# Point Set Topology. A. Topological Spaces and Continuous Maps

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1 Point Set Topology A. Topological Spaces and Continuous Maps Definition 1.1 A topology on a set X is a collection T of subsets of X satisfying the following axioms: T 1.,X T. T2. {O α α I} T = α IO α T. T3. O, O T = O O T. A topological space is a pair (X, T )wherexis a set and T is a topology on X. When the topology T on X under discussion is clear, we simply denote (X, T ) by X. Let (X, T ) be a topological space. Members of T are called open sets. (T 3) implies that a finite intersection of open sets is open. 1. Let X be a set. The power set 2 X of X is a topology on X and is called the discrete topology on X. The collection I = {,X} is also a topology on X and is called the indiscrete topology on X. 2. Let (X, d) be a metric space. Define O X to be open if for any x O, there exists an open ball B(x, r) lying inside O. Then, T d = {O X O is open} { } is a topology on X. T d is called the topology induced by the metric d. 3. Since R n is a metric space with the usual metric: d((x 1,..., x n ), (y 1,..., y n )) = n (x i y i ) 2, R n has a topology U induced by d. This topology on R n is called the usual topology. 4. Let X be an infinite set. Then, T = {,X} {O X X\O is finite} is a topology on X. T is called the complement finite topology on X. 5. Define O R to be open if for any x O, there exists 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 topology on R. NotethatU T but U T. We shall denote this topological space by R l. 1 i=1

2 Proposition 1.2 Let (X, T ) be a topological space and A a subset of X. Then, the collection T A = {O A O T}is a topology on A. T A is called the subspace topology on A and (A, T A ) is called a subspace of (X, T ). 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 (1) and X are closed in X. (2) If E,F are closed in X, thene F is closed in X. (3) If {E α α I} is a collection of closed sets in X, then α IE α 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 set in 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 collection of all finite intersections of members of S is a basis for T. Hence, any open subset of X is a union of finite intersections of members in S. 1. {(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) beametricspaceandt d the topology on X induced by d. The collection B of all open balls of X is a basis for T d. 3. {[a, b) a<b}is a basis for the lower limit topology on R l. What would be a subbasis for the lower limit topology on R l? 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 a collection of subsets of X such that B1. U B U = X, B2. for any U 1,U 2 Band x U 1 U 2, there exists U Bsuch that x U U 1 U 2. Then the collection T B of all unions of members of B is a topology on X and B is a basis for T B. Proof T 1andT2 are clearly satisfied. B2 implies that if U α,u β B, then U α U β T B.LetO 1 = U α and let O 2 = U β be in T B,whereU α,u β B.Then 2

3 O 1 O 2 = (U α U β )isint.thatbis a basis for T B follows from the definition of T B. 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 T S of all unions of finite intersections of members of S is a topology on X and S is a subbasis for T S. Proof Similar to 1.7. Definition 1.9 A subset N containing a point x in X is called a neighbourhood of x if there exists an 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 neighbourhood U of x inside A. (2) x X is called a limit point of A if (O \{x}) A, for any neighbourhood O of x. (3) A o = {x A x is an interior point of A} is called the interior of A. (4) A = A {x X xis a limit point of A} is called the closure of A. Proposition 1.11 (1) A o is an open set and A o A. (2) X \ A =(X\A) o and X \ A o = X \ A. (3) A is a closed set and A A. (4) A is open A = A o. (5) A is closed A = A. (6) A o is the largest open set contained in A. (7) A is the smallest closed set containing A. (8) A o = {O O is open and O A}. (9) A = {E E is closed and E A}. (10) A oo = A o and A = A. (11) (A B) o = A o B o. (12) A B = A B. (13) Let B A X. Then B is closed in A if and only if B = E A for some closed set E in X. (14) Let B A X and let B A be the closure of B in A. ThenB A =B A. (15) Let B A X and let B o A be the interior of B in A. If A is open, then B o A = B o. 3

