mh404 point set topology

MH404 Point Set Topology

Wong Yan Loi

Contents

1 Topology and Continuous Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Product and Sum Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Separation Axioms and Compactness . . . . . . . . . . . . . . . . . . . . . . . . 74 Countability, Separability and paracompactness . . . . . . . . . . . . . . . . . . . 125 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Identifications and Adjunction Spaces . . . . . . . . . . . . . . . . . . . . . . . . 227 Function spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1

1 Topology and Continuous Map 2

1 Topology and Continuous Map

Definition 1.1 A topology on a set X is a collection T of subsets of X satisfying the followingaxioms:

( T 1 ) φ , X ∈ T .( T 2 ) {Oα | α ∈ I} ⊆ T =⇒

⋃α ∈ I Oα ∈ T .

( T 3 ) O, O′ ∈ T =⇒ O ∩O′ ∈ T .

A topological space is a pair (X, T ) where X is a set and T is a topology on X . When thetopology T on X under discussion is clear, we simply denote (X, T ) by X . Let (X, T ) be atopological space. Members of T are called open sets. ( T 3 ) implies that a finite intersection ofopen sets is open.

Examples1. Let X be a set. The power set 2X of X is a topology on X and is called the discrete topologyon X . The collection I = {φ , X} is also a topology on X and is called the indiscrete topologyon X .

2. Let (X, d) be a metric space. Define O ⊆ X to be open if for any x ∈ O, there exists anopen ball B(x, r) lying inside O. Then Td = { O ⊆ X | O is open } ∪ {φ} is a topology onX . Td is called the topology induced by the metric d.

3. Since Rn is a metric space with the usual metric :

d( (x1, ..., xn), (y1, ..., yn) ) =

√√√√ n∑i=1

(xi − yi)2 ,

Rn has a topology U induced by d. This topology on Rn is called the usual topology .

4. Let X be an infinite set. Then T = { φ,X } ∪ {O ⊆ X | X−O is finite } is a topologyon X . T is called the complement finite topology on X .

5. Define O ⊆ R to be open if for any x ∈ O, there exists a δ > 0 such that [x , x+δ) ⊆ O.Then the collection T ′ of all such open sets is a topology on R. T ′ is called the lower limit topol-ogy on R. Note that U ⊂ T ′ but U 6= T ′. We shall denote this topological space by Rl .

Proposition 1.2 Let (X, T ) be a topological space and A be a subset of X .Then the collection TA = { O ∩ A | O ∈ T } is a topology on A. TA is called the subspacetopology on A and (A , TA) is called a subspace of (X , T ).

Proof Exercise.

Let (X, T ) be a topological space.


Definition 1.3 E ⊆ X is said to be closed if X − E is open. ( i.e. X − E ∈ T )

Proposition 1.4(a) φ and X are closed in X.(b) If E , F are closed in X , then E ∪ F is closed in X .(c) If { Eα | α ∈ I } is a collection of closed sets in X , then

⋂α ∈ I Eα is closed in X .

Proof Exercise.

Definition 1.5 A basis for the topology T on X is a subcollection B of T such that any open setin X is a union of members of B.

Definition 1.6 A subbasis for the topology T on X is subcollection S of T such that the collec-tion of all finite intersections of members of S is a basis for T . Hence any open subset of X is aunion of finite intersections of members in S.

Examples1. { (a, b) | a < b } is a basis for the usual topology on R. { (a,∞) , (−∞, b) | a, b ∈ R }is a subbasis for the usual topology on R.

2. Let (X, d) be a metric space and Td the topology on X induced by d. The collection B of allopen balls of X is a basis for Td.

3. { [a, b) | a < b } is a basis for the lower limit topology on R. What would be a subbasis forthe lower limit topology on R ?

4. If (X, T ) is a topological space, then T itself is a basis for T .

Proposition 1.7 Let X be a set and B be a collection of subsets of X such that( B1 )

⋃U∈B U = X

( B2 ) for any U1 , U2 ∈ B and x ∈ U1∩U2 , there existsU ∈ B such that x ∈ U ⊆ U1∩U2.Then the collection TB of all unions of members of B is a topology on X and B is a basis for TB.

Proof ( T 1 ) and ( T 2 ) are clearly satisfied. ( B2 ) implies that if Uα, Uβ ∈ B, thenUα ∩ Uβ ∈ TB. Let O1 = ∪Uα and O2 = ∪Uβ be in TB, where Uα , Uβ ∈ B. ThenO1 ∩O2 =

⋃(Uα ∩ Uβ) is in T . That B is a basis for TB follows from the definition of TB.

Proposition 1.8 Let X be a set and S be a collection of subsets of X such that X =⋃V ∈S V.

Then the collection TS of all unions of finite intersections of members of S is a topology on Xand S is a subbasis for TS .


Proof Similar to 1.7.

Definition 1.9 A subset N containing a point x in X is called a neighborhood of x if there existsan open set O in X such that x ∈ O ⊆ N .

Definition 1.10 Let A be a subset of X .(1) x ∈ A is called an interior point of A if there exists a neighborhood U of x inside A.(2) x ∈ X is called a limit point of A if for any neighborhood O of x , O ∩A 6= φ.(3) Ao = { x ∈ A | x is an interior point of A } is called the interior of A.(4) A = { x ∈ X | x is a limit point of A } is called the closure of A.

Proposition 1.11(1) Ao is an open set and Ao ⊆ A .(2) X −A = (X −A)o and X −Ao = X −A.(3) A is a closed set and A ⊇ A .(4) A is open ⇐⇒ A = Ao.(5) A is closed ⇐⇒ A = A .(6) Ao is the largest open set contained in A.(7) A is the smallest closed set containing A.(8) Ao =

⋃{ O | O is open and O ⊆ A }.

(9) A =⋂{ E | E is closed and E ⊇ A }.

(10) Aoo= Ao and A = A.

(11) (A ∩B)o = Ao ∩ Bo .(12) A ∪B = A ∪B .(13) Let B ⊆ A ⊆ X. Then B is closed in A ⇐⇒ B = E ∩A for some closed set E in X.(14) Let B ⊆ A ⊆ X and let BA be the closure of B in A. Then BA

= B ∩A.(15) Let B ⊆ A ⊆ X and let B◦A be the interior of B in A. If A is open, then B◦A = B◦.

Proof Exercise.

Definition 1.12 Let (X, T ) and (Y, T ′) be topological spaces and f : X −→ Y be a map.Let x ∈ X. f is said to be continuous at x if for any open neighborhood U of f(x) , thereexists an open neighborhood O of x such that f [O] ⊆ U . f is said to be continuous on X if fis continuous at each point of X .

Proposition 1.13 Let (X, T ) and (Y, T ′) be topological spaces and f : X −→ Y be a map.The following statements are equivalent .

(1) f is continuous on X .(2) For any open set U ⊆ Y , f−1[U ] is open in X .

2 Product and Sum Topologies 5

(3) For any closed set E ⊆ Y f−1[E] is closed in X .(4) For any A ⊆ X , f [A] ⊆ f [A].

Proof Exercise.

Proposition 1.14 Let f : X −→ Y and f : Y −→ Z be maps. If f and g are continuous,then f ◦ g is continuous.

Proof This follows easily by 1.13 (2).

Definition 1.15 Let (X, T ) and (Y, T ′) be topological spaces and f : X −→ Y a map.(1) f is called an open map if for any open set U in X , f [U ] is open in Y .(2) f is called a closed map if for any closed set E in X , f [E] is closed in Y .(3) f is called a homeomorphism if f is bijective , continuous and f−1 is continuous.

Proposition 1.16 Let f : X −→ Y be a bijective continuous map. The following statementsare equivalent.

(1) f−1 : X −→ Y is a homeomorphism.(2) f : X −→ Y is an open map.(3) f : X −→ Y is a closed map.

Proposition 1.17 ( The Combination Principle) Let A1 , ... , An be a finite collection of closedsubsets of X such that

⋃ni=1Ai = X. Then f : X → Y is continuous if and only if f | Ai

is continuous for each i = 1, ... , n .

Proof The ’only if’ part is obvious. Suppose f | Ai is continuous for each i = 1, ... , n . LetE be a closed subset of Y . Then f−1[E] =

⋃ni=1(f

−1[E]∩Ai) =⋃ni=1(f | Ai)−1[E] . Since

f | Ai is continuous, (f | Ai)−1[E] is closed in Ai . As Ai is closed in X , (f | Ai)−1[E]is closed in X. Hence f−1[E] is closed in X . This shows that f is continuous on X.

2 Product and Sum Topologies

Definition 2.18 Let {Xλ}λ ∈ I be a family of topological spaces and pα :∏λ ∈ I Xλ −→ Xα

be the natural projection. The product topology on∏λ ∈ I Xλ is defined to be the one generated

by the subbasis {p−1α [Oα] | Oα is open in Xα, α ∈ I}.

Note that when∏λ ∈ I Xλ is given the product topology the projection pα is continuous and open

2 Product and Sum Topologies 6

. In the case that the product is finite the collection {∏α ∈ I Oα | Oα is open in Xα } is a

basis for the product topology on∏λ ∈ I Xλ

Proposition 2.19 Let f : X −→∏λ ∈ I Xλ be a map. Then f is continuous if and only if

pα ◦ f is continuous for all α ∈ I .

Proof Exercise.

The following results are immediate consequences.

Proposition 2.20 The diagonal map d : X −→ X ×X, defined by d(x) = (x, x), and thetwisting map τ : X × Y −→ Y ×X, defined by τ(x, y) = (y, x) , are continuous.

Proposition 2.21 Given a family of maps { fλ : Xλ → Yλ }λ∈I , the product map∏λ ∈ I fλ :

∏λ ∈ I Xλ −→

∏λ ∈ I Yλ , defined by (

∏λ ∈ I fλ) (xλ) = (fλ(xλ)), is

continuous.

Definition 2.22 Let {Xα} be a family of topological spaces. The topological sum tXα isthe topological space whose underlying set is ∪Xα equipped with the topology { O ⊆ ∪Xα |O ∩ Xα is open in Xα for each α } .

Note that when tXα is given the sum topology the inclusion iα : Xα −→ tXα is continu-ous. In general it may not be possible to fit together the topologies on a family of spaces {Xα}to obtain a topology on the union restricting to the original topology on each Xα. For example,if two spaces Xα and Xβ overlap they may not induce the same topologies on Xα ∩ Yβ . Afamily {Xα} of spaces is said to be compatible if the two topologies on Xα ∩Xβ are identicalfor all pair α, β .

Proposition 2.23 Let {Xα} be a compatible family of spaces such that Xα∩Xβ is open in Xα

and Xβ , for each α, β. Then Xα has the subspace topology from tXα .

Proof Exercise.

For example if {Xα} is a family of disjoint topological spaces, then each Xα is a subspace oftXα .

Proposition 2.24 If {fα : Xα −→ Y } is a family of maps such that fα and fβ agree on

3 Separation Axioms and Compactness 7

Xα ∩Xβ for all α , β , then there is a unique continuous map tfα : tXα −→ Y such that(tfα) | Xα = fα, for all α .

Proof Exercise.

3 Separation Axioms and Compactness

Definition 3.25 let X be a topological space.

(1) X is T0 if for each pair of distinct points, at least one has a neighborhood not containingthe other.

(2) X is T1 or Frechet if for any distinct x , y ∈ X , there exist open neighborhoods U ofx and V of y such that y is not in U and x is not in V .

(3) X is T2 or Hausdorff if for any distinct x , y ∈ X , there exist open neighborhoods Uof x and V of y such that U ∩ V = φ .

(4) X is said to be regular if for any closed set E and any point x not in E , there exist anopen neighborhood U of E and an open neighborhood V of x such that U ∩ V = φ .

(5) X is said to be normal if for any two disjoint closed sets E and F , there exist an openneighborhood U of E and an open neighborhood V of F such that U ∩ V = φ .

(6) X is T3 if X is T1 and regular.

(7) X is T4 if X is T1 and normal.

It is easy to see that X is T1 if and only if every singleton set in X is closed. Hence we haveT4 =⇒ T3 =⇒ T2 =⇒ T1 =⇒ T0 .

Examples

1. Let X = {0, 1} and T = {φ,X, {0}} . Then ( X , T ) is T0 but not T1 .

2. Every metric space is T4 .

3. Let X be an infinite set with the complement finite topology. Then X is T1 but not T2 .


4. Let K = { 1n | n ∈ Z+ } and let B = { (a, b) ⊆ R | a < b } ∪ { (a, b)−K | a < b } .

Then B is a basis for a topology on R . Clearly R with this topology is T2 but it is not regular.

5. Rl is T4

6. Rl ×Rl is regular but not normal. In general product of normal spaces may not be normal.

Urysohn’s Lemma If A and B are nonempty disjoint closed subsets in a normal space X, thenthere exists a continuous function f : X −→ [0, 1] such that f [A] = 0 and f [B] = 1.

Definition 3.26 A cover of a topological space X is a collection C = {Oα}α∈I of subsets ofX such that ∪α∈IOα = X . It is said to be an open cover if each Oα is open in X . If theindex set I is finite ( or countable ), then {Oα}α∈I is called a finite ( or countable ) cover. Asubcover of a cover C of X is a subcollection S of C such that S is a cover of X.

Definition 3.27 A space X is said to be compact if every open cover of X has a finite subcover.A ⊆ X is said to be compact if A with the subspace topology is a compact space.

Examples1. Any finite space is compact.2. [a, b] is compact.3. An infinite set provided with the complement finite topology is a compact space.4. R is not compact. Neither does Rl .

Proposition 3.28 A ⊆ X is compact if and only if every cover of A by open subsets of X hasa finite subcover.

Proposition 3.29(1) A closed subspace of a compact space is compact.(2) A compact subset of a Hausdorff space is closed.(3) A continuous image of compact space is compact.(4) A bijective continuous map from a compact space to a T2 space is a homeomorphism.

Proof Exercise.

Proposition 3.30 Let {Xλ}λ∈I be a collection of spaces. If an infinite number of Xλ arenon-compact, then any compact subset in

∏λ∈I Xλ has empty interior.


Proof Exercise.

Proposition 3.31 Let X be compact. ThenX is T2 ⇐⇒ X is T3 ⇐⇒ X is T4 .

Proof Exercise.

Tychonoff Theorem A product of compact spaces is compact.

Proof See Munkres, Topology .

Let X be the closed interval [a, b] and for each α ∈ I let Xα be a copy of X . Then∏α∈I Xα is compact. In particular [a, b]n is compact. Moreover we have the following result.

Heine-Borel Theorem Let A be a subset of Rn . Then A is compact if and only if A isclosed and bounded.

Extreme Value Theorem Let X be compact and f : X → R be a continuous function.Then f attains its maximum and minimum on X .

Lebesgue Covering Lemma Let (X , d) be a compact metric space and {Oα} be an opencover of X . Then there exists a positive number δ , called a Lebesgue number of the cover,such that each open ball of radius ≤ δ is contained in at least one Oα .

Definition 3.32 A space X is said to be locally compact if every point of X has a compactneighborhood.

Examples1. A compact space is locally compact.2. R is locally compact.( for any x ∈ R , [x− 1, x+ 1] is a compact neighborhood of x.)3. Rn is locally compact.4. Rl is not locally compact. ( Exercise )5. A discrete space is locally compact.6. Q is not locally compact.


Proposition 3.33 Let X be a locally compact Hausdorff space. Then for any x ∈ X, thecollection of all compact neighborhoods of x is a local basis at x. ( That is any neighborhood of xcontains a compact neighborhood of x . )

Proof Let U be an open neighborhood of x and E be a compact neighborhood of x . ThenE is regular. Since U ∩ E is an open set in E containing x , there exists a set V open inE such that x ∈ V ⊆ V

E ⊆ U ∩ E ⊆ U . V is open in E implies that V = E ∩ V ′

for some V ′ open in X . This shows that V is a neighborhood of x and hence VE is also a

neighborhood of x . Since VE is a closed subset of the compact subset E , V

E is compact.

Corollary 3.34 A locally compact Hausdorff space is regular.

Note that a locally compact Hausdorff space may not be normal. ( see Royden, Real AnalysisP.169 )

Proposition 3.35(1) A closed subspace of a locally compact space is locally compact.

(2) Let f : X −→ Y be a continuous open surjection. If X is locally compact, then Y islocally compact.

(3)∏λ∈I Xλ is locally compact if and only if there exists a finite subset F of I such that Xλ

is locally compact ∀λ ∈ I and Xλ is compact ∀λ ∈ I − F .

Proof We leave (1) and (2) as exercises. Let’s prove (3).Suppose

∏λ∈I Xλ is locally compact . Since the projection pλ is an open map , Xλ =

pλ[∏λ∈I Xλ ] is locally compact by (2). If infinitely many Xλ ’s are non-compact, then by

1.30 each compact subset of∏λ∈I Xλ has empty interior. Hence a point in

∏λ∈I Xλ cannot have

a compact neighborhood.Conversely, let F be a finite subset of I such that Xλ is compact ∀λ ∈ I − F . Let(xλ) ∈

∏λ∈I Xλ . For each λ ∈ F , there exists a compact neighborhood Eλ of xλ .

Then by Tychonoff Theorem ,∏λ∈I Oλ where Oλ = Eλ if λ ∈ F and Oλ = Xλ if

λ ∈ I − F is a compact neighborhood of (xλ) .

Definition 3.36(1) A ⊆ X is said to be dense in X if A = X .

(2) A ⊆ X is said to be nowhere dense in X ifo

A = φ .(3) X is said to be of 1st category if it is a countable union of nowhere dense subsets of X .

It is of 2nd Category if it is not of 1st Category.


Theorem 3.37 Let X be a locally compact Hausdorff space. Then the intersection of a countablecollection of open dense subsets of X is dense in X .

Baire Category Theorem Let X be a non-empty locally compact Hausdorff space. Then Xis of 2nd Category.

Proof Let { En | n ∈ Z+ } be a countable collection of nowhere dense subsets of X . Theneach X − En is open and dense in X . Hence by 1.37

⋂∞n=1( X − En ) is dense in X .

Therefore

X − (∞⋃n=1

En) =∞⋂n=1

(X − En) ⊇∞⋂n=1

( X − En ) 6= φ .

Definition 3.38 A compactification of a space X is a pair (X∗ , i) where X∗ is a compactspace and i : X −→ X∗ is a homeomorphism of X onto a dense subspace of X∗ .

Theorem 3.39Let (X , T ) be a topological space and let ∞ be a point not in X .Then T ∗ = T ∪ {U ∪ {∞} | U ⊆ X and X − U is a compact closed subset of X } is atopology on X∗ = X ∪ {∞} . Furthermore :

(a) (X∗ , T ∗) is compact.(b) If X is non-compact, then (X∗ , i), where i : X → X∗ is the inclusion, is a compacti-fication of X .(c) If X is compact, then ∞ is an isolated point of X∗ .(d) X∗ is Hausdorff if and only if X is locally compact and Hausdorff.

Proof Exercise.

Definition 3.40 The space X∗ = X ∪ {∞} in 3.39 is called the 1-point compactification of X .

Examples

1. The 1-point compactification of the subspace Z+ of positive integers of R is homeomorphicto the subspace {0} ∪ { 1

n | n ε Z+ } . ( Exercise )

4 Countability, Separability and paracompactness 12

2. The n-sphere is the subspace Sn = { (x1, ..., xn+1) ∈ Rn+1 | x21 + ...+ x2n+1 = 1 } ofRn+1 . It is a closed and bounded subset of Rn+1 . Therefore Sn is compact.

The 1-point compactification of Rn is the n-sphere Sn . A homeomorphism between Sn

and Rn∗

is provided by the stereographic projection.Let η = (0, ...0, 1) ∈ Rn+1 be the north pole of Sn and let pn : Rn+1 → R be the projectiononto the last factor of Rn+1 . Identify Rn as Rn × {0} ⊆ Rn+1 . Then the stereographicprojection s : Sn − {η} −→ Rn is given by s(x) = η + 1

1−pn(x)(x− η) . One can easilycheck that s is a homeomorphism. Now extend s to s(x) : Sn −→ 1-point compactificationRn

∗of Rn by

s(x) =

{s(x) if x ∈ Sn − η∞ if x = η

Then s is continuous at η . Since Sn is compact and Rn∗

is Hausdorff, s is a homeomorphism.

4 Countability, Separability and paracompactness

Definition 4.41 A space X is said to be separable if X contains a countable dense subset.

Examples

1. Rn is separable. ( Qn is a countable dense subset.)

2. If (X, T ) is separable and T ′ is a topology on X such that T ⊆ T ′, then (X, T ′) is separable.

3. Rl is separable. ( Q is a countable dense subset.)

Proposition 4.42(1) A continuous image of separable space is separable.(2) An open subspace of a separable space is separable.(3) A product of countably many separable spaces is separable.

Proof Easy exercise.

Definition 4.43 A space X is said to be 2nd countable if it has a countable basis.

Examples


1. R is 2nd countable. ( {(a, b) | a < b, a, b ∈ Q} is a countable basis. )

2. Similarly Rn is 2nd countable.

3. Rl is not 2nd countable. ( Let B be a basis for Rl. For each x ∈ Rl, pick Bx ∈ B s.t.x ∈ Bx ⊆ [x, x+1). Note that x = inf Bx. Then the function f : R −→ B given by f(x) = Bxis injective. This shows that B is not countable. )

Definition 4.44 A space X is said to have a countable basis at x ∈ X if there is a countablecollection B of open neighborhoods of x such that any open neighborhood of x contains a memberof B. A space X is said to be 1st countable if it has a countable basis at each of its point.

Examples

1. If X is 2nd countable, then X is 1st countable. Hence Rn is 1st countable.

2. Rl is 1st countable. ( {[x, x+ 1n) | n ∈ Z

+} is a countable basis at x.)

3. The discrete or indiscrete topology on X is 1st countable.

4. Every metric space is 1st countable. ( {B(x, 1n) | n ∈ Z+} is a countable basis at x.)

5. Let X be an uncountable set and T be the complement finite topology on X. Then (X, T ) isnot 1st countable. ( Exercise )

6. Let I be an uncountable set. For each λ ∈ I, let Xλ be a copy of {0, 1} with the discretetopology. Note that Xλ is 1st countable and is even 2nd countable. But

∏λ∈I Xλ is not 1st

countable. ( Exercise )

Proposition 4.45(1) A subspace of a 1st (2nd) countable space is 1st (2nd) countable.(2) A countable product of 1st (2nd) countable spaces is 1st (2nd) countable.(3) Let f : X −→ Y be a surjective open continuous map. If X is 1st (2nd) countable , then Yis 1st (2nd) countable.

Proof (1) and (3) are very easy. Let’s prove (2) for 1st countability. Let {Xi | i ∈ Z+} be acountable family of 1st countable spaces. Let (xi) ∈

∏i∈Z+ Xi. For each i ∈ Z+, let Bi be a

countable basis at xi. Then B = {∏i∈Z+ Oi | Oi ∈ Bi and Oi = Xi ∀i 6∈ J where J is a

finite subset of Z+ } is a countable basis at (xi).

Remarks


1. An arbitrary product of 1st (2nd) countable spaces may not be 1st (2nd) countable .( see example 6 above.)

2. Consider id : R −→ Rl. It is continuous. Note that R is 2nd countable but Rl is not.

Definition 4.46 A space X is called Lindelof if every open cover of X has a countable subcover.

Theorem 4.47 A 2nd countable space X is Lindelof.Proof Let B = {Ui | i ∈ Z+} be a countable basis ofX. Let {Oα} be an open cover ofX. EachOα is a union of members of B. Therefore there exists a countable open cover {Uαi | i ∈ Z+}of X by basic open sets in B such that each Uαi lies in some Oα. Now for each i ∈ Z+, chooseOαi ⊇ Uαi . Then {Uαi | i ∈ Z+} is a countable subcover of {Oα}.

Examples

1. Rn is Lindelof.

2. Compact spaces are Lindelof.

3. Rl is Lindelof. ( see Munkres p.192 ) Note that Rl is not 2nd countable. Therefore theconverse of 1.47 is not true in general.

4. A discrete space on an uncountable set is not Lindelof.

Theorem 4.48 If X is 2nd countable, then X is separable.

Proof Let B = {Oi | i ∈ Z+} be a countable basis of X. For each i ∈ Z+, let xi ∈ Oi. ThenA = {xi | i ∈ Z+} is a countable dense subset of X.

Note that the converse of this theorem is not true. For example take X = Rl.

Theorem 4.49 If X is a separable metric space, then X is 2nd countable.

Proof Let A be a countable dense subset ofX. We shall prove that B = {B(a, r) | r ∈ Q+, a ∈A} is a countable basis of X. It suffices to show that any open ball is a union of members of B.Let x ∈ B(p, r). Since A is dense, there exists a a ∈ A s.t. d(a, x) < 1

2(r − d(x, p)). Picka q ∈ Q+ s.t. d(a, x) < q < 1

2(r − d(x, p)). Then x ∈ B(a, q) ⊆ B(p, r). This proves theassertion.

Theorem 4.50 If X is a Lindelof metric space, then X is 2nd countable.


Proof We shall prove X is separable. For each j ∈ Z+, {B(x, 1j ) | x ∈ X} is an open coverof X. Since X is Lindelof, it has a countable subcover Bj = {B(xi,j ,

1j ) | i ∈ Z+}. Let

A = {xi,j | i, j ∈ Z+} be the collection of all the centers. Then A is a countable subset of X.Let B(p, r) be an open ball. Pick an jo ∈ Z+ s.t. 1

jo< r. Then Bjo = {B(xi,jo ,

1jo) | i ∈ Z+}

covers X. Therefore p ∈ B(xio,jo ,1jo) for some io. Then d(p, xio,jo) <

1jo< r implies that

xio,jo ∈ B(p, r). Hence A ∩B(p, r) 6= ∅. That is A is dense in X.

Definition 4.51 A locally finite family of subsets of a topological space X is a family such thateach point of X has a neighborhood meeting only finitely many members of the family.A refinement F of a cover C of X is a cover of X such that each member of F is contained in somemember of C.A space X is said to be paracompact if every open cover of X has a locally finite open refinement.

Examples

1. A compact space is paracompact.

2. R is paracompact. ( Let U be an open cover ofR. For each closed interval [N,N+1], N ∈ Z,we have by compactness a finite cover {UN1 , ..., UNn} by members of U . Take as a refinement{UNi ∩ (N,N + 1)}. This does not cover points in Z. To remedy this, we add in a small openneighborhood ( of diameter less than 1 and small enough to be contained in some member of U )of each N ∈ Z. We then have the required locally finite open refinement.)

3. Rn is paracompact. ( similar proof as above )

4. Rl is paracompact. ( Exercise )

Proposition 4.52 A paracompact Hausdorff space is regular.

Proof Let U be an open neighborhood of a point p. We shall construct a neighborhood V of psuch that V ⊆ U. For each q 6∈ U, choose open disjoint neighborhoods Vq, Uq of p, q respec-tively. Then {Uq | q ∈ X \ U} ∪ {U} is an open cover of X and thus has a locally finite openrefinement. Let M be a neighborhood of p meeting only finitely many members of the refinementand letW1, ...,Wn be those not contained in U. Then there exist q1, ..., qn ∈ X \U s.t. Wi ⊆ Uqi .let V = M ∩

⋂ni=1 Vqi . It remains to show that V ⊆ U. Let W be the union of all members of

the above refinement not contained in U. Then V ∩W ⊆ M ∩ (⋂ni=1 Vqi) ∩ (

⋃ni=1 Uqi) = ∅.

Thus V ⊆ X \W. Since W ⊇ X \ U, V ⊆ X \W ⊆ U.

Proposition 4.53 A paracompact regular space is normal.( Exercise )

5 Connectedness 16

Proposition 4.54 A Lindelof regular space is normal.

Proof Exercise.

Theorem 4.55 If X is paracompact and separable, then X is Lindelof.

Proof Let {Uα} be an open cover of X. Let {Vβ} be a locally finite open refinement of {Uα}.We may assume each Vβ 6= ∅. Because X has a countable dense subset {xi | i ∈ Z+}, {Vβ} isat most a countable family. ( Each Vβ contains at least one xi. If {Vβ} is an uncountable family,then there is some xi contained in uncountably many Vβ. This contradicts the local finiteness of{Vβ}. ) For each Vβ, choose Uα(β) ⊇ Vβ. Then {Uα(β)} is a countable subcover of {Uα}.

Remark

A product a two paracompact ( Lindelof ) spaces may not be paracompact ( Lindelof ). For exam-ple, take Rl ×Rl.

Urysohn’s Metrization Theorem A 2nd countable T3 is metrizable.

5 Connectedness

Definition 5.56 A space X is said to be connected if it is not the union of two non-empty disjointopen subsets of X . A ⊆ X is connected if it is connected as a subspace of X .

Examples

1. Any indiscrete space is connected.

2. R is connected. ( Exercise. )

3. Rl is not connected.

4. The subspace Q of all rational numbers of R is not connected.

5. Rn is connected.

6. A ⊆ R is connected if and only if A is an interval.

5 Connectedness 17

7. A compact connected T2 space cannot be a union of countably many but more than one disjointclosed subsets.

Proposition 5.57 The following statements are equivalent :(1) X is connected .(2) There is no proper non-empty subset of X which is both open and closed in X .(3) Every continuous map from X to {0, 1} is constant, where {0, 1} is the two point space withthe discrete topology.

Proof Exercise.

Proposition 5.58 Let f : X −→ Y be continuous and A be a connected subset of X . Thenf [A] is connected.

Proof This follows directly from 1.57(3).

Proposition 5.59 R is connected.

Proof A non-empty proper open subset O of R is a disjoint union of open intervals. Then one ofthese open intervals has an endpoint a not in O. Hence O is not closed.

Proposition 5.60(1) If A is a connected subset of X , then A is also a connected subset of X .

(2) If A is a connected subset of X and B is subset of X such that A ⊆ B ⊆ A, then B isconnected.

Proof Exercise.

Let A be the set { ( x, sin( 1x) ) ∈ R2 | x > 0 }. Then A and A ∪ {(0, 0)} are connected

subsets of R2.

Corollary 5.61 A ⊆ R is connected if and only if A is an interval or a singleton set.

Proof An interval I of R is a subset of R having at least two points and satisfying the conditionthat for any a, b ∈ I , the line segment joining a and b is also contained in I . Then I is an intervalif and only if I is of the form (a, b), (a, b], [a, b) or [a, b] where −∞ ≤ a < b ≤ ∞. Nowthe forward implication is clear. Conversely if A is 5an open interval, then it is homeomorphic toRwhich is connected by 5.59. The rest of different types of intervals are also connected by 5.60(2).

5 Connectedness 18

Intermediate Value Theorem(1) If f : X −→ R is continuous and X is connected, then f [X] is an interval.(2) Let f : [a, b] −→ R be continuous. Then f assumes all the values between f(a) and f(b) .

Proposition 5.62 [a, b) is not homeomorphic to (a, b) .

Proposition 5.63 Let {Aα}α∈J be a family of connected subsets ofX such that⋂α∈J Aα 6= φ .

Then⋃α∈J Aα is connected.

Proof Use 5.57(3).

Definition 5.64 Two points a, b in a space X are said to be connected, written a ∼ b, if thereis a connected subspace of X containing a and b.

Proposition 5.65 If every pair of points in X are connected, then X is connected.

Proof Let U be both open and closed in X . Suppose U 6= φ, X. Then there exista ∈ U, b ∈ X − U. By assumption, there exists a connected subset C containing a and b.But then U ∩C is a proper non-empty both open and closed subset of C. This contradicts the factthat C is connected.

Proposition 5.66 ∼ is an equivalence relation.

Proof Exercise.

Definition 5.67 The connected component of a point p ∈ X , denoted by C(p) , is the equiva-lence class of ∼ containing p.

Proposition 5.68(1) C(p) is the largest connected set containing p .(2) C(p) is closed.

Proof(1) C(p) =

⋃{ C : C is connected and p ∈ C }. By 1.63 C(p) is connected.

(2) C(p) is connected and contains p . Hence by (1) , C(p) ⊆ C(p).

Proposition 5.69 If X and Y are connected, then X × Y is also connected.

5 Connectedness 19

Proof Let (a, b), (x, y) ∈ X × Y. Since X × {b} and {x} × Y are connected, we have(a, b) ∼ (x, b) ∼ (x, y). By 1.65 X × Y is connected.

In fact one can make use 1.69 to prove that arbitrary product of connected spaces is connected.This is left as an exercise.

Corollary 5.70 Rn, In and Sn are connected.

Proof Let p ∈ Sn. As Sn − {p} is homeomorphic to Rn, Sn − {p} is connected. HenceSn = Sn − {p} is connected.

Definition 1.71 X is said to be locally connected at p ε X if for any neighborhood U of p, thereexists a connected neighborhood V of p contained in U.X is locally connected if it is locally connected at each of its points.

Note that X is locally connected if and only if the collection of all open connected subsets of Xis a basis for the topology on X.

Examples1. Let A be the set { (x, sin( 1x)) ∈ R

2 | x > 0 }. Then A ∪ {(0, 0)} is not locally connectedbut connected.

2. Let A = (0, 1) ∪ (2, 3). Then A is locally connected but not connected.

3. Rn and Sn are locally connected.

4. Rl is not locally connected.( Let U be any neighborhood of x. Hence U ⊇ [x, x + 2δ) for some δ > 0. Then[U ∩ (∞, x + δ)]

⋃[U ∩ [x + δ,∞)] = U is a disjoint union of two non-empty open sub-

sets of U . Therefore any neighborhood of x is not connected. )

5. Discrete spaces are locally connected.

6. Q is not locally connected.

Proposition 5.72 If a space is locally connected at p ∈ X, then p is an interior point of C(p).

Proof By assumption p has a connected neighborhood U . Therefore p ∈ U ⊆ C(p).

Corollary 5.73 If X is locally connected, then C(p) is an open subset of X.

5 Connectedness 20

Proposition 5.74 Every open subspace of a locally connected space is locally connected.

Proposition 5.75 X is locally connected if and only if the connected components of every opensubspace of X are open in X .

Proof Suppose X is locally connected. Let O be an open subspace of X. By 1.58 and 1.59O is locally connected and each connected component of O is open in O. Since O is open inX, each connected component of O is open in X. Conversely let U be an open neighborhoodof p ∈ X. By assumption the connected component CU (p) in the subspace U is open in X.Hence CU (p) is open in X. Therefore CU (p) is a connected open neighborhood of p containedin U. This shows that X is locally connected.

Corollary 5.76 Let N be a neighborhood of a point x in a locally connected space X. Thenthere exists a connected open neighborhood C of x lying inside N.

Proposition 5.77 Let {Xα}α∈I be a family of locally connected spaces such that all but at mostfinitely many are connected. Then

∏α∈I Xα is locally connected.

Proof Let F be a finite subset of I such that Xα is not connected for all α ∈ F. Letx ∈

∏α∈I Xα and O be a basic open neighborhood of x. Note that O =

∏α∈I Oα where

Oα is open in Xα and Oα = Xα for all α ∈ I − J with J some finite subset of I. For eachα ∈ F ∪J, there exists an open connected neighborhood Vα of Xα such that Vα ⊆ Xα. LetU =

∏α∈I Uα where Uα = Vα if α ∈ F ∪ J, and Vα = Xα if α ∈ I − F ∪ J. Then U

is a product of connected sets and hence it is connected. Also U is a basic open neighborhood ofx lying in O.Note that the converse of this result is also true since local connectedness is invariant under contin-uous open surjection.

Definition 5.78 A path joining a pair of points a and b of a space X is a continuous mapα : I → X such that α(0) = a and α(1) = b, where I denotes the closed unit in-terval [0, 1].

It can be verified easily that the relation of being joined by a path is an equivalence relation. Theequivalence classes of this relation are called the path components of X. The path component of apoint x in X is denoted by P (x). P (x) is the largest path connected set containing x. Howeverunlike C(x) , P (x) may not be closed.

Definition 5.79 A space X is said to be path connected if it has exactly one path component. (That is any pair of points in X can be joined by a path. )

Proposition 5.80 A path connected space is connected.

5 Connectedness 21

Proposition 5.81 Let f : X −→ Y be a continuous surjection. If X is path connected, thenY is path connected.

Proof Exercise.

Examples1. Rn and Sn are path connected.( It is easy to check that Rn and Rn − {0} are path connected. Let σ : Rn+1 − {0} −→ Sn

be the map given by σ(x) = x‖x‖ . Then by 5.65, Sn is path connected. )

2. Let A be the set { (x, sin( 1x) ) | x > 0 }⋃{(0, 0)}. Then A is not path connected.

Note that A − {(0, 0)} is a path component which is not closed in A.

Proposition 5.82 Let {Xα}α∈I be a family of spaces. Then∏α∈I Xα is path connected if

and only if each Xα is path connected.

Proof Exercise.

Definition 5.83 A space is said to be locally path connected at x ∈ X if each neighborhood ofx contains a path connected neighborhood. X is said to be locally path connected if it is locallypath connected at each of its points.Note that X is locally path connected if and only if the collection of all path connected open sets ofX is a basis for the topology on X.

Examples1. Rn is locally path connected since every open ball in Rn is path connected. Similarly Sn islocally path connected.

2. (0,1) ∪ (2,3) is locally path connected but not path connected.

3. For each positive integer n , let Ln be the set of all points on the line segment joining thepoints (0,1) and ( 1

n ,0) in R2 and also let L∞ = {(0, y) | 0 ≤ y ≤ 1} . Then the subspaceA = L∞ ∪ {

⋃∞n=1 Ln } in R2 is not locally path connected but path connected.

4. If X is locally path connected, then X is locally connected.

5. If L∞ of A in 3. is replaced by L∗∞ = { (0, 1− 1n) | n ∈ Z

+}, then A is locally connectedat (0,1) but not locally path connected at (0,1).

6 Identifications and Adjunction Spaces 22

Proposition 5.84 Every open subspace of a locally path connected space is locally path connected.

Proposition 5.85 X is locally path connected if and only if the path components of every opensubspace of X are open.

Proposition 5.86∏α∈I Xα is locally path connected if and only if all Xα are locally path con-

nected and all but at most finitely many are also path connected.

Proposition 5.87 If X is locally path connected and connected, then X is path connected.

Proof By 5.85 a path component of X is open and hence is also closed. Since X is connected,there is exactly one path component.

6 Identifications and Adjunction Spaces

Definition 6.88 A continuous surjective map p : (X, T ) −→ (Y, T ′) is called an identificationor a quotient map if p−1[O] is open in X if and only if O is open in Y.

Proposition 6.89 Let (X, T ) be a topological space and Y be a set. Let p : X −→ Y be asurjective map. Then Tp = {U ⊆ Y | p−1[U ] is open in X} is a topology on Y. Furthermorep : (X, T ) −→ (Y, Tp) is an identification.

Proof Direct verification.

The topology Tp on Y defined in 6.89 is called the identification or quotient topology induced byp. In fact it is the largest topology on Y that makes p continuous.

Definition 6.90 Let f : X −→ Y be a map. A subsetO in X is said to be saturated with respectto f if f−1[f [O]] = O.

If p : X −→ Y is a map, then any set of the form p−1[U ] is saturated with respect to p.Hence a continuous surjective map p : (X, T ) −→ (Y, T ′) is an identification if and only for anyopen set O ⊆ X saturated with respect to p, p[O] is open in Y.

Proposition 6.91 Let p : X −→ Y be continuous.(1) If p is an open surjection ( or a closed surjection ) , then p is an identification.(2) If there exists a continuous s : Y −→ X such that p ◦ s = id |Y , then p is an identification.


Proof Exercise. Note that the converses of (1) and (2) are not true.

Proposition 6.92 Let p : X −→ Y be an identification and g : Y −→ Z be a surjective map.Then g ◦ p is an identification if and only if g is an identification.

Proof(=⇒) One can easily see that the continuity of g ◦ p implies the continuity of g. Let O ⊆ Z besuch that g−1[O] is open in Y. Because p and g are continuous, (g ◦ p)−1[O] = p−1[g−1[O]] isopen in X. Since g ◦ p is an identification, O is open in Z. Hence g is an identification.

(⇐=) Let O ⊆ Z be such that (g ◦ p)−1[O] is open in X. Since p and g are identifications, Ois open in Z. Therefore g ◦ p is an identification.

Let (X, T ) be a space and R be an equivalence relation defined on X. For each x ∈ X, let [x]be the equivalence class containing x. Denote the quotient set by X/R. Let p : X −→ X/Rgiven by p(x) = [x] be the natural projection. The set X/R with the identification topology Tpinduced by p is called the quotient space of X by R.

Let A be a subset of X. Define an equivalence relation RA on X by xRAy if and only ifx, y ∈ A. Then the quotient space X/RA is the space X with A identified to a point [A] and isusually denoted by X/A. Note that E ⊆ X is saturated ( with respect to the projection p ) if andonly if E ⊇ A or E is disjoint from A.

Corollary 6.93 Let X,Y be spaces with equivalence relations R and S respectively, and letf : X −→ Y be a relation preserving, ( i.e. x1Rx2 =⇒ f(x1)Sf(x2) ) continuous map.Then the induced map f∗ : X/R −→ Y/S given by f∗([x]) = [f(x)] is continuous. Further-more, f∗ is an identification if f is an identification.

Proposition 6.94 Let Y and A be closed subspaces of X, then Y/(Y⋂A) is a subspace of X/A.

( In fact Y/(Y⋂A) is homeomorphic to a subspace of X/A. )

Proof Consider the commutative diagram

-

-

? ?

Y

Y/Y⋂A X/A

X

p p

i

i

Clearly i is a continuous injection. Thus it suffices to show that i : Y/Y⋂A −→ pi[Y ] is an

open map. Then Y/Y ∩A is homeomorphic to the subspace pi[Y ] = i[Y/Y⋂A] of X/A. Let


U ⊆ Y/Y⋂A be open. Then p−1[U ] is open in Y. Therefore there exists a W open in X such

that Y⋂W = p−1[U ]. Since p−1[U ] is saturated with respect to p, either p−1[U ]

⋂A = Y

⋂A

or p−1[U ]⋂A = ∅. In the first case, i[U ] = p[Y ]

⋂p[[X \Y ]

⋃W ]. Thus, since (X \Y )

⋃W

is open and saturated, i[U ] is open in p[Y ]. In the second case, i[U ] = p[Y ]⋂p[(X \A)

⋂W ]

which is also open in p[Y ] since X \A is saturated and open in X.

Examples1. Let X be a space with a preferred base point ∗, then the reduced suspension SX is the quotientspace (X × I)/(X × {0, 1}

⋃{∗} × I) where I = [0, 1] is the unit interval.

2. The reduced cone on X, CX is the quotient space (X × I)/(X × {0}⋃{∗} × I). If {∗}

is a closed subspace ( e.g. X is T1 ), X × {0}⋃{∗} × I is closed in X × I. Hence by 6.94

X × {1} ∼= X is a subspace of CX.

Let p : Q −→ Q/Z be the natural projection which is an identification when Q/Z is providedwith the quotient topology. The map p× idQ : Q×Q −→ (Q/Z)×Q is not an identification.In general a product of two identifications may not be an identification. (c.f. A product of twoclosed maps may not be closed. ) However we have the following results.

Proposition 6.95 Let f : X −→ Y and g : T −→ Z be identifications. Suppose T is compactand Z is regular. Then f × g : X × T −→ Y × Z is an identification

Proposition 6.96 If f : X −→ Y is an identification and Z is locally compact and regular, thenf × 1Z : X × Z −→ Y × Z is an identification.(This result can also be deduced from the exponential law in function space topology.)

Let X and Y be topological spaces and A be a subset of X. Let f : A −→ Y be a continuousmap. Define an equivalence relation ∼ on Y tX by :

a ∼ b if and only if

a = b for a, b ∈ X, or a, b ∈ Yf(a) = f(b) for a, b ∈ Aa = f(b) for a ∈ Y, b ∈ Af(a) = b for a ∈ A, b ∈ Y

Definition 6.97 The quotient space (Y t X)/ ∼ is called an adjunction space and is denoted byY⋃f X. The construction means that X is attached to Y by means of f : A −→ Y.

Suppose θ : X −→ Z and φ : Y −→ Z satisfy θ |A= φ ◦ f, then a map θ ∪f φ :Y⋃f X −→ Z is defined by


(θ ∪f φ)(x) =

{θ(x) x ∈ Xφ(x) x ∈ Y

Note that this is well-defined since if x ∼ y then θ(x) = φ(y). Let p : Y tX −→ Y⋃f X be the

quotient map. Clearly, (θ ∪f φ) ◦ p is continuous and hence by 1.92 θ ∪f φ is continuous.

Proposition 6.98 Let A be closed in X. Then the map i : Y −→ Y ∪f X, defined byi(y) = p(y), is an injective homeomorphism and i[Y ] is closed in Y ∪f Y.

Proof Let B ⊆ Y be closed. Then p−1[i[B]] = B t f−1[B] is closed in Y tX. Hence i[B] isclosed in Y ∪f X. Therefore i is a closed injective map. In particular, i[Y ] is a closed subspaceof Y ∪f X.

Proposition 6.99 Let A be closed in X. Then p |X\A is an injective homeomorphism onto anopen subspace of Y ∪f X.

Proof p |X\A is obviously continuous and bijective. Let O be open in X \A. Then O is openand saturated in X. Since p is an identification, p[O] is open in Y ∪f X.

Proposition 6.100 If X and Y are normal and A ⊆ X is closed, then Y ∪f X is normal.

Proof Let P andQ be disjoint closed sets in Y ∪fX. Then P1 = P ∩ i[Y ] andQ1 = Q∩ i[Y ] areclosed in i[Y ]. Since Y is normal, i[Y ] is normal by 1.83. Therefore there exist neighborhoodsR of P1 and S of Q1 such that R,S are open in i[Y ] and they have disjoint closures. Note thatsince i[Y ] is closed in Y ∪f X , R and S have the same closures in i[Y ] and in Y ∪f X. LetP2 = p−1[P ∪R]∩X and Q2 = p−1[Q∪ S]∩X. Then P2 and Q2 are disjoint and closed in X .Since X is normal, there exist disjoint open neighborhoods M and N of P2 and Q2 respectively.ThenD = p[M \A]∪R andE = p[N \A]∪S are disjoint open neighborhoods of P andQ respec-tively ( They are open since , for example p−1[D] = R t (M \A) ∪ f−1[R] =M \ (A \ f−1[R])which is open in Y tX ) .

Proposition 6.101 The adjunction space construction preserves connectedness, path- connected-ness and compactness. When A ⊆ X is closed, the T1 property is preserved.

Proof Since X and Y are connected ( path-connected ) , their continuous images are alsoconnected ( path-connected). As they have non-empty intersection and their union is Y ∪f X,Y ∪f X is connected ( path-connected) . Since Y ∪f X is the continuous image of the compactspace Y tX, it is compact. Finally, let y ∈ Y. Then since Y is closed in Y ∪f X, so is {y}.Similarly if x ∈ X \A, then p−1[{x}] is closed in X. Hence {x} is closed in Y ∪f X since p isan identification.

7 Function spaces 26

Corollary 5.102 If X and Y are normal Hausdorff and A ⊆ X is closed, then Y ∪f X is alsonormal Hausdorff.

7 Function spaces

Let X and Y be topological spaces and let Y X be the set of all continuous maps from X to Y.For any subset C ⊆ X and U ⊆ Y, define N(C,U) = {f : X −→ Y | f [C] ⊆ U}.

Examples

1. If X is discrete, then Y X is homeomorphic to the product space∏α∈X Yα, where each Yα is a

copy of Y.

2. Let X be a compact space and Y be a metric space. Then the compact-open topology on Y X

equals the topology of uniform convergence on Y X .

Definition 7.103 The topology on Y X generated by the subbasis {N(C,U) | C is compact in Xand U is open in Y } is called the compact-open topology on Y X .

Lemma 7.104 If X is Hausdorff and S is a subbasis for the topology on Y, then C = {N(C, V ) |C is compact in X and V ∈ S } is a subbasis for the compact-open topology on Y X .

Proof C is subbasis for a topology on Y X . It is clear that this topology is a subcollection of thecompact-open topology on Y X . Therefore it suffices to show that for any compact subset C ⊆ X,any open subset U ⊆ Y and any f ∈ N(C,U), there exist V1, ..., Vm ∈ S and compact subsetsC1, ..., Cm ⊆ X such that f ∈

⋂mi=1N(Ci, Vi) ⊆ N(C,U). Let x ∈ C. Then U is an open

neighborhood of f(x). Hence there exist V x1 , ..., V

xn(x) ∈ S such that f(x) ∈

⋂n(x)i=1 V

xi ⊆ U. f

is continuous implies that there exists an open neighborhood Wx of x such that f [Wx] ⊆⋂n(x)i=1 V

xi .

Since C is compact and Hausdorff, it is regular and thus there is an open neighborhood Sx ofx in C such that Sx ⊆ Wx ∩ C ⊂ Wx. ( Note that Sx = Sx

C is compact. ) Hencef ∈ N(Sx,

⋂n(x)i=1 V

xi ) =

⋂n(x)i=1 N(Sx, V

xi ). SinceC is compact, the open cover {Sx | x ∈ C}

has a finite subcover {Sx1 , ..., Sxn}. Then f ∈⋂nr=1[

⋂n(xr)i=1 N(Sxr , V

xri )] ≡ B. It remains

to show that B ⊆ N(C,U). Let g ∈ B and x ∈ C. Then x ∈ Sxr for some r. Henceg(x) ∈ g[Sxr ] ⊆

⋂n(xr)i=1 V xr

i ⊆ U. Therefore g[C] ⊆ U and g ∈ N(C, V ).

Lemma 7.105 If Y is Hausdorff, then Y X is Hausdorff.

Proof Exercise.

Given f ∈ ZX×Y the function e(f) : X −→ ZY is defined by e(f)(x)(y) = f(x, y), forx ∈ X, y ∈ Y.


Lemma7.106 e(f) is continuous.

Proof We shall prove that inverse images of subbasic open sets in ZY are open in X. Letx ∈ e(f)−1[N(C, V )], where C is compact in Y and V is open in Z. Hence f [{x} × C] ⊆ V or{x} × C ⊆ f−1[V ]. Since f is continuous, f−1[V ] is an open neighborhood of {x} × C. Bycompactness ofC, there exist open setsAi ⊆ X andBi ⊆ Y such that {x}×C ⊆

⋃ni=1Ai×Bi ⊆

f−1[V ], and x ∈⋂ni=1Ai. It remains to show that

⋂ni=1Ai ⊆ e(f)−1[N(C, V )]. Let

x′ ∈⋂ni=1Ai and y ∈ C. Then y ∈ Br, for some r and (x′, y) ∈ Ar × Br ⊆ f−1[V ].

Thus e(f)(x′)[C] ⊆ V and x′ ∈ e(f)−1[N(C, V )].

Definition 7.107 The function e : ZX×Y −→ (ZY )X given by e(f)(x)(y) = f(x, y), forx ∈ X, y ∈ Y is called the exponential function.

Lemma 7.108 If X is Hausdorff, then e : ZX×Y −→ (ZY )X is continuous.

Proof By the first lemma, {N(C,N(K,V )) | C is compact in X, K is compact in Y, V is openin Z } is a subbasis for the topology on (ZY )X . Since e−1[N(C,N(K,V ))] = N(C ×K,V ) isopen in (ZY )X , e is continuous.

Lemma 7.109 If Y is locally compact and regular, then e is bijective.

Proof It is easy to see that e is injective. To show e is surjective, let f ′ : X −→ ZY . Letf : X × Y −→ Z be the map defined by f(x, y) = f ′(x)(y). It suffices to show that f is contin-uous. Then e(f) = f ′. Let V be an open neighborhood of f(x, y) in Z. Since f ′(x) : Y −→ Zis continuous and Y is locally compact and regular, there exists a compact neighborhood C of ysuch that f ′(x)[C] ⊆ V. Hence f ′(x) ∈ N(C, V ). Since f ′ is continuous, there exists an openneighborhood W of x such that f ′[W ] ⊆ N(C, V ). Then W × C is a neighborhood of (x, y) andf [W × C] ⊆ V.

Lemma 7.110 If X, Y are Hausdorff, then e is an injective homeomorphism onto its image.

Proof It has been shown in previous lemmas that e is injective and continuous. It thus sufficesto show that for C compact in X × Y and V open in Z, e[N(C, V )] is open in e[ZX×Y ]. Letf ′ ∈ e[N(C, V )] and let f ∈ N(C, V ) be such that e(f) = f ′. LetA andB be the projections ofCin X,Y respectively. A and B are compact and Hausdorff and hence are regular. Since f is con-tinuous, for each point p = (x, y) ∈ C, there exist closed compact neighborhoodsRp of x inA andSp of y inB such that f [Rp×Sp] ⊆ V. AsC is compact, the open cover {Rop×Sop p ∈ C} has a fi-nite subcover {Rop1×S

op1 , ..., R

opn×S

opn}. Hence f ′[Rpi ][Spi ] ≡ f [Rpi×Spi ] ⊆ V, i = 1, ..., n.

Therefore f ′ ∈ W ≡ e[ZX×Y ] ∩ {⋂ni=1N(Rpi , N(Spi , V ))}, which is open in e[ZX×Y ]. It

remains to show that W ⊆ e[N(C, V )]. Let g′ ∈ W and g ∈ ZX×Y be such that e(g) = g′.Given any p = (x, y) ∈ C, p ∈ Rpr × Spr for some r. Then g(p) = g′(x)(y) ∈ V. Thusg ∈ N(C, V ) and hence g′ ∈ e[N(C, V )].

Theorem 7.111 ( Exponential Law ) Let X,Y be Hausdorff and Y be locally compact. Thene : ZX×Y −→ (ZY )X is a homeomorphism.


Proof This follows from the previous lemmas.

Proposition 7.112 Let g : Y −→ Z be a map. Then the maps(i) g∗ : XZ −→ XY defined by gX(f) = f ◦ g and(ii) g∗ : Y X −→ ZX defined by gX(f) = g ◦ fare continuous.

Proof Exercise.

Proposition 7.113 (X × Y )Z ∼= XZ × Y Z .

Proof The map φ : XZ×Y Z −→ (X×Y )Z given by φ(f, g) = (f×g)◦dZ is clearly bijective.Here dZ is the diagonal map on Z. Since φ−1 = (pX∗ × pY ∗) ◦ dZ , φ is continuous. For anycompact C in Z and open sets U ⊆ X, V ⊆ Y, φ[N(C,U)×N(C, V )] = N(C,U × V ). Henceφ is open and hence a homeomorphism.

References

[1] J. Dugundji, Topology , Prentice-Hall.

[2] J. R. Munkres, Topology, Prentice-Hall.

mh404 point set topology

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