Lecture Notes Topology (2301631) Phichet Chaoha Department of ...

Lecture Notes
Topology (2301631)
Phichet Chaoha
Department of Mathematics and Computer Science,
Faculty of Science, Chulalongkorn University,
Bangkok 10330, Thailand.
[2001-03-05-06-08-09-11-12]

Contents
Preliminaries 5
Sets and Maps 5
Countable and Uncountable Sets 5
Ordered Sets 6
Chapter 1. Basic Concepts in Topology 7
1. Spaces and Subspaces 7
2. Basis for a Topology 11
3. Limit Points 14
4. Continuous Maps 17
5. Sequences and Nets 21
6. Product Topology 23
7. Quotient Topology 26
8. Metric Topology 29
Chapter 2. Countability and Separation Axioms 33
1. The Countability Axioms 33
2. The Separation Axioms 35
Chapter 3. Connectedness 41
Chapter 4. Compactness 47
Chapter 5. Function Spaces 59
Chapter 6. Complete Metric Spaces 65
Bibliography 73
3

Preliminaries
Sets and Maps
We assume that the reader is familiar with definitions and basic operations of
sets as well as of maps (or functions) between sets. We will use X c and P(X) to
denote the complement (with respect to some universe) and the powerset of the
set X, respectively. For a collection A of sets, ∪ A and ∩ A are the union and
intersection of sets in A, respectively.
Conventionally, the sets of natural numbers, integers, rational numbers, real
numbers and complex numbers will be denoted by N, Z, Q, R and C, respectively.
The properties listed below are easy to verify.
(1) Let {A α } α∈Λ be a collection of subsets of X.
(a) X − ∪ α∈Λ A α = ∩ α∈Λ (X − A α).
(b) X − ∩ α∈Λ A α = ∪ α∈Λ (X − A α).
(2) Let f : X → Y be a map, A ⊆ X and B ⊆ Y .
(a) f(X − A) ⊇ f(X) − f(A). The equality holds when f is 1-1.
(b) f −1 (Y − B) = X − f −1 (B).
(c) f −1 (f(A)) ⊇ A. The equality holds when f is 1-1.
(d) f(f −1 (B)) ⊆ B. The equality holds when f is onto.
(3) Let f : X → Y be a map, {A α } α∈Λ a collection of subsets of X, and
{B β } β∈∆ a collection of subsets of Y .
(a) f( ∪ α∈Λ A α) = ∪ α∈Λ f(A α).
(b) f( ∩ α∈Λ A α) ⊆ ∩ α∈Λ f(A α). The equality holds when f is 1-1.
(c) f −1 ( ∪ β∈∆ B β) = ∪ β∈∆ f −1 (B β ).
(d) f −1 ( ∩ β∈∆ B β) = ∩ β∈∆ f −1 (B β ).
Countable and Uncountable Sets
Definition 0.1. A set A is countable if there is a surjection f : N → A, or
equivalently an injection g : A → N. Thus, a countable set is either finite or
countably infinite. A set that is not countable is called an uncountable set.
Example 0.2. N × N is countable because f : N × N → N given by f(m, n) =
2 m 3 n is injective.
Example 0.3. Since g : N × N → Q + given by g(m, n) = m n
is surjective
and, from the previous example, there is a surjection h : N → N × N, then the
composition g ◦ h : N → Q + is clearly surjective; i.e., Q + is countable.
Theorem 0.4. A subset of a countable set is countable.
Theorem 0.5. A countable union of countable sets is countable.
5

6 Topology (2301631)
Proof. Let {A α } α∈Λ be a countable collection of countable sets. Then there
are surjections i : N → Λ and f α : N → A α for each α ∈ Λ. It follows that the
function F : N × N → ∪ α∈Λ A α defined by
is a surjection.
F (m, n) = f i(m) (n)
Example 0.6. Z × Z and Q are countable.
Theorem 0.7. A finite product of countable sets is countable.
Example 0.8. If B i is a countable set for each i ∈ N, the set ∪ n∈N
countable.
Ordered Sets
□
n∏
B i is also
Definition 0.9. A relation ≤ on a set A is called a preorder if it is reflexive
and transitive. When ≤ is a preorder on A, we will call (A, ≤) a preordered set.
Definition 0.10. A relation ≤ on a set A is called a partial order if it is
reflexive, antisymmetric and transitive. When ≤ is a partial order on A, we will
call (A, ≤) a partially ordered set or a poset.
It is not difficult to see that a preorder ≤ on a set A induces a partial order on
A/ ∼ , where ∼ is the equivalence relation on A defined by a ∼ b iff a ≤ b and b ≤ a.
Theorem 0.11 (Zorn’s lemma). Let (A, ≤) be a nonempty partially ordered
set. If every chain in A has an upper bound in A, then A has a maximal element.
Definition 0.12. A partially ordered set (A, ≤) is said to be totally ordered if
every a, b ∈ A, we have a ≤ b or b ≤ a.
Definition 0.13. A totally ordered set (A, ≤) is said to be well ordered if every
nonempty subset of A has a smallest element.
Definition 0.14. For a well ordered set (A, ≤) and α ∈ A, the section of A by
α defined to be the set S α = {x ∈ A | x < α}.
Theorem 0.15. There exists a well ordered set L having the largest element
Ω such that the section S Ω is uncountable but every other section is countable.
Moreover, each countable subset of S Ω has an upper bound in S Ω .
i=1

CHAPTER 1
Basic Concepts in Topology
1. Spaces and Subspaces
Definition 1.1. A topological space (or simply a space) consists of a set X
together with a subcollection τ X of P(X) satisfying the following properties :
(1) ∅ ∈ τ X ,
(2) X ∈ τ X ,
(3) ∪ A ∈ τ X for an arbitrary subcollection A of τ X ,
(4) ∩ A ∈ τ X for a finite subcollection A of τ X .
The collection τ X is called a topology on X and an element U ∈ τ X is said to
be open (in X).
We usually write τ instead of τ X when there is no ambiguity. Most of the time,
we will assume that X ≠ ∅ and write a space X without specifying its topology τ X .
The reader should keep in mind that a space X always comes with a topology on
it.
Although ∅ is trivially a space, we will usually assume that a space in this note
is nonempty.
Example 1.2. The collection of all open subsets of R forms a topology on R.
Example 1.3. Let X = {a, b} (where a ≠ b). Then {∅, {a, b}}, {∅, {a}, {a, b}},
{∅, {b}, {a, b}} {∅, {a}, {b}, {a, b}} are all possible topologies on X.
Exercise 1.4. Find all possible topologies on the set {a, b, c} (where a, b, c are
all distinct).
Example 1.5. Some interesting topologies on a given set X.
(1) The discrete topology : τ dis = P(X).
(2) The indiscrete topology : τ indis = {∅, X}.
(3) The cofinite topology : τ cf = {∅, X} ∪ {A ⊆ X | A c is finite}.
(4) The cocountable topology : τ cc = {∅, X} ∪ {A ⊆ X | A c is countable}.
7

8 Topology (2301631)
Exercise 1.6. In R with the cofinite topology, prove that any two nonempty
open sets always intersect. Is this still true for R equipped with the cocountable
topology?
Exercise 1.7. For a collection {τ α } α∈Λ of topologies on a set X, show that
the intersection ∩ α∈Λ τ α is still a topology on X. What about ∪ α∈Λ τ α?
Definition 1.8. A subset A of a space X is said to be
• closed if A c = X − A is an open set,
• G δ if A is a countable intersection of open sets,
• F σ if A is a countable union of closed sets, or equivalently, if A c is a G δ -set
in X.
Example 1.9. In R (with all usual open subsets), Q = ∪ q∈Q {q} is clearly F σ,
and hence Q c is G δ . However, both of them are neither open nor closed.
Definition 1.10. Let (X, τ X ) be a space, x ∈ X and A ⊆ X. A subset U of
X is called a neighborhood of x if x ∈ U ∈ τ X . Similarly, a subset V of X is called
a neighborhood of A if A ⊆ V ∈ τ X .
Definition 1.11. A space X is said to be Hausdorff if any two distinct points
of X have disjoint neighborhoods.

Phichet Chaoha 9
Example 1.12. The space {a, b} (where a ≠ b) equipped with the topology
{∅, {a}, {a, b}} is not Hausdorff.
Definition 1.13. Let d : R n × R n → R be the euclidean metric defined by
∑
d(x, y) = √ n (x i − y i ) 2 ,
i=1
for all x, y ∈ R n . The open ball centered at x ∈ R n of radius r > 0 is defined to be
B(x; r) = {y ∈ R n | d(x, y) < r}.
A subset U of R n is said to be open if for each x ∈ U, there exists r > 0 such that
B(x; r) ⊆ U.
It is straightforward to verify that the collection of all such open sets (notice that
both ∅ and R n are open) forms a topology on R n . This topology is called the usual
topology on R n . Note also that R n with the usual topology is Hausdorff.

10 Topology (2301631)
Exercise 1.14. Prove that a closed [open] set in R n (with the usual topology)
is a G δ -set [F σ -set].
Definition 1.15. Let τ and τ ′ be two topologies on a set X. We say that τ ′
is finer than τ (or equivalently, τ is coarser than τ ′ ) if τ ⊆ τ ′ .
Example 1.16. τ dis is finer than τ indis . What can we say about τ cf and τ cc ?
Exercise 1.17. Let X be a space and Y ⊆ X.
{U ∩ Y | U ∈ τ X } forms a topology on Y .
Prove that the collection
Definition 1.18. Let X be a space and Y ⊆ X. The subspace topology of Y
(in X) is defined to be
τ Y = {U ∩ Y | U ∈ τ X }.
In this case, we say that (Y, τ Y ) is a subspace of X.
Example 1.19. The subspace topology of Z (in R) is discrete.
Exercise 1.20. Let Y be a subspace of X and A ⊆ Y . Prove that A is closed
in Y iff A = C ∩ Y for some closed set C in X.
Theorem 1.21. Let Y be an open [closed] subspace of X. If A is open [closed]
in Y , then A is also open [closed] in X.
Proof. Let A be open [closed] in Y . Then A = B ∩ Y for some open [closed]
subset B of X. Since Y itself is open [closed] in X and the interesection is finite,
it follows that A is also open [closed] in X.
□

Phichet Chaoha 11
2. Basis for a Topology
Definition 1.22. For a set X and B ⊆ P(X), we call B a basis for a topology
on X if it satisfies the following conditions :
(1) For each x ∈ X, there exists B x ∈ B such that x ∈ B x .
(2) For each B 1 , B 2 ∈ B and x ∈ B 1 ∩ B 2 , there exists B 3 ∈ B such that
x ∈ B 3 ⊆ B 1 ∩ B 2 .
Exercise 1.23. Let B be a basis for a topology on a set X. Prove that X = ∪ B
and {U ⊆ X | ∀x ∈ U∃B x ∈ B, x ∈ B x ⊆ U} forms a topology on X.
Definition 1.24. Let B be a basis for a topology on a set X. We define the
topology generated by B to be
Clearly, B ⊆< B >.
< B >= {U ⊆ X | ∀x ∈ U∃B x ∈ B, x ∈ B x ⊆ U}.
Lemma 1.25. Let B be a basis for a topology on a set X. Every element of
< B > is a union of elements of B, and < B > is the coarsest (smallest) topology
on X containing B.
Proof. Let U ∈< B >. For each x ∈ U, let B x ∈ B be chosen from the the
definition above so that x ∈ B x ⊆ U. Then U = ∪ x∈U B x; i.e., U is a union of
elements of B. Moreover, if τ is a topology on X such that B ⊆ τ, then a union of
elements of B is clearly in τ. Therefore, < B >⊆ τ as desired.
□
Remark 1.26. The previous lemma implies that τ =< τ > for any topology τ
on X. Hence, U ∈ τ iff for each x ∈ U, there exists U x ∈ τ such that x ∈ U x ⊆ U.
Also, if B ⊆ B ′ , we clearly have < B >⊆< B ′ >. The converse is not true, but we
have the following theorem instead.
Theorem 1.27. Let B and B ′ be bases for topologies on a set X. We have
< B >⊆< B ′ > iff for each B ∈ B and x ∈ B, there exists B ′ ∈ B ′ such that
x ∈ B ′ ⊆ B.

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Proof. (⇒) Trivial.
(⇐) Let U ∈< B > and x ∈ U. Then there is B ∈ B such that x ∈ B ⊆ U. By
assumption, we can find B ′ ∈ B ′ such that x ∈ B ′ ⊆ B. Therefore, x ∈ B ′ ⊆ U;
i.e., U ∈< B ′ >.
□
Corollary 1.28. Let (X, τ) be a space and a basis B such that B ⊆ τ. Then
τ is generated by B iff for each U ∈ τ and x ∈ U, there exists B ∈ B such that
x ∈ B ⊆ U.
Example 1.29. The usual topology on R n is generated by the collection of
open balls {B(x; r) | x ∈ R n , r > 0}. In particular, the usual topology on R is
generated by {(a, b) | a < b}.
Definition 1.30. The lower limit topology on R is the topology generated by
{[a, b)|a < b}. We simply write R l to denote R equipped with the lower limit
topology.
Exercise 1.31. Prove that the lower limit topology on R is strictly finer than
the usual topology.
Definition 1.32. Let (X, ≤) be a totally ordered set with |X| > 1. An interval
in X is one of the following sets :
(a, b) = {x ∈ X | a < x < b}
[a, b) = {x ∈ X | a ≤ x < b}
(a, b] = {x ∈ X | a < x ≤ b}
[a, b] = {x ∈ X | a ≤ x ≤ b}
The order topology on X is the topology generated by the basis consisting of all
intervals of the following forms :
• (a, b),
• [x min , b) if X has the smallest element x min ,
• (a, x max ] if X has the largest element x max .
Exercise 1.33. Consider the set R 2 = R × R whose element is denoted by
a × b, and the dictionary order (≼) on R 2 defined by
a 1 × b 1 ≼ a 2 × b 2 iff (a 1 × b 1 = a 2 × b 2 ) or (a 1 × b 1 ≺ a 2 × b 2 ),
where a 1 × b 1 ≺ a 2 × b 2 iff a 1 < a 2 or (a 1 = a 2 and b 1 < b 2 ).
Is the dictionary order topology on R × R the same as the usual topology on
R 2 ? If not, compare them.

Phichet Chaoha 13
Exercise 1.34. For a set X and S ⊆ P(X) with ∪ S = X, prove that the
collection of all finite intersections of elements of S forms a basis for a topology on
X.
Definition 1.35. For a set X and S ⊆ P(X), we will call S a subbasis for a
topology on X if ∪ S = X.
If S is a subbasis for a topology on X, we will define the topology generated
by S (as a subbasis), denoted by < S >, to be the collection of all unions of finite
intersections of elements of S.
Exercise 1.36. Let (X, ≤) be a totally ordered set with |X| > 1 and for each
a ∈ X, let S a = {x ∈ X | x < a} and T a = {x ∈ X | x > a}.
Prove that the collection S = {S a | a ∈ X} ∪ {T a | a ∈ X} forms a subbasis for the
order topology on X.

14 Topology (2301631)
3. Limit Points
Let X be a (nonempty) space and ∅ ̸= A ⊆ X.
Definition 1.37. A point x ∈ X is called a limit point of A if, for each
neighborhood U of x, we have U ∩ A − {x} ̸= ∅.
The set of all limit points of A is called the derived set of A and denoted by A ′ . If
A = A ′ , we will call A a perfect set.
Example 1.38. In R (with the usual topology), we have (0, 1] ′ = [0, 1], Q ′ =
R, { 1 n | n > 0}′ = {0}, and N ′ = ∅.
Exercise 1.39. Let (L, ≤) be a well ordered set having the largest element Ω
together with the following properties :
(1) the section S Ω is uncountable,
(2) the section S α is countable for any α < Ω,
(3) each countable subset of S Ω has an upper bound in S Ω .
In L equipped with the order topology, prove that Ω is a limit point of S Ω .
Definition 1.40. The interior of A (in X), denoted by Int X (A) or A ◦ , is the
largest open subset of X contained in A, or equivalently
Int X (A) = ∪ {U ⊆ X | U ⊆ A and U is open}.
Definition 1.41. The closure of A (in X), denoted by Cl X (A) or A, is the
smallest closed subset of X containing A, or equivalently
Cl X (A) = ∩ {C ⊆ X | A ⊆ C and C is closed}.
Definition 1.42. The boundary (or frontier) of A is defined to be
∂A = A ∩ (X − A).
Remark 1.43. From the definition above, we clearly have
• A is open in X iff A = Int X (A).
• A is closed in X iff A = Cl X (A).
Example 1.44. In R (with the usual topology), we clearly have (0, 1] ◦ = (0, 1),
(0, 1] = [0, 1] and ∂(0, 1] = {0, 1}.

Phichet Chaoha 15
{ 1 n
Exercise 1.45. What are the interior, the closure and the boundary of Q and
| n ∈ N}? Justify your answers.
Exercise 1.46. Let X be a space and A ⊆ X. Are the following pairs of sets
comparable? If so, compare them.
(1) (A ◦ ) and (A) ◦ .
(2) (∂A) and ∂(A).
(3) ∂(A ◦ ) and (∂A) ◦ .
Theorem 1.47. For a subspace Y of X such that A ⊆ Y , we have
Cl Y (A) = Cl X (A) ∩ Y.
Proof. (⊆) Since Cl X (A) is closed in X, then Cl X (A) ∩ Y is closed in Y .
Since A ⊆ Cl X (A) ∩ Y , it follows that Cl Y (A) ⊆ Cl X (A) ∩ Y .
(⊇) Since Cl Y (A) is closed in Y , we have Cl Y (A) = C ∩ Y for some closed
subset C of X containing A. Therefore, Cl X (A) ⊆ C, and hence Cl X (A) ∩ Y ⊆
C ∩ Y = Cl Y (A).
□
Example 1.48. Consider Y = (0, 2] as a subspace of R. Then,
Cl Y ((0, 1]) = Cl R ((0, 1]) ∩ Y = [0, 1] ∩ (0, 2] = (0, 1]
which is closed in Y , but not in R.
Theorem 1.49. Let X be a space, A ⊆ X and x ∈ X. Then
x ∈ A iff U ∩ A ≠ ∅ for each neighborhood U of x.
Proof. We will show that x /∈ A iff U ∩ A = ∅ for some neighborhood U of x.
(⇒) Suppose x /∈ A and let U = X − A. Then, it is easy to see that U is a
neighborhood of x such that U ∩ A = ∅ as desired.
(⇐) Suppose there is a neighborhood U of x such that U ∩ A = ∅. Then
C = X − U is closed, A ⊆ C and x /∈ C. Since A ⊆ C, it follows that x /∈ A. □
Remark 1.50. If the topology on X is generated by a basis B, we clearly have
the followings :
• x ∈ A ′ iff B ∩ A − {x} ≠ ∅ for each B ∈ B containing x.
• x ∈ A iff B ∩ A ≠ ∅ for each B ∈ B containing x.
Exercise 1.51. Let X be a space and A ⊆ X. Prove that A = A ∪ A ′ .
Remark 1.52. From Exercise 1.39 and the previous exercise, we may represent
L by S Ω .
Corollary 1.53. Let X be a space and A ⊆ X. Then A is closed iff A ′ ⊆ A.
Proof. A is closed ⇔ A = A ⇔ A ′ ⊆ A.
Exercise 1.54. Let A and B be subsets of a space X. Prove that
(1) A ◦ ∩ ∂A = ∅.
(2) A ◦ ∪ ∂A = A.
(3) A = A.
(4) A ⊆ B ⇒ A ⊆ B.
(5) A ∪ B = A ∪ B.
(6) A ∩ B ⊆ A ∩ B. When does the other containment fail?
□

16 Topology (2301631)
Definition 1.55. Let X be a space and A ⊆ X. We say that A is dense (in
X) if A = X. On the other hand, A is nowhere dense if X − A is dense in X.
Example 1.56. In R, Q is dense while Z is nowhere dense.
Definition 1.57. Subsets A and B of a space X are said to be separated if
A ∩ B = A ∩ B = ∅.
Example 1.58. In R, the intervals (0, 1) and [1, 2) are disjoint but not separated,
while (0, 1) and (1, 2) are separated.