4 Proof Exercise. Definition 1.12 Let (X, T )and(y,t ) be topological spaces and f : X Y be amap.letx X.fis said to be continuous at x if for any open neighbourhood U of f(x), there exists an open neighbourhood O of x such that f[o] U. f is said to be continuous on X if f is 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] isopeninx. (3) For any closed set E Y, f 1 [E] isclosedinx. (4) For any A X, f[a] f[a]. 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] isopeniny. (2) f is called a closed map if for any closed set E in X, f[e] isclosediny. (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 mapping. The following statements are equivalent. (1) f 1 : Y X 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 A 1,...,A n be a finite collection of closed subsets of X such that n i=1 A i = X. Thenf:X Yis continuous if and only if f A i is continuous for each i =1,...,n. Proof The only if part is obvious. Suppose f A i is continuous for each i =1,...,n.LetEbe a closed subset of Y.Thenf 1 [E]= n i=1 (f 1 [E] A i )= ni=1 (f A i ) 1 [E]. Since f A i is continuous, (f A i ) 1 [E] isclosedina i.asa i 4

5 is closed in X, (f A i ) 1 [E]isclosedinX. Hence f 1 [E] isclosedinx. This shows that f is continuous on X. B. Product and Sum Topologies Definition 1.18 Let {X λ } λ I be a collection of topological spaces and let p α : λ I X λ X α be the natural projection. The product topology on λ I X λ is defined to be the one generated by the subbasis {p 1 α [O α] O α is open in X α,α I}. Note that when λ I X λ is given the product topology the projection p α is continuous and open. In the case that the product is finite the collection { α I O α O α is open in X α } is a basis for the product topology on λ I X λ. Proposition 1.19 Let f : X λ I X λ be a map. Then f is continuous if and only if p α f is continuous for all α I. The following results are immediate consequences. Proposition 1.20 The diagonal map d : X X X, defined by d(x) =(x, x), and the twisting map τ : X Y Y X, defined by τ(x, y) =(y, x), are continuous. Proposition 1.21 Given a family of maps {f λ : X λ Y λ } λ I, the product map λ I f λ : λ I X λ λ I Y λ, defined by ( λ I f λ )(x λ )=(f λ (x λ )), is continuous. Definition 1.22 Let {X α } be a family of topological spaces. The topological sum X α is the topological space whose underlying set is X α equipped with the topology {O X α O X α is open in X α for each α}. Note that when X α is given the sum topology the inclusion i α : X α X α is continuous. In general it may not be possible to fit together the topologies on a family of spaces {X α } to obtain a topology on the union restricting to the original topology on each X α. For example, if two spaces X α and X β overlap they may not induce the same topologies on X α Y β. A family {X α } of spaces is said to be compatible if the two topologies on X α X β are identical for all pair α, β. Proposition 1.23 Let {X α } be a compatible family of spaces such that X α X β is open in X α and X β, for each α, β. Then X α has the subspace topology from X α. 5

6 Proof Exercise. For example if {X α } is a family of disjoint topological spaces, then each X α is a subspace of X α. Proposition 1.24 If {f α : X α Y } is a family of maps such that f α and f β agree on X α X β for all α, β, then there is a unique continuous map f α : X α Y such that ( f α ) X α = f α, for all α. Proof Exercise. C. Separation Axioms and Compactness Definition 1.25 Let X be a topological space. (1) X is T 0 if for each pair of distinct points, at least one has a neighbourhood not containing the other. (2) X is T 1 or Fréchet if for any distinct x, y X, there exist open neighbourhoods U of x and V of y such that y is not in U and x is not in V. (3) X is T 2 or Hausdorff if for any distinct x, y X, there exist open neighbourhoods U of x and V of y such that U V =. (4) X is said to be regular if for any closed set E and any point x not in E, there exist an open neighbourhood U of E and an open neighbourhood V of x such that U V =. (5) X is said to be normal if for any two disjoint closed sets E and F,thereexist an open neighbourhood U of E and an open neighbourhood V of F such that U V =. (6) X is T 3 if X is T 1 and regular. (7) X is T 4 if X is T 1 and normal. It is easy to see that X is T 1 if and only if every singleton set in X is closed. Hence we have T 4 = T 3 = T 2 = T 1 = T Let X = {0, 1} and T = {,X,{0}}. Then(X, T )ist 0 but not T Every metric space is T Let X be an infinite set with the complement finite topology. 4. Then X is T 1 but not T 2. 6

7 5. Let K = { 1 n 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 T 2 but it is not regular. 6. R l is T R l R l is regular but not normal. In general, a product of normal spaces may not be normal. 8. A subspace of a Hausdorff space is Hausdorff and a subspace of a regular space is regular. However, a subspace of a normal space may not be normal. 9. A product of Hausdorff spaces is Hausdorff and a product of regular spaces is regular. Urysohn s Lemma If A and B are nonempty disjoint closed subsets in a normal space X, then there exists a continuous function f : X [0, 1] such that f[a] =0 and f[b] =1. Proof See [2] p.207. Definition 1.26 A cover of a topological space X is a collection C = {O α } α I of subsets of X such that α I O α = X. ItissaidtobeanopencoverifeachO α is open in X. If the index set I is finite (or countable), then {O α } α I is called a finite (or countable) cover. A subcover of a cover C of X is a subcollection S of C such that S is a cover of X. Definition 1.27 AspaceXis 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. 1. Any finite space is compact. 2. An indiscrete space is compact. 3. [a, b] iscompact. 4. An infinite set provided with the complement finite topology is a compact space. 5. R is not compact. Neither does R l Proposition 1.28 A X is compact if and only if every cover of A by open subsets of X has a finite subcover. 7

8 Proposition 1.29 (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 T 2 space is a homeomorphism. Proposition 1.30 Let {X λ } λ I be a collection of spaces. If an infinite number of X λ are non-compact, then any compact subset in λ I X λ has empty interior. Proof Exercise. Proposition 1.31 Let X be compact. Then X is T 2 if and only if X is T 4. Proof Exercise. Tychronoff s Theorem Proof See [2] p.229. A product of compact spaces is compact. Let X be the closed interval [a, b] and for each α I let X α be a copy of X. Then α I X α is compact. In particular [a, b] n is compact. Moveover we have the following result. Heine-Borel Theorem Let A be a subset of R n.thenais compact if and only if A is closed 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 {O α } an open cover 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 O α. Definition 1.32 AspaceXis said to be locally compact if every point of X has a compact neighbourhood. 1. A compact space is locally compact. 2. R is locally compact. (For any x R, [x 1,x+ 1] is a compact neighbourhood of x). 3. R n is locally compact. 8

9 4. R l is not locally compact. (Exercise. Note that [a, b) isclosedinr l.) 5. A discrete space is locally compact. 6. Q is not locally compact. Proposition 1.33 Let X be a locally compact Hausdorff space. Then for any x X the collection of all compact neighbourhoods of x is a local basis at x. (That is any neighbourhood of x contains a compact neighbourhood of x.) Proof Let U be an open neighbourhood of x and E be a compact neighbourhood of x. ThenEis regular. Since U E is an open set in E containing x, there exists asetv open in E 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 neighbourhood of x and hence V E is also a neighbourhood of x. Since V E is a closed subset of the compact subset E, V E is compact. Corollary 1.34 A locally compact Hausdorff space is regular. Note that a locally compact Hausdorff space may not be normal. (See Royden, Real Analysis p.169.) Proposition 1.35 (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 is locally compact. (3) λ I X λ is locally compact if and only if there exists a finite subset F of I such that X λ is locally compact λ I and X λ is compact λ I \ F. Proof We leave (1) and (2) as exercises. Note that (1) is not true if the subspace is not closed. For example Q R is not locally compact. Let s prove (3). Suppose λ I X λ is locally compact. Since the projection p λ is an open map, X λ = p λ [ λ I X λ ] is locally compact by (2). If infinitely many X λ s are noncompact, then by 1.30 each compact subset of λ I X λ has empty interior. Hence apointin λ IX λ cannot have a compact neighbourhood. Conversely, let F be a finite subset of I such that X λ is compact λ I \ F.Let (x λ ) λ IX λ.foreachλ F, there exists a compact neighbourhood E λ of x λ. Then by Tychonoff s theorem, λ I O λ where O λ = E λ if λ F and O λ = X λ if λ I \ F is a compact neighbourhood of (x λ ). Definition 1.36 (1) A X is said to be dense in X if A = X. (2) A X is said to be nowhere dense in X if o A=. 9

10 (3) X is said to be of First Category if it is a countable union of nowhere dense subsets of X. It is of Second Category if it is not of First Category. Theorem 1.37 Let X be a locally compact Hausdorff space. Then the intersection of a countable collection of open dense subsets of X is dense in X. Baire Category Theorem Let X be a non-empty locally compact Hausdorff space. Then X is of Second Category. Proof Let {E n n Z + } be a countable collection of nowhere dense subsets of X. Then each X \ E n is open and dense in X. Hence by 1.37, n=1 (X \ E n )is dense in X. Therefore X \ ( E n ) = n=1 (X \ E n ) n=1 (X \ E n ). n=1 Definition 1.38 A compactification of a space X is a pair (X,i)whereX is a compact space and i : X X is a homeomorphism of X onto a dense subspace of X. Theorem 1.39 Let (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 a topology on X = X { }. Furthermore (1) (X, T )iscompact. (2) If X is non-compact, then (X,i), where i : X X is the inclusion, is a compactification of X. (3) If X is compact, then is an isolated point of X. (4) X is Hausdorff if and only if X is locally compact and Hausdorff. Proof Exercise. Definition 1.40 The space X = X { } in 1.39 is called the 1-point compactification of X. 1. The 1-point compactification of the subspace Z + of positive integers of R is homeomorphic to the subspace {0} { 1 n n Z+ }. (Exercise). 2. The n-sphere is the subspace S n = {(x 1,...,x n+1 ) R n+1 x x2 n+1 =1} of R n+1. It is a closed and bounded subset of R n+1. Therefore S n is compact. 10

11 The 1-point compactification of R n is the n-sphere S n. A homeomorphism between S n and R n is provided by the stereographic projection. Let η =(0, 0,1) R n+1 be the north pole of S n and let p n : R n+1 R be the projection onto the last factor of R n+1. Identify R n as R n {0} R n+1. Then the stereographic projection s : S n \{η} R n is given by s(x) =η+ 1 1 p n(x) (x η). One can easily check that s is a homeomorphism. Now extend s to s(x) :S n 1-point compactification R n of R n by s(x) = { s(x) if x S n \{η} if x = η. Then s is continuous at η. Since S n is compact and R n is Hausdorff, s is a homeomorphism. D. Countability, Separability and paracompactness Definition 1.41 A space X is said to be separable if X contains a countable dense subset. 1. R n is separable. (Q n 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. R l is separable. (Q is a countable dense subset.) Proposition 1.42 (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 Exercise. Definition 1.43 AspaceXis said to be 2nd countable if it has a countable basis. 1. R is 2nd countable. ({(a, b) a < b, a, b Q}is a countable basis.) 2. Similarly R n is 2nd countable. 3. R l is not 2nd countable. (Let B be a basis for R l. For each x R l,pick B x Bsuch that x B x [x, x + 1). Note that x =infb x. Then the function f : R B given by f(x) =B x is injective. This shows that B is not countable.) 11

12 Definition 1.44 AspaceXis said to have a countable basis at x X if there is a countable collection B of open neighbourhoods of x such that any open neighbourhood of x contains a member of B. A space X is said to be 1st countable if it has a countable basis at each of its point. 1. If X is 2nd countable, then X is 1st countable. Hence R n is 1st countable. 2. R l is 1st countable. ({[x, x + 1 n ) 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, 1 n ) 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 ) is not 1st countable. (Exercise.) 6. Let I be an uncountable set. For each λ I, letx λ be a copy of {0, 1} with the discrete topology. Note that X λ is 1st countable and is even 2nd countable. But λ I X λ is not 1st countable. (Exercise.) Proposition 1.45 (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 Y is 1st (2nd) countable. Proof (1) and (3) are very easy. Let s prove (2) for 1st countability. Let {X i i I} be a countable family of 1st countable spaces. Let (x i ) i I X i.foreach i I,letB i be a countable basis at x i. Then B = { i IO i O i B i and O i = X i i J where J is a finite subset of I} is a countable basis at (x i ). Remarks 1. An arbitrary product of 1st (2nd) countable spaces may not be 1st (2nd) countable. (See example 6 above.) 2. Consider id : R R l. It is continuous. Note that R is 2nd countable but R l is not. Definition 1.46 AspaceXis called Lindelöf if every open cover of X has a countable subcover. Theorem 1.47 A 2nd countable space X is Lindelöf. Proof Let B = {U i i I} be a countable basis of X. Let{O α }be an open cover of X. Each O α is a union of members of B. Therefore there exists a countable 12

13 open cover {U αi i I} of X by basic open sets in B such that each U αi lies in some O α. Now for each i I, chooseo αi U αi.then{u αi i I}is a countable subcover of {O α }. 1. R n is Lindelöf. 2. Compact spaces are Lindelöf. 3. R l is Lindelöf. (See Munkres p.192.) Note that R l is not 2nd countable. Therefore the converse of 1.47 is not true in general. 4. A discrete space on an uncountable set is not Lindelöf. Theorem 1.48 If X is 2nd countable, then X is separable. Proof Let B = {O i i I} be a countable basis of X. Foreachi I,letx i O i. Then A = {x i i I} is a countable dense subset of X. Note that the converse of this theorem is not true. For example take X = R l. Theorem 1.49 If X is a separable metric space, then X is 2nd countable. Proof Let A be a countable dense subset of X. 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 such that d(a, x) < 1 (r d(x, p)). Pick a positive rational number q such 2 that d(a, x) <q< 1 (r d(x, p)). Then x B(a, q) B(p, r). This proves the 2 assertion. Theorem 1.50 If X is a Lindelöf metric space, then X is 2nd countable. Proof We shall prove that X is separable. For each j Z +, {B(x, 1) x X} j is an open cover of X. Since X is Lindelöf, it has a countable subcover B j = {B(x i,j, 1) i j Z+ }. Let A = {x i,j i, j Z + } be the collection of all the centers. Then A is a countable subset of X. LetB(p, r) beanopenball.pickan j o Z + such that 1 j o <r.thenb jo ={B(x i,jo, 1 j o ) i Z + } covers X. Therefore p B(x io,jo, 1 j o ) for some i o.thend(p, x io,jo ) < 1 j o <rimplies that x io,jo B(p, r). Hence A B(p, r). ThatisAis dense in X. Definition 1.51 (1) A locally finite family of subsets of a topological space X is a family such that each point of X has a neighbourhood meeting only finitely many members of the family. (2) A refinement F of a cover C of X is a cover of X such that each member of F is contained in some member of C. 13

14 (3) A space X is said to be paracompact if every open cover of X has a locally finite open refinement. 1. A compact space is paracompact. 2. R is paracompact. (Let U be an open cover of R. For each closed interval [N,N +1],N Z, we have, by compactness, a finite cover {U N1,...,U Nn } by members of U. Take as a refinement {U Ni (N,N +1)}. Thisdoesnot cover points in Z. Toremedythis,weaddinasmallopenneighbourhood(of diameter less than 1 and small enough to be contained in some member of U) of each N in Z. We then have the required locally finite open refinement.) 3. R n is paracompact. (Similar proof as above.) 4. R l is paracompact. (Exercise) Proposition 1.52 A paracompact Hausdorff space is regular. Proof Let U be an open neighbourhood of a point p. We shall construct a neighbourhood V of p such that V U. Foreachq U, choose open disjoint neighbourhoods V q,u q of p, q respectively. Then {U q q X \ U} {U}is an open cover of X and thus has a locally finite open refinement. Let M be a neighbourhood of p meeting only finitely many members of the refinement and let W 1,...,W n be those not contained in U. Then there exist q 1,...,q n X \ U such that W i U qi. Let V = M n i=1 V qi. 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 ( n i=1 V qi ) ( n i=1 U qi )=. Thus V X \W. Since W X \U, we have V X \ W U. Proposition 1.53 A paracompact regular space is normal. Proposition 1.54 A Lindelöf regular space is normal. Theorem 1.55 If X is paracompact and separable, then X is Lindelöf. Proof Let {U α } be an open cover of X. Let{V β }be a locally finite open refinement of {U α }. We may assume each V β. Because X has a countable dense subset {x i i Z + }, {V β } is at most a countable family. (Each V β contains at least one x i. If {V β } is an uncountable family, then there is some x i contained in uncountably many V β. This contradicts the local finiteness of {V β }.) For each V β, choose U α(β) V β.then{u α(β) }is a countable subcover of {U α }. 14

15 Note that a product of two paracompact (Lindelöf) spaces may not be paracompact (Lindelöf). For example, take R l R l. Urysohn s Metrization Theorem A 2nd countable T 3 space is metrizable. Proof See [2] p.349. E. Connectedness Definition 1.56 AspaceXis said to be connected if it is not the union of two non-empty disjoint open subsets of X. A X is connected if it is connected as a subspace of X. 1. Any indiscrete space is connected. 2. R is connected. (Exercise) 3. R l is not connected. 4. The subspace Q of all rational numbers of R is not connected. 5. R n is connected. 6. A R is connected if and only if A is an interval. 7. A compact connected T 2 space cannot be a union of countably many but more than one disjoint closed subsets. Proposition 1.57 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 with the discrete topology. Proposition 1.58 Let f : X Y be continuous and A be a connected subset of X. Then f[a] is connected. Proof This follows directly from 1.57(3). Proposition 1.59 R is connected. Proof A non-empty proper open subset O of R is a disjoint union of open intervals. Then one of these open intervals has an endpoint a not in O. Hence O is not closed. 15

16 Proposition 1.60 (1) If A is a connected subset of X, thenais 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 is connected. Let A be the set {(x, sin( 1 )) x>0}. Then A and A {(0, 0)} are connected x subsets of R 2. Corollary 1.61 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 condition that for anya, b I, the line segment joining a and b is also contained in I. ThenIis an interval if and only if I is of the form (a, b), (a, b], [a, b) or[a, b] where a<b. Now the forward implication is clear. Conversely if A is an open interval, then it is homeomorphic to R which is connected by The rest of different types of intervals are also connected by 1.60(2). Intermediate Value Theorem (1) If f : X R is continuous and X is connected, then f[x] isaninterval. (2) Let f :[a, b] R be continuous. Then f assumes all the values between f(a) and f(b). Proposition 1.62 [a, b) is not homeomorphic to (a, b). Proposition 1.63 Let {A α } α I be a family of connected subsets of X such that α I A α. Then α IA α is connected. Proof Use 1.57(3). Definition 1.64 Two points a, b in a space X are said to be connected, written a b, if there is a connected subspace of X containing a and b. Proposition 1.65 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,X. Then there exist a U, b X \ U. By assumption, there exists a connected subset C containing a and b. ButthenU Cis a proper non-empty both open and closed subset of C. This contradicts the fact that C is connected. Proposition 1.66 is an equivalence relation. Definition 1.67 The connected component of a point p X, denoted by C(p), is the equivalence class of containing p. 16

17 Proposition 1.68 (1) C(p) is the largest connected set containing p. (2) C(p) isclosed. Proof (1) C(p) = {C:C is connected and p C}. By1.63C(p) is connected. (2) C(p) is connected and contains p. Hence by (1), C(p) C(p). Proposition 1.69 If X and Y are connected, then X Y is also connected. 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 prove that arbitrary product of connected spaces is connected. This is left as an exercise. Corollary 1.70 R n, [a, b] n and S n are connected. Proof Let p S n. As S n \{p} is homeomorphic to R n,s n \{p} is connected. Hence S n = S n \{p}is connected. Definition 1.71 X is said to be locally connected at p X if for any neighbourhood U of p, there exists a connected neighbourhood 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 X is a basis for the topology on X. 1. Let A be the set {(x, sin( 1 )) x>0}.thena {(0, 0)} is not locally connected x but connected. 2. Let A =(0,1) (2, 3). Then A is locally connected but not connected. 3. R n and S n are locally connected. 4. R l is not locally connected. (Let U be any neighbourhood 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 subsets of U. Therefore any neighbourhood of x is not connected.) 5. Discrete spaces are locally connected. 6. Q is not locally connected. Proposition 1.72 If a space is locally connected at p X, thenpis an interior point of C(p). 17

18 Proof By assumption p has a connected neighbourhood U. Therefore p U C(p). Corollary 1.73 If X is locally connected, then C(p) is an open subset of X. Proposition 1.74 connected. Every open subspace of a locally connected space is locally Proposition 1.75 X is locally connected if and only if the connected components of every open subspace of X are open in X. Proof Suppose X is locally connected. Let O be an open subspace of X. By1.73 and 1.74 O is locally connected and each connected component of O is open in O. Since O is open in X, each connected component of O is open in X. Conversely let U be an open neighbourhood of p X. By assumption the connected component C U (p) in the subspace U is open in X. Hence C U (p) isopeninx. Therefore C U (p) is a connected open neighbourhood of p contained in U. This shows that X is locally connected. Corollary 1.76 Let N be a neighbourhood of a point x in a locally connected space X. Then there exists a connected open neighourhood C of x lying inside N. Proposition 1.77 Let {X α } α I be a family of locally connected spaces such that all but at most finitely many are connected. Then α I X α is locally connected. Proof Let F be a finite subset of I such that X α is not connected for all α F. Let x α I X α and O a basic open neighbourhood of x. NotethatO= α IO α where O α is open in X α and O α = X α for all α I \ J with J some finite subset of I. Foreachα F J, there exists an open connected neighbourhood V α of X α such that V α O α.letu= α IU α where U α = V α if α F J and V α = X α if α I \ (F J). Then U is a product of connected sets and hence it is connected. Also U is a basic open neighbourhood of x lying in O. Note that the converse of this result is also true since local connectedness is invariant under continuous open surjection. Definition 1.78 A path joining a pair of points a and b ofaspacexis a continuous map α : I X such that α(0) = a and α(1) = b, whereidenotes the closed unit interval [0, 1]. It can be verified easily that the relation of being joined by a path is an equivalence relation. The equivalence classes of this relation are called the path components of X. The path component of a point x in X is denoted by P (x). P (x) isthelargest path connected set containing x. However unlike C(x), P (x) may not be a closed set. 18

19 Definition 1.79 AspaceXis said to be path connected if it has exactly one path component. (That is any pair of points in X canbejoinedbyapath.) Proposition 1.80 A path connected space is connected. Proposition 1.81 Let f : X Y be a continuous surjection. If X is path connected, then Y is path connected. 1. R n and S n are path connected. (It is easy to check that R n and R n \{0}are path connected. Let σ : R n+1 \{0} S n be the map given by σ(x) = x. x Then by 1.81 S n is path connected.) 2. Let A be the set {(x, sin( 1 x )) x>0} {(0, 0)}. ThenAis not path connected. Note that A \{(0, 0)} is a path component which is not closed in A. Proposition 1.82 Let {X α } α I be a family of spaces. Then α I X α is path connected if and only if each X α is path connected. Definition 1.83 A space is said to be locally path connected at x X if each neighbourhood of x contains a path connected neighbourhood. X is said to be locally path connected if it is locally path 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 of X is a basis for the topology on X. 1. R n is locally path connected since every open ball in R n is path connected. Similarly S n is locally path connected. 2. (0, 1) (2, 3) is locally path connected but not path connected. 3. For each positive integer n, letl n be the set of all points on the line segment joining the points (0, 1) and ( 1 n, 0) in R2 and let L = {(0,y) 0 y 1}. Then the subspace A = L n=1 L n in R 2 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 1 n ) n Z+ },thenais locally connected at (0, 1) but not locally path connected at (0, 1). Proposition 1.84 Every open subspace of a locally path connected space is locally path connected. 19

20 Proposition 1.85 X is locally path connected if and only if the path components of every open subspace of X are open. Proposition 1.86 α I X α is locally path connected if and only if all X α are locally path connected and all but at most finitely many are also path connected. Proposition 1.87 If X is locally path connected and connected, then X is path connected. Proof By 1.85 a path component of X is open and hence is also closed. Since X is connected, there is exactly one path component. F. Identifications and Adjunction Spaces Definition 1.88 A continuous surjective map p :(X, T ) (Y,T ) is called an identification or a quotient map if p 1 [O] isopeninxif and only if O is open in Y. Proposition 1.89 Let (X, T ) be a topological space and let Y be a set. Let p : X Y be a surjective map. Then T p = {U Y p 1 [U] isopeninx}is a topology on Y. Furthermore p :(X, T ) (Y,T p )isanidentification. Proof Direct verification. The topology T p on Y defined in 1.89 is called the identification or quotient topology induced by p. In fact it is the largest topology on Y that makes p continuous. Definition 1.90 Let f : X Y be a map. A subset O in X is said to be saturated with respect to 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 )isanidentification if and only for any open set O X saturated with respect to p, p[o] isopeniny. Proposition 1.91 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,thenpis an identification. Note that the converses of (1) and (2) are not true. Also the restriction of an identification may not be an identification. 20

21 Proposition 1.92 Let p : X Y be an identification and g : Y Z a surjective map. Then g p is an identification if and and only if g is an identification. Proof Suppose that g p is an identification. One can easily see that the continuity of g p implies the continuity of g. LetO Zbe such that g 1 [O] isopeniny. Because p and g are continuous, (g p) 1 [O] =p 1 [g 1 [O]] is open in X. Since g p is an identification, O is open in Z. Hence g is an identification. Conversely, suppose that g is an identification. Let O Z be such that (g p) 1 [O] isopen in X. Since p and g are identifications, O is open in in Z. Therefore g p is an identification. Let (X, T )be a space and Rbe 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/R given by p(x) =[x] be the natural projection. The set X/R with the identification topology T p induced by p is called the quotient space of X by R. Let A be a subset of X. Define an equivalence relation R A on X by xr A y if and only if x, y A. Then the quotient space X/R A is the space X with A identified to a point [A] and is usually denoted by X/A. Note that E Xis saturated (with respect to the projection p) if and only if E A or E is disjoint from A. Corollary 1.93 Let X, Y be spaces with equivalence relations R and S respectively, and let f : X Y be a relation preserving, (i.e. x 1 Rx 2 = f(x 1 )Sf(x 2 )) continuous map. Then the induced map f : X/R Y/S given by f ([x]) = [f(x)] is continuous. