + 1 ∈/
n∈N n∈N
subcover, and so, ω1 is not Lindelöf. ■

Proposition 256: For each α ∈ ω1 , the collection {(β, α] | β < α} is a countable local base for α.

Corollary 257: The space ω1 + 1 is compact.

Proof: Exercise. ■

Theorem 258: Closed and unbounded (Club) subsets of ω1 are not disjoint.

Proof: Suppose that E, F ⊆ ω1 are closed and unbounded. Choose x0 ∈ E and x1 ∈ F such that

x0 < x1 . For n ∈ N, choose x2n ∈ E such that x2n−1 < x2n and choose x2n+1 ∈ F such that x2n <

x2n+1 . Then {xn }n∈N is a strictly increasing sequence. By a previous proposition, sup({xn }n∈N ) is an

accumulation point of the sequence {xn }n∈N . Since E and F are closed, sup({xn }n∈N ) ∈ E ′ ⊆ E = E

and sup({xn }n∈N ) ∈ F ′ ⊆ F = F . Therefore, sup({xn }n∈N ) ∈ E ∩ F . ■

Corollary 259: The space ω1 is normal.

Proof: Exercise. ■

Exercises

156. Prove corollary 248.

157. Prove corollary 250.

158. (Principle of Transfinite Induction) Let A be a class and suppose that for each x ∈ Ord,

if whenever y ∈ Ord and y < x implies that y ∈ A, then x ∈ A. Prove that Ord ⊆ A.

Section 8.3: Cardinals

Definition 260: If X is a set, then |X| = card(X) = first({α ∈ Ord | |α| = |X|}).

Example 261:

(a) Elements of ω are cardinal numbers.

54
(b) ω, ω1 , ω2 , . . . , ωω are cardinals.

Definition 262: Let κ, λ be cardinals.

(i) κ + λ = |(κ × {0}) ∪ (λ × {1})|;

(ii) κ · λ = |κ × λ|;

(iii) κλ = {f | f : λ → κ}.

Proposition 263: For any set X, |P (X)| = 2|X| .

Proposition 264: For any set X, |X| < |P (X)|.

Conjecture 265 (Continuum Hypothesis): ω1 = 2ω .

Definition 266: If κ is a cardinal, then κ+ is the first cardinal strictly larger than κ.

Example 267: ω + = ω1

Conjecture 268 (Generalized Continuum Hypothesis): For any infinite cardinal κ, κ+ = 2κ .

Notation 269:

(i) ℵ0 = ω;

(ii) ℵβ+1 = (ℵβ )+ ;

S
(iii) If α is a limit ordinal, then ℵα = ℵβ .
β<α
Theorem 270: Let κ, λ be infinite cardinals.

(i) κ + λ = max{κ, λ};

(ii) κ · λ = max{κ, λ};

(iii) If κ ≤ λ, then κλ = 2λ ;

(iv) (2λ )κ = 2λ·κ .

Definition 271: A subset S ⊆ ω1 is stationary if S ∩ C ̸= ϕ for any club set C ⊆ ω1 .

Lemma 272 (Pressing Down Lemma): Let S ⊆ ω1 be a stationary set and suppose that

f : S → ω1 such that f (α) < α for every α ∈ S. Then there is a β ∈ ω1 such that f −1 (β) is

stationary.

Exercises

159. Show that the limit ordinals in ω1 form a stationary set.

160. Show that any function f : ω1 → R is not strictly increasing.

161. Use the definition of cardinal arithmetic to compute the following.

(a) 2 + 3; (b) 2 · 3; (c) 23 .

162. Compute each of the following.

(a) ω1 + ω3 ; (b) ω1 · ω3 ; (c) 2ω .

163. Assuming that the Generalized Continuum Hypothesis is true, what is 2ω6 ?

55
164. Suppose that U is an open cover of ω1 consisting of basic open sets. Prove that there is a

γ ∈ ω1 and an unbounded set S ⊆ ω1 such that (γ, α] ∈ U for all α ∈ S.

165. Let L denote the set of limit ordinals in ω1 .

(a) If U ⊆ ω1 is open and L ⊆ U , prove that ω1 \ U is countable.

(b) Show that L cannot be written as the intersection of a countable collection of open sets in ω1 .

(c) Conclude that ω1 is not perfect.

Section 8.4: Products of Ordinals

Note 273: The space (ω1 + 1) × (ω + 1) is compact and Hausdorff, and hence normal.

Example 274 (Tychonoff Plank): The space [(ω1 +1)×(ω +1)]\{(ω1 , ω)}, called the Tychonoff

Plank, is not normal.

Proof: Set E = {ω1 } × ω and F = (ω1 × {ω}) \ {(ω1 , ω)}. To see that the Tychonoff Plank is not

normal, suppose that U and V are open sets with E ⊆ U and F ⊆ V . It remains to show that

U ∩ V ̸= ϕ. For each n ∈ ω, let αn ∈ ω1 such that (αn , ω1 ] × {n} ⊆ U . For each α ∈ ω1 , let nα ∈ ω

such that {α} × [n, ω] ⊆ V . Set γ = sup({αn | n ∈ ω}) + 1. Then for each n ∈ ω, [γ, ω1 ] × {n} ⊆ U .

Therefore, {γ} × [nγ , ω] ⊆ U ∩ V . So, U ∩ V ̸= ϕ, as desired. ■

Example 275: The space [(ω1 + 1) × (ω1 + 1)] \ {(ω1 , ω1 )} is not normal.

Proof: Set E = {(α, α) ∈ α ∈ ω1 } and F = ω1 × {ω1 }. To see that [(ω1 + 1) × (ω1 + 1)] \ {(ω1 , ω1 )}

is not normal, suppose that U and V are open with E ⊆ U and F ⊆ V . It remains to show that

U ∩ V ̸= ϕ. For each α ∈ ω1 , there is a βα < α such that (βα , α] × (βα , α] ⊆ U . For each γ ∈ ω1 ,

there is a βγ ∈ ω1 such that {γ} × [βγ , ω1 ] ⊆ V . By the Pressing Down Lemma, there is a δ ∈ ω1

and a stationary set S such that βα = δ for every α ∈ S. Then for any α ∈ S with α > βδ + 1, we

obtain that ([βα , α] × {α}) ∩ ({δ + 1} × [βδ × ω1 ]) ⊆ U ∩ V . So, U ∩ V ̸= ϕ, as desired. ■

Exercises

166. Is the space (ω1 + 1) × ω Lindelöf?

56
 Chapter 9: Nets

Section 9.1: Definition and Examples

Definition 276: A set Λ with a binary relation ≼ is a directed set if the following conditions are

satisfied.

(i) ≼ is reflexive;

(ii) ≼ is transitive;

(iii) for all λ1 , λ2 ∈ Λ, there is a λ3 ∈ Λ such that λ1 ≼ λ3 and λ2 ≼ λ3 .

Example 277: Any lattice is a directed set.

Definition 278: A net in a set X is a function f : Λ → X where Λ is a directed set.

Notation 279: We sometimes write xα for f (α).

Example 280: Since N is a directed set, any sequence is a net.

Definition 281: Let (xα )α∈Λ be a net on a set X and A ⊆ X.

(i) The net is eventually in A if there is a β ∈ Λ such that xα ∈ A for all α ≥ β.

(ii) The net is frequently in A if for every β ∈ Λ, there is an α ≥ β such that xα ∈ A.

Note 282: If a net is eventually in a set, then it is frequently in the set.

Definition 283: A net (xα ) in a topological space X converges to a point x ∈ X if whenever

U ⊆ X is open with x ∈ U , there is a λ ∈ Λ, xβ ∈ U for all α ≥ λ.

Example 284: Let X be a topological space and x ∈ X. Also, let U denote the collection of open

sets that contain x. Define a binary relation ≼ on U by U ≼ V if and only if V ⊆ U . For each

U ∈ U , choose xU ∈ U . Then (xU )U ∈U is a net in X that converges to x.

Section 9.2: Nets and Topologies

Theorem 285: Let X be a topological space and A ⊆ X. Then x ∈ A if and only if there is a net

in A that converges to x.

Proof: (=⇒) Suppose that x ∈ A. Set U = {U ⊆ X | U is an open neighborhood of x}. For each

open U ∈ U , choose xU ∈ A ∩ U . By the example above, the net {xU }U ∈U converges to x.

(⇐=) Clear. ■

Theorem 286: Let X and Y be topological spaces and f : X → Y . Then f is continuous at x ∈ X

if and only if whenever (xα ) is a net in X that converges to x, f (xα ) converges to f (x).

Proof: (=⇒) Suppose that f is continuous at x and {xα }α∈Λ ⊆ X is a net converging to x. To

see that {f (xα )}α∈Λ converges to f (x), let U be an open neighborhood of f (x). Then f −1 (U ) is an

57
open neighborhood of x. Since {xα }α∈Λ converges to x, there is a β ∈ Λ such that xα ∈ f −1 (U ) for

every α with β ≤ α. Then, for every α with β ≤ α, f (xα ) ∈ f (f −1 (U )) ⊆ U .

(⇐=) We prove the contrapositive. Suppose that f is not continuous at x. Let U = {U ⊆ X |

U is an open neighborhood of x}. Since f is not continuous at x, there is an open neighborhood

V of f (x) such that f (U ) ⊈ V for all U ∈ U . For each U ∈ U , choose an xU ∈ U such that

f (xU ) ∈ / V for all U ∈ U , {xU }U ∈U converges to x but {f (xU )}U ∈U does

/ V . Then, since f (xU ) ∈

not converge to f (x), as desired. ■

Definition 287: Let X be a set. A subnet of a net f : Λ → X is a composition g ◦ f where

g : M → Λ is increasing and cofinal and M is a directed set.

Notation 288: We usually write xαµ for f (g(µ)).

Example 289: Suppose that X is a nonempty set and f : N → X is a sequence in X. Define

g : R → N by g(r) = ⌊r⌋. Then f ◦ g : R → X is a subnet of f . In other words, the net (xnr )r∈R is

a subnet of the net (xn )n∈N

Definition 290: Let (xα )α∈Λ be a net in a topological space X and suppose that x ∈ X. Then x

is a cluster point of the net if for every open neighborhood U of x and every β ∈ Λ, there is an

α ≥ β such that xα ∈ U . In other words, the net (xα )α∈Λ is frequently in each open neighborhood

of x.

Note 291: In the definition above, it is permissible that x = xβ .

Proposition 292: Let ⟨X, d⟩ be a metric space, {xn }n∈N ⊆ X, and x ∈ X. If x is an accumulation

point of the sequence {xn }n∈N , then there is a subsequence that converges to x.

Proof: Since ⟨X, d⟩ is T1 , every open neighborhood of x contains infinitely many elements of the

sequence. Choose n1 ∈ N such that xn1 ∈ B(x, 1). For k ≥ 2, choose nk ∈ N such that nk−1 < nk
 
1
and xnk ∈ B x, \ {xn1 , xn2 , . . . , xnk−1 }. Then {xnk }∞
k=1 is the desired subsequence. ■
k
Note 293: The converse of the proposition is not true.

Example 294: Consider the sequence {(−1)n }∞

n=1 in R with the usual topology. Then the sequence

has no accumulation points but there are many convergent subsequences.

Proposition 295: If a net converges to a point in a topological space X, then every subnet of the

net also converges to the point.

Proof: Exercise. ■

Note 296: The above proposition does not hold for cluster points.

Lemma 297: A net in a topological space has a cluster point x if and only if there is a subnet that

converges to x.

Proof: Let (xα )α∈Λ be a net in a topological space X and x ∈ X.

58
(=⇒) Suppose that x is a cluster point of the net. Set M = {(α, U ) | α ∈ Λ and xα , x ∈ U }. Define

a relation ⪯ on M by (α, U ) ⪯ (β, V ) if α ≤ β and V ⊆ U . It is readily verified that ⪯ is a direction

on M . Define g : M → Λ by g[(α, U )] = α. Then g is increasing. Also, since x is a cluster point, g is

cofinal. It remains to show that (xα(β,U ) )(β,U )∈M converges to x. Let W be an open neighborhood

of x and δ ∈ Λ. Since x is a cluster point, we may choose γ ∈ Λ such that γ ≥ δ and xγ ∈ W .

Then for any (β, U ) ∈ M with (γ, W ) ⪯ (β, U ), xα(β,U ) = xβ ∈ U ⊆ W by the construction of g,

(β, U ) ∈ M , and the relation ⪯. Therefore, (xα(β,U ) )(β,U )∈M converges to x.

(⇐=) Suppose that (xαµ )µ∈M is a subnet that converges to x. To see that x is a cluster point of

(x, α)α∈Λ , let U be an open neighborhood of x and β ∈ Λ. Since (αµ )µ∈M is cofinal in Λ, there is

a λ1 ∈ M such that αλ1 ≥ β. Also, since (xαµ )µ∈M converges to x, there is a λ2 ∈ M such that

xαµ ∈ U for all µ ≥ λ2 . Since M is a directed set, we may choose a λ ∈ M such that λ1 ≤ λ and

λ2 ≤ λ. Since (αµ )µ∈M is increasing, β ≤ αλ1 ≤ αλ and since αλ2 ≤ αλ , xαλ ∈ U , as desired. ■

Theorem 298: The following are equivalent for any topological space X.

(i) X is compact.

(ii) Every net in X has a cluster point.

(iii) Every net in X has a convergent subnet.

Proof: ((i) =⇒ (ii)) Suppose that X is a compact topological space and (xα )α∈Λ is a net in X.

Toward a contradiction, suppose that (xα )α∈Λ has no cluster point in X. Then for every x ∈ X,

there is an open neighborhood Ux of x and αx ∈ Λ such that xα ∈ / U for all α ≥ αx . Since X is

n
Uxi . Since Λ is a directed set and {αxi }ni=1
S
compact, there are x1 , x2 , . . . , xn ∈ X such that X ⊆
i=1
n
S
is a finite set, we may choose a β ∈ Λ such that αxi ≤ β for all 1 ≤ i ≤ n. Then xβ ∈
/ Uxi = X,
i=1
a contradiction. Therefore, (xα )α∈Λ has a cluster point.

((ii) =⇒ (i)) We prove the contrapositive. Suppose that U be an open cover of X with no finite

subcover. Let F denote the collection of all finite subcollections of U ordered by subcollection

inclusion. For each V ∈ F choose an xV ∈ X \ ∪V . Then (xV )V ∈F is a net in X. To see that

(xV )V ∈F has no cluster point, let x ∈ X. Choose a U ∈ U with x ∈ U . Then for any V ∈ F

with {U } ⊆ V , xV ∈
/ U , which means that x is not a cluster point of (xV )V ∈F . Since x was chosen

arbitrarily, the net (xV )V ∈F has no cluster point in X, as desired.

((ii) ⇐⇒ (iii)) Previous lemma. ■

Q
Theorem 299: A net (xα )α∈Λ converges to x in a product space Xγ if and only if πγ [(xα )α∈Λ ]
γ∈Γ
converges to πγ (x) for all γ ∈ Γ.

Proof: (=⇒) Recall that the projection maps are continuous.

59
(⇐=) Suppose that the projection maps πγ [(xα )α∈Λ ] converge to πγ (x) for all γ ∈ Γ. To see that

(xα )α∈Λ converges to x, let U be a basic open neighborhood of x. Then there are γ1 , γ2 , . . . , γn ∈ Γ
n
πγ−1
T
and U1 is open in Xγ1 , U2 is open in Xγ2 , . . . , Un is open in Xγn such that U = i
(Ui ) and
i=1
πγi (x) ∈ Ui for every 1 ≤ i ≤ n. Recall that πγi [(xα )α∈Λ ] is a net that converges to πγi (x) for every

1 ≤ i ≤ n by the assumption. So, for each 1 ≤ i ≤ n, there is a βi ∈ Λ such that πγi (xα ) ∈ Ui for

all α ≥ βi by the definition of convergence. Since Λ is a finite directed set, we may choose a β ∈ Λ

such that βi ≤ β for all 1 ≤ i ≤ n. Then for all 1 ≤ i ≤ n and all α ≥ β, πγi (xα ) ∈ Ui , which means

that for all α ≥ β, xα ∈ U . Therefore, the net (xα )α∈Λ converges to x. ■

Q
Corollary 300: If x is a cluster point of a net (xα )α∈Λ in a product space Xγ , then the image
γ∈Γ
of the projection map πγ (x) is a cluster point of the net πγ [(xα )α∈Λ ] for all γ ∈ Γ.

Proof: By the proposition above, there is a subnet (xαµ )µ∈M that converges to x. Then by the
 
theorem above, πγ (xαµ )µ∈M converges to πγ (x) for all γ ∈ Γ. Therefore, πγ (x) is a cluster point

of πγ [(xα )α∈Λ ] in Xγ for all γ ∈ Λ. ■

Note 301: The converse of the proposition above is not true.

Example 302: Consider the sequence, and hence net, S = {(1, 0), (0, 1), (2, 0), (0, 2), (3, 0), (0, 3), . . .}

in R2 with the usual topology. Then S has no cluster point. However, 0 is a cluster point of both

π1 (S) and π2 (S).

Proposition 303: Let X be a topological space and x ∈ X. If a net in X converges to x, then

every subnet converges to x.

Proof: Exercise. ■

Note 304: The proposition above does not hold for cluster points.
 
1 1 1
Example 305: Consider the sequence, and hence net, S = 1, , 3, , 5, , . . . in R with the usual
2 4 6
topology. Then 0 is a cluster point of S but is not a cluster point of every subnet.

Theorem 306 (Tychonoff ’s Theorem): Let {Xα }α∈Λ be an indexed collection of topological
Q
spaces. Then Xα is compact if and only if Xα is compact for all α ∈ Λ.
α∈Λ
Proof: (=⇒) Recall that the projection maps are continuous and continuous maps preserve com-

pactness.
Q
(⇐=) Let Xα be a compact space for every α ∈ Λ and suppose that (fγ )γ∈Γ is a net in Xα .
α∈Λ
Q
Set P = {(∆f ) | ∆ ⊆ Λ and f ↾∆ is a cluster point of the net (fγ ↾∆ ) in Xδ }. Order P by
δ∈∆
(∆, f ) ⪯ (ε, g) if ∆ ⊆ ε and f ⊆ g. In other words, g ↾∆ = f . Since each Xα is compact, P is

not empty. It is readily verified that (P, ⪯) is a poset. Now suppose that C is a chain in P . Set
S S Q
C= ∆ and h = f . To see that h ↾C is a cluster point of (fγ ↾C )γ∈Γ , let U = Uc
(∆,f )∈C (∆,f )∈C c∈C
be a basic open set of h ↾C and β ∈ Γ. Then there are c1 , . . . , cn ∈ C such that Uc = Xc for all

60
c∈/ {c1 , . . . , cn }. Since C is a chain, there is (M, j) ∈ C such that c1 , . . . , cn ∈ M . Then there is
Q Q
a ρ ≥ β such that fρ ↾M ∈ Uc . Since c1 , . . . , cn ∈ M , fρ ↾C ∈ Uc . Therefore, (C, h) is an
c∈M c∈C
upper bound of C . So P has a maximal element (Z, k) by Zorn’s Lemma. Toward a contradiction,

suppose that Z ̸= Λ. Let α∗ ∈ Λ \ Z. Since k is a cluster point of the net (fγ ↾Z )γ∈Γ , it is the limit
∗
 the net (fγµ (α ))µ∈ξ has a cluster point y ∈ Xα∗ .
of a subnet (fγµ ↾Z )µ∈ξ . Since Xα∗ is compact,

k(α) if α ∈ Z

∗
Q
Define ℓ : Z ∪ {α } → Xα by ℓ(α) =
α∈Z∪{α∗ }
if α = α∗ .

y


Xα and (Z, k) ≺ (Z ∪ {α∗ }, l), contra-

Q
Then ℓ has a cluster point of (fγ ↾Z∪{α∗ } )γ∈Γ in
α∈Z∪{α∗ }
dicting the maximality of (Z, k). Therefore, Z = Λ, which means that k is a cluster point of the net
Q
(fγ )γ∈Γ , concluding the proof that the product space Xα is compact. ■
α∈Λ
Example 307: Suppose that X is a nonempty set and f : N → X is a sequence in X. Define

g : [1, ∞) → N by g(r) = ⌊r⌋. Then f ◦ g : [1, ∞) → X is a subnet of f . In other words, the net

(xnr )r∈[1,∞) is a subnet of the net (xn )n∈N .

Exercises

167. Let E = {f ∈ RR | f (x) = {0, 1}, and f (x) = 0 finitely often} and let g ∈ RR be identically 0.

(a) Show that g ∈ E in the product topology on RR .

(b) Find a net (fλ )λ∈Λ ⊆ E that converges to g.

168. Find a net (fλ )λ∈Λ ⊆ Ω0 that converges to ω1 ∈ Ω.

169. Let M be a metric space.

(a) Show that a map P : Ω0 → M defined by P (α) = xα is a net.

(b) Show that a net (xα )α∈Λ converges to x ∈ M if and only if xα is eventually equal to x.

170. Let (xλ )λ∈Λ be a net in a topological space and for each λ0 ∈ Λ, set Tλ0 = {xλ | λ ≥ λ0 }.

Show that y is a cluster point of (xλ )λ∈Λ if and only if y ∈ Tλ for every λ ∈ Λ.

171. Prove proposition 282.

Section 9.3: Ultranets

Proposition 308: Let (xα )α∈Λ be a net on a set X. Then the following are equivalent.

(i) For every A ⊆ X, the net is eventually in A or eventually in X \ A.

(ii) For every A ⊆ X, if the net is frequently in A, then it is eventually in A.

Proof: ((i) =⇒ (ii)) We will use (i) to prove the contrapositive of (ii). Suppose that A ⊆ X and

(xα )α∈Λ is not eventually in A. By (i), the net is eventually in X \ A. Then there is a β ∈ Λ such

that xα ∈ X \ A for all α ∈ β. So the net is not frequently in A.

61
((ii) =⇒ (i)) Note that for any A ⊆ X, the net is either eventually in A or frequently in X \ A. In

which case, it is eventually in X \ A. ■

Definition 309: A net satisfying one of the properties in the proposition above is called an ultranet.

Example 310 (Trivial Ultranet): Suppose that X is a set, x ∈ X, and Λ is a directed set. For

each α ∈ Λ, let xα = x. Then (xα )α∈Λ is an ultranet called the trivial ultranet.

Proposition 311: If (xα )α∈Λ is an ulranet on a set X and f : X → Y , then (f (xα ))α∈Λ is an

ultranet on Y .

Proof: Let B ⊆ Y . Since X \ f −1 (B) = f −1 (Y \ B) and (xα )α∈Λ is an ultranet in X, the net

is either eventually in f −1 (B) or f −1 (Y \ B). So, the net (f (xα )α∈Λ is either eventually in B or

eventually in Y \ B. ■

Proposition 312: A subnet of an ultranet is an ultranet.

Proof: Exercise. ■

Proposition 313: An ultranet converges to its cluster points.

Proof: Exercise. ■

Theorem 314: Every net has a subnet that is an ultranet.

Proof: Will be proved in the next chapter. ■

Theorem 315 (Tychonoff ’s Theorem): Let {Xα }α∈Λ be an indexed collection of topological
Q
spaces. Then Xα is compact if and only if Xα is compact for all α ∈ Λ.
α∈Λ
Proof: Exercise. ■

Exercises

172. Prove proposition 283.

173. Prove proposition 284.

174. Use ultranets to prove Tychonoff’s Theorem.

62
 Chapter 10: Filters

Section 10.1: Definitions and Examples

Definition 316: A nonempty collection F of nonempty subsets of a set X is a proper filter on

X if F is closed under finite intersection and supersets. In other words, each of the following is

satisfied.

(i) Whenever E, F ∈ F , then E ∩ F ∈ F .

(ii) Whenever E ∈ F and E ⊆ F ⊆ X, then F ∈ F .

Note 317: A filter F is improper if and only if F = P (X).

Note 318: Unless otherwise stated, the term “filter” will mean “proper filter.”

Definition 319: A nonempty collection C of nonempty subsets of a set X is a filter base (filter-

base) on X if for all C1 , C2 ∈ C , there is a C3 ∈ C such that C3 ⊆ C1 ∩ C2 .

Example 320: The collection {(a, ∞) | a ∈ R} is a filter base on R but not a filter since it is not

closed under supersets.

Theorem 321: Suppose that X is a set and C is a filter base on X. Then the collection

{F ⊆ X | there is a C ∈ C with C ⊆ F } is a filter on X.

Proof: Let C be a filter base on X and set F = {F ⊆ X | there is a C ∈ C with C ⊆ F }. By

the way F is defined, it is closed under supersets. Now suppose that F1 , F2 ∈ F . Then there are

C1 , C2 , C3 ∈ C such that C1 ⊆ F1 , C2 ⊆ F2 , and C3 ⊆ C1 ∩ C2 . Then C3 ⊆ F1 ∩ F2 , which means

that F1 ∩ F2 ∈ F . ■

Definition 322: The filter F in the theorem above is called the filter generated by C .

Definition 323: The filter on R generated by {(a, ∞) | a ∈ R} is called the Fréchet filter.

Example 324: Let X be a set and A ⊆ X. Then the set {F ⊆ X | A ⊆ F } is a filter on X

generated by the filter base {A}.

Example 325 (Neighborhood Filter): Let X be a topological space and x ∈ X. Then the set

{U ⊆ X | U is open and x ∈ U } is a filter base. The filter generated by the filter base above is

called the neighborhood filter at x.

Definition 326: A filter F is free if ∩F = ϕ and fixed if ∩F ̸= ϕ.

Note 327: The Fréchet filter is a free filter.

Note 328: The filters in previous examples are fixed filters.

Definition 329: Let F , G are filters and F ⊆ G . Then G is finer than F and F is coarser than

63
G.

Section 10.2: Filters and Topologies

Definition 330: Let F be a filter on a topological space X and x ∈ X.

(i) F converges to x if for every open neighborhood U of x, U ∈ F .

(ii) F clusters at x, or has x as a cluster point, if for every open neighborhood U of x and every

F ∈ F , U ∩ F ̸= ϕ.

Note 331: Let F be a filter on a topological space X and x ∈ X.

(i) If F converges to x, then F clusters at x.

(ii) F clusters at x if and only if x ∈ ∩{F | F ∈ F }.

Example 332: Note that the collection {A ⊆ R | Q ⊆ A} is a filter on R. Then the filter clusters

at every element of R with the usual topology since Q is dense. However, the filter has no points of

convergence since every point has a neighborhood that is not in the filter.

Theorem 333: Let F be a filter on topological space X and x ∈ X. Then F clusters at x if and

only if there is a finer filter G such that G converges to x.

Proof: (=⇒) Suppose that F clusters at x. Set C = {U ∩F | U is an open neighborhood of x and F ∈

F }. For C1 = U1 ∩F1 , C2 = U2 ∩F2 ∈ C , C1 ∩C2 = (U1 ∩F1 )∩(U2 ∩F2 ) = (U1 ∩U2 )∩(F1 ∩F2 ) ∈ C .

Hence, C is a filter base on X. Let G be the filter generated by C . Then F ⊆ G and every open

neighborhood of x is in G . Thus, G converges to x.

(⇐=) Suppose that H is a finer filter than F that converges to x. To see that F clusters at x,

let U be an open neighborhood of x and F ∈ F . Then U ∈ H and F ∈ H , which means that

U ∩ F ̸= ϕ since H is a filter. Therefore, H clusters at x, as desired. ■

Theorem 334: Let X be a topological space, A ⊆ X, and x ∈ X. Then x ∈ A if and only if there

is a filter F on X such that A ∈ F and F converges to x.

Proof: (=⇒) Suppose that x ∈ A. Set C = {U ∩ A | U is an open neighborhood of x}. Then C

is a filter base on X. Let F be the filter generated by C . Then F contains A and every open

neighborhood of x. So, F converges to x.

(⇐=) Suppose that there is a filter F on X such that A ∈ F and F converges to x. Then every

open neighborhood of x is in F and since F is a filter, every open neighborhood of x meets A Thus,

x ∈ A, as desired. ■

Note 335: Let F be a filter on a set X and f : X → Y . For each E, F ⊆ F , f (E∩F ) ⊆ F (E)∩f (F ),

which means that the collection {f (F ) | F ∈ F } is a filter base on Y .

Notation 336: In the above, f (F ) denotes the filter generated by the filter base.

64
Theorem 337: Let X and Y be topological spaces, f : X → Y , and x ∈ X. Then the following

are equivalent.

(i) f is continuous at x.

(ii) Whenever F is a filter on X converges to x, then f (F ) converges to f (x).

Proof: ((i) =⇒ (ii)) Let f be continuous at x and F be a filter on X that converges to x. To see

that f (F ) converges to f (x), let V be an open neighborhood of f (x). Since f is continuous at x,

there is an open neighborhood U of x such that f (U ) ⊆ V . Since F converges to x, U ∈ F , which

means that V ∈ f (F ).

((ii) =⇒ (i)) To see that f is continuous at x, let V be an open neighborhood of f (x). Set

B = {U ⊆ X | U is an open neighborhood of x} and let F be the filter generated by B. Then

F converges to x by the construction of B. By (ii), f (F ) converges to f (x), which means that

V ∈ f (F ), and so, f (U ) ⊆ f (F ) ⊆ V for some U ∈ B and some F ∈ F . Therefore, f is continuous

at x. ■

Theorem 338: A filter F on a product space

Q Q
Xα converges to x ∈ Xα if and only if
α∈Λ α∈Λ
πα (F ) → πα (x).

Proof: (=⇒) Recall that the projection maps are continuous.

Q
(⇐=) Let Uα be a basic open neighborhood of x. Then there are α1 , . . . , αn ∈ Λ such that
α∈Λ
/ {α1 , . . . , αn }. For each 1 ≤ i ≤ n, there is a Fi ∈ F such that παi (Fi ) ⊆
Uα = Xα for all α ∈
n
Fi ∈ F . Note that F ̸= ϕ and παi (F ) ⊆ Uαi for each 1 ≤ i ≤ n. Then
T
Uαi . Set F =
i=1
n
πα−1 Uα ∈ F , as desired.
T Q Q
F ⊆ i
(Uαi ) = Uα . Then ■
i=1 α∈Λ α∈Λ

Section 10.3: Ultrafilters

Definition 339: A filter on a set X is an ultrafilter if there is no strictly finer filter on X.

Example 340: Let X be a set and x ∈ X. Then the collection {F ⊆ X | x ∈ F } is an ultrafilter

on X that converges to x.

Proof: Let G be a filter on X and F ⊆ G . To see that F = G , let F ∈ G . Note that {x} ∈ G

since {x} ∈ F and F ⊆ G . If x ∈

/ F , then {x} ∩ F = ϕ, which is not possible since G is a filter. So

x ∈ F , which means that F ∈ F . ■

Theorem 341: Let F be a filter on a set X. Then F is an ultrafilter on X if and only if for every

A ⊆ X, A ∈ F or X \ A ∈ F .

Proof: (=⇒) Let F be an ultrafilter on X and suppose that A ⊆ X and X \ A ∈

/ F . Since F

is a filter, no subset of X \ A is an element of F . So A meets every element of F . Then the

collection {A ∩ F | F ∈ F } is a filter base on X. Let G be the filter generated by the filter base

65
{A ∩ F | F ∈ F }. Then F ⊆ G and A ∈ G . Since F is an ultrafilter and F ⊆ G , F = G , which

means that A ∈ F .

(⇐=) Suppose that F is a filter on X such that for every A ⊆ X, A ∈ F or X \ A ∈ F . To see

that F is an ultrafilter, let G be a filter on X such that F ⊆ G and G ∈ G . Then X \ G ∈

/G ⊇F

/ F , G ∈ F by the assumption. Therefore, G ⊆ F , and so, F is an ultrafilter, as

Since X \ G ∈

desired. ■

Theorem 342 (Ultrafilter Lemma): Every filter is contained in an ultrafilter.

Proof: Let F is a filter on a set X and P be the collection of all finer filters than F on X ordered

by subcollection inclusion. Then (P, ⊆) is a poset by previous arguments. Let C be a chain in P

and set G = ∪C. Clearly, H ⊆ G for every H ∈ C. It remains to show that G is a filter on X.

Let G1 , G2 ∈ G . Since C is a chain, there is a K ∈ C such that G1 , G2 ∈ K . Since K is a filter,

G1 ∩ G2 ∈ K ⊆ G . Finally, let G ∈ G and G ⊆ H ⊆ X. Then there is an E ∈ C such that G ∈ E .

Since E is a filter, H ∈ E ⊆ G . Hence, G is a filter. By Zorn’s Lemma, P has a maximal element

which is an ultrafilter containing F . ■

Example 343: Suppose that X is a set, A ⊆ X, and A ̸= ϕ. Set F = {F ⊆ X | A ⊆ F } and recall

that F is a filter with filter base {A}. Then for any x ∈ A, the collection {F ⊆ X | x ∈ F } is an

ultrafilter containing F .

Theorem 344: Let F be an ultrafilter on a set X and f : X → Y . Then f (F ) is an ultrafilter on

Y.

Proof: To see that f (F ) is an ultrafilter on Y , let A ⊆ Y . Since preimages preserve complements

and F is an ultrafilter, f −1 (A) ∈ F or f −1 (Y \ A) = X \ f −1 (A) ∈ F . Then one of f [f −1 (A)] or

f [f −1 (Y \ A)] ∈ f (F ) which means that A or Y \ A is in f (F ). Therefore, F is an ultrafilter. ■

Section 10.4: Filters and Nets

Definition 345: Let (xα )α∈Λ be a net in a set X and λ ∈ Λ. Then the set Tλ = {xα | λ ≤ α} is

the tail following λ or the λ-tail.

Note 346: Let (xα )α∈Λ is a net in a set X. Since Λ is a directed set, the set {Tλ | λ ∈ Λ} is a filter

base on X.

Definition 347: Let (xα )α∈Λ be a net in a set X. The filter generated by {Tλ | λ ∈ Λ} is the filter

on X generated by the net (xα )α∈Λ .

Definition 348: Let F be a filter on a set X. Set ΛF = {(x, F ) | x ∈ F ∈ F }. Since F is a filter,

ΛF is directed by the relation (x, F ) ≤ (y, G) if and only if G ⊆ F . The net (x)(x,F )∈ΛF is the net

in X generated by F .

66
Theorem 349:

(i) A filter on a topological space X converges to ℓ if and only if the net generated by the filter

converges to ℓ.

(ii) A net in a topological space X converges to ℓ if and only if the filter generated by the net

converges to ℓ.

Proof:

(i) (=⇒) Suppose that F is a filter on X that converges to ℓ. To see that the net (x)(x,F )∈ΛF

converges to ℓ, let U be an open neighborhood of ℓ. Then U ∈ F . Now suppose that (y, G) ∈ ΛF

such that ℓU ≤ (y, G). Then y ∈ G ⊆ U . So the net (x)(x,F )∈ΛF converges to ℓ.

(⇐=) Suppose that the net (x)(x,F )∈ΛF converges to ℓ. To see that F converges to ℓ, let U be an

open neighborhood of ℓ. Since the net (x)(x,F )∈ΛF converges to ℓ, there is a (z, H) ∈ ΛF such that

y ∈ U whenever (y, F ) ∈ ΛF and (z, H) ≤ (y, F ).

Claim: H ⊆ U .

Proof (Claim): Toward a contradiction, suppose that H ⊈ U . Let w ∈ H \ U . Then (z, H) ≤

(w, H) and w ∈
/ U , a contradiction. ■Claim

Since H ∈ F , H ⊆ U , and F is a filter, U ∈ F as desired.

(ii) (=⇒) Suppose that (xα )α∈Λ is a net in X and F is the filter generated by (xα )α∈Λ . First

suppose that the net converges to ℓ. To see that F converges to ℓ, let U be an open neighborhood of

ℓ. Since the net converges to ℓ, there is a β ∈ Λ such that the β-tail Tβ ⊆ U . So U ∈ F as desired.

(⇐=) Suppose that F converges to ℓ. To see that the net converges to ℓ, let U be an open

neighborhood of ℓ. Then U ∈ F , which means that Tβ ⊆ U for some β ∈ Λ. Therefore, the net

converges to ℓ. ■

Theorem 350:

(i) A filter is an ultrafilter if and only if the net generated by the filter is an ultranet.

(ii) A net is an ultranet if and only if the filter generated by the net is an ultrafilter.

Proof:

(i) Let F be a filter on a set X.

(=⇒) Suppose that F is an ultrafilter. To see that (x)(x,F )∈ΛF is an ultranet, let A ⊆ X. Since F is

an ultrafilter, either A ∈ F or X \ A ∈ F . Without loss of generality, suppose that A ∈ F . Choose

an a ∈ A. Then for any (x, F ) ∈ ΛF with (a, A) ≤ (x, F ), x ∈ F ⊆ A. So the net (x)(x,F )∈ΛF is

eventually in A. Therefore, the net generated by F is an ultranet.

(⇐=) Suppose that the net (x)(x,F )∈ΛF is an ultranet. To see that F is an ultrafilter, suppose that

A ⊆ X. Since (x)(x,F )∈ΛF is an ultranet, the net is either eventually in A or eventually in X \ A.

67
Without loss of generality, suppose that the net is eventually in A. Then there is a (y, G) ∈ ΛF such

that x ∈ A whenever (x, F ) ∈ ΛF and (y, G) ≤ (x, F ). Note that for every x ∈ G, (y, G) ≤ (x, G).

Then G ⊆ A. Since G ∈ F , G ⊆ A, and F is a filter, A ⊆ F .

(ii) Let (xα )α∈Λ be a net in a set X and F be the filter generated by (xα )α∈Λ .

(=⇒) Suppose that (xα )α∈Λ is an ultranet. To see that F is an ultrafilter, let A ⊆ X. Then

(xα )α∈Λ is eventually in A or eventually in X \ A. Without loss of generality, suppose that (xα )α∈Λ

is eventually in A. Then there is a β ∈ Λ such that Tβ ⊆ A, which means that A ∈ F , as desired.

(⇐=) Suppose that F is an ultrafilter. To see that the net (xα )α∈Λ is an ultranet, let A ⊆ X. Then

since F is an ultrafilter, A ∈ F or X \ A ∈ F . Without loss of generality, suppose that A ∈ F .

Since A ∈ F , there is a β ∈ Λ such that Tβ ⊆ A by the way F is defined. This means that (xα )α∈Λ

is eventually in A, as desired. ■

Theorem 351: Every net has a subnet that is an ultranet.

Proof: Let (xα )α∈Λ be a net in a set X and F be an ultrafilter that contains the filter generated

by the net (xα )α∈Λ . Set M = {(α, F ) ∈ Λ × F | xα ∈ F } and put an ordering ≤M on M defined by

(α, F ) ≤M (β, G) if α ≤ β and G ⊆ F .

Claim 1: ≤M is a direction on M .

Proof (Claim 1): That ≤M is reflexive and transitive follows immediately from the fact that Λ is a

directed set and superset inclusion is a partial order relation. Now suppose that (α, F ), (β, G) ∈ M .

Since Λ is a directed set, there is a γ ∈ Λ such that α ≤ γ and β ≤ γ. By the definition of F ,

Tγ ∩ F ∩ G ̸= ϕ. Choose a δ ∈ Λ such that λ ≤ δ and xδ ∈ F ∩ G. Then (α, F ) ≤M (δ, F ∩ G) and

(β, G) ≤M (δ, F ∩ G). ■Claim 1

Claim 2: Every element of F meets every tail of (xα )α∈Λ .

Proof (Claim 2): If β ∈ Λ, then Tβ ∩ F ̸= ϕ for F ∈ F . This means there is a γ ∈ Λ with β ≤ γ

and xγ ∈ F . Then f (γ, F ) = γ ≥ β. ■Claim 2

By claim 2, f is increasing and cofinal. Define g : M → ΛF by g[(α, F )] = (xα , F ).

Claim 3: Every tail of (xα )α∈Λ meets every element of F .

Proof (Claim 3): Suppose that (x, F ) ∈ ΛF . Then there is a β ∈ Λ such that Tβ ∩ F ̸= ϕ. Choose

γ ≥ β such that xγ ∈ F . Then g[(γ, F )] = (xγ , F ) ≥ (x, F ) since F ⊆ F . ■Claim 3

By claim 3, g is increasing and cofinal. So the net (xα )(α,F )∈M is a subnet of both the net (xα )α∈Λ

and the net (x)(x,F )∈ΛF . Since F is an ultrafilter, the net (x)(x,F )∈ΛF is an ultranet. By a previous

proposition, a subset of an ultranet is an ultranet. So, the net (xα )(α,F )∈M is an ultranet which is

a subnet of (xα )α∈Λ . ■

68
 Chapter 11: Quotient Spaces

Section 11.1: The Quotient Topology

Proposition 352: Let X be a topological space, Y be a set, and f : X → Y be onto. Then the set

{U ⊆ Y | f −1 (U ) is open in X} is a topology on Y .

Proof: Set τY = {U ⊆ Y | f −1 (U ) is open in X}. Since f −1 (Y ) = X and f −1 (ϕ) = ϕ, Y, ϕ ∈ τY .

Now suppose that U ⊆ τY . Then for each U ∈ U , f −1 (U ) is open in X. So, f −1 (∪U ) =

f −1 ( f (U ), which is open in X. So, ∪U ∈ τY . Finally, suppose that V ⊆ τY is

S S −1
U) =
U ∈U U ∈U
finite. For each V ∈ V , note that f −1 (V ) is open in X. So, f −1 (∩V ) = f −1 (
T T −1
V)= f (V ),
V ∈V V ∈V
which is open in X. Therefore, ∩V ∈ τY . ■

Definition 353: The topology defined in the above proposition is called the quotient topology

on Y induced by f and f is called a quotient map.

Note 354: Since preimages of functions preserve complements, a set is closed in the quotient

topology if and only if its preimage is closed in the domain space.

Theorem 355: Let X and Y be topological spaces and suppose that f : X → Y is a continuous

and open surjection. Then the topology on Y is the quotient topology on Y induced by f .

Proof: Let τY be the topology on Y and τq be the quotient topology induced by f . First, suppose

that U ∈ τY . Since f is continuous, note that f −1 (U ) is open in X, which means that U ∈ τq by

the definition of τq . Now suppose that U ∈ τq . Then by the definition of τq , note that f −1 (U ) is

open in X. Since f is onto and open, f [f −1 (U )] = U ∈ τY . ■

Theorem 356: Let X and Y be topological spaces and suppose that f : X → Y is continuous and

closed. Then the topology on Y is the quotient topology on Y induced by f .

Proof: Mimic the proof above using the note above. ■

Definition 357: Let X be a set and D ⊆ P (X). Then D is a decomposition (partition) of X

if it is pairwise disjoint and ∪D = X.

Theorem 358: Let X be a topological space and D be a decomposition of X. Let τ = {U ⊆ D |

∪U is open in X}. Then τ is a topology on D.

Proof: First, since ∪D = X and ∪ϕ = ϕ, note that ϕ, D ∈ τ . Now suppose that γ ⊆ τ . Then

U)= (∪U ), which is open in X. Finally, suppose that U1 , . . . , Un ∈ τ . Then

S S
∪(∪γ) = ∪(
U ∈γ U ∈γ
n n
Ui ) = (∪Ui ), which is open in X. Therefore, τ is a topology on D.
T T
∪( ■
i=1 i=1
Definition 359: The pair (D, τ ) in the above theorem is called a decomposition space of X.

69
Theorem 360: Let X be a space and D be a decomposition of X. Define f : X → D by f (x) = A

if and only if x ∈ A. Then the decomposition topology is the quotient topology induced by f .

Proof: Let τd be the decomposition topology on D and τq be the quotient on D induced by f .

Then for any U ⊆ D, f −1 (U ) = {x ∈ X | f (x) ∈ U } = {x ∈ X | f (x) = U for some U ∈ U } =

{x ∈ X | x ∈ U for some U ∈ U } = ∪U . So, U ∈ τd if and only if U ∈ τq . ■

Definition 361: Suppose that X is a topological space and ∼ is an equivalence relation on X. The

identification space X / ∼ is the decomposition space of the equivalence classes of ∼.

Section 11.2: Examples

Example 362: Let S be the unit circle in R2 .

(a) Define f : [0, 2π] → S by f (x) = (cos(x), sin(x)). Note that f is continuous and onto. To see

that f is closed, let E ⊆ [0, 2π] be closed. Since [0, 2π] is compact, E is compact. Also, since f

is continuous, f (E) is compact. Furthermore, since S is Hausdorff, f (E) is closed. So, the usual

topology on S is the quotient topology induced by f . ■

Note that the map f is not open.

(b) Define an equivalence relation ∼ on [0, 2π] by x ∼ y if x = y and 0 ∼ 2π. Then the identification

space [0, 2π] / ∼ is formed by identifying 0 and 2π, and is homeomorphic to S. ■

(c) The above implies that S is a quotient space of [0, 2π] induced by f (t) = (cos t, sin t). The

corresponding decomposition space is D = {{t} | t ∈ (0, 2π)} ∪ {{0, 2π}}. Define an equivalence

relation ∼ on [0, 2π] by t ∼ t and 0 ∼ 2π. So the unit circle is created from the interval [0, 2π] by

gluing 0 to 2π.

Example 363: Set X = [0, 2π] × [0, 2π] and let Y be the quotient space formed by identifying each

point (0, y) with (2π, y) for each y ∈ [0, 2π]. Then the space Y is the cylinder S × [0, 2π], where S

is the unit circle. The quotient map f : X → Y is f (x, y) = (cos(x), sin(x), y).

Example 364: Set X = [0, 2π] × [0, 2π] and let Y be the quotient space formed by identifying each

point (0, y) with (2π, y) for each y ∈ [0, 2π] and identifying each point (x, 0) with (x, 2π). Then the

space Y is the torus S × S, where S is the unit circle. The first identification creates a cylinder. The

second identification joins the ends of the cylinder to create the torus. The quotient map f : X → Y

if f (x, y) = ((cos(x), sin(x)), (cos(y), sin(y))).

Example 365: Set X = [0, 2π] × [0, 2π] and let Y be the quotient space formed by identifying each

point (x, 0) with (2π − x, 2π) for each x ∈ [0, 2π]. Then Y is the Möbius strip.

Example 366: Set X = [0, 2π] × [0, 2π] and let Y be the quotient space formed by identifying

each point (0, y) with (2π, y) for each y ∈ [0, 2π] and each point (x, 0) with (2π − x, 2π) for each

70
x ∈ [0, 2π]. Then Y is the Klein bottle (see the cover of Willard).

Example 367: Let Γ be the Moore-Niemytzki Plane and X be the identification space formed by

identifying all points of Q × {0} with (0, 0). Since Γ is regular, X is Hausdorff. However, X is not

regular since [(0, 0)] cannot be separated from {[x, 0] | x ∈ R \ Q}. ■

Definition 368: Let X be a topological space.

(i) Consider X × [0, 1]. The cone over X, denoted ΛX, is constructed by identifying all points

(x, 1) ∈ X × [0, 1] as a single point.

(ii) Consider X × [−1, 1]. The suspension over X, denoted ΣX, is constructed by identifying all

points (x, 1) ∈ X × [−1, 1] and all points (x, −1) ∈ X × [−1, 1] as single points.

Note 369: If X is a topological space, D is a decomposition of X, and A ⊆ X, then D induces a

decomposition DA on A.

Example 370: Let Γ be the Moore-Niemytzki Plane and X be the identification space formed by

identifying all points of Q × {0} with (0, 0). Since Γ is regular, X is Hausdorff. However, X is not

regular since [(0, 0)] cannot be separated from {[x, 0] | x ∈ R \ Q}. ■

Chapter 12: Continua

Section 12.1: Definitions and Examples

Definition 371: A compact connected Hausdorff space is called a continuum.

Proposition 372: Any product of continua is a continuum.

Proof: This follows from the fact that a product of connected spaces is connected, a product of

Hausdorff spaces is Hausdorff, and a product of compact spaces is compact. ■

Definition 373: A collection {Kα | α ∈ Λ} of subsets of a topological space is directed by inclusion

if whenever α, β ∈ Λ, then there is a γ ∈ Λ such that Kγ ⊆ Kα ∩ Kβ .

Theorem 374: If K = {Kα | α ∈ Λ} is a nonempty directed collection of subsets of a topological

T
space, then Kα is a continuum.
α∈Λ
T
Proof: Choose an α0 ∈ Λ. Since Kα0 is a continuum and Kα ⊆ Kα0 , we may suppose without
α∈Λ
loss of generality that X = Kα0 . So suppose that X is a compact Hausdorff space. Then each Kα
T
is closed in X, which means that Kα is closed in X, and thus, compact. So it suffices to prove
α∈Λ
T
that Kα is connected. Attempting a contradiction, suppose that {H, K} is a disconnection of
α∈Λ

71
T T
Kα . Since Kα is closed in X, H, K ⊆ X are closed and disjoint. Since X is compact and
α∈Λ α∈Λ
Hausdorff, it is normal, which means that there are disjoint open sets U, V ⊆ X such that H ⊆ U

and K ⊆ V . Note that for each α ∈ Λ, Kα \ (U ∪ V ) ̸= ϕ. Set F = {Kα \ (U ∪ V ) | α ∈ Λ}. Note

that F is nonempty. Since K = {Kα | α ∈ Λ} is directed, F has the finite intersection property.

Since X is compact, ∩F ̸= ϕ. Hence,

   
T T
∩F = ∩{Kα \ (U ∪ V ) | α ∈ Λ} = Kα \ (U ∪ V ) ⊆ Kα \ (H ∪ K) = ϕ.
α∈Λ α∈Λ
T
This is a contradiction since ∩F ̸= ϕ. Therefore, Kα is connected. ■
α∈Λ
Corollary 375: The intersection of any decreasing sequence of continua is a continuum.

Section 12.2: Subcontinua and Irreducible Continua

Definition 376: A subcontinuum of a continuum K is a subset of K which is a continuum in the

subspace topology.

Definition 377: A continuum K contained in a topological space X is called irreducible about

a subset A ⊆ X if A ⊆ K and no proper subcontinuum of K contains A.

Note 378: In the above, if A = {a, b}, then K is irreducible between a and b.

Theorem 379: If K is a continuum and A ⊆ K is nonempty, then there is a subcontinuum C ⊆ K

that is irreducible about A.

Proof: Suppose that K is a continuum and that A ⊆ K is nonempty. Let K denote the set of

subcontinua of K which contain A. Put an ordering ≤ on K as follows: K1 ≤ K2 if and only if

K2 ⊆ K1 . By the theorem above, the intersection of any linearly ordered subcollection of K is a

member of K . This means that every chain in K has an upper bound in K . Hence, by Zorn’s

Lemma, there is a maximal element C of K . Since C is maximal, it is irreducible about A. ■

Section 12.3: Cut and Noncut Points of Continua

Definition 380: Let X be a connected T1 -space.

(i) A cut point of X is a point p ∈ X such that X \ {p}.

(ii) If p ∈ X is not a cut point, then p is called a noncut point.

(iii) A cutting of X is a triple {p, U, V }, where p ∈ X is a cut point and {U, V } is a disconnection

of X \ {p}.

Example 381: In the interval [0, 1], the points 0 and 1 are noncut points and all other points are

cut points.

Theorem 382: If K is a continuum and {p, U, V } is a cutting of K, then U ∪ {p} and V ∪ {p} are

continua.

Proof: Suppose that K is a continuum and {p, U, V } is a cutting of K. Define f : K → U ∪ {p} by

72
 

x if x ∈ U ∪ {p};

f (x) =

p
 if x ∈ V.
Note that f ↾U ∪{p} is the identity function and f ↾V ∪{p} is a constant function, which means that

both functions are continuous. Also, note that the sets U ∪ {p} and V ∪ {p} are closed sets and

K = (U ∪ {p}) ∪ (V ∪ {p}), which means that f is continuous. Since U ∪ {p} is the continuous image

of a continuum, it is compact and connected, which means that U ∪ {p} is a continuum. To see that

V ∪ {p} is a continuum, apply a similar argument to the one above. ■

Theorem 383: If K is a continuum and {p, U, V } is a cutting of K, then U and V contain a noncut

point.

Proof: Toward a contradiction, suppose that each x ∈ U is a cut point. In other words, suppose

that {x, Ux , Vx } is a cutting of K for every x ∈ U . If Ux ∩ (V ∪ {p}) ̸= ϕ and Vx ∩ (V ∪ {p}) ̸= ϕ,

then Ux and Vx disconnect V ∪ {p} since Ux ∪ Vx = K \ {x} and x ∈ U . Hence, one of Ux and Vx is

contained in U = K \ (V ∪ {p}). Suppose that Ux ⊆ U for all x ∈ U . Since Ux ∪ {x} ⊆ U , V ⊆ Vx

for each x ∈ U .

Claim 1: If x ∈ U and y ∈ Ux , then Uy ∪ {y} ⊆ Ux .

Proof (Claim 1): Note that Uy and Vy are mutually separated in K and Vx ∪{x} ⊆ Uy ∪Vy , which

means that Vx ∪ {x} ⊆ Uy or Vx ∪ {x} ⊆ Vy . However, it is not the case that Vx ∪ {x} ⊆ Uy since

Uy ∩Vy = ϕ and V ⊆ Vx ∩Vy . Hence, Vx ∪{x} ⊆ Vy , and so, Ux = K \(Vx ∪{x}) ⊇ K \Vy = Uy ∪{y}.

■Claim 1

Let U = {Ux ∪ {x} | x ∈ U } and P = {C | C ⊆ U and C is linearly ordered by subset inclusion}.

Note that P ̸= ϕ since for each x ∈ U , {Ux ∪ {x}} ∈ P. Suppose K = {Cα | α ∈ Λ} ⊆ C is linearly

ordered.
S
Claim 2: The set Cα ∈ P.
α∈Λ
Cα . Since Cα ⊆ U for each α ∈ Λ, Cα ⊆ U . So suppose that
S S
Proof (Claim 2): Consider
α∈Λ α∈Λ
S S
Ux1 ∪ {x1 } ∈ Cα and Ux2 ∪ {x2 } ∈ Cα . Choose α1 , α2 ∈ Λ such that Ux1 ∪ {x1 } ∈ Cα1 and
α∈Λ α∈Λ
Ux2 ∪ {x2 } ∈ Cα2 . Since K is linearly ordered, either Cα1 ⊆ Cα2 or Cα2 ⊆ Cα1 , and both Cα1 and

Cα2 are linearly ordered. Thus, either (Ux1 ∪ {x1 }) ⊆ (Ux2 ∪ {x2 }) or (Ux2 ∪ {x2 }) ⊆ (Ux1 ∪ {x1 }).
S S
Hence, Cα is linearly ordered, and so, Cα ∈ P. ■Claim 2
α∈Λ α∈Λ
By Zorn’s Lemma, P has a maximal element C = {Ux ∪ {x} | x ∈ U ∗ } for some U ∗ ⊆ U . By a

previous theorem, ∩C is a nonempty continuum and ∩C ⊆ U . Let q ∈ ∩C . Since Uq ̸= ϕ, let

r ∈ Uq and Ur ∪ {r} ⊆ Uq and (Uq ∪ {q}) ⊆ (Ux ∪ {x}) for every x ∈ U ∗ . Hence, C ∪ {Ur ∪ {r}}

is linearly ordered, and so, by the maximality of C , r ∈ U ∗ . This means that q ∈ Ur ∪ {r} ⊆ Uq , a

73
contradiction. Hence, U does not consist entirely of cut points. ■

Corollary 384: Every continuum with more than one point has at least two noncut points.

Proof: Let K be a continuum with more than one point. If K has no cut points, then each of the

points are noncut points and there are more than one of them. If K has a cut point, then by the

previous theorem, it has at least two noncut points. ■

Section 12.4: Metrizable Continua

Question 385: Are there other metrizable continua with exactly two noncut points besides closed

bounded intervals?

Theorem 386: A metrizable continuum with exactly two noncut points is homeomorphic to the

interval [0, 1].

Proof: Let K be a metrizable continuum with exactly two noncut points a and b and suppose that

D ⊆ K is a countable and dense subset which does not contain a and b. Note that D has no smallest

or largest element and if p, q ∈ D with p < q, there is a r ∈ D such that p < r < q. Also, D is order

isomorphic, and hence homeomorphic, to the dyadic rationals P in (0, 1). Let f : D → P be an order

isomorphism. Note that if p ∈ K \ {a, b}, then p is a cut point, which means K = Ap ∪ Bp , where if

x ∈ Ap and y ∈ Bp , then x < y. It follows that f (Ap ∩ D) and f (Bp ∩ D) form a Dedekind cut of

the dyadic rationals, and thus, uniquely determine an element F (p) of (0, 1). Then define F (a) = 0

and F (b) = 1, and so, f is extended to an order isomorphism, and hence, a homeomorphism from

K onto [0, 1]. ■

Theorem 387: If K is a metrizable continuum such that K \{x, y} is disconnected for any x, y ∈ K,

then K is homeomorphic to the unit circle.

Proof: Let K be a metrizable continuum such that K \ {x, y} is disconnected for any x, y ∈ K.

Claim 1: K has no cut points.

Proof (Claim 1): If {p, U0 , V0 } is a cutting of K, then since U0 ∪ {p} and V0 ∪ {p} are continuua,

there are noncut points y ∈ U0 ∪ {p} and z ∈ V0 ∪ {p}. This implies that [(U0 ∪ {0}) \ {y}] ∩ [(V0 ∪

{p}) \ {z}] ̸= ϕ and [(U0 ∪ {0}) \ {y}] ∪ [(V0 ∪ {p}) \ {z}] = K \ {y, z} is connected, a contradiction.

Hence, K has no cut points. ■Claim 1

Let a, b ∈ K with a ̸= b. Then K \ {a, b} = U ∪ V , where U, V ⊆ K are disjoint nonempty open

subsets. Set U ∗ = U ∪ {a, b} and V ∗ = V ∪ {a, b}.

Claim 2: The sets U ∗ and V ∗ are connected.

Proof (Claim 2): Suppose that U ∗ = S ∪ T , where S, T ⊆ U ∗ are disjoint, nonempty, and open

subsets. If a, b ∈ S, then T is open in U , and hence, open in K, which is a contradiction since T

74
is closed in U ∗ . Hence, one of a and b is in S and the other is in T . Without loss of generality,

suppose that a ∈ S and b ∈ T . By an argument similar to the one above, S \ {a} is a clopen subset

of K \ {a}, which is a contradiction. Therefore, U ∗ and V ∗ are connected. ■Claim 2

Claim 3: The points a and b are noncut points of U ∗ .

Proof (Claim 3): If S and T disconnect U ∗ \ {a} and b ∈ S, then by arguments similar to the one

above, T is a clopen subset of K \ {a}, which is a contradiction. ■Claim 3

A similar argument to the one in claim 3 implies that both a and b are noncut points of V ∗ .

Claim 4: Each of U ∗ and V ∗ have exactly two noncut points.

Proof (Claim 4): Toward a contradiction, suppose not. Then either both U ∗ and V ∗ have a third

noncut point or one of U ∗ and V ∗ has a third noncut point and the other does not.

Case 1: Both U ∗ and V ∗ have a third noncut point.

Suppose that p ∈ U ∗ and q ∈ V ∗ are noncut points, different from both a and b. Then the sets

U ∗ \ {p} and V ∗ \ {q} are connected, are not disjoint, and their union is K \ {p, q}, which is a

disconnected set, a contradiction.

Case 2: One of U ∗ and V ∗ has a third noncut point and the other does not.

Without loss of generality, suppose that U ∗ has a third noncut point p. Then if q ∈ V , there is a

cutting {q, A, B} of V ∗ , where A and B are connected, a ∈ A, and b ∈ B. This means that U ∗ \ {p},

A, and B form a chain of connected sets whose union is K \ {x, y}, a contradiction.

Therefore, both U ∗ and V ∗ have exactly two noncut points. ■Claim 4

Thus, both U ∗ and V ∗ are metric continua with two noncut points, a and b, and U ∗ ∩ V ∗ = {a, b}.

Hence, K = U ∗ ∪ V ∗ is homeomorphic to the unit circle. ■

Exercises

175. Verify that the spaces below are continua.

(a) The interval [0, 1];

(b) The unit circle S1 ;

(c) The torus S1 × S1 .

176. Suppose that f : X → Y is a homeomorphism of X onto Y . Prove that for each x ∈ X, the

restriction f ↾X\{x} : (X \ {x}) → (Y \ {f (x)}) is a homeomorphism.

177. Prove that the sets R and [0, ∞) are not homeomorphic.

178. For n ∈ N \ {1}, prove that the sets R and Rn are not homeomorphic.

179. Prove that any locally connected compact Hausdorff space is the union of finitely many

continua.

75
Definition: Let X and Y be ordered spaces. A function f : X → Y is an order isomorphism if

f is one-to-one and x < y if and only if f (x) < f (y).

180.

(a) Prove that every order isomorphism is a homeomorphism reltive to the order topologies on X

and Y .

(b) Let P denote the set of dyadic rationals in (0, 1). Prove that P has no largest or smallest

elements. Then show that if p, q ∈ P with p < q, then p < r < q for some r ∈ P .

(c) Prove that any countable linearly ordered set D with the properties in (b) is order isomorphic

to P .

181.

(a) Let K be a continuum in a topological space X and let U ⊆ X be an open subset that meets

both K and X \ K. Prove that every component of U ∩ K meets Fr(U ).

(b) Prove that no continuum can be written as a countable union of disjoint closed sets.

76
 Chapter 13: Mysior’s Space

Example 388 (A. Mysior Proc. AMS. Apr. 1981): Set X = [0, ∞) × [0, 2] ∪ {x∗ }, where

x∗ ∈
/ R × [0, 2]. For each x ∈ [0, ∞) × {0}, let x′ ∈ R such that x ∈ (x′ , 0). Also, for each

x ∈ [0, ∞)×{0}, let Vx = {(x′ , t) ∈ [0, ∞)×[0, 2] | 0 ≤ t ≤ 1}∪{(s, t) ∈ [0, ∞)×[0, 2] | x′ ≤ s ≤ x′ +1

and t = s − x′ }. For each x ∈ [0, ∞) × (0, 2], define Bx = {{x}}, and for each x ∈ [0, ∞) × {0}, define

Bx = {Vx \ F | F ⊆ [0, ∞) × (0, 2] and |F | < ℵ0 }. Define Bx∗ = {{x∗ } ∪ ([a, ∞) × [0, 2] | a ∈ [0, ∞)}

and let B = Bx .
S
x∈X
Claim 1: The collection B is a base for a topology on X.

Proof (Claim 1): Since {x∗ } ∪ ({[0, ∞) × [0, 2]) ∈ B, ∪B = X. Now suppose that B1 , B2 ∈ B

and B1 ∩ B2 ̸= ϕ. Then B1 ∩ B2 is either a singleton in [0, ∞) × (0, 2], an element of Bx for some

x ∈ [0, ∞) × {0}, or an element of Bx∗ . ■Claim 1

Let τ be the topology on X generated by B.

Claim 2: The space X is T1 .

Proof (Claim 2): Suppose x, y ∈ X with x ̸= y. If x ∈ [0, ∞) × (0, 2], let U = {x}. If

x ∈ [0, ∞) × {0}, let U = Vx \ {y}. Finally, if x = x∗ , let U = {x∗ } ∪ ([a, ∞) × [0, 1]), where y ∈ (s, t)

and a > s. In any case, x ∈ U and y ∈

/ U , and so, X is T1 . ■Claim 2

Claim 3: If x ̸= x∗ , then the elements of Bx are closed sets.

Proof (Claim 3): If x ∈ [0, ∞) × (0, 1], then Bx = {{x}} and {x} is closed since X is T1 . Now

suppose x ∈ [0, ∞) × {0} and let Bx ∈ Bx . To see that Bx is closed, suppose y ∈

/ B. We will

show there is a By ∈ By such that Bx ∩ By = ϕ. If y ∈ [0, ∞) × (0, 2], then let By = {y}. If

y ∈ [0, ∞) × {0}, note that there is a z ∈ [0, ∞) × {0} such that Vx ∩ Vy ⊆ {z}. Let B = Vy \ {z}.

Then ϕ = By ∩ Vx ⊇ By ∩ Bx . Finally, if y = x∗ , then let By = {x} ∪ [x′ + 3, ∞) × [0, 2]. Then

ϕ = By ∩ Vx ⊇ By ∩ Bx . ■Claim 3

Claim 4: The space X is T3 .

Proof (Claim 4): Since X is T1 , we need only show that X is regular. Suppose x ∈ X and

U ⊆ X is open. If x ̸= x∗ , then by claim 3, there is a B ∈ Bx such that x ∈ B = B ⊆ U .

So suppose that x = x∗ . Then there is an a ≥ 0 such that {x∗ } ∪ ([a, ∞) × [0, 2]) ⊆ U . Now,

x∗ ∈ {x∗ } ∪ ([a + 3, ∞) × [0, 2]) ⊆ {x∗ } ∪ ([a + 3, ∞) × [0, 2]) ⊆ {x∗ } ∪ ([a, ∞) × [0, 2]). ■Claim 4

Claim 5: Let n ∈ N. Also, let E ⊆ (n − 1, n) × {0} and F ⊆ (n, n + 1) × {0} such that E and F

are infinite and one of E and F is uncountable. Then E and F cannot be separated by open sets.

Proof (Claim 5): Without loss of generality, suppose that F is uncountable. Let A ⊆ E be

77
countable and infinite and suppose that U and V are open sets with E ⊆ U and F ⊆ V . For each
S
a ∈ A, there is a finite Fa ⊆ [0, ∞) × (0, 2] such that Va \ Fa ⊆ U . Then Fa is countable. Since
a∈A
S
F is uncountable, there is an x ∈ F such that ({x} × [0, 2]) ∩ [ Fa ] = ϕ. Also, since A is infinite,
a∈A
S
(Vx \ F ) ∩ [ Va \ Fa ] ̸= ϕ for any finite F ⊆ [0, ∞) × (0, 2]. Thus, U ∩ V ̸= ϕ. ■Claim 5
a∈A
Claim 6: Suppose f : X → [0, 1] is continuous and f (x) = 0 for every x ∈ [0, 1] × {0}. Then the

set {x ∈ [n − 1, n] × {0} | f (x) ̸= 0} is countable for all n ∈ N.

Proof (Claim 6): We will proceed by induction on n. For each n ∈ N, let P (n) be the statement

“The set {x ∈ [n − 1, n] × {0} | f (x) ̸= 0} is countable.” If n = 1, then the set {x ∈ [n − 1, n] × {0} |

f (x) ̸= 0} = ϕ, which is countable. Hence, P (1) is true. So suppose that P (k) is true for some

k ∈ N. Toward a contradiction, suppose that the set {x ∈ [k, k + 1] | f (x) ̸= 0} is uncountable. Then

there is an r > 0 such that the set {x ∈ [k, k + 1] × {0} | f (x) > r} is uncountable. Then f −1 ([0, r))

contains all but countably many elements of [k − 1, k] × {0}, f −1 ((r, 1]) contains uncountably many

elements of [k, k + 1] × {0} and f −1 ([0, r)) and f ((r, 1]) are disjoint open sets. This contradicts claim

5. Therefore, the set P (k + 1) is true, and the claim is proved by the Principle of Mathematical

Induction. ■Claim 6

Claim 7: Suppose f : X → [0, 1] is continuous and f (x) = 0 for every x ∈ [0, 1] × {0}. Then

f (x∗ ) = 0.

Proof (Claim 7): Toward a contradiction, suppose that f (x∗ ) > 0. Set r = 1 ∗
2 f (x ). Then

f −1 ((r, 1]) is an open set containing x∗ and disjoint from f −1 (0), contradicting claim 6. ■Claim 7

By claim 7, the space X is not Tychonoff, and hence, is not completely regular. ■

78
 Chapter 14: Topological Groups

Definition 389: Suppose that G is a topological space and a group such that the inverse and

multiplicative operators are continuous. Then G is a topological group.

Note 390: If U is open in a topological group G, then the set U −1 = {u−1 | u ∈ U } is open in G.

Notation 391: If G is a topological group and A, B ⊆ G, then AB = {ab | a ∈ A, b ∈ B}.

Example 392: The pair ⟨R+ , ·⟩ is a topological group.

1
Proof: To see that the inverse operator is continuous, recall that the function f (x) = is continuous
x
on R+ . To see that the multiplication operator is continuous, let I ⊆ R+ be an open interval. Let

(x, y) be in the preimage of I under the multiplication operator. Without loss of generality, suppose
 r 
ε ε ε
that I = (xy − ε, xy + ε) for some ε > 0. Let δ > 0 such that δ < min , , . Suppose
|3x| |3y| 3
(r, s) ∈ (x − δ, x + δ) × (y − δ, y + δ). Then,

|rs − xy| = |(r − x + x)(s − y + y) − xy|

= |(r − x)(s − y) + (r − x)y + (s − y)x + xy − xy|

≤ |(r − x)(s − y)| + |(r − x)y| + |(s − y)x|

= |r − x| · |s − y| + |r − x| · |y| + |s − y| · |x|

< δ 2 + δ|y| + δ|x|

ε ε ε
< + + = ε.
3 3 3
So, the multiplication operator is continuous, and thus, the pair ⟨R+ , ·⟩ is a topological group. ■

Theorem 393: For any topological space G that is also a group, the following are equivalent.

(i) G is a topological group;

(ii) For any a, b ∈ G and any open U ⊆ G with ab−1 ∈ U , there are open sets V, W ⊆ G such that

a ∈ V , b ∈ W , and V W −1 ⊆ U .

Proof: ((i) =⇒ (ii)) Suppose that U ⊆ G is open, a, b ∈ G, and ab−1 ∈ U . Since multiplication

is continuous, there are open V, W −1 ⊆ G such that the pair (a, b−1 ) ∈ V × W −1 and V W −1 ⊆ U .

Since b−1 ∈ W −1 , b ∈ W .

((ii) =⇒ (i)) To see that the inverse operator is continuous, let U ⊆ G be open. We must show that

U −1 is open in G. Let a ∈ U −1 . Then ea ∈ U −1 , where e is the identity element. So, by (ii), there

are open V, W ⊆ G such that e ∈ V , a ∈ W , and V W −1 ⊆ U . Then (V W −1 )−1 = W V −1 ⊆ U −1 .

Since e ∈ V −1 , W ⊆ W V −1 ⊆ U −1 . So, a ∈ W ⊆ U −1 . Thus, the inverse operator is continuous.

To see that the multiplication operator is continuous, let U ⊆ G be open. Let U ∗ ⊆ G × G

be the preimage of U under multiplication. To see that U ∗ is open, let (a, b−1 ) ∈ U ∗ . Then

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ab−1 ∈ U . So, by (ii), there are open V, W ⊆ G such that a ∈ V , b ∈ W , and V W −1 ⊆ U .

Then (a, b−1 ) ∈ V × W −1 ⊆ U ∗ . So, U ∗ is open, which means that the multiplication operator is

continuous. Therefore, G is a topological group. ■

Theorem 394: If G is a topological group, then for every a ∈ G, the mapping Ta : G → G defined

by Ta (x) = xa is a homeomorphism.

Proof: Define f : G → G × {a} by f (x) = (x, a). Then f is a homeomorphism. Then Ta is

continuous since it is the composition of f and the multiplication operator restricted to G × {a}. To

see that Ta is one-to-one, suppose that x, y ∈ G with Ta (x) = Ta (y). Then xa = ya, which means

that x = y. To see that Ta is onto, note that Ga = G. Finally, to see that Ta is a homeomorphism,

note that Ta−1 = Ta−1 , which is continuous. ■

Corollary 395: Every topological group is topologically homeomorphic.

Proof: Let G be a group and a, b ∈ G. By the theorem above, Ta−1 b is a homeomorphism and

Ta−1 b (a) = b. ■

Corollary 396: Suppose G is a topological group and a ∈ G.

(i) If U ⊆ G is open, then U a is open.

(ii) If E ⊆ G is closed, then Ea is closed.

Proof: Since Ta is a homeomorphism, Ta (U ) = U a is open and Ta (E) = Ea is closed. ■

Corollary 397: Suppose that G is a topological group, U ⊆ G is open, and A ⊆ G. Then U A is

open.
S
Proof: Note that U A = U a. ■
a∈A
Lemma 398: Suppose that G is a topological group, x ∈ G, and A ⊆ G. Then Ax = Ax.

Proof: Since Ax is a closed set and Ax ⊆ Ax, Ax ⊆ Ax. Now suppose that p ∈ A. To see

that px ∈ Ax, let U ⊆ G be open with px ∈ U . Then p ∈ U x−1 , which is open. Since p ∈ A,

U x−1 ∩ A ̸= ϕ. Then U ∩ Ax ̸= ϕ. Therefore, px ∈ Ax. ■

Theorem 399: Every topological group is regular.

Proof: Let G be a topological group. To see that G is regular, suppose that U ⊆ G is open and

a ∈ U . Then e ∈ Ua−1 , which is open, where e is the identity element of G. Since e−1 e ∈ U a−1 ,

there are open sets V1 , V2 ⊆ G such that e ∈ V1 , e ∈ V2 , and V1−1 V2 ⊆ U a−1 . Set V = V1 ∩ V2 . To

see that V ⊆ U a−1 , let p ∈ V . Since e ∈ V , p ∈ V p, which means that V p ∩ V ̸= ϕ. So, there are

b, c ∈ V such that bp = c. So, p = b−1 c ∈ V −1 V ⊆ Ua−1 . So, a ∈ V a ⊆ V a = V a ⊆ U a−1 a = U . ■

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