Phichet Chaoha 17
Let X and Y be (nonempty) spaces.
4. Continuous Maps
Definition 1.59. A map f : X → Y is said to be continuous at x ∈ X if, for
any neighborhood V of f(x), there is a neighborhood U of x such that f(U) ⊆ V .
We simply say that f is continuous if it is continuous at each x ∈ X.
Example 1.60. For a selfmap of R n (with the usual topology), the definition
of continuity above is exactly the ε − δ definition in analysis.
Theorem 1.61. A map f : X → Y is continuous iff f −1 (V ) is open in X for
each open subset V of Y .
Proof. (⇒) Let V ⊆ Y be an open set and x ∈ f −1 (V ). Since f is continuous
at x and V is a neighborhood of f(x), there is a neighborhood U of x such that
f(U) ⊆ V . Hence, U ⊆ f −1 (f(U)) ⊆ f −1 (V ) which implies that f −1 (V ) is open in
X.
(⇐) Let x ∈ X and V a neighborhood of f(x). It follows from the assumption
that f −1 (V ) is a neighborhood of x. So, by letting U = f −1 (V ), we clearly have
f(U) = f(f −1 (V )) ⊆ V and this proves the continuity of f.
□

18 Topology (2301631)
Remark 1.62. Definition 1.59 is generally known as the local definition of
continuity while Theorem 1.61 is usually taken to be the global definition. Moreover,
using the global definition, one can easily prove that f : X → Y is continuous iff
f −1 (C) is closed in X for each closed subset C of Y .
Example 1.63. Let τ and τ ′ be topologies on X. It is easy to verify that the
identity map 1 X : (X, τ) → (X, τ ′ ) is always continuous iff τ is finer than τ ′ .
Remark 1.64. If the topology of Y is generated by a basis B, it is easy to see
that a map f : X → Y is continuous iff f −1 (B) is open for each B ∈ B. This is
because f −1 ( ∪ α B α) = ∪ α f −1 (B α ). Note also that f −1 (B) may not be a basis
element of X.
Exercise 1.65. If the topology of Y is generated by a subbasis S, prove that
a map f : X → Y is continuous iff f −1 (S) is open in X for each S ∈ S.
Theorem 1.66. A map f : X → Y is continuous iff f(A) ⊆ f(A) for any
subset A of X.
Proof. (⇒) Let A ⊆ X, y ∈ f(A) and V be a neighborhood of y. Then, there
exists x ∈ A such that f(x) = y. Hence, x ∈ f −1 (V ). By the continuity of f, it
follows that f −1 (V ) is a neighborhood of x. Therefore, f −1 (V ) ∩ A ≠ ∅ and hence
∅ ̸= f(f −1 (V ) ∩ A) ⊆ f(f −1 (V )) ∩ f(A) ⊆ V ∩ f(A).
This implies y ∈ f(A).
(⇐) Let C be a closed subset of Y and A = f −1 (C). Since
f(A) ⊆ f(A) = f(f −1 (C)) ⊆ C = C,
then we have A ⊆ f −1 (f(A)) ⊆ f −1 (C) = A.
desired.
Therefore, A is closed in X as
□
Example 1.67. It is easy to verify that a constant map, an inclusion and a
restriction of a continuous map are continuous.
Theorem 1.68. If f : X → Y and g : Y → Z are continuous maps, then so is
g ◦ f.
Proof. Let W be an open subset of Z.
Then g −1 (W ) is open in Y by the continuity of g. It follows that (g ◦ f) −1 (W ) =
f −1 (g −1 (W )) is open in X by the continuity of f. This proves the continuity of
g ◦ f as desired.
□
Exercise 1.69. Let X be a space and A ⊆ X. Prove that the subspace
topology on A is the smallest (coarsest) topology that makes the inclusion i : A ↩→
X continuous.

Phichet Chaoha 19
Exercise 1.70. Let f : X → Y be a map. If X = ∪ α U α where each U α is
open in X and f| Uα is continuous, prove that f is continuous. Is it still true if we
do not assume that each U α is open?
Now, suppose we have two maps f : A → Y and g : B → Y that agree on the
intersection A ∩ B. We clearly have a well-defined map f ∪ g : A ∪ B → Y given by
{
f(x) if x ∈ A,
f ∪ g(x) =
g(x) if x ∈ B.
Notice that the previous exercise already guarantees the continuity of f ∪ g
whenever both A and B are open subspaces of X. The next theorem also gives the
similar result when X is a union of two closed subspaces.
Theorem 1.71 (Pasting Lemma). Let X = A ∪ B, where A and B are closed
in X. If f : A → Y and g : B → Y are continuous maps that agree on A ∩ B, then
h = f ∪ g : A ∪ B → Y is continuous.
Proof. Let C be a closed subset of Y . Since
h −1 (C) = f −1 (C) ∪ g −1 (C),
the continuity of f and g implies that the sets f −1 (C) and g −1 (C) are closed in
A and B, respectively. Also, since A and B are closed in X, so are f −1 (C) and
g −1 (C). Therefore, h −1 (C) = f −1 (C) ∪ g −1 (C) is also closed in X.
□
The pasting lemma can also be extended to a finite union of closed subspaces
of X, but for an infinite union, we have the following counterexample.
Example 1.72. For each r ∈ R, let f r : {r} → R be defined by
{
1 ; r ∈ Q
f r (r) =
0 ; r /∈ Q.
Clearly, each f r is continuous and f = ∪ r∈R f r is well-defined. It is easy to see that
f is the characteristic map on Q. Therefore, f is not continuous.
Definition 1.73. A map f : X → Y is called a homeomorphism if f is a
continuous bijection whose inverse map is also continuous. We will say that X is
homeomorphic to Y , written X ∼ = Y , if there is a homeomorphism from X to Y .
Remark 1.74. It is easy to verify that being homeomorphic is an equivalence
relation.
Definition 1.75. We say that a property P of a space is a topological property
or a topological invariant if P is invariant under homeomorphisms.
Example 1.76. For topologies τ and τ ′ on a set X such that τ is strictly finer
than τ ′ , the identity map 1 X : (X, τ) → (X, τ ′ ) is a continuous bijection, but not a
homeomorphism!
Definition 1.77. A continuous injection f : X → Y is called an imbedding if
the induced bijection f : X → f(X) is a homeomorphism, where f(X) is considered
as a subspace of Y .

20 Topology (2301631)
Example 1.78. Let S 1 = {(x, y) | x 2 +y 2 = 1} be a subspace of R 2 . The continuous
bijection f : [0, 1) → R 2 given by f(t) = (cos 2πt, sin 2πt) is not an imbedding
because f −1 : S 1 → [0, 1) is not continuous. Notice that (f −1 ) −1 ([0, 1 2 )) = f([0, 1 2 ))
is not open in S 1 . However, f| (0,1) is an imbedding.
Exercise 1.79. Prove that f : N ∪ {0} → R given by f(n) = (−1) n n 2 is an
imbedding. What about g : N ∪ {0} → R given by g(0) = 0 and g(n) = 1 n
for
n > 0?

Phichet Chaoha 21
5. Sequences and Nets
Definition 1.80. A sequence in a space X is a map (x n ) : N → X. We usually
denote (x n )(i) by x i .
Definition 1.81. The sequence (x n ) in X is said to converge to x ∈ X, abbreviated
by (x n ) → x, if for each neighborhood U of x, there exists N ∈ N such
that x n ∈ U for all n ≥ N. When (x n ) → x, we will call x a limit of (x n ).
Example 1.82. In the space {a, b}, where a ≠ b, with the indiscrete topology,
the constant sequence (a) converges to both a and b. Hence, a limit of a convergent
sequence may not be unique.
Theorem 1.83. A convergent sequence in a Hausdorff space has a unique limit.
Proof. Let (x n ) be a convergent sequence in X. Suppose (x n ) → x and
(x n ) → x ′ . If x ≠ x ′ , there exist disjoint neighborhoods U and V of x and x ′ ,
respectively. Hence, there is a large enough N ∈ N such that x n ∈ U ∩ V for all
n ≥ N (can you see why?). This is impossible because U ∩ V = ∅. Hence, we must
have x = x ′ .
□
Definition 1.84. Let (x n ) and (y n ) be sequences in a space X. We say that
(y n ) is a subsequence of (x n ) if there is a strictly increasing map f : N → N such
that y i = x f(i) for all i ∈ N.
Exercise 1.85. Let (y n ) be a subsequence of a convergent sequence (x n ) in a
Hausdorff space. Prove that if (y n ) converges to y ∞ ∈ X, then so is (x n ). What
happens if (x n ) is not assumed to be convergent?
Theorem 1.86 (Sequence Lemma). Let X be a space, x ∈ X and A ⊆ X. If
there is a sequence in A converging to x, then x ∈ A.
Proof. Let (a n ) be a sequence in A converging to x, and U a neighborhood
of x. Then, by the definition of convergence, there exists N ∈ N such that a n ∈ U
for all n ≥ N. In particular, a N ∈ U ∩ A. Hence, U ∩ A ≠ ∅; i.e, x ∈ A. □
The converse of the above theorem is not true in general as we will see in the
next example.
Example 1.87. The space S Ω does not satisfy the converse of the sequence
lemma because there is no sequence in S Ω converging to Ω. To see this, let (x n ) be
a sequence in S Ω . Since C = {x n | n ∈ N} is a countable subset of S Ω , it has an
upper bound in S Ω , says α. Now it is easy to see that C ∩ (α, Ω] = ∅ and hence
(x n ) does not converge to Ω.

22 Topology (2301631)
Definition 1.88. A directed set is a preordered set (I, ≤) such that for any
α, β ∈ I, we have α ≤ γ and β ≤ γ for some γ ∈ I.
Example 1.89. Clearly, (P(S), ⊆) and (N, |) are directed sets. Moreover, if
N (x) denotes the set of all neighborhoods of x ∈ X, both (N (x), ⊆) and (N (x), ⊇)
are also directed sets.
Definition 1.90. A net in a space X is simply a map (x α ) : I → X where I
is a directed set.
Example 1.91. A sequence in a space X is simply a net in X whose directed
set is (N, ≤).
Definition 1.92. A net (x α ) : I → X is said to converge to x ∈ X, abbreviated
by (x α ) → x, if for each neighborhood U of x, there exists γ ∈ I such that x α ∈ U
for all α ≥ γ.
Exercise 1.93. Prove that a convergent net in a Hausdorff space has a unique
limit.
Theorem 1.94. Let X be a space, x ∈ X and A ⊆ X. There is a net in A
converging to x iff x ∈ A.
Proof. (⇒) Let (a α ) : (I, ≤) → A be a net in A converging to x, and U a
neighborhood of x. Then, by the definition of convergence, there exists γ ∈ I such
that a α ∈ U for all α ≥ γ. In particular, a γ ∈ U ∩ A. Hence, U ∩ A ≠ ∅; i.e, x ∈ A.
(⇐) Suppose x ∈ A. Let I be the poset of neighborhoods of x partially ordered
by inverse inclusions; i.e., U ≤ V iff U ⊇ V . Then for each U ∈ I, we have U∩A ≠ ∅.
By the axiom of choice, we can form a net (a U ) in A such that a U ∈ U ∩ A for each
U ∈ I. Now, it is clear from the construction that (a U ) → x.
□
Exercise 1.95. Let X and Y be spaces, x ∈ X and f : X → Y a map.
(1) Prove that if f is continuous at x, then for each sequence (x n ) in X
converging to x, the sequence (f(x n )) converges to f(x).
(2) Show that the converse of the above statement fails in general.
(3) Prove that a map f : X → Y is continuous at x iff for each net (x α ) in X
converging to x, the net (f(x α )) converges to f(x).
Definition 1.96. A subset I ′ of a directed set I is said to be cofinal in I if
for each α ∈ I, we have α ≤ α ′ for some α ′ ∈ I ′ .
Proposition 1.97. A cofinal subset of a directed set is also a directed set.
Proof. Suppose I ′ be a cofinal subset of a directed set I. Let α ′ , β ′ ∈ I ′ .
Since I ′ ⊆ I and I is directed, we clearly have α ′ ≤ γ and β ′ ≤ γ for some γ ∈ I.
Since I ′ is cofinal in I, there exists γ ′ ∈ I ′ such that γ ≤ γ ′ . Hence, α ′ ≤ γ ′ and
β ′ ≤ γ ′ as required.
□
Definition 1.98. A net (y β ) : J → X is said to be a subnet of a net (x α ) :
I → X if (y β ) = (x α ) ◦ f for some order-preserving map f : J → I where f(J ) is
cofinal in I.
Example 1.99. A subsequence is clearly a subnet.
Exercise 1.100. Let X be a Hausdorff space, (x α ) : I → X a convergent net
in X and (y β ) : J → X a subnet of (x α ). Prove that if (y β ) converges to a point
z ∈ X, then so does (x α ).

Phichet Chaoha 23
6. Product Topology
Let Λ be an index set. In this section, we will introduce some topologies on a
cartesian product ∏ α∈Λ X α, where each X α is a topological space. Conventionally,
an element of ∏ α∈Λ X α is written as (x α ) α∈Λ , and for a collection {f α : A →
X α } α∈Λ of maps, (f α ) α∈Λ denotes the map f : A → ∏ α∈Λ X α defined by f(a) =
(f α (a)) α∈Λ for each a ∈ A.
Definition 1.101. Let {X α } α∈Λ be a collection of spaces. We define the box
topology on the set ∏ α∈Λ X α to be the topology generated by the basis
{ ∏ α∈Λ
U α | U α is open in X α }.
We also define the product topology on the set ∏ α∈Λ X α to be the topology
generated by the basis
{ ∏ α∈Λ
U α | U α is open in X α and U α ≠ X α for finitely many α’s}.
The set ∏ α∈Λ X α together with the product topology will be simply called the
product space.
Example 1.102. In R ω = ∏ n∈N R, the set ∏ n∈N
(−n, n) is open in the box
topology, but not in the product topology.
Remark 1.103. The box topology is generally finer than the product topology.
However, when Λ is finite, the box and the product topologies are the same.
Exercise 1.104. Let A and B be subsets of a space X. Prove that A × B =
A × B in the product topology.
The proofs of the following two theorems are straightforward, and hence left to
reader.
Theorem 1.105. Suppose for each α ∈ Λ, the topology on X α is generated by
a basis B α . Then the collection { ∏ α∈Λ B α | B α ∈ B α } forms a basis for the box
topology on the set ∏ α∈Λ X α.
Similarly, the collection
{ ∏ α∈Λ
B α | B α ∈ B α ∪ {X α } and B α ≠ X α for finitely many α’s}
forms a basis for the product topology on the set ∏ α∈Λ X α.
Theorem 1.106. If A α is a subspace of X α for each α ∈ Λ, then ∏ α∈Λ A α is
also a subspace of ∏ α∈Λ X α when both products are given the box or the product
topology.
Remark 1.107. For simplicity, we sometimes write U α1 ×· · ·×U αn × ∏ α≠α 1 ,...,α n
X α
to represent the subset ∏ α∈Λ U α of ∏ α∈Λ X α such that U α = X α for all α /∈
{α 1 , . . . , α n }.
Theorem 1.108. For each β ∈ Λ, the projection map
π β : ∏ α∈Λ
X α → X β
is continuous when the product is given the box or the product topology.

24 Topology (2301631)
Proof. First note that it suffices to prove only the case where the product
topology is given because the box topology is finer. Let β ∈ Λ and U β an open
subset of X β . Since π −1
β
(U β) = U β × ∏ α≠β X α which is clearly open in the product
topology, then π β is continuous as desired.
□
Exercise 1.109. Show that the product topology on ∏ α∈Λ X α is the coarsest
topology that makes the projection map π β : ∏ α∈Λ X α → X β continuous for each
β ∈ Λ. In other words, the collection
∪
{π −1
β (U β) | U β is open in X β }
β∈Λ
forms a subbasis for the product topology on the set ∏ α∈Λ X α.
Exercise 1.110. If X α is Hausdorff for each α ∈ Λ, then ∏ α∈Λ X α is also
Hausdorff in the box and the product topologies.
Theorem 1.111. Let ∏ α∈Λ X α be given the product topology. The map f =
(f α ) α∈Λ : A → ∏ α∈Λ X α is continuous iff f α : A → X α is continuous for each
α ∈ Λ.
Proof. (⇒) Trivial by composing f with the projection maps.
(⇐) Let B be a basis element in ∏ α∈Λ X α with respect to the product topology;
i.e., B = U α1 × · · · × U αn × ∏ α≠α 1 ,...,α n
X α for some open sets U αi ⊆ X αi . Clearly,
f −1 (B) = ∩ n
i=1 f α −1
i
(U αi ) and each fα −1
i
(U αi ) is open by assumption. Therefore,
f −1 (B) is open.
□
Example 1.112. The previous theorem is not true if the topology on ∏ α∈Λ X α
is taken to be the box topology. Consider R ω = ∏ n∈N R. Then ∆ : R → Rω defined
by ∆(x) = (x, x, . . . ) is clearly continuous (being continuous in each component)
when R ω is given the product topology. However, ∆ is not continuous when R ω is
given the box topology because the set
(
∆ −1 (−1, 1) × (− 1 2 , 1 2 ) × (−1 3 , 1 ) ∞∩
3 ) × . . . = (− 1 n , 1 n ) = {0}
is not open in R.
Exercise 1.113. For a sequence (x n ) in the product space ∏ α∈Λ X α, prove
that (x n ) converges to x iff the sequence (π α (x n )) converges to π α (x) for all α ∈ Λ.
We end this subsection by stating the universal property of product spaces.
∏
Theorem 1.114. Let {X α } α∈Λ be a collection of spaces. The product space
α∈Λ X α together with the collection {π β : ∏ α∈Λ X α → X β } β∈Λ of projection
maps satisfies the following universal property :
Given a space A with a collection {p β : A → X β } β∈Λ of continuous maps, there
exists a unique continuous map f : A → ∏ α∈Λ X α such that p β = π β ◦ f for each
β ∈ Λ; i.e., we have the following commutative diagram for each β ∈ Λ :
A
♣ ♣ ♣ ♣ ♣ ♣
p β
✲
X β
n=1
∃!f
❘
∏
α∈Λ
X α
✒ π β

Phichet Chaoha 25
Moreover, if P is a space together with a collection {π ′ β : P → X β} β∈Λ satisfying
the above universal property, then P must be homeomorphic to the product
space ∏ α∈Λ X α.
Proof. Let A be a space and {p β : A → X β } β∈Λ a collection of continuous
maps. Let f = (p α ) α∈Λ : A → ∏ α∈Λ X α and β ∈ Λ. Clearly, p β = π β ◦ f. Since
each p β is continuous, then f is continuous by the previous theorem. Now, suppose
f ′ : A → ∏ α∈Λ X α has the property that p β = π β ◦ f ′ for each β ∈ Λ. Then,
f = (π β ◦ f) β∈Λ = (p β ) β∈Λ = (π β ◦ f ′ ) β∈Λ = f ′
as desired.
Finally, suppose P together a the collection {π
β ′ : P → X β} β∈Λ satisfies the
above universal property. Then, there exist continuous maps f : P → ∏ α∈Λ X α
and g : ∏ α∈Λ X α → P such that π
β ′ = π β ◦ f and π β = π
β ′ ◦ g. It follows that
π ′ β = π′ β ◦ (g ◦ f) and π β = π β ◦ (f ◦ g). Again, since ∏ α∈Λ X α satisfies the universal
property, the identity map id : ∏ α∈Λ X α → ∏ α∈Λ X α must be the only continuous
map satisfying π β = π β ◦ id. It follows that f ◦ g = id. Similarly, g ◦ f must be the
identity map on P . Therefore, f is a homeomorphism whose inverse is g. □

26 Topology (2301631)
7. Quotient Topology
Definition 1.115. Let X and Y be spaces. A surjective map p : X → Y is
called a quotient map (or an identification) if for each V ⊆ Y , V is open in Y iff
p −1 (V ) is open in X.
Remark 1.116. Every quotient map is always continuous, but not vice versa.
Note also that, the condition in the definition above is equivalent to ”C is closed in
Y iff p −1 (C) is closed in X”, but NOT to ”U is open in X iff p(U) is open in Y ”.
Definition 1.117. Let X and Y be spaces. A map f : X → Y is said to be
open if for each open subset U of X, the set f(U) is open in Y . A closed map is
defined in the similar manner.
Theorem 1.118. Every continuous surjective map that is either open or closed
is a quotient map.
Proof. Let f : X → Y be a continuous surjective map that is open, and
V ⊆ Y . By the continuity of f, if V is open in Y , then f −1 (V ) is open in X.
Conversely, assume that f −1 (V ) is open in X. Since f is open and surjective,
then V = f(f −1 (V )) is open in Y . The case where f is closed can be proved
similarly.
□
Exercise 1.119. Find (and justify your answers) :
(1) a surjective continuous map that is not a quotient map.
(2) a quotient map that is neither open nor closed.
Exercise 1.120. Let {X α } α∈Λ be a collection of spaces. Show that for each
β ∈ Λ, the projection map π β : ∏ α∈Λ X α → X β is always open, but it may not be
closed.
Definition 1.121. Let (X, τ X ) be a space, A a set and p : X → A a surjective
map. The quotient topology τ p on A induced by p is the unique topology on A
which makes p a quotient map; i.e.,
τ p = {U ⊆ A | p −1 (U) ∈ τ X }.
Example 1.122. Consider the surjective map p : R → {a, b, c} defined by
⎧
⎪⎨ a ; x < 0
p(x) = b ; x = 0
⎪⎩
c ; x > 0.
Then the quotient topology on {a, b, c} induced by p is {∅, {a, b, c}, {a}, {c}, {a, c}}.
Definition 1.123. Let X be a space, A a partition of X into disjoint subsets
and p : X → A the surjective map that sends each point in X to the element of
A containing it. The set A together with the quotient topology induced by p is
usually called a quotient space (or an identification space) of X. Moreover, if ∼ is
the equivalence relation on X induced by A, we will usually denote the quotient
space A by X/ ∼.
Remark 1.124. Since a subset V of the quotient space X/ ∼ is just a collection
of equivalence classes, then V is open in X/ ∼ iff ∪ V is open in X.
Example 1.125. Let I = [0, 1], and I 2 = I × I.

Phichet Chaoha 27
(1) The 1-sphere (S 1 ) := I/ ∼, where 0 ∼ 1.
(2) The 2-sphere (S 2 ) := I 2 / ∼, where (x, 0) ∼ (x ′ , 1) ∼ (0, y) ∼ (1, y ′ ), for
each x, x ′ , y, y ′ ∈ I.
(3) The Mobius strip := I 2 / ∼, where (0, y) ∼ (1, 1 − y), for each y ∈ I.
(4) The 2-torus (T 2 ) := I 2 / ∼, where (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y), for
each x, y ∈ I.
(5) The Klien bottle := I 2 / ∼, where (x, 0) ∼ (x, 1) and (0, y) ∼ (1, 1 − y),
for each x, y ∈ I.
(6) The real projective space (RP 2 ) := I 2 / ∼, where (x, 0) ∼ (1 − x, 1) and
(0, y) ∼ (1, 1 − y), for each x, y ∈ I.

28 Topology (2301631)
Example 1.126. Let L = R × {−1, 1}. Define an equivalence relation ∼ on L
by (x, −1) ∼ (x, 1) for all x ≠ 0. Then, the quotient space L/ ∼ is called the line
with double points. It is easy to see that L is Hausdorff, but L/ ∼ is not!
Definition 1.127. Let p : X → Y be a map and y ∈ Y . We will call the set
p −1 ({y}) the fiber of p over y.
Theorem 1.128. Let p : X → Y be a quotient map, Z a space, and g :
X → Z a continuous map. If g is constant on each fiber of p, then there exists a
unique continuous map f : Y → Z such that g = f ◦ p; i.e., we have the following
commutative diagram :
X
❅ ❅❘ p
g
✲ Z
♣ ♣ ♣ ♣
✒
∃!f
Y
Proof. Since g is constant on each fiber of p, we can simply define f : Y → Z
by f(y) = g(a) for each y ∈ Y and some point a in the fiber of p over y. Clearly,
g = f ◦ p. For each open subset V of Z, the set g −1 (V ) = p −1 (f −1 (V )) is open by
the continuity of g, and hence f −1 (V ) must be open in Y since p is a quotient map.
Therefore, f is continuous. Now suppose f ′ : Y → Z is a map such that g = f ′ ◦ p.
Then, for each y ∈ Y , we have f(y) = g(a) = f ′ ◦ p(a) = f ′ (y) where a is any point
in the fiber over y.
□
Example 1.129. Let p : I 2 → S 2 and q : I 2 → T 2 be the quotient maps
from the previous example. Then, by the previous theorem, there exists a unique
continuous map f : T 2 → S 2 such that p = f ◦ q. Morever, it is easy to see that f
is surjective.
Exercise 1.130. Prove that there is a surjective continuous map from the
Mobius strip onto the Klien bottle.

Phichet Chaoha 29
8. Metric Topology
Definition 1.131. A metric on a set X ≠ ∅ is a map d : X ×X → R satisfying
the following properties for all x, y, z ∈ X :
(1) d(x, y) = d(y, x).
(2) d(x, y) = 0 iff x = y.
(3) d(x, z) ≤ d(x, y) + d(y, z).
The pair (X, d) is called a metric space.
Remark 1.132. From the conditions above, we have
0 = d(x, x) ≤ d(x, y) + d(y, x) = d(x, y) + d(x, y) = 2d(x, y).
Therefore, d(x, y) ≥ 0; i.e. d : X × X → R + 0 .
Exercise 1.133. Let d be a metric on the set X. Prove that the map defined
by d(x, y) = min{d(x, y), 1}, for any x, y ∈ X, is also a metric on X. In fact, this
is still true if we replace 1 by any k ∈ R + .
Definition 1.134. For a metric space (X, d), the metric d defined as in the
previous example is called the standard bounded metric.
Definition 1.135. Let (X, d) be a metric space, x ∈ X and r > 0. We define
the open ball centered at x of radius r to be
B d (x; r) = {y ∈ X | d(x, y) < r}.
Definition 1.136. For a metric space (X, d), we define the metric topology
induced by d, denoted by τ d , to be the topology on X whose basis is the set of all
open balls.
Example 1.137. For any set X ≠ ∅, the map d : X × X → R + 0
{
0 ; x = y
d(x, y) =
1 ; x ≠ y
given by
is a metric on X. This metric induces the discrete topology on X and hence it is
called the discrete metric.
Definition 1.138. A space X is said to be metrizable if its topology is induced
by a metric.
Example 1.139. The euclidean space R n is metrizable because its topology
(the usual topology) is induced by the euclidean metric d.
Exercise 1.140. For a metric space (X, d), prove that the map d : (X, τ d ) ×
(X, τ d ) → R + 0 is continuous.
Theorem 1.141. For a metric d on the set X, the metric topology induced by
d is the coarsest topology on X which makes d continuous.
Proof. The map d : (X, τ d ) × (X, τ d ) → R + 0 is continuous by the previous
exercise. Now, let τ be a topology on the set X which makes the map d : (X, τ) ×
(X, τ) → R + 0 continuous. It suffices to show that τ contains all open balls. For x ∈
X and r > 0, the continuity of d implies that the restriction d| {x}×X : {x} × X →
R + 0 is continuous. Let d x : (X, τ) → R + 0 be the following composition :

30 Topology (2301631)
d x
X ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣✲♣ ♣ R + 0
◗
ι ◗◗ ✑ ✑✸
d {x}×X
{x} × X
where ι(y) = (x, y) for each y ∈ X. Clearly, d x is continuous and hence B d (x; r) =
d −1
x ([0, r)) is open in X; i.e., B d (x; r) ∈ τ. □
Theorem 1.142. The converse of the sequence lemma holds in a metric space.
Proof. Let (X, d) be a metric space, A ⊆ X and a ∈ A. Then, we have
B d (a; 1 n )∩A ≠ ∅ for each n ∈ N. By the axiom of choice, we obtain a sequence (a n)
in A such that a n ∈ B d (a; 1 n ) ∩ A for each n ∈ N. To show that (a n) converges to a,
we let U be an neighborhood of a. Since the metric topology on X is generated by
open balls, there exits ϵ > 0 such that B d (a; ϵ) ⊆ U. By choosing N ∈ N such that
1
N < ϵ, we have a n ∈ B d (a; 1 n ) ∩ A ⊆ B d(a; 1 N ) ∩ A ⊆ B d(a; ϵ) ∩ A ⊆ U ∩ A ⊆ U for
all n ≥ N. This proves that (a n ) → a as desired.
□
Theorem 1.143. Let f : X → Y be a map of spaces, where X satisfies the
converse of the sequence lemma. Then, f is continuous iff for any x ∈ X and any
sequence (x n ) converging x, the sequence (f(x n )) converges to f(x).
Proof. (⇒) Follows from Exercise 1.95(1).
(⇐) Let A ⊆ X and y ∈ f(A). Then y = f(x) for some x ∈ A. Since X
satisfies the converse of the sequence lemma, there exists a sequence (a n ) in A
converging to x. By the assumption, the sequence (f(a n )) (in f(A)) converges
to f(x). Therefore, by the sequence lemma, y = f(x) ∈ f(A). It follows that
f(A) ⊆ f(A) which implies the continuity of f.
□
Corollary 1.144. The previous theorem is true when X is metrizable.
Example 1.145. The space S Ω is not metrizable because it does not satisfy
the converse of the sequence lemma.
Definition 1.146. Two metrics d 1 and d 2 on a set X are said to be (topologically)
equivalent if d 1 and d 2 induce the same metric topology on X.
Exercise 1.147. Let d and ρ be metrics on a set X. If there exist positive real
numbers a, b and c such that for any x, y ∈ X,
prove that d and ρ are equivalent.
ad(x, y) ≤ bρ(x, y) ≤ cd(x, y),
Exercise 1.148. On R n , we can define the square metric ρ and the taxicab
metric d t by
• ρ(x, y) = max{|x 1 − y 1 |, . . . , |x n − y n |}
• d t (x, y) = ∑ n
i=1 |x i − y i |
for any x, y ∈ R n . Prove that ρ and d t are really metrics on R n , and both are
equivalent to the euclidean metric d. [Hint : use the previous exercise]
Unfortunately, for the metrizability of R ω , we do not have obvious generalizations
of the d or d t because of the convergence issue. However, we can modify ρ to
get the bounded version by letting
ρ(x, y) = sup{d(x i , y i ) | i ∈ N}

Phichet Chaoha 31
where d(a, b) = min{|a − b|, 1} denotes the standard bounded metric on R.
Definition 1.149. The metric ρ as defined above is called the uniform metric
on R ω , and the metric topology induced by ρ is called the uniform topology.
Exercise 1.150. Prove that the uniform topology on R ω is finer than the
product topology, and coarser than the box topology.
To obtain the product topology on R ω , we need to modify ρ to get a new metric
D as in the following theorem.
Theorem 1.151. The map D : R ω × R ω → R defined by
D(x, y) = sup{ d(x i, y i )
| i ∈ N}
i
is a metric and it induces the product topology on R ω .
Proof. The fact that D is a metric is quite easy to verify.
To show that the metric topology induced by D is finer than the product
topology, we let U be an open set in the product topology and x = (x 1 , x 2 , . . . ) ∈ U.
Then there is a basis element B in the product topology such that x ∈ B ⊆ U;
says B = B 1 × B 2 × · · · × B n × ∏ i>n R for some n, where each B i is open in R.
Then for each i = 1, . . . , n, there exists 0 < ϵ i ≤ 1 such that (x i − ϵ i , x i + ϵ i ) ⊆ B i .
Now, let ϵ = min{ ϵ i
i
|i = 1, 2, . . . , n} and we claim that B D (x, ϵ) ⊆ B. To see this,
let y ∈ B D (x, ϵ). Then for each i = 1, 2, . . . , n, we have |x i − y i | = d(x i , y i ) ≤
iD(x, y) < iϵ ≤ ϵ i ≤ 1. Hence, y i ∈ (x i − ϵ i , x i + ϵ i ) ⊆ B i for each i = 1, 2, . . . , n;
i.e., y ∈ B.
On the other hand, we will show that the product topology on R ω is finer than
the metric topology induced by D. Let U be an open set in the metric topology
induced by D and x = (x 1 , x 2 , . . . ) ∈ U. Then B D (x, ϵ) ⊆ U for some ϵ > 0. Now,
choose N ∈ N such that N > 1 ϵ . We claim that V = ∏ N
i=1 (x i−ϵ, x i +ϵ)× ∏ i>N R ⊆
B D (x, ϵ). Let y ∈ V . Clearly, for 1 ≤ i ≤ N, we have d(x i,y i )
i
≤ |x i−y i |
i
< ϵ i ≤ ϵ.
Also, since d is bounded above by 1, we clearly have d(x i,y i )
i
≤ 1 N
< ϵ for a i ≥ N.
It follows that
D(x, y) ≤ max{ d(x 1, y 1 )
1
That is y ∈ B D (x, ϵ) as desired.
, d(x 2, y 2 )
2
, . . . , d(x N , y N )
, 1 N N } < ϵ.
□

CHAPTER 2
Countability and Separation Axioms
1. The Countability Axioms
Definition 2.1. Let X be a space and x ∈ X. A collection B x of open subsets
of X is called a (local) basis at x if for each neighborhood U of x, there exists
B ∈ B x such that x ∈ B ⊆ U.
Definition 2.2. A space X is called
(1) first-countable if it has a countable basis at each point.
(2) second-countable if it has a countable basis.
(3) separable if it has a countable dense subset.
Theorem 2.3. If X is second-countable, then it is first-countable and separable.
Proof. Let X be second-countable. Then it is clear from the definition that
X is also first-countable. Now, let B be a countable basis for X. By the axiom of
choice, we can form a subset D of X by choosing one element from each set in B.
It follows that D is the desired countable dense subset of X.
□
Example 2.4. R n is clearly second-countable by taking the collection all products
of intervals with rational endpoints as a basis. Hence, it is first-countable by
the previous theorem. Moreover, it is separable since Q n is its countable dense
subset. Hence, R n satisfies all countability axioms.
Example 2.5. R l is first-countable and separable, but not second-countable.
It is easy to see that the intervals [x, x + 1 n ) form the basis at x ∈ R l, and clearly
Q is dense in R l . To see that R l is not second-countable, let B be a basis for R l .
Then, for x, x ′ ∈ R l , there exists B x , B x ′ ∈ B such that x ∈ B x ⊆ [x, x + 1) and
x ′ ∈ B x ′ ⊆ [x ′ , x ′ + 1). Clearly, inf(B x ) = x because {x} ⊆ B x ⊆ [x, x + 1), and
similarly, inf(B x) ′ = x ′ . Hence, if x ≠ x ′ , we must have B x ≠ B x ′. So, B contains
an uncountable subcollection {B x | x ∈ R l }, and hence B is uncountable.
Exercise 2.6. Let X be a second-countable space. Prove that
(1) Every collection of disjoint open sets in X is countable.
(2) The subspace topology on an uncountable subset of X cannot be discrete.
Theorem 2.7. A subspace of a first-countable [second-countable] space is firstcountable
[second-countable]. A countable product of first-countable [second-countable]
spaces is first-countable [second-countable].
Proof. The first statement is clear since we can construct a basis element
of a subspace simply by intersecting a basis element of the whole space with the
subspace. For a countable product, we use the fact that a countable union of
countable sets and a finite product of countable sets are countable. For example,
33

34 Topology (2301631)
let {X n } n∈N be a countable collection of second countable spaces and B n a countable
basis of X n . Then
B = ∪ (
∏ n
B i × ∏ )
{X i }
n∈N i=1 i>n
can be shown to be a countable basis for the product space ∏ n∈N X n. Notice the
use of the product topology here!
□
Exercise 2.8. Prove that a countable product of a separable space is also
separable.
Example 2.9. R ω satisfies all countability axioms by the previous theorem and
exercise.
The next example shows that a subspace of a separable space may not be
separable.
Example 2.10. Since R l is separable (Q is also dense in R l ), then from the
previous exercise, so is R 2 l
. However, the subspace L = {(x, y) | x + y = 1} of
R 2 l
is not separable because the topology on L is discrete (i.e., for each (x, y) ∈ L,
we have ([x, x + 1) × [y, y + 1)) ∩ L = {(x, y)} which is open in L). Therefore, the
uncountable space L cannot have a countable dense subset.
As we have seen in Theorem 2.3, the separability is generally weaker than
second-countability. However, these two concepts are equivalent when the space is
metrizable.
Theorem 2.11. A metric space is first-countable, and it is separable iff it is
second-countable.
Proof. Let X be a metric space. Then, for each x ∈ X, the collection
{B(x; 1 n
) | n ∈ N} forms a countable basis at x. Now, if X is separable, we let D be a
countable dense subset of X. To see that the collection {B(p; 1 n
) | p ∈ D and n ∈ N}
forms a countable basis for X, let U be an open subset of X and x ∈ U. Then there
is n ∈ N such that B(x; 1 1
n
) ⊆ U. Since D is dense in X, there exists p ∈ D∩B(x;
2n ).
Now it is easy to verify that x ∈ B(p; 1
2n ) ⊆ B(x; 1 n
) ⊆ U. The converse follows
directly from Theorem 2.3.
□
Corollary 2.12. R l is not metrizable.
Theorem 2.13. The converse of the sequence lemma holds for a first-countable
space (e.g. a metric space).
Proof. Let X be a first-countable space, A ⊆ X and a ∈ A. Let B a =
{B 1 , B 2 , . . . } be a countable basis at a. For each n ∈ N, let B n ′ = B 1 ∩ · · · ∩ B n .
Clearly, each B n ′ is open and B n+1 ′ ⊆ B n ′ ⊆ B n . Since a ∈ A, we can form a
sequence (a n ) in A by choosing a n ∈ B n ′ ∩ A for each n. To see that the sequence
(a n ) converges to a, we let U be a neighborhood of a. Since B a is a basis at a, there is
N ∈ N such that a ∈ B N ⊆ U. Then for n ≥ N, we have a n ∈ B n ′ ⊆ B N ′ ⊆ B N ⊆ U
as desired.
□
Corollary 2.14. The space S Ω is not first-countable.
Exercise 2.15. Is S Ω separable? Justify your answer.
Exercise 2.16. Which one of our three countability axioms does S Ω satisfy?
Justify your answer.

Phichet Chaoha 35
Definition 2.17. Let X be a space.
2. The Separation Axioms
(1) X is a T 0 −space if for any two distinct points in X, we can find a neighborhood
of one of the point not containing the other.
(2) X is a T 1 −space or a quasi-separated space if for any two distinct points
in X, we can find a neighborhood of each point not containing the other.
(3) X is a T 2 −space or a separated space or a Hausdorff space if any two
distinct points of X have disjoint neighborhoods.
(4) X is a T 3 −space or a regular space if it is T 1 and any point and any closed
subset not containing that point have disjoint neighborhoods.
(5) X is a T 3 1 −space or a completely regular space if it is T 1 and any closed
2
subset A of X and any point x /∈ A, there exists a continuous map f :
X → [0, 1] such that f(x) = 0 and A ⊆ f −1 ({1}). (Note that if A = ∅,
then f is the zero map)
(6) X is a T 4 −space or a normal space if it is T 1 and any two disjoint closed
subsets have disjoint neighborhoods.
(7) X is a T 5 −space or a completely normal space if it is T 1 and any two
separated subsets A, B of X (i.e. A ∩ B = A ∩ B = ∅) have disjoint
neighborhoods.
Remark 2.18. We need X to be T 1 in the definition of T i for i ≥ 3 because
the space {a, b} with indiscrete topology satisfies the other part of the definition,
but it is not even Hausdorff!
Example 2.19. X = {a, b} with the topology {∅, {a}, {a, b}} is clearly T 0 , but
not T 1 .
Exercise 2.20. Prove that X is T 0 iff {x} ̸= {y}, for any distinct points x, y
in X.
Theorem 2.21. A space X is T 1 iff every singleton is closed.
Proof. (⇒) Suppose X is T 1 and let x ∈ X. For each y ≠ x, we can find a
neighborhood U y not containing x. Then X − {x} = ∪ y≠x U y is open, and hence
{x} is closed.
(⇐) Suppose that every singleton is closed. Then, for distinct points x, y ∈ X,
the open sets X − {y} and X − {x} are the desired neighborhoods of x and y,
respectively.
□
Corollary 2.22. A space X is T 1 iff every finite subset of X is closed.
Remark 2.23. Using the previous theorem, it is not difficult to see that a
T i -space is also a T i−1 -space for i = 1, 2, 3, 4, 5. Moreover, it is straightforward to
verify that a T 3 1 -space is a T 3-space. Leter on, we will also see that a T
2 4 -space is
a T 3 1 -space.
2
Theorem 2.24. A space X is Hausdorff iff the diagonal
is closed in X × X.
∆ = {(x, x) | x ∈ X}

36 Topology (2301631)
Proof. Suppose X is T 2 . Let (x, y) ∈ X × X − ∆; i.e., x ≠ y. Then there are
neighborhoods U x and U y of x and y, respectively, such that U x ∩ U y = ∅. This
implies that V = U x × U y is open in X × X and (x, y) ∈ V ⊆ X × X − ∆. Thus,
∆ is closed in X × X. We can prove the other direction simply by reversing the
above argument.
□
Exercise 2.25. If f, g : X → Y are continuous maps and Y is Hausdorff, then
the coincidence set
C(f, g) = {x ∈ X | f(x) = g(x)}
is closed in X. In particular, for a continuous self-map f of a Hausdorff space X,
the fixed point set
is closed in X.
F (f) = C(f, id X ) = {x ∈ X | f(x) = x}
Theorem 2.26. Let D be a dense subset of X. If f : D → Y is a continuous
map and Y is Hausdorff, then there is at most one continuous extension of f to X.
Proof. Suppose g, h : X → Y are continuous extensions of f; i.e., g| D = h| D .
Let x ∈ X and suppose that g(x) ≠ h(x). Since Y is Hausdorff, there exist disjoint
neighborhoods U ′ and V ′ of g(x) and h(x), respectively. By continuity of g and h,
the sets U = g −1 (U ′ ), V = h −1 (V ′ ) and W = U ∩ V are all neighborhoods of x.
Since D is dense in X, there exists d ∈ D ∩ W . Since both h and g are extensions
of f, we have g(d) = f(d) = h(d) ∈ U ′ ∩ V ′ . This contradicts to the fact that
U ′ ∩ V ′ = ∅. Therefore, g(x) = h(x).
□
Remark 2.27. The previous theorem only guarantees the uniqueness, not the
existence, of an extension. However, we will see later on that if f is ”uniformly”
continuous, an extension always exists.
Theorem 2.28. A T 1 -space X is regular iff for each point x ∈ X and a neighborhood
U of x, there exists a neighborhood V of x such that V ⊆ V ⊆ U.
Proof. (⇒) Assume that X is regular. Let x ∈ X and U a neighborhood of
x. Then, by regularity, there exist disjoint neighborhoods V of x and W of X − U.
So, V ⊆ X − W and X − W is a closed set contained in U. Therefore, we have
x ∈ V ⊆ V ⊆ X − W = X − W ⊆ U as desired.
(⇐) Assume that, for each x ∈ X and a neighborhood U of x, there exists a
neighborhood V of x such that x ∈ V ⊆ V ⊆ U. Now, let x ∈ X and A be a closed
subset of X not containing x. By setting U = X −A which is clearly a neighborhood
of x, there exists a neighborhood V of X such that x ∈ V ⊆ V ⊆ U = X − A.
Hence, X − V is a neighborhood of A such that V ∩ (X − V ) = ∅.
□
Theorem 2.29. A T 1 -space X is normal iff for each closed set A in X and an
open set U containing A, there exists a neighborhood V of A such that V ⊆ V ⊆ U.
Proof. The proof is similar to the proof of the above theorem.
Exercise 2.30. Prove that every well-ordered set with the order topology is
normal. In particular, S Ω and S Ω are normal.
Exercise 2.31. Prove that a space X is completely normal iff every subspace
of X is normal.
□

Phichet Chaoha 37
Lemma 2.32. A metric space is normal.
Proof. Let A and B be disjoint closed subsets of metric space X. For each
a ∈ A and each b ∈ B, let ϵ a > 0 and ϵ b > 0 be such that
B(a; ϵ a ) ∩ B = ∅ and B(b; ϵ b ) ∩ A = ∅.
Now, let U = ∪ a∈A B(a; ϵ a
2
), and V = ∪ b∈B B(b; ϵ b
2
). Clearly, U and V are open sets
containing A and B, respectively. To see that U ∩V = ∅, suppose there is z ∈ U ∩V .
Then, z ∈ B(a; ϵ a
2
) ∩ B(b; ϵ b
2
) for some a ∈ A and b ∈ B. WLOG, we may assume
that ϵ b ≤ ϵ a . So, we have d(a, b) ≤ d(a, z) + d(z, b) < ϵa 2 + ϵ b
2
≤ ϵa 2 + ϵa 2 = ϵ a which
is impossible because B(a; ϵ a ) ∩ B = ∅.
□
Corollary 2.33. A metric space is completely normal.
Proof. Since every subspace of a metric space X is also a metric space, and
hence normal, then X itself is completely normal.
□
Theorem 2.34. A second-countable regular space is normal.
Proof. Let A and B be disjoint closed subsets of a second-countable regular
space X, and B a countable basis of X. By regularity, we can find countable
subcollections {U 1 , U 2 , . . . } and {V 1 , V 2 , . . . } of B such that A ⊆ ∪ ∞ i=1 U i, B ⊆
∪ ∞ i=1 V i, U n ∩ B = ∅ and V n ∩ A = ∅ for all n ∈ N.
Since ∪ ∞ i=1 U i and ∪ ∞ i=1 V i may not be disjoint, then for each n ∈ N, we let
U n ′ = U n − ∪ n i=1 V i and V n ′ = V n − ∪ n i=1 U i. Clearly, U n’s ′ and V n’s ′ are open and
pairwise disjoint. Now, it is rather straightforward to check that U ′ = ∪ n≥1 U n ′ and
V ′ = ∪ n≥1 V n ′ are disjoint neighborhoods of A and B, respectively.
□
Theorem 2.35 (Urysohn Lemma). Let A and B be disjoint closed subsets of a
normal space X, and [a, b] a closed in interval in R. Then there exists a continuous
map f : X → [a, b] such that A ⊆ f −1 ({a}) and B ⊆ f −1 ({b}).
Proof. WLOG, we can assume that A ≠ ∅ ̸= B, a = 0 and b = 1.
We will construct a continuous map f : X → [0, 1] with the property that
f(A) = {0} and f(B) = {1}. First let P = Q ∩ [0, 1] and fix a representation
of P in such a way that the first two elements are 1 and 0 (for example, P =
{1, 0, 1 2 , 1 3 , 2 3 , . . . }). Let P n be the subset of the first n elements of P according to
the representation we have just fixed.
First let U 1 = X − B and, by normality, let U 0 be an open set such that
A ⊆ U 0 ⊆ U 0 ⊆ U 1 . Hence, we have defined U i for all i ∈ P 2 . Inductively, we
assume that for some n ≥ 2, U i is defined for all i ∈ P n so that U i ⊆ U j for i < j.
Now, suppose P n+1 = P n ∪ {r}. Since r /∈ {0, 1}, we can find in P n the immediate
predecessor and the immediate successor of r (according to the usual ordering of
R), says p and q, respectively. Again, by normality, there exists an open set U r
such that U p ⊆ U r ⊆ U r ⊆ U q ; i.e., U i is now defined for all i ∈ P n+1 . Hence, we
obtain the collection {U i } i∈P with the desired property. Moreover, we can extend
the definition of U i to all i ∈ Q simply by letting
{
∅ for i ∈ Q ∩ (−∞, 0)
U i =
X for i ∈ Q ∩ (1, ∞).
Finally, we define f : X → [0, 1] by
f(x) = inf{i | x ∈ U i }

38 Topology (2301631)
for all x ∈ X. Clearly, we have f(A) = {0} and f(B) = {1} as desired. For the
continuity of f, we leave it as the next exercise.
□
Exercise 2.36. Verify the continuity of the map f in the theorem above.
Remark 2.37. From the theorem above, we usually says that A and B can
be separated by a continuous map. The reason we do not write f(A) = {0} and
f(B) = {1} is because A or B may be empty.
Corollary 2.38. A normal space is always completely regular.
Recall : (Weierstrass M-test)
Let (f n : X → R) be a sequence of maps. If, for each n ≥ 1, there exists
M n ∈ R such that |f n (x)| ≤ M n for all x ∈ X, and the series ∑ ∞
n=1 M n converges,
then the series of maps ∑ ∞
n=1 f n converges uniformly to a map f : X → R.
Theorem 2.39 (Tietze Extension Theorem). Let A be a closed subset of a
normal space X. Then any continuous map f : A → [a, b] can be extended to a
continuous map g : X → [a, b].
Proof. WLOG, we may assume that A ≠ ∅, a = −1 and b = 1.
First, we claim that for any r > 0 and a continuous map f : A → [−r, r], there
exists a continuous map g : X → [− r 3 , r 2r
3
] such that |f(a) − g(a)| ≤
3
for all a ∈ A.
To prove this claim, we divide [−r, r] into 3 intervals, namely I 1 = [−r, − r 3 ],
I 2 = [− r 3 , r 3 ] and I 3 = [ r 3 , r]. Let B = f −1 (I 1 ) and C = f −1 (I 3 ). Then B and
C are disjoint closed subsets of X because f is continuous and A is closed in X.
By the Urysohn lemma, there exists a continuous map g : X → [− r 3 , r 3
] such that
B ⊆ g −1 ({− r 3 }) and C ⊆ g−1 ({ r 3
}). Also, it is not difficult to check (by considering
3 cases where a ∈ B, a ∈ C and a /∈ B ∪ C) that |f(a) − g(a)| ≤ 2r
3
for all a ∈ A.
Therefore, we have proved the claim.
Now, apply the above claim to f : A → [−1, 1] in the condition of this theorem.
We get g 1 : X → [− 1 3 , 1 3 ] such that |f(a) − g 1(a)| ≤ 2 3
for all a ∈ A. Apply the
claim again to (f − g 1 ) : A → [− 2 3 , 2 3 ], we get g 2 : X → [−( 1 3 )( 2 3 ), ( 1 3 )( 2 3
)] such that
|f(a) − g 1 (a) − g 2 (a)| ≤ ( 2 3 )2 for all a ∈ A. Inductively, by applying the claim to
the map (f − g 1 − g 2 − · · · − g n ) : A → [−( 2 3 )n , ( 2 3 )n ], we get a continuous map
g n+1 : X → [−( 1 3 )( 2 3 )n , ( 1 3 )( 2 3 )n ] such that
n+1
∑
|f(a) − g i (a)| = |f(a) − g 1 (a) − g 2 (a) − · · · − g n (a) − g n+1 (a)| ≤ ( 2 3 )n+1 . . . (∗)
i=1
for all a ∈ A.
Since, for each n ≥ 1 and x ∈ X, |g n (x)| ≤ ( 1 3 )( 2 3 )n−1 and ∑ ∞
n=1 ( 1 3 )( 2 3 )n−1 converges,
Weierstrass M-test implies that the sequence of maps ( ∑ n
i=1 g i) converges
uniformly on X, says to a g. Clearly, g is continuous (since each ∑ n
i=1 g i is continuous)
and |g(x)| ≤ 1 (by comparing with the geometric series ∑ ∞
n=1 ( 1 3 )( 2 3 )n−1 ).
Moreover, f(a) = g(a) for all a ∈ A (by *). Hence, g : X → [−1, 1] is the desired
continuous extension of f.
□
Exercise 2.40. Prove that if A be a closed subset of a normal space X, then
any continuous map f : A → R can be extended to a continuous map g : X → R.
Theorem 2.41. For a T 1 space X, the following statements are equivalent :
(1) X is normal.

Phichet Chaoha 39
(2) For any disjoint closed subsets A and B of X, there exists a continuous
map f : X → [0, 1] such that A ⊆ f −1 ({0}) and B ⊆ f −1 ({1}).
(3) For any closed subsets A of X, a continuous map f : A → [0, 1] can be
extended to a continuous map g : X → [0, 1].
Proof. (1) ⇒ (2) : Urysohn lemma.
(1) ⇒ (3) : Tietze extension theorem.
(2) ⇒ (1) : Let A, B be disjoint closed subsets of X. By (2), there exists
a continuous map f : X → [0, 1] such that A ⊆ f −1 ({0}) and B ⊆ f −1 ({1}).
Then, U = f −1 [0, 1 2 ) and V = f −1 ( 1 2
, 1] for disjoint neighborhoods of A and B,
respectively.
(3) ⇒ (2) : Let A, B be disjoint closed subsets of X. By the pasting lemma,
we have a continuous map f : A ∪ B → [0, 1] defined by f(x) = 0 if x ∈ A, and
f(x) = 1 if x ∈ B. Then, by (3), f can be extended the desired map.
□
The following theorem summarizes some other properties of T i -spaces, the
proofs are straightforward and hence left to the reader.
Theorem 2.42. Let i ≤ 3 1 2 .
(1) A subspace of a T i -space is also a T i -space.
(2) A product T i -spaces is a T i -space.
Proof. Exercise.
Example 2.43 (See [4] for details). The product space S Ω ×S Ω is regular since
both S Ω and S Ω are normal (being well-ordered). However, S Ω × S Ω is not normal.
In fact, we will see later on that S Ω ×S Ω is normal (being compact and Hausdorff).
Hence, S Ω × S Ω also serves as an example of a subspace of a normal space which is
not normal.
Exercise 2.44. Prove that a closed subspace of a normal space is always normal.
We end this chapter by stating the Urysohn metrization theorem. The proof
can be found in [4].
Theorem 2.45 (Urysohn Metrization Theorem). A second-countable regular
space is metrizable.
Exercise 2.46. Give an example of a second-countable Hausdorff space that
is not metrizable.
□

CHAPTER 3
Connectedness
Definition 3.1. A separation (or a disconnection) of a space X is a pair {U, V }
of disjoint nonempty open subsets of X such that X = U ∪ V .
Remark 3.2. If {U, V } is a separation of X, then U and V are both open and
closed in X at the same time.
Definition 3.3. A space X is said to be connected if it has no separation.
Otherwise, it is said to be disconnected. Similary, a subset A of X is said to
be connected if it is connected as a subspace of X. Otherwise, it is said to be
disconnected.
Example 3.4. A discrete space X with |X| ≥ 2 is always disconnected. An
empty space and a one-point space are always connected.
Exercise 3.5. Prove that, for any −∞ ≤ a < b ≤ ∞, an open interval (a, b)
is connected. In particular, R is connected.
Exercise 3.6. Prove that X is disconnected iff there exists a continuous surjective
map f : X → Y where Y is any discrete space with 2 points.
Theorem 3.7. A space X is connected iff ∅ and X are the only subsets of X
that are both open and closed in X.
Proof. Follows directly from the definition.
Theorem 3.8. Let X be a space and Y ⊆ X. The pair {U, V } is a separation
of Y iff U and V are nonempty separated sets in X such that U ∪ V = Y .
Proof. (⇒) Let {U, V } be a separation of Y . Then, U and V are disjoint
nonempty open subsets of Y such that U ∪ V = Y . It remains to show that
U and V are separated in X. Since both U and V are also closed in Y , we have
Cl Y (U) = U and Cl Y (V ) = V . It follows that U ∩V = (U ∩Y )∩V = U ∩(Y ∩V ) =
U ∩ Cl Y (V ) = U ∩ V = ∅, and similarly, U ∩ V = ∅ as desired.
(⇐) Now, let U and V be nonempty separated sets in X such that U ∪ V = Y .
Clearly, U and V are disjoint subsets of Y . Since Cl Y (U) ∩ V ⊆ U ∩ V = ∅, it
follows that Cl Y (U) ⊆ Y − V = U; i.e., U is closed in Y . Similarly, V is also closed
in Y . Therefore, both U and V are open, and hence forms a separation of Y . □
Example 3.9. In R, the pair {[−1, 0), (0, 1]} forms a separation of the subspace
Y = [−1, 0) ∪ (0, 1], while the pair {[−1, 0], (0, 1]} does not form a separation of the
subspace Z = [−1, 1].
Lemma 3.10. If {U, V } is a separation of X and Y is a nonempty connected
subspace of X, then either Y ⊆ U or Y ⊆ V .
41
□

42 Topology (2301631)
Proof. Since U and V are both open and closed in X, it follows that U ∩ Y
and V ∩ Y are both open and closed in Y . Since Y is connected, either U ∩ Y = Y
or V ∩Y = Y (and the other one will be ∅). Therefore, either Y ⊆ U or Y ⊆ V . □
Theorem 3.11. Let {X α } α∈Λ be a collection of connected spaces such that
X α ∩ X β ≠ ∅ for each α, β ∈ Λ. Then ∪ α∈Λ X α is connected.
Proof. Suppose ∪ α∈Λ X α is disconnected with a separation {U, V }. We claim
that either X α ⊆ U for all α ∈ Λ, or X α ⊆ V for all α ∈ Λ. The claim will imply
that either V = ∅ or U = ∅, which is contradiction. Hence, ∪ α∈Λ X α must be
connected. To prove the claim, suppose X β ⊆ U and X γ ⊆ V for some β, γ ∈ Λ.
Since X β ∩ X γ ≠ ∅, it follows that U ∩ V ≠ ∅ which is a contradiction. Therefore,
we have proved the claim.
□
Corollary 3.12. The union of a collection of connected spaces that have a
point in common is connected.
Theorem 3.13. If A is a connected subset of X and A ⊆ B ⊆ A, then B is
also connected.
Proof. Suppose B is disconnected with a separation {U, V }. Then U and V
are nonempty separated sets in X such that U ∪V = B. Since A is connected, then
either A ⊆ U or A ⊆ V . WLOG, assume that A ⊆ U. Then, A ⊆ U and hence
V = B ∩ V ⊆ A ∩ V ⊆ U ∩ V = ∅ which is a contradition. Therefore, B must be
connected.
□
Corollary 3.14. If A is a connected subset of X, then so is A.
Example 3.15. The intervals [a, b], [a, b) and (a, b] are connected.
Theorem 3.16. If X is a connected space and f : X → Y is a continuous map,
then f(X) is also connected.
Proof. WLOG, assume that f is surjective; i.e., f(X) = Y . Suppose Y is
disconnected with a separation {U, V }. Then it is easy to verify that the pair
{f −1 (U), f −1 (V )} forms a separation of X which contradicts to the fact that X is
connected. Therefore, Y must be connected.
□
Corollary 3.17. Connectedness is a topological property.
Corollary 3.18 (Intermediate Value Theorem). If f : [a, b] → R is a continuous
map and y 0 is a point between f(a) and f(b), then there exists x 0 ∈ (a, b) such
that f(x 0 ) = y 0 .
Proof. Let y 0 be a point between f(a) and f(b). Suppose y 0 /∈ f([a, b]).
Then, f : [a, b] → R − {y 0 } is clearly continuous. Since {(−∞, y 0 ), (y 0 , ∞)} forms
a separation of R − {y 0 } and f([a, b]) is connected by the previous theorem, then
either f([a, b]) ⊆ (−∞, y 0 ) or f([a, b]) ⊆ (y 0 , ∞) which contradicts to the fact that
y 0 is between f(a) and f(b). It follows that y 0 = f(x 0 ) for some x 0 ∈ [a, b]−{a, b} =
(a, b).
□
Theorem 3.19. An arbitrary product of connected spaces is connected.

Phichet Chaoha 43
Proof. First, we will prove that X ×Y is connected if X and Y are connected.
Fix (a, b) ∈ X ×Y and consider T x = (X ×{b})∪({x}×Y ). Clearly, T x is connected
for all x ∈ X because X×{b} ∼ = X, {x}×Y ∼ = Y and (X×{b})∩({x}×Y ) = {(x, b)}.
Now, since X × Y = ∪ x∈X T x and (a, b) ∈ T x for all x ∈ X, it follows that X × Y
is also connected, and hence so is a finite product of connected spaces.
To prove this theorem for an arbitrary product, let X = ∏ α∈Λ X α be a product
of connected spaces X α ’s. WLOG, we may assume that each X α has more than
one element. Fix a point b = (b α ) ∈ X, and let X(α 1 , . . . , α n ) denote the subspace
of X containing all points (x α ) such that x α = b α for α /∈ {α 1 , . . . , α n }. It is not
difficult to see that X(α 1 , . . . , α n ) ∼ = X α1 × · · · × X αn and hence it is connected
from the previous paragraph. Now, let Y = ∪ α 1 ,...,α n ∈Λ X(α 1, . . . , α n ) ⊆ X. Since
b ∈ X(α 1 , . . . , α n ) for any α 1 , . . . , α n ∈ Λ, then Y is connected. Note that Y ≠ X
because, for example, the point (x α ) with x α ≠ b α for all α ∈ Λ is not in Y .
However, we can show that Y = X. For a point (x α ) ∈ X and a basis neighborhood
(in the product topology) ∏ α∈Λ U α of (x α ), we have U α ≠ X α for finitely many
α’s, says α 1 , . . . , α n . Let (y α ) ∈ X be defined by
{
x α if α ∈ {α 1 , . . . α n },
y α =
b α if α /∈ {α 1 , . . . α n }.
It is easy to see that
(y α ) ∈ X(α 1 , . . . , α n ) ∩ ∏ α∈Λ
U α ⊆ Y ∩ ∏ α∈Λ
U α .
Therefore, Y = X and hence, by Corollary 3.14, X is also connected as desired.
Definition 3.20. Let x and y be points in a space X. A path (in X) from x
to y is a continuous map p : [0, 1] → X such that p(0) = x and p(1) = y. A loop
(in X) at x is simply a path from x to x.
Definition 3.21. A space X is said to be path connected if each pair of points
in X can be joined by a path in X.
Example 3.22. Both R n and D n = {x ∈ R n : ||x|| ≤ 1} are path connected
for n ≥ 0, while S n = ∂D n+1 is path connected for n > 0.
Exercise 3.23. Prove that a continuous image of a path connected space is
always path connected, and use this fact to show that R cannot be homeomorphic
to R 2 .
Exercise 3.24. Let X and Y be path connected spaces. Prove that X × Y is
path connected, and if X ∩ Y ≠ ∅, then X ∪ Y is also path connected.
Theorem 3.25. A path connected space is always connected.
Proof. Let X be a path connected space. Suppose X is not connected with
a separation {U, V }. Since a path in X is a continuous map, then the image of
any path in X (which is connected) must lie entirely in either U or V . That is we
cannot join a point in U and a point in V by a path. This contradicts to the fact
that X is path connected. Hence, X must be connected.
□
Example 3.26. In R 2 , let
A = (I × {0}) ∪ ({ 1 | n > 0} × I)
n
□

44 Topology (2301631)
and the comb space
C = A ∪ ({0} × I).
Clearly, both A and C are path connected. Moreover, since A is connected and
C = A, then C is also connected. However, the deleted comb space
C ∗ = C − ({0} × (0, 1))
is connected (A ⊆ C ∗ ⊆ A = C) but not path connected.
Exercise 3.27. The topologist’s sine curve (denoted by TSin) is a subspace of
R 2 defined by
TSin = Cl R 2({(x, sin( 1 )) | 0 < x ≤ 1}).
x
Prove that TSin is connected, but not path connected.
Definition 3.28. For a space X, we can define an equivalence relation (verify!)
on X by x ∼ y if x and y belong to the same connected subspace of X. The
equivalence classes of this relation are call the (connected) components of X.
Theorem 3.29. The components of X are connected disjoint subspaces of X
whose union is X and each connected subspace of X intersects only one of them.
Proof. Since ∼ is an equivalence relation, then the components of X are
disjoint and their union is X. Next, we will show that each connected subspace A
of X intersects only one of the components. Suppose A intersects the components
C 1 and C 2 . Then there exist x 1 ∈ A ∩ C 1 and x 2 ∈ A ∩ C 2 , which implies x 1 ∼ x 2
because x 1 and x 2 belong to the same connected subspace A of X. Since C 1 and
C 2 are equivalence classes, we must have C 1 = C 2 .
To show that the component C is connected, pick a point x 0 ∈ C. For each
x ∈ C, we have x ∼ x 0 and hence there exists a connected subspace A x of X
containing both x and x 0 . It is not difficult to see that C ⊆ ∪ x∈C A x. Also, since
for each x ∈ C, A x is connected and x 0 ∈ A x ∩ C, then A x ⊆ C (by the above
paragraph). Therefore, C = ∪ x∈C A x, and hence C is connected because x 0 ∈ A x
for all x ∈ X.
□
Remark 3.30. The previous theorem clearly implies that if A is a connected
subspace of X and C is a component of X such that A ∩ C ≠ ∅, the we must have
A ⊆ C.
Definition 3.31. For a space X, we can define an equivalence relation (verify!)
on X by x ∼ p y if there is a path in X from x to y. The equivalence classes of this
relation are called the path components of X.
Theorem 3.32. The path components of X are path connected disjoint subspaces
of X whose union is X and each path connected subspace of X intersects
only one of them.
Proof. Similar to the previous theorem.
Example 3.33. The deleted comb space C ∗ has only one component, but two
path components. Moreover, C ∗ ∪ {(0, q) | q ∈ Q ∩ (0, 1)} also has one component,
but countably many path components.
Exercise 3.34. Prove that a component of X is always closed in X. Can we
say the same thing for path component? Why?
□

Phichet Chaoha 45
Definition 3.35. A space X is said to be locally connected at x if each neighborhood
of x contains a connected neighborhood of x. We say that X is locally
connected if it is locally connected at each of its points.
Definition 3.36. A space X is said to be locally path connected at x if each
neighborhood of x contains a path connected neighborhood of x. We say that X is
locally path connected if it is locally path connected at each of its points.
Example 3.37. The subspace [−1, 0) ∪ (0, 1] of R is locally connected and
locally path connected, but it is neither connected nor path connected. On the
other hand, TSin is neither locally connected nor locally path connected.
Theorem 3.38. A space X is locally connected iff for each open subset U of
X, each component of U is open in X
Proof. Suppose X is locally connected. Let U be an open subset of X, C
the component of U and x ∈ C. Then by local connectedness (at x), U contains a
connected neighborhood V of x. Since C is a component of U and V is a connected
subset of U, we must have V ⊆ C. Therefore, C is open in X.
Conversely, assume that each component of an open subset of X is also open
in X. Let x ∈ X, U a neighborhood of x and C the component of U containing x.
By assumption, C is open in X and hence it is the desired connected neighborhood
of x contained in U.
□
Theorem 3.39. A space X is locally path connected iff for each open subset U
of X, each path component of U is open in X
Proof. Similar to the previous theorem.
Remark 3.40. The previous two theorems immediately imply that a (path)
component of a locally (path) connected space is always open.
Theorem 3.41. A connected and locally path connected space is path connected.
Proof. Let X be a connected, locally path connected space. Fix x 0 ∈ X
and let P be the path component of X containing x 0 . Then P ≠ ∅. Since X is
locally path connected, the path connected component P must be open in X by the
previous theorem. Moreover, if y ∈ X − P , the path component Q of X containing
y must be open in X. Since y ∈ Q ⊆ X − P , the set X − P must be open in X.
Now, if X − P ≠ ∅, then {P, X − P } will clearly form a separation of X which
contradicts to the assumption. Therefore, X − P = ∅; i.e., P = X.
□
Exercise 3.42. Prove that an open connected subset of R n is always path
connected.
Exercise 3.43. Discuss the connectedness, path connectedness, local connectedness
and local path connectedness of Q and R − Q.
Definition 3.44. A space X is said to be totally disconnected if only connected
subspaces of X are one-point sets.
Example 3.45. Q is totally disconnected. For if a connected subspaces A of
Q contains distinct points p < q, we can find an irrational number s such that
p < s < q and hence {A ∩ (−∞, s), A ∩ (s, ∞)} will form a separation of A which
is a contradiction.
□

CHAPTER 4
Compactness
Definition 4.1. Let X be a space and C a collection of subsets of X. We say
that C is a covering of X (or C covers X) if ∪ C = X. A subcollection C that still
covers X is called a subcovering of C. If each element of C is open in X, we will
call C an open covering. Moreover, a space X is said to be compact if every open
covering has a finite subcovering.
Definition 4.2. When X is a subspace of Y , we define a [open] covering of X
in Y to be a collection of [open] subsets of Y whose union contains X. Therefore, a
(open) covering of a space X in the previous definition is simply a [open] covering
of X in X.
Theorem 4.3. Let X and Y be spaces such that X ⊆ Y . Then X is compact
iff every open covering of X in Y has a finite subcovering.
Proof. Suppose X is compact and let C be an open covering of X in Y .
It follows that the collection {A ∩ X | A ∈ C} forms an open covering of X (in
X). Then, by compactness of X, it has a finite subcovering, says {A 1 ∩ X, A 2 ∩
X, . . . , A n ∩ X}. Now, it is clear that {A 1 , A 2 , . . . , A n } is a finite subcovering of C.
Conversely, assume that every open covering of X in Y has a finite subcovering.
Let D be an open covering of X (in X). For each D ∈ D, there is an open subset U D
of Y such that D = U D ∩ X. Then, by axiom of choice, we have an open covering
{U D |D ∈ D} of X in Y and, by assumption, it reduces to a finite subcovering, says
{U D1 , U D2 , . . . , U Dn }. Now, it is easy to verify that {D 1 , D 2 , . . . , D n } is a finite
subcovering of D. Therefore, X is compact.
□
Example 4.4. Any finite discrete space is compact, but an infinite discrete
space is not.
Example 4.5. R is not compact because {(n, n+2) | n ∈ Z} is an open covering
of R that does not have a finite subcovering.
Example 4.6. A finite union of compact spaces is compact.
Exercise 4.7. Show that { 1 n | n ∈ N} is compact, but { 1 n
| n ∈ N} is not.
Remark 4.8. There are some other types of compactness : a space X is called
• countably compact if every countable open covering of X has a finite subcovering.
• limit point compact if every infinite subset of X has a limit point.
• sequentially compact if every sequence in X has a convergent subsequence.
It is immediate from the definitions that compactness implies countable compactness.
In general, one can show that (see [4] for details):
47

48 Topology (2301631)
compact
⇓
sequentially compact ⇒ countably compact ⇒ limit point compact
When the space is metrizable, we also have the following implications :
limit point compact ⇒ sequentially compact ⇒ compact
Therefore, the notions of compactness above are all equivalent for metric spaces,
and hence any closed interval [a, b] in R is compact since it is limit point compact
by the famous Bolzano-Wierestrass theorem.
Exercise 4.9. Show that the space X = N × {1, 2} indiscrete is not compact,
but limit point compact.
Exercise 4.10. Prove that a space X is compact iff every net in X has a
convergent subnet.
Theorem 4.11. A closed subspace of a compact space is compact.
Proof. Let C be a closed subspace of a compact space X and C a collection
of open subsets of X such that C ⊆ ∪ C. Then the collection D = C ∪ {X − C} is
clearly an open covering of X. Since X is compact, there is a finite subcollection
D ′ of D such that X = ∪ D ′ . Let C ′ = D ′ − {X − C}. Clearly, C ′ is finite and
∪
C ′ = ∪ (D ′ − {X − C}) ⊇ ( ∪ D ′ ) − (X − C) = X − (X − C) = C
Hence, C ′ is a finite subcollection of C whose union still covers C; i.e., C is compact.
□
Lemma 4.12. If A is a compact subspace of a Hausdorff space X and x ∈ X−A,
then there exist neighborhoods U and V of x and A, respectively, such that U∩V = ∅.
Proof. Since X is Hausdorff, for each a ∈ A, there exist neighborhoods U a
and V a of x and a, respectively, such that U a ∩ V a = ∅. Clearly, {V a } a∈A is an open
covering of A, and since A is compact, we obtain a finite subcovering {V ai } n i=1 .
Then, by letting U = ∩ n i=1 U a i
and V = ∪ n i=1 V a i
, it is easy to see that both are
open in X. Moreover, if there is y ∈ U ∩ V , we will have y ∈ U ak ∩ V ak for some
k = 1, . . . , n. This contradicts to the fact that U ak ∩ V ak = ∅. Hence, we must have
U ∩ V = ∅.
□
Theorem 4.13. A compact subspace of a Hausdorff space is closed.
Proof. Let A be a compact subspace of a Hausdorff space X, and x /∈ A. It
follows directly from the previous lemma that there is a neighborhood U of x such
that U ∩ A = ∅. Therefore, A is closed in X.
□
Corollary 4.14. An arbitrary intersection of compact subspace of a Hausdorff
space is also compact.
Proof. Let K be a collection of compact subspaces of a Hausdorff space X.
Since X is Hausdorff, each K ∈ K is also closed in X by the previous theorem.
Hence, ∩ K is also closed in each K ∈ K. Since each K ∈ K is compact, by
Theorem 4.11, ∩ K is then compact.
□
Theorem 4.15. A compact Hausdorff space is normal.

Phichet Chaoha 49
Proof. Let X be a compact Hausdorff space. Let A and B be disjoint
closed subsets of X. By Theorem 4.11, the sets A and B are compact. Also,
by Lemma 4.12, for each a ∈ A, there are disjoint neighborhoods U a of a and V a of
B. Since {U a } a∈A forms an open covering of A and A is compact, there is a finite
subcollection {U ai } n i=1 that covers A. By letting U = ∪n i=1 U a i
and V = ∩ n i=1 V a i
, it
is easy to check that U and V are disjoint neighborhoods of A and B, respectively.
Therefore, X is normal.
□
Theorem 4.16. If X is a compact space and f : X → Y is a continuous map,
then f(X) is also compact.
Proof. Suppose X is a compact space and f : X → Y is a continuous map.
Let {U α } α∈Λ be an open covering of f(X). Then, {f −1 (U α )} α∈Λ clearly forms
an open covering of X and, by compactness, it has a finite subcovering, says
{f −1 (U αi )} n i=1 . Hence, {U α i
} n i=1 is the desired finite subcollection of {U α} α∈Λ
that still covers f(X).
□
Example 4.17. A quotient space of a compact space is always compact.
Corollary 4.18. Compactness is a topological property.
Theorem 4.19. A bijective continuous map from a compact space onto a Hausdorff
space is a homeomorphism.
Proof. To show that f −1 is continuous, let C be a closed subset of X. Since X
is compact, then so is C (by Theorem 4.11). Since f is continuous, by the previous
theorem, f(C) is compact in Y . Finally, since Y is Hausdorff, it follows directly by
Theorem 4.13 that f(C) is closed in Y .
□
Example 4.20. Let C = {(x, y) : x 2 + y 2 = 1} ⊆ R 2 and f : I → C be given
by f(t) = (cos 2πt, sin 2πt). Clearly, f is continuous. Also, let p : I → I /∼ = S 1 be
the quotient map where ∼ is given by 0 ∼ 1. So, f is constant on each fiber of p
and hence it induces a continuous map ˜f : S 1 → C such that ˜f ◦ p = f. It is easy
to see that ˜f is bijective. Since S 1 is compact and C is Hausdorff, ˜f is in fact a
homeomorphism; i.e., S 1 ∼ = C.
Theorem 4.21 (Extreme Value Theorem). If X is a compact space and f :
X → R is a continuous map, then f attains its minimum and maximum on X.
Proof. Suppose that f does not attain its minimum on X; i.e., for each
z ∈ f(X), there exists y ∈ f(X) such that y < z. Then the collection {(y, ∞) | y ∈
f(X)} clearly forms an open covering of f(X). Since f(X) is compact, we obtain
a finite subcovering of f(X), says {(y 1 , ∞), (y 2 , ∞), . . . , (y n , ∞)}. Hence,
f(X) ⊆ (y min , ∞) where y min = min{y 1 , y 2 , . . . , y n }. This is clearly a contradition
since y min ∈ f(X) but y min /∈ (y min , ∞). Therefore, f must attain its minimum.
Similarly, we can show that f attains its maximum as well.
□
Definition 4.22. Let (X, d) be a metric space, ∅ ̸= A ⊆ X and x ∈ X. We
define the diameter of A, denoted by diam(A), and the distance between x and A,
denoted by d(x, A), as follows :
diam(A) = sup{d(a, b) | a, b ∈ A},
d(x, A) = inf{d(x, a) | a ∈ A}.

50 Topology (2301631)
If {d(a, b) | a, b ∈ A} is not bounded above, we simply write diam(A) = ∞. Otherwise,
we will say that A is bounded.
Exercise 4.23. Let A ≠ ∅ subset of a metric space (X, d). Prove that A is
bounded iff A ⊆ B d (x; R) for some x ∈ X and R > 0.
Exercise 4.24. Let C ≠ ∅ be a closed subset of a metric space (X, d) and
x /∈ C. Prove that d(x, C) > 0.
Theorem 4.25 (Lebesgue Lemma). If C is an open covering of a compact
metric space (X, d), then there is δ > 0 such that, for every subset A of X with
diam(A) < δ, we have A ⊆ C for some C ∈ C.
Proof. Let C be an open covering of a compact metric space (X, d). If X ∈ C,
we are done. Assume that X /∈ C. Since X is compact, C has a finite subcovering,
says {C 1 , C 2 , . . . , C n }. Now, define f : X → R by
f(x) = 1 n
n∑
d(x, X − C i )
i=1
for each x ∈ X. Since X /∈ C, then each X − C i is nonempty and hence f is
well-defined. It is not difficult to see that f is also continuous. Then f attains
its minimum, says δ. Note also that for each x ∈ X, there exists k = 1, 2, . . . , n
such that x ∈ C k and, since X − C k is closed, the previous exercise implies that
d(x, X − C k ) > 0. It follows that f(x) > 0 for all x ∈ X. Hence, δ > 0.
Finally, let A be any subset of X with diam(A) < δ, and a ∈ A. Clearly,
A ⊆ B(a; δ) and hence
f(a) = 1 n
n∑
d(a, X − C i ) ≤ d(a, X − C m ),
i=1
where d(a, X −C m ) = max{d(a, X −C 1 ), . . . , d(a, X −C n )} (depending on a). Since
δ is the minimum of f, we have f(a) ≥ δ and hence d(a, X − C m ) ≥ δ. This implies
B(a; δ) ⊆ C m and hence A ⊆ C m ∈ C as desired.
□
Definition 4.26. Let (X, d X ), (Y, d Y ) be metric spaces and f : X → Y a map.
We say that f is uniformly continuous if for each ϵ > 0, there is δ > 0 such that
for each x 1 , x 2 ∈ X, if d X (x 1 , x 2 ) < δ, we have d Y (f(x 1 ), f(x 2 )) < ϵ.
Clearly, uniform continuity implies continuity. However, when X is compact,
the two notions are equivalent.
Theorem 4.27 (Uniform Continuity Theorem). Let (X, d X ) and (Y, d Y ) be
metric spaces where X is compact. Any continuous map f : X → Y is uniformly
continuous.
Proof. Let ϵ > 0. Since {B(y, ϵ 2
) | y ∈ Y } is an open covering of Y and
f is continuous, then C = {f −1 (B(y, ϵ 2
)) | y ∈ Y } forms an open covering of X.
Since X is compact, there exists a Lebesgue number δ > 0 for C. Then for any
x 1 , x 2 ∈ X such that d X (x 1 , x 2 ) < δ, there exists y 0 ∈ Y such that {x 1 , x 2 } ⊆
f −1 (B(y 0 , ϵ 2 )) because diam{x 1, x 2 } < δ. It follows that f(x 1 ), f(x 2 ) ∈ B(y 0 , ϵ 2 )
and hence d Y (f(x 1 ), f(x 2 )) < ϵ. Therefore, f is uniformly continuous. □

Phichet Chaoha 51
Lemma 4.28 (Tube Lemma). Let X, Y be spaces such that Y is compact. If N
is an open subset of X × Y containing a slice {x 0 } × Y of X × Y , then there is a
neighborhood U of x 0 such that U × Y ⊆ N.
Proof. Let X, Y be spaces such that Y is compact. Let N be an open subset
of X × Y containing {x 0 } × Y of X × Y . For each y ∈ Y , let U y × V y be a basis
neighborhood of (x 0 , y) such that U y × V y ⊆ N. Then, {U y × V y } y∈Y is an open
covering of {x 0 } × Y such that ∪ y∈Y (U y × V y ) ⊆ N. Since Y is compact, then
so is {x 0 } × Y , and hence there is a finite subcollection {U yi × V yi } n i=1 that covers
{x 0 } × Y . Let U = ∩ n
i=1 U y i
, which is clearly a neighborhood of x 0 . Therefore, we
have {x 0 } × Y ⊆ U × Y = U × ( ∪ n
i=1 V y i
) = ∪ n
i=1 (U × V y i
) ⊆ ∪ n
∪
i=1 (U y i
× V yi ) ⊆
y∈Y (U y × V y ) ⊆ N.
□
Theorem 4.29. A finite product of compact spaces is compact.
Proof. By using the induction on the number of the factors, it suffices to
prove for a product of two compact spaces X and Y . Let C be an open covering of
X × Y . Then, for each x ∈ X, since Y is compact, there is a finite subcollection
C
∪x f of C that covers {x} × Y . Now, by applying the tube lemma to the open set
C
f
x containing {x} × Y , we get a neighborhood U x of x such that U x × Y ⊆ ∪ Cx.
f
Since X is compact, we can reduce the covering {U x } x∈X to a finite subcollection
{U xi } n i=1 that still covers X. Now, ∪ n
i=1 Cf x i
is a finite subcollection of C that covers
X × Y since
n∪
n∪
n∪
X × Y ⊆ ( U xi ) × Y = (U xi × Y ) ⊆ ( ∪ Cx f i
) = ∪ n∪
( Cx f i
).
i=1
Therefore, X × Y is compact.
i=1
i=1
Example 4.30. A cube [a, b] n is always compact.
Lemma 4.31. Let X be a subspace of a metric space (Y, d). If X is compact,
then X is closed and bounded.
Proof. Let X be a compact subspace of (Y, d) and fix y 0 ∈ Y . Then X is
closed by Theorem 4.13. Since {B d (y 0 ; r) | r > 0} is an open covering of X, we then
have X ⊆ B(y 0 ; R) for some R > 0 by compactness. Hence, X is also bounded. □
Theorem 4.32. A subspace X of R n is compact iff X is closed and bounded
(with respect to the euclidean metric on R n ).
Proof. Let X be a subspace of R n .
(⇒) Follows directly from the previous lemma.
(⇐) Suppose X is closed and bounded. Then X ⊆ B(x 0 ; R) for some x 0 ∈ R n
and R > 0. WLOG, we may assume that x 0 = 0 (otherwise, we may replace R
by R + d(x 0 , 0)). Since B(0; R) ⊆ [−R, R] n , we have X ⊆ [−R, R] n . Since the
cube [−R, R] n is compact by the previous theorem and X is closed, it follows from
Theorem 4.11 that X is compact.
□
Example 4.33. A closed ball B(x; R) and a sphere ∂B(x; R) in R n are compact.
In particular D n and S n are compact for each n ≥ 0.
Definition 4.34. Let X be a set and C a collection of subsets of X. We say
that C has the finite intersection property if every finite subcollection of C has a
nonempty intersection.
i=1
□

52 Topology (2301631)
Example 4.35. The collection {[−n, n] | n ∈ N} of closed subsets of R has the
finite intersection property and so does the collection {[n, ∞) | n ∈ N}.
Theorem 4.36. A space X is compact if and only if every collection of closed
subsets of X having the finite intersection property has a nonempty intersection.
Proof. (⇒) Assume that X is a compact space. Let C be a collection of
closed subsets of X having the finite intersection property. For each C ∈ C, let
A C = X − C which is open in X. Then, by DeMorgan’s law, we have
∩
A C .
C∈C
C = ∩ C∈C(X − A C ) = X − ∪ C∈C
If ∩ C∈C C = ∅, the collection {A C} C∈C will then form an open covering of X and it
can be reduced to a finite subcovering {A Ci } n i=1 by compactness of X. This implies
(again, by DeMorgan’s law) ∩ n
i=1 C i = ∅, which contradicts to the fact that C has
the finite intersection property.
(⇐) Assume that every collection of closed subsets of X having the finite intersection
property has a nonempty intersection. Let {U α } α∈Λ be an open covering
of X. It follows that the collection C = {X − U α } α∈Λ of closed subsets of X has
an empty intersection and hence it does not have the finite intersection property.
Then, there exists a finite subcollection {X − U αi } n i=1 (of C) whose intersection is
empty. Hence, by DeMorgan’s law, {U αi } n i=1 is the desired finite subcollection of
{U α } α∈Λ that still covers X. □
Example 4.37. From the previous example, it is easy to see that ∩ {[−n, n] | n ∈
N} ̸= ∅ while ∩ {[n, ∞) | n ∈ N} = ∅. Hence, R is not compact.
Exercise 4.38. Prove that a space X is compact if and only if, for every
collection A of subsets of X having the finite intersection property, ∩ A∈A A ≠ ∅.
Lemma 4.39. Let X be a set and A a collection of subsets of X having the
finite intersection property, then there exists a maximal collection D of subsets of
X with respect to the finite intersection property such that A ⊆ D.
Proof. Consider the set
F = {B | A ⊆ B ⊆ P(X) and B has the finite intersection property}
which is partially ordered by set inclusion. Clearly, F ≠ ∅. Moreover, for each
chain C ⊆ F, we clearly have A ⊆ ∪ C ⊆ P(X) and ∪ C has the finite intersection
property. For if {C i } n i=1 is a finite subcollection of ∪ C, then for each i, there exists
B i ∈ C so that C i ∈ B i and hence {C i } n i=1 ⊆ ∪ n
i=1 B i. Since ∪ n
i=1 B i = B k ∈ C
for some k ∈ {1, . . . , n} and B k have the finite intersection property, we must have
∩ {Ci } n i=1 = ∩ n
i=1 C i ≠ ∅. Therefore, ∪ C is an upper bound of C in F and the
existence of D follows directly from Zorn’s lemma.
□
Exercise 4.40. Let X be a set and D a maximal collection of subsets of X
with respect to the finite intersection property. Prove that any finite intersections
of elements of D belongs to D and any S ⊆ X that intersects every elements of D
belongs to D.
Theorem 4.41 (Tychonoff’s Theorem). An arbitrary product of compact spaces
is compact.

Phichet Chaoha 53
Proof. Let X = ∏ α∈Λ X α where X α is compact, and A a collection of subsets
∩
of X having the finite intersection property. By Exercise 4.38, we need show that
A∈A
A ≠ ∅. According to the previous lemma, let D be a maximal collection of
subsets of X with respect to the finite intersection property such that A ⊆ D. The
result will then follow once we can show that ∩ D∈D D ≠ ∅.
Let us fix α ∈ Λ for a moment and consider {π α (D)|D ∈ D}, where π α :
X → X α is the projection map. This collection clearly has the finite intersection
property since D does. Then, by compactness of X α and Exercise 4.38, there exists
x α ∈ ∩ D∈D π α(D). Note also that for each D ∈ D and a neighborhood U α of x α ,
since U α ∩ π α (D) ≠ ∅, then there is y ∈ D such that π α (y) ∈ U α ∩ π α (D) and hence
y ∈ πα
−1 (U α ) ∩ D. In other words, πα
−1 (U α ) intersects every element of D. Hence,
the previous exercise implies that πα −1 (U α ) ∈ D.
Now, let x = (x α ) α∈Λ and a basis neighborhood U = ∩ n
i=1 π−1 α i
(U αi ) of x.
Since U is a finite intersection of elements of D, we have U ∈ D by the previous
exercise. Since D has the finite intersection property, then U ∩ D ≠ ∅ for each
D ∈ D. It follows that x ∈ D for each D ∈ D, and hence x ∈ ∩ D∈D
D. Therefore,
D ≠ ∅ as desired.
□
∩
D∈D
Definition 4.42. A space X is said to be locally compact at x if there is a
compact subset of X containing a neighborhood of x. If X is locally compact at
each of its points, we simply say that X is locally compact.
Remark 4.43. Local compactness is a topological property, and a compact
space is always locally compact.
Example 4.44. R n is locally compact.
Exercise 4.45. Show that Q is not locally compact.
Definition 4.46. A compactification of a space X is a pair ( ̂X, ι) where ̂X is
a compact Hausdorff space and ι is an imbedding of X into ̂X as a dense subspace
(i.e., ι(X) = ̂X).
Remark 4.47. When there is no ambiguity, we may identify X with the subspace
ι(X), and simply regard ̂X as a compactifaction of X without specifying
ι. Moreover, if a space X has a compactification, it must be at least completely
regular since a subspace of a completely regular space is always completely regular
(but not true for a normal space!).
Definition 4.48. A compactification ( ̂X, ι) of a space X is said to be a onepoint
compactification if ̂X − ι(X) is a singleton.
Example 4.49. ([0, 1], (0, 1) ↩→ [0, 1]) is a compactification of (0, 1) while
(S 1 , f| (0,1) ), as in Example 1.78, is a one-point compactification of (0, 1).
Theorem 4.50. Let X be a non-compact space. Then X has a one-point compactification
iff it is locally compact Hausdorff.
Proof. (⇒) Let ( ̂X, ι) be a one-point compactification of X, says ̂X − ι(X) =
{∞}. Since ̂X is Hausdorff, then so is the subspace ι(X). Let x ∈ ι(X). Since
̂X is Hausdorff, there exist disjoint neighborhoods U and V of x and ∞ in ̂X,
respectively. Then ̂X − V is closed in ̂X, and hence compact. Since ̂X − V ⊆ ι(X),

54 Topology (2301631)
then ̂X − V is also a compact subset of ι(X) containing the neighborhood U of x;
i.e., ι(X) is locally compact at x. Therefore, X ∼ = ι(X) is locally compact Hausdorff.
(⇐) Let ̂X = X ∪ {∞}. We will call a subset U of ̂X open if U is an open
subset of X or U = ̂X − K for some compact subset K of X. It is easy to see
that the definition of open sets above in fact gives us a topology on ̂X since a finite
union and an arbitrary intersection (in a Hausdorff space) of compact sets are also
compact. Moreover, X is dense in ̂X because a neighborhood of ∞ must be of the
form ̂X − K and hence intersects X (since X itself is not compact!). To show that
̂X is Hausdorff, let x and y be distinct points in ̂X. Since X itself is Hausdorff,
we may assume that x ∈ X and y = ∞. By local compactness at x, there is a
compact subset K of X containing a neighborhood U of x. Then V = ̂X − K is
clearly a neighborhood of ∞ that is disjoint from U. For the compactness of ̂X,
we let C be an open covering of ̂X. Then there is an element A of C containing ∞.
Since A is open in ̂X, it must be of the form ̂X − K for some compact subset K of
X. Since C also covers K, there exists a finite subcollection {C 1 , C 2 , . . . , C n } that
still covers K. Now, it is easy to see that {A, C 1 , C 2 , . . . , C n } is the desired finite
subcollection of C that covers ̂X. Therefore, ̂X is compact, and ( ̂X, X ↩→ ̂X) is a
one-point compactification of X.
□
Definition 4.51. Two compactifications ( ̂X, ι) and ( ̂X ′ , ι ′ ) are said to be
homeomorphic if there is a homeomorphism h : ̂X → ̂X′ that makes the following
diagram commutative :
̂X
♣ ♣ ♣ ♣ ♣ ♣
h♣ ♣ ♣ ♣ ♣✲♣ ♣ ̂X′
❅■ ι ❅
X.
✒
ι ′
Theorem 4.52. A one-point compactification of a space X is unique up to a
homeomorphism.
Proof. Suppose ( ̂X, ι) and ( ̂X ′ , ι ′ ) are one-point compactifications of X, says
̂X − ι(X) = {∞} and ̂X ′ − ι ′ (X) = {ξ}. Define h : ̂X → ̂X′ by h(∞) = ξ and
h(x) = ι ′ ◦ ι −1 (x) for all x ∈ ι(X) . Clearly, h is a bijection and ι ′ = h ◦ ι. To
see that h is a homeomorphism, it suffices to show that h is continuous since a
compactification is always compact Hausdorff. Let U be an open subset of ̂X′ .
If ξ /∈ U, then h −1 (U) = ι ◦ (ι ′ ) −1 (U) is also an open subset of ι(X). Since
ι(X) = ̂X − {∞} is open in ̂X, h −1 (U) is open in ̂X. Otherwise, if ξ ∈ U, then
K = ̂X ′ − U ⊆ ι ′ (X) is closed in ̂X ′ and hence compact. It follows that h −1 (K) is
compact and hence closed in ̂X. Therefore, h −1 (U) = h −1 ( ̂X ′ − K) = ̂X − h −1 (K)
is open in ̂X, and h is continuous.
□
Remark 4.53. Since a one-point compactification a space X is unique up to a
homeomorphism, we usually call it ”the” one-point compactification of X.
Exercise 4.54. Find the one-point compactifications of N and R 2 .
your answers.
Justify
Theorem 4.55. Let X be a Hausdorff space. Then X is locally compact iff for
each x ∈ X and a neighborhood U of x, there is a neighborhood V of x such that V
is compact and V ⊆ U.

Phichet Chaoha 55
Proof. Let X be a Hausdorff space.
(⇒) Suppose X is locally compact. Let x ∈ X and U a neighborhood of x
(in X). If X itself is compact, let ̂X = X. Otherwise, let ̂X be the one-point
compactification of X (by identifying X with ι(X)). Notice that X is open in ̂X
for both cases. Thus, U is also open in ̂X. Since ̂X is regular (it is even normal by
Theorem 4.15), there is a neighborhood V of x in ̂X such that x ∈ V ⊆ Cl ̂X
V ⊆ U.
Being closed in ̂X, Cl ̂X
V is clearly compact. Moreover, since U ⊆ X, we have
U ∩ X = U, V ∩ X = V and V = Cl ̂X
V ∩ X = Cl ̂X
V . Therefore, x ∈ V ⊆ V ⊆ U
and V is compact as desired.
(⇐) Follows directly from the definition.
□
Theorem 4.56 (Imbedding Theorem). If X is a T 0 -space and Λ is a collection
of continuous maps from X to [0, 1] satisfying that, for each x ∈ X and a neighborhood
U of x, there exists f ∈ Λ such that f(x) > 0 and X − U ⊆ f −1 ({0}) (i.e.,
f vanishes outside U), then the map F : X → [0, 1] Λ defined by
is an imbedding of X into [0, 1] Λ .
F (x) = (f(x)) f∈Λ
Proof. Clearly, F is continuous since each f ∈ Λ is continuous. To show
that F is injective, suppose x ≠ y. Since X is T 0 , we may assume that there is a
neighborhood U of x not containing y. By the requirement above, there is f 0 ∈ Λ
such that f 0 (x) > 0 while f 0 (y) = 0. Hence, F (x) ≠ F (y). Now, we need to show
that the bijection F : X → F (X) is a homeomorphism. Since F : X → F (X) is
continuous, we only have show that F is open. Let U be a nonempty open subset
of X, y ∈ F (U) and pick x ∈ U such that F (x) = y. Since U is a neighborhood
of x, there is f 0 ∈ Λ such that f 0 (x) > 0 and f 0 (X − U) = {0}. By letting
W = F (X) ∩ π −1
f 0
((0, 1]) ⊆ [0, 1] Λ where π f0 : [0, 1] Λ → [0, 1] is the projection
map, W is clearly open in F (X) and y = F (x) ∈ W . To see that W ⊆ F (U),
let w ∈ W . Then w = F (a) for some a ∈ X and π f0 (w) > 0. It follows that
0 < π f0 (w) = π f0 (F (a)) = f 0 (a), and hence, a ∈ U (since f vanishes outside U).
Therefore, w ∈ F (U) as desired.
□
Remark 4.57. The collection Λ having the requirement as above is usually
said to separate a point from a closed set not containing it.
Corollary 4.58. A space X is completely regular iff it can be imbedded into
[0, 1] Λ for some Λ.
Proof. (⇒) Assume that X is completely regular. Let Λ be the collection of
all continuous maps from X to [0, 1]. Let x ∈ X and U a neighborhood of x. Then
by complete regularity of X, there exists a continuous map f : X → [0, 1] such that
f(x) = 0 and X − U ⊆ f −1 ({1}). Clearly, 1 − f ∈ Λ, (1 − f)(x) = 1 > 0 and
X − U ⊆ (1 − f) −1 ({0}). Therefore, Λ satisfies the requirement in the imbedding
theorem, and hence, X can be embedded into [0, 1] Λ .
(⇐) Since [0, 1] Λ is compact Hausdorff, then it is normal and hence completely
regular. It follows that X is completely regular because it is homeomorphic to a
subspace of [0, 1] Λ .
□

56 Topology (2301631)
Exercise 4.59. If X is a completely regular space and F is a collection of
all bounded continuous maps from X to R, prove that X can be imbedded into
∏
f∈F [a f , b f ], where f(X) ⊆ [a f , b f ] for each f ∈ F.
For an imbedding h : X → K of X into a compact Hausdorff space K, the
subspace h(X) of K is clearly compact (because it is closed in the compact space
K) and Hausdorff (because it is a subspace of a Hausdorff space K). Moreover,
h(X) contains h(X) as a dense subspace. Therefore, (h(X), h) is a compactification
of X, and we call it the compactification induced by h. The above corollary together
with the Tychonoff’s theorem now imply :
Corollary 4.60. If X is a completely regular space, then X has a compactification.
When a space X has a compactification ( ̂X, ι), it is natural to ask whether a
given continuous map f : X → R can be extended to a continuous map F : ̂X → R
in the sense that the following diagram commutes :
X
❅ ❅❘ ι
f
̂X
✲ R
♣ ♣ ♣ ♣
✒ F
This is certainly not true for an arbitrary map f. In fact, f must be bounded
since the domain of F is compact. However, sin( 1 x
) : (0, 1) → R is bounded, but it
can not be extended to either compactification S 1 or [0, 1] of (0, 1). The following
examples give us other compactifications of (0, 1) such that the map sin( 1 x
) can be
extended.
Example 4.61. The compactification of (0, 1) induced by the imbedding
h(x) = (x, sin( 1 )) ∈ [0, 1] × [−1, 1]
x
is clearly TSin. Note also that the map sin( 1 x
) : (0, 1) → R can be extended to
TSin as follows :
sin( 1 x
(0, 1) ) ✲ R
❅
h❅ ❅❘ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣
✒
π 2
TSin
Example 4.62. Let ̂X be a compactification induced by the imbedding
h(x) = (x, sin( 1 x ), cos( 1 x ))
of (0, 1) into [0, 1] × [−1, 1] 2 . Then, the maps x, sin( 1 x ) and cos( 1 x
) can all be
extended to ̂X because of the following commutative diagrams :
sin( 1 x )
cos( 1 x )
x
(0, 1) ✲ R (0, 1) ✲ R (0, 1) ✲ R
❅ ❅ ❅
h ❅❘ ♣ ♣ ♣ ♣ ♣ ♣
✒
π1
h ❅❘ ♣ ♣ ♣ ♣ ♣ ♣
✒
π2
h ❅❘ ♣ ♣ ♣ ♣ ♣ ♣
✒
π3
̂X ̂X ̂X

Phichet Chaoha 57
Exercise 4.63. Let f : D → Y be a continuous map, ι : D → X an imbedding
such that ι(D) = X and Y a Hausdorff space. If F is a continuous extension of f
in the sense that the following diagram commutes :
f
D ✲ Y
then F must be unique.
❅ι
❅❘
✒ F
X,
Theorem 4.64. Let X be a completely regular space. There exists a compactification
( ̂X, ι) of X such that every bounded continuous map f : X → R extends to
a unique continuous map from ̂X to R.
Proof. Let F be the set of all bounded continuous maps from X to R.
By Exercise 4.59, we obtain an imbedding of X into a compact Hausdorff space
∏
f∈F [a f , b f ], says h : X → ∏ f∈F [a f , b f ]. Let ( ̂X = h(X), ι = h) be the compactification
of X induced by h. Then, for each f 0 ∈ F, the projection π f0 : ̂X → [a f0 , b f0 ]
is clearly the unique by the previous exercise (since h(X) is dense in ̂X and R is
Hausdorff) continuous extension of f 0 to ̂X.
□
Lemma 4.65. Let X be a completely regular space and let ( ̂X, ι) be a compactification
of X satisfying the extension property in Theorem 4.64. Then, any
continuous map f from X to a compact Hausdorff space K extends uniquely and
continuously to ̂X.
Proof. Let K be a compact Hausdorff space and f : X → K a continuous
map. Then we can imbed K (since K is completely regular) into the product
space [0, 1] Λ , for some Λ, via h : K ∼= → h(K) ⊆ [0, 1] Λ . Now, for each α ∈ Λ, the
composition π α ◦h◦f : X → [0, 1] is bounded and can be continuously extended to ̂X
by the extension property, says to g α : ̂X → [0, 1]. It is not hard to check that we get
the continuous map G = (g α ) α∈Λ : ̂X → [0, 1] Λ which extends h ◦ f : X → [0, 1] Λ .
Moreover, by the continuity of G, we have
G( ̂X) = G(ι(X)) ⊆ G(ι(X)) = h ◦ f(X) ⊆ h(K) = h(K).
So, G really maps X into the subspace h(K) of [0, 1] Λ . Therefore, we can define
F = h −1 ◦ G : ̂X → K which is the desired extension of f.
f
X ✲ K h ✲ [0, 1] Λ
ι
π α
❄
❄
̂X ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣✯♣ ♣
G
♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣
✒
F
♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣✲♣ ♣ [0, 1]
g α
As usual, the uniqueness of F follows directly from the fact that K is Hausdorff.
□
Theorem 4.66. For a completely regular space X, any two compactifications
of X satisfying the extension property in Theorem 4.64 are homeomorphic.

58 Topology (2301631)
Proof. Let ( ̂X, ι) and ( ̂X ′ , ι ′ ) be two such compactifications of X. Then, by
the previous lemma, ι : X → ̂X and ι ′ : X → ̂X ′ extend continuously to f : ̂X ′ → ̂X
and g : ̂X → ̂X ′ , respectively, as follows :
ι
X ✲ ̂X X ✲ ̂X′
❅ ❅
ι
′ ❅❘ ♣ ♣ ♣ ♣
✒ f
ι ❅❘ ♣ ♣ ♣ ♣
✒ g
̂X ′
̂X
Then, ι = f ◦g ◦ι and ι ′ = g ◦f ◦ι ′ ; i.e., f ◦g and g ◦f are also continuous extensions
of ι and ι ′ , respectively.
ι
X ✲ ̂X X ✲ ̂X′
ι ′
ι ′
❅
ι ❅❘
̂X
✒ f◦g
❅ ❅❘ ι
′
̂X′
✒ g◦f
As usual, since both ̂X and ̂X ′ are Hausdorff, f ◦ g and g ◦ f must be unique.
Therefore, we must have f ◦ g = id ̂X
and g ◦ f = id ̂X′; i.e., f and g are homeomorphisms.
□
Definition 4.67. The Stone- ˘Cech compactification β(X) of X is the (up to a
homeomorphism) is the compactification satisfying the extension property in Theorem
4.64.
Exercise 4.68. Let X be a completely regular space. Show that X is connected
iff β(X) is connected.

CHAPTER 5
Function Spaces
For spaces X and Y , let Y X denote the set of all maps from X to Y , and
C[X, Y ] the set of all continuous maps from X to Y . In this chapter, we will
introduce some topologies on Y X .
Definition 5.1. Let X and Y be spaces. For each x ∈ X and an open subset
U of Y , let
S(x, U) = {f ∈ Y X | f(x) ∈ U}.
The point-open topology on Y X is the topology generated by the sets S(x, U) as a
subbasis.
Remark 5.2. Since a basis element is a finite intersection of subbasis elements,
then a basis neighborhood (w.r.t. the point-open topology) of f ∈ Y X consists of
g ∈ Y X that is ”close” to f at finitely many points. The point-open topology
is sometimes called the topology of pointwise convergence because of the following
exercise.
Exercise 5.3. Prove that a sequence (f n ) in Y X converges to f ∈ Y X in the
point-open topology iff for each x ∈ X, (f n (x)) converges to f(x) in Y .
Example 5.4. Let I = [0, 1]. In the topology of pointwise convergence, the
subspace C[I, R] of R I is not closed. This is because the sequence (f n ) in C[I, R],
where f n (x) = x n , converges pointwise to the map
{
0 for 0 ≤ x < 1,
f(x) =
1 for x = 1.
which is not in C[I, R].
Definition 5.5. Let X and Y be spaces. For a compact subset K of X and
an open subset U of Y , let
S(K, U) = {f ∈ Y X | f(K) ⊆ U}.
The compact-open topology on Y X is the topology generated by the sets S(K, U)
as a subbasis.
Remark 5.6. It is clear from the definitions above that the compact-open
topology is finer than the point-open topology. Even though the compact-open
topology can be defined on Y X , we usually consider this topology on the subset
C[X, Y ].
Theorem 5.7. Let X and Y be spaces where X is locally compact Hausdorff.
If C[X, Y ] is given the compact-open topology, then the evaluation map ev : X ×
C[X, Y ] → Y , defined by ev(x, f) = f(x), is continuous.
59

60 Topology (2301631)
Proof. Let (x, f) ∈ X × C[X, Y ], we will show that ev is continuous at (x, f).
Let V be a neighborhood of f(x). Since f is continuous, f −1 (V ) is a neighborhood
of x. Since X is locally compact Hausdorff, there is a neighborhood U of x such
that x ∈ U ⊆ U ⊆ f −1 (V ) and U is compact. Clearly, f(U) ⊆ f(f −1 (V )) ⊆ V , and
hence W = U × S(U, V ) is a neighborhood of (x, f). Moreover, for (x ′ , f ′ ) ∈ W ,
we have ev(x ′ , f ′ ) = f ′ (x ′ ) ∈ V ; i.e, W ⊆ ev −1 (V ).
□
Exercise 5.8. Let X, Y, Z be spaces where Y is locally compact Hausdorff.
Prove that the composition
C[X, Y ] × C[Y, Z] → C[X, Z]
is continuous when the compact-open topology is used throughout.
as sets.
Lemma 5.9 (Exponential Law). For any sets X, Y and Z, we have
Z X×Y ∼ = (Z Y ) X
Proof. Define ϕ : Z X×Y → (Z Y ) X by
for each f ∈ Z X×Y , x ∈ X and y ∈ Y .
Also, define ψ : (Z Y ) X → Z X×Y by
ϕ(f)(x)(y) = f(x, y)
ψ(g)(x, y) = g(x)(y)
for each g ∈ (Z Y ) X , x ∈ X and y ∈ Y .
It is easy to check that ϕ is bijective with ϕ −1 = ψ.
Exercise 5.10. Let X, Y and Z be spaces, and give C[Y, Z] the compact-open
topology. If f : X × Y → Z is continuous, then ϕ(f) in the above lemma is a
continuous map from X to C[Y, Z].
Theorem 5.11 (Adjunction Formula). Let X, Y, Z be spaces where Y is locally
compact Hausdorff. If C[Y, Z] has the compact-open topology, then the set of all
continuous maps from X to C[Y, Z] is isomorphic to the set of all continuous maps
from X × Y to Z; i.e.,
as sets.
C[X × Y, Z] ∼ = C[X, C[Y, Z]]
Proof. From the previous exercise, we see that ϕ decends to the function
ϕ : C[X × Y, Z] → C[X, C[Y, Z]].
Then, the result will follow once we can show that ψ also decends to the function
ψ : C[X, C[Y, Z]] → C[X × Y, Z]
which is the inverse of ϕ. However, for each g ∈ C[X, C[Y, Z]], it is not difficult to
verify that ψ(g) is the composition
X × Y g×1 Y
→ C[Y, Z] × Y → ev Z.
Since Y is locally compact Hausdorff and C[Y, Z] is given the compact-open topology,
the evaluation map ev is continuous and hence ψ(g) ∈ C[X × Y, Z] as desired.
□
□

Phichet Chaoha 61
Exercise 5.12. Let X, Y, Z be spaces where X is Hausdorff. If the compactopen
topology is used throughout, then
C[X, Y × Z] ∼ = C[X, Y ] × C[X, Z]
(as spaces).
When Y is a metric space, we can define the uniform topology on Y X as follows.
Definition 5.13. Let X be a space (or even a set), (Y, d) a metric space and
d be the standard bounded metric (i.e., d = min{d, 1}). Define
ρ(f, g) = sup{d(f(x), g(x)) | x ∈ X}
for f, g ∈ Y X . It is easy to check that ρ is also a metric on Y X , and it is called the
uniform metric corresponding to the metric d. The topology on Y X induced by ρ
is then called the uniform topology.
If F is a subset of Y X such that the set {d(f(x), g(x)) | x ∈ X} is bounded for
any f, g ∈ F, we can also define another metric on F by
ρ(f, g) = sup{d(f(x), g(x)) | x ∈ X}.
(For examples, when F = B[X, Y ] - the set of all bounded map from X to Y ,
or when F = C[X, Y ] when X is compact) This metric is called the sup metric
corresponding to the metric d. Under this circumstance, one can show that ρ is in
fact the standard bounded metric of ρ; i.e., ρ = min{ρ, 1}, and hence both metric
are equivalent.
Definition 5.14. Let X be a space, (Y, d) a metric space and (f n ) a sequence
in Y X . We say that (f n ) converges uniformly to f : X → Y if for each ϵ > 0, there
is N ∈ N such that
d(f n (x), f(x)) < ϵ
for all n ≥ N and x ∈ X.
Exercise 5.15. Let X be a space and (Y, d) a metric space. Prove that a
sequence (f n ) in Y X converges uniformly to f ∈ Y X iff (f n ) converges to f in the
uniform topology.
Theorem 5.16 (Uniform Limit Theorem). Let X be a space and (Y, d) a metric
space. If a sequence (f n ) in C[X, Y ] converges uniformly to a map f, then f ∈
C[X, Y ].
Proof. Suppose (f n ) converges uniformly to f. Let x 0 ∈ X, and we will show
that f is continuous at x 0 . Let ϵ > 0, there exists N ∈ N such that d(f N (x), f(x)) <
ϵ/3 for each x ∈ X. Since f N is continuous at x 0 , there is a neighborhood U of x 0
such that d(f N (x), f N (x 0 )) < ϵ/3 for any x ∈ U. Then, for x ∈ U, we have
d(f(x), f(x 0 )) ≤ d(f(x), f N (x)) + d(f N (x), f N (x 0 )) + d(f N (x 0 ), f(x 0 )) < ϵ
which implies the continuity of f at x 0 as desired.
Corollary 5.17. Let X be a space and (Y, d) a metric space. The subspace
C[X, Y ] of Y X is closed in the uniform topology.
Proof. Let Y X be equipped with the uniform topology and f ∈ Y X a limit
point of C[X, Y ]. Since Y X is metrizable in the uniform topology, then f is the
limit of a sequence in C[X, Y ]. By the uniform limit theorem, we have f ∈ C[X, Y ].
Therefore, C[X, Y ] is closed in Y X .
□
□

62 Topology (2301631)
Exercise 5.18. Let X be a space and (Y, d) a metric space. Let B[X, Y ]
denote the set of all bounded maps from X to Y . Prove that if a sequence (f n ) in
B[X, Y ] converges uniformly to a map f, then f ∈ B[X, Y ]. Hence, the subspace
B[X, Y ] of Y X is also closed in the uniform topology.
Definition 5.19. Let X be a space and (Y, d) a metric space. For each f ∈ Y X ,
a compact subset K of X and ϵ > 0, let
B K (f, ϵ) = { g ∈ Y X | sup{d(f(x), g(x)) | x ∈ K} < ϵ } .
Then B = {B K (f, ϵ) | f ∈ Y X , ϵ > 0, K a compact subset of X} forms a basis
for a topology on Y X . We will call the topology generated by B the topology of
compact convergence or the topology of uniform convergence on compact sets.
Exercise 5.20. Let X be a space and (Y, d) a metric space. Prove that a
sequence (f n ) in Y X converges uniformly to f on each compact subset K of X iff
(f n ) converges to f ∈ Y X in the topology compact convergence.
Definition 5.21. A space X is said to be compactly generated if it satisfies
the following condition : a set A is open [closed] in X if A ∩ K is open [closed] in
K for each compact subset K of X.
Theorem 5.22. Every locally compact space is compactly generated.
Proof. Let X be a locally compact space and A ⊆ X. Suppose A ∩ K is open
in K for any compact subset K of X. To show that A is open in X, let x ∈ A. By
local compactness at x, we can find a compact set K and an open set U in X such
that x ∈ U ⊆ K. By assumption, A∩K is open in K and hence A∩U = (A∩K)∩U
is open in U. Since U is open in X, so is U ∩ A and hence U ∩ A is a neighborhood
of x such that U ∩ A ⊆ A. Therefore, A is open in X.
□
Theorem 5.23. Every first countable space are compactly generated.
Proof. Let X be a first countable space and B ⊆ X. Suppose B ∩ K is closed
in K for any compact subset K of X. To show that B is closed in X, let x ∈ B.
Then there is a sequence (x n ) in B converging to x. Now, K = {x n | n ∈ N} ∪ {x}
is clearly compact and hence B ∩ K is closed in K by assumption. Since (x n ) is
also a sequence in B ∩ K converging to x, we must have x ∈ B ∩ K ⊆ B. Therefore
B is closed in X.
□
Lemma 5.24. Let X be a compactly generated space and Y a space. Then a
map f : X → Y is continuous iff for each compact subset K of X, the restriction
f| K is continuous.
Proof. (⇒) Trivial.
(⇐) Suppose f is continuous on each compact subset of X. To show that f
is continuous, let A be an open subset of Y . Then, for each compact subset K of
X, the set f −1 (A) ∩ K = (f| K ) −1 (A) is open in K since f| K is continuous. Since
X is compactly generated, it follows that f −1 (A) is open in X. Therefore, f is
continuous as desired.
□
Theorem 5.25. Let X be a compactly generated space and (Y, d) a metric space.
The subspace C[X, Y ] of Y X is closed in the topology of compact convergence.

Phichet Chaoha 63
Proof. Let Y X be equipped with the topology of compact convergence, f a
limit point to C[X, Y ] and K a compact subset of X. Then, for each n ∈ N, we
can pick f n ∈ B K (f, 1 n ) ∩ C[X, Y ] and hence obtain a sequence (f n) in C[X, Y ].
It is straightforward to verify that (f n ) converges uniformly to f on K. Then,
the uniform limit theorem implies that f| K is continuous. Therefore, f itself is
continuous by the previous lemma.
□
Theorem 5.26. Let X be a space and (Y, d) a metric space.
following inclusions of topologies on Y X :
We have the
uniform topology ⊇ topology of compact convergence ⊇ point-open topology.
The first two coincide when X is compact, and the second two coincide when X is
discrete.
Proof. For a basis neighborhood B K (f, ϵ) of f ∈ Y X in the topology of compact
convergence, we cleary have B ρ (f, δ) ⊆ B K (f, ϵ), where δ = min{ϵ, 1}. Hence,
the uniform topology is generally finer than the topology of compact convergence.
Moreover, When X is compact, for a basis neighborhood B ρ (f, ϵ) of f in the uniform
topology, we also have B X (f, δ) ⊆ B ρ (f, ϵ), where δ = min{ϵ, 1}.
Also, for a basis neighborhood ∩ n
i=1 S(x i, U i ) of f ∈ Y X in the topology of
point-open convergence, we clearly have B K (f, ϵ) ⊆ ∩ n
i=1 S(x i, U i ), where ϵ =
min{ϵ 1 , ϵ 2 , . . . , ϵ n }, B d (f(x i ), ϵ i ) ⊆ U i and K = {x 1 , x 2 , . . . , x n }. Hence, the topology
of compact convergence is finer than the point-open topology. Moreover, when
X is discrete, for a basis neighborhood B K (f, ϵ) of f in the topology of compact
convergence, we have K = {x 1 , x 2 , . . . , x n } (since a compact subset of a discrete
space must be finite) and hence ∩ n
i=1 S(x i, U i ) ⊆ B K (f, ϵ), where U i = B d (f(x i ), ϵ)
for all i = 1, 2, . . . , n.
□
Exercise 5.27. Let X be a space and (Y, d) a metric space. Prove that,
on C[X, Y ], the topology of compact convergence and the compact-open topology
coincide. Hence, the topology of compact convergence on C[X, Y ] depends on the
metric topology induced by d instead of the metric d.
Corollary 5.28. When X is compact and (Y, d) is a metric space, the uniform
topology on C[X, Y ] depends on the metric topology induced by d instead of the
metric d.
Proof. Follows directly from the previous theorem and example.
Example 5.29. When X = {1, 2, . . . , n} and Y = R with the euclidean metric,
we clearly have Y X = C[X, Y ] = R n . Since X is compact and discrete, all four
topologies defined above coincide.
Example 5.30. When X = N and Y = R with the euclidean metric, we clearly
have Y X = C[X, Y ] = R ω . Since X is discrete, the uniform topology is finer than
the other three.
□

CHAPTER 6
Complete Metric Spaces
Definition 6.1. A sequence (x n ) in a metric space (X, d) is called a Cauchy
sequence if for any ϵ > 0, there exists N ∈ N such that d(x m , x n ) < ϵ for all
m, n ≥ N.
Remark 6.2. A convergent sequence in a metric space X is always Cauchy.
Definition 6.3. A metric space X is said to be complete if every Cauchy
sequence in X converges.
Theorem 6.4. A metric space (X, d) is complete if every Cauchy sequence in
X has a convergent subsequence.
Proof. Let (x n ) be a Cauchy sequence in (X, d) and (x nk ) its subsequence
converging to x. Note that n k ≥ k for all k ∈ N. Since (x nk ) converges to x
and (x n ) is Cauchy, then for ϵ > 0, there exists a big enough K ∈ N such that
d(x, x nk ) < ϵ/2 and d(x nk , x k ) < ϵ/2 for all k ≥ K. By the triangle inequality, we
have d(x, x k ) ≤ d(x, x nk ) + d(x nk , x k ) < ϵ for all k ≥ K. Therefore, (x n ) converges
to x as desired.
□
Corollary 6.5. For any k ∈ N, R k is complete (with respect to the euclidean
metric).
Proof. Let d denote the euclidean metric on R k , and let (x n ) be a Cauchy
sequence in (R k , d). Then there exists N ∈ N such that d(x m , x n ) ≤ 1 for all
m, n ≥ N and hence, we have
d(x n , 0) ≤ M = max{d(x 1 , 0), . . . , d(x N−1 , 0), d(x N , 0) + 1}
for all n ∈ N. It follows that (x n ) is a sequence in the (sequentially) compact metric
space B(0; M). Therefore, (x n ) has a convergent subsequence, and by the previous
theorem, (R k , d) is complete.
□
Corollary 6.6. A compact metric space is complete.
Proof. Since compactness is equivalent to sequential compactness for metric
spaces, the result then follows directly from the previous theorem.
□
Example 6.7. Q and R − Q are not complete with respect to the euclidean
metric on R.
Exercise 6.8. Prove that a subspace A of a complete metric space (X, d) is
complete iff A is closed in X.
Remark 6.9. It should be pointed out here that completeness is not topologically
invariant. For example, the interval (0, 1) is not complete (by considering the
Cauchy sequence ( 1 n
)), but it is homeomorphic to R. The following example shows
that the completeness of a metric space (X, d) really depends on the metric d.
65

66 Topology (2301631)
x
Example 6.10. Let d(x, y) = ∣
1+|x| − y
1+|y|
∣ for x, y ∈ R. It is straightforward
to verify that d is a metric on R and it is equivalent to the euclidean metric on
R. Since the sequence (n) is Cauchy with respect to d but it does not converge, it
follows that (R, d) is not complete.
Exercise 6.11. Let d and d ′ be two metrics on X. Suppose further that there
are positive real numbers A, B and C such that
Ad(x, y) ≤ Bd ′ (x, y) ≤ Cd(x, y)
for any x, y ∈ X. Prove that a sequence (x n ) in X is Cauchy with respect to d iff
it is Cauchy with respect to d ′ . Hence, (X, d) is complete iff (X, d ′ ) is complete.
Notice also that d and d ′ are equivalent.
The previous example immediately implies that R k is complete with respect to
the square metric and the taxicab metric.
Theorem 6.12. R ω is complete with respect to the metric
D(x, y) = sup{ d(π i(x), π i (y))
| i ∈ N}
i
as defined in Theorem 1.151.
Proof. Let (x n ) be a Cauchy sequence in (R ω , D). Notice that, for each i ∈ N
and ϵ > 0, there exists N i ∈ N such that D(x m , x n ) < min{ϵ,1}
i
for all m, n ≥ N i .
So we have d(π i (x m ), π i (x n )) = d(π i (x m ), π i (x n )) ≤ iD(x m , x n ) < min{ϵ, 1} ≤ ϵ
for all m, n ≥ N i . It follows that (π i (x n )) is also a Cauchy sequence in (R, d) and
hence converges to some a i ∈ R. To see that (x n ) converges to a = (a 1 , a 2 , . . . ), let
B be a basis neighborhood of a. WLOG, we may write B = ∏ i∈N U i where U i is a
neighborhood of a i for all i ∈ N, and U i = R for all i > m for some m ∈ N. Then
for each i = 1, 2, . . . , m, there exists N i ∈ N such that π i (x n ) ∈ U i . It follows that
x n ∈ B for all n ≥ max{N 1 , N 2 , . . . , N m }. Therefore, (x n ) → a and hence (R ω , D)
is complete as desired.
□
Theorem 6.13. For a space X and a complete metric space (Y, d), the space
Y X is complete with respect to the uniform metric ρ corresponding to d.
Proof. Let (f n ) be a Cauchy sequence in (Y X , ρ). For each x ∈ X, we can
show that (f n (x)) is Cauchy in (Y, d) and hence converges to some y x ∈ Y by
completeness. Now, define f ∈ Y X by f(x) = y x for each x ∈ X. To show that
(f n ) → f, we let ϵ > 0. Since (f n ) is Cauchy in (Y X , ρ), there exists N ∈ N such
that for m, n ≥ N and x ∈ X,
d(f m (x), f n (x)) ≤ ρ(f m , f n ) < ϵ 2 .
By letting m → ∞, we have d(f(x), f n (x)) ≤ ϵ 2
for any n ≥ N and x ∈ X. It
follows that ρ(f, f n ) ≤ ϵ 2 < ϵ for any n ≥ N. Therefore, (f n) → f and hence
(Y X , ρ) is complete. □
Corollary 6.14. Let X be a space and (Y, d) a complete metric space. Then,
the subspaces B[X, Y ] and C[X, Y ] of (Y X , ρ) are complete.
Proof. According to the previous chapter, both B[X, Y ] and C[X, Y ] are
closed in (Y X , ρ). It follows from the previous theorem that both subspaces are
complete.
□

Phichet Chaoha 67
Example 6.15. R ω , B[X, R n ] and C[X, R n ] are complete with respect to the
uniform metric ρ. Moreover, it is not difficult to show that both B[X, R n ] and
C[X, R n ] (when X is compact) are complete with respect to ρ.
Definition 6.16. Let (X, d X ) and (Y, d Y ) be metric spaces. A map f : X → Y
is called
• an isometry if d Y (f(x), f(y)) = d X (x, y) for each x, y ∈ X.
• a contraction if there exists α ∈ [0, 1) such that
d Y (f(x), f(y)) ≤ αd X (x, y)
for each x, y ∈ X.
• contractive if d Y (f(x), f(y)) < d X (x, y) for each x, y ∈ X and x ≠ y.
• nonexpansive if d Y (f(x), f(y)) ≤ d X (x, y) for each x, y ∈ X.
Clearly, we have the following implications :
isometry ⇒ nonexpansive ⇐ contractive ⇐ contraction.
Example 6.17. Translations, rotations and reflections of R n are isometries.
Exercise 6.18. Prove that a nonexpansive map is always uniformly continuous.
Exercise 6.19. Prove that an isometry is always an imbedding.
Exercise 6.20. Prove that f(x) = sin(x) on [0, 1] is contractive but not a
contraction.
Theorem 6.21. Any metric space can be isometrically imbedded into a complete
metric space.
Proof. Let (X, d) be a metric space and x 0 ∈ X. We will show that X can
be imbedded into the metric space (B[X, R], ρ) which is complete by the above
example.
For each a ∈ X, we define ϕ a : X → R by
ϕ a (x) = d(x, a) − d(x, x 0 )
for each x ∈ X. By the triangle inequality, we clearly have
|ϕ a (x)| = |d(x, a) − d(x, x 0 )| ≤ d(a, x 0 )
for each x ∈ X; i.e., ϕ a ∈ B[X, R].
Since for each a, b ∈ X, we clearly have the inequality
d(a, b) ≤ ρ(ϕ a , ϕ b ) = sup |ϕ a (x) − ϕ b (x)| = sup |d(x, a) − d(x, b)| ≤ d(a, b)
x∈X
x∈X
which imples ρ(ϕ a , ϕ b ) = d(a, b). Therefore, the map a ↦→ ϕ a is the desired isometry.
□
Definition 6.22. A completion of a metric space X is a pair ( ˜X, ι) of a complete
metric space ˜X and an isometric imbedding ι : X → ˜X such that ι(X) = ˜X.
Notice that, for a given isometric imbedding h : X → Y of a metric space X
into a complete metric space Y , we immediately obtain a completion of X simply by
letting ˜X = h(X) and ι = h. We will call this completion (h(X), h) the completion
induced by h.
Theorem 6.23. A completion of a metric space is unique upto isometric homeomorphisms.

68 Topology (2301631)
Proof. Let ( ˜X, ι) and ( ˜X ′ , ι ′ ) be completions of a metric space X. Then,
we clearly have an isometric homeomorphism ι ′ ◦ ι −1 : ι(X) → ι ′ (X), and after
composing with the inclusion of ι ′ (X) into ι ′ (X) = ˜X ′ , we obtain an isometric
imbedding f : ι(X) → ˜X ′ . Notice that for a Cauchy sequence (x n ) in ι(X), the
sequence (f(x n )) is also Cauchy in ˜X ′ since f is an isometry. This allows us to
define the desired isometric homeomorphism [exercise below] f : ˜X → ˜X ′ by
f(x) = lim
n→∞ f(x n)
for each x ∈ ˜X and any sequence (x n ) converging to x.
Exercise 6.24. Verify that the map f in the theorem above is a well-defined
(independent from the sequence (x n )) isometric homeomorphism such that the
following diagram commutes :
˜X
❅■ ι ❅
f
∼ =
X.
✲
✒
ι ′
Exercise 6.25. Let (X, d) be a metric space and CS(X) the set of all Cauchy
sequences in (X, d). Define a relation ∼ on CS(X) as follows :
˜X′
(x n ) ∼ (y n ) ⇔ lim
n→∞ d(x n, y n ) = 0.
(1) Prove that ∼ is an equivalence relation.
(2) Prove that for any K ∈ N, we always have (x n ) ∼ (x K+n ).
Now, let ˜X = CS(X)/ ∼ and for any [(x n )], [(y n )] ∈ ˜X,
D([(x n )], [(y n )]) = lim
n→∞ d(x n, y n ).
(3)Prove that D is well-defined and it is a metric on ˜X.
Define ι : (X, d) → ( ˜X, D) by ι(x) = [(x)].
(4) Prove that ι is an isometric imbedding.
(5) Prove that ι(X) is dense in ˜X.
(6) Prove that that ( ˜X, D) is complete.
Theorem 6.26 (Banach’s Fixed Point Theorem). Let (X, d) be a complete
metric space and f : X → X a contraction. Then f has a unique fixed point, says
x 0 , and the sequence (f n (x)) converges to x 0 for any x ∈ X.
Proof. Since f : X → X is a contraction, there exists α ∈ [0, 1) such that
d(f(x), f(y)) ≤ αd(x, y) for any x, y ∈ X. Let x ∈ X. If f(x) = x, then x 0 = x.
Suppose f(x) ≠ x. To obtain x 0 , we first show that the sequence (f n (x)) is Cauchy.
Observe that
• For m, n ∈ N with m > n, we have
d(f m (x), f n (x)) ≤ αd(f m−1 (x), f n−1 (x)) ≤ · · · ≤ α n d(f m−n (x), x).
• For k ∈ N, we have
d(f k (x), x) ≤ Σ k i=1d(f i (x), f i−1 (x)) ≤ Σ k i=1α i−1 d(f(x), x)
= 1−αk
1
1−α
d(f(x), x) ≤
1−αd(f(x), x).
□

Let ϵ > 0 and N ∈ N be such that α N <
assume that m > n and we have
( α
d(f m (x), f n (x)) ≤ α n d(f m−n n
(x), x) ≤
1 − α
Phichet Chaoha 69
ϵ(1−α)
d(f(x),x)
. Then for m, n ≥ N, we may
)
d(f(x), x) < ϵ.
Hence, (f n (x)) is Cauchy and converges by the completeness of (X, d). Let
x 0 = lim n→∞ f n (x). Since f is continuous, x 0 is a fixed point of f by
f(x 0 ) = f( lim
n→∞ f n (x)) = lim
n→∞ f(f n (x)) = lim
n→∞ f n+1 (x) = x 0 .
Finally, suppose x ′ 0 is another fixed point of f. Since f is a contraction with
α < 1, we have d(x 0 , x ′ 0) = d(f(x 0 ), f(x ′ 0)) ≤ αd(x 0 , x ′ 0) which is impossible unless
d(x 0 , x ′ 0) = 0. Therefore, we must have x 0 = x ′ 0.
□
Definition 6.27. A metric space (X, d) is said to be totally bounded if for each
ϵ > 0, X can be covered by finitely many ϵ-balls; i.e., there exists x 1 , x 2 , . . . , x n ∈ X,
such that X = ∪ n
i=1 B d(x i , ϵ).
Theorem 6.28. A totally bounded metric space is always bounded.
Proof. Let (X, d) be a totally bounded metric space. Then, we can cover X by
finitely many unit balls, says X = ∪ n
i=1 B d(x i , 1). Let M = max{d(x i , x j ) | i, j =
1, 2, . . . , n}. For any x, y ∈ X, we clearly have x ∈ B d (x i , 1) and y ∈ B d (x j , 1) for
some i, j = 1, 2, . . . , n. Hence, d(x, y) ≤ d(x, x i ) + d(x i , x j ) + d(x j , y) ≤ M + 2; i.e.,
(X, d) is bounded.
□
Example 6.29. The converse of the previous theorem is not true. Consider an
infinite set X together with the discrete metric d. Clearly, (X, d) is bounded since
d(x, y) ≤ 1 for all x, y ∈ X. However, it is impossible to cover X by finitely many
open balls of radius 1 2 .
Example 6.30. A compact metric space is always totally bounded.
Theorem 6.31. A metric space is compact iff it is complete and totally bounded.
Proof. (⇒) Follows directly from Corollary 6.6 and the previous example.
(⇐) Let (X, d) be a metric space that is complete and totally bounded. Then,
it suffices to show that X is sequentially compact. Let (x n ) be a sequence in X.
WLOG, we may assume that the set A 0 = {x n | n ∈ N} is infinite. Since X is
totally bounded, we can cover X by finitely many open balls of radius 1, and hence,
there exists a ball B 1 of radius 1 such that A 1 = B 1 ∩ A 0 is infinite. Now, assume
that an infinite set A k and an open ball B k of radius 1 k
are defined for some k ≥ 1,
1
we will construct an infinite set A k+1 and a ball B k+1 of radius
k+1
as follows
1
: cover X by finitely many open balls of radius
k+1 and use the fact that A k is
1
infinite to obtain an open ball B k+1 of radius
k+1 such that A k+1 = B k+1 ∩ A k is
infinite. Now, for each k ∈ N, let J k = {n | x n ∈ A k }. Clearly, each J k is an infinite
subset of N and J 1 ⊇ J 2 ⊇ J 3 ⊇ . . . Choose n 1 ∈ J 1 and for k > 1, we can always
choose n k ∈ J k such that n k > n k−1 because J k is infinite. By this construction, it
is easy to see that, for all i, j ≥ k, we have d(x ni , x nj ) < 2 k because x n i
, x nj ∈ B k .
That is (x nk ) forms a Cauchy subsequence of (x n ). Therefore, by completeness,
(x nk ) is a convergent subsequence of (x n ). □

70 Topology (2301631)
We will end this chapter by considering the well-known Ascoli’s theorem which
characterizes compact subsets of the uniform space (C[X, R n ], ρ) when X is compact.
Since the sup metric ρ also induces the uniform topology in this case, we will
always consider C[X, R n ] in the sup metric throughout.
Definition 6.32. Let X be a space, (Y, d) a metric space and F ⊆ C[X, Y ].
For x 0 ∈ X, the set F is equicontinuous at x 0 if for each ϵ > 0, there exists a
neighborhood U of x 0 such that
for all x ∈ U and all f ∈ F.
equicontinuous at each x 0 ∈ X.
d(f(x), f(x 0 )) < ϵ
We simply say that F is equicontinuous if it is
Exercise 6.33. Show that F = {f n : I → I | f n (x) = x n , n ∈ N} is not
equicontinuous at 1.
Lemma 6.34. Let X be a compact space, (Y, d) a compact metric space and
F ⊆ (C[X, Y ], ρ). Then F is equicontinuous iff F is totally bounded.
Proof. (⇒) Suppose F is equicontinuous and let ϵ > 0. Then for each x ∈ X,
there is a neighborhood U x of x such that d(f(x), f(y)) < ϵ 3 for any y ∈ U x and
f ∈ F. By the compactness of X, we then can cover X by U x1 , U x2 , . . . U xn for
some x 1 , x 2 , . . . , x n ∈ X. Also, by the compactness of Y , we can cover Y by finitely
many open sets V 1 , V 2 , . . . , V m with diam(V i ) < ϵ 3
. Consider the set
I = {α : {1, . . . , n} → {1, . . . , m} | ∃f ∈ F∀i ∈ {1, . . . , n}, f(x i ) ∈ V α(i) }
which is clearly finite. For each α ∈ I, we fix a map f α ∈ F such that f α (x i ) ∈ V α(i)
for all i ∈ {1, 2, . . . , n}.
Now, we claim that F ⊆ ∪ α∈I B ρ(f α , ϵ). Let f ∈ F. Then we obtain a function
α : {1, . . . , n} → {1, . . . , m} such that f(x i ) ∈ V α(i) for all i ∈ {1, 2, . . . , n}. That is
α ∈ I (notice that f and f α may not be the same map). Moreover, for each x ∈ X,
we have x ∈ U xi for some i ∈ {1, 2, . . . , n} and hence
d(f(x), f α (x)) ≤ d(f(x), f(x i )) + d(f(x i ), f α (x i )) + d(f α (x i ), f α (x)) < ϵ.
It follows that ρ(f, f α ) = max{d(f(x), f α (x)) | x ∈ X} < ϵ ; i.e., f ∈ B ρ (f α , ϵ).
(⇐) Suppose F is totally bounded. Let ϵ > 0 and x ∈ X. Then there exist
f 1 , f 2 , . . . , f n ∈ F such that F ⊆ ∪ n
i=1 B ρ(f i , ϵ 3
). For each i = 1, 2, . . . , n, by
the continuity of each f i at x, we can find a neighborhood U i of x such that
d(f i (x), f i (y)) < ϵ 3 for any y ∈ U i. So U = ∩ n
i=1 U i is a neighborhood of x. For any
y ∈ U and f ∈ F, we must have f ∈ B ρ (f i , ϵ 3
) for some i = 1, 2, . . . , n and hence
d(f(x), f(y)) ≤ d(f(x), f i (x)) + d(f i (x), f i (y)) + d(f i (y), f(y)) < ϵ.
Therefore, F is equicontinuous at x.
Exercise 6.35. Let X be a compact space and F ⊆ (C[X, R n ], ρ). Prove that
if F is bounded, then F ⊆ C[X, Y ] for some compact subset Y of R n .
Theorem 6.36 (Classical Ascoli’s Theorem). Let X be a compact space and
F ⊆ (C[X, R n ], ρ). Then F is compact iff it is closed, bounded and equicontinuous.
Proof. (⇒) Since F is compact, it is closed and totally bounded by the previous
theorem. Hence, F is also bounded. Since F ⊆ C[X, Y ] for some compact
□

Phichet Chaoha 71
subset Y of R n by the previous exercise, F must be equicontinuous by the previous
lemma.
(⇐) Since F is closed and (C[X, R n ], ρ) is complete, F is complete with respect
to ρ. Since F is bounded with respect to ρ, the previous exercise implies that
F ⊆ C[X, Y ] for some compact subset Y of R n . Finally, since F is equicontinuous,
it must be totally bounded by the previous lemma.
□
Definition 6.37. Let X be a space and (Y, d) a metric space. A subset F of
C[X, Y ] is said to be pointwise bounded if for each x ∈ X, the subset
is bounded.
F x = {f(x) | f ∈ F}
Exercise 6.38. Let X be a compact space and F ⊆ (C[X, R n ], ρ). Prove :
(1) If F is equicontinuous and pointwise bounded, then F is bounded.
(2) If F is equicontinuous, then so is F.
(3) F has compact closure iff it is pointwise bounded and equicontinuous.
Theorem 6.39 (Arzela’s Theorem). Let X be compact and (f n ) a sequence
in (C[X, R n ], ρ). If the collection F = {f n | n ∈ N} is pointwise bounded and
equicontinuous, then (f n ) has a uniformly convergent subsequence.
Proof. Since F is pointwise bounded and equicontinuous, the previous exercise
implies that F is compact in (C[X, R n ], ρ) and hence sequentially compact.
Since (f n ) a sequence in F and ρ induces the uniform topology, (f n ) has a uniform
convergent subsequence.
□
Example 6.40. The family F = {f n : I → I | f n (x) = x n , n ∈ N} ⊆ C[I, R]
is not equicontinuous by Arzela’s theorem since it is pointwise bounded (can you
see why?) and the sequence (f n ) converges to a function f that is not continuous
(hence it cannot have a uniform convergent subsequence).

Bibliography
[1] S. W. Davis, Topology, McGraw-Hill, 2005.
[2] M. A. Khamsi and W. A. Kirk, An introduction to metric spaces and fixed point theory, John
Wiley & Sons, Inc., 2001.
[3] J. G. Hocking and G. S. Young, Topology, Dover, 1998.
[4] J. R. Munkres, Topology (second edition), Prentice Hall, Upper Saddle River, NJ, 2000.
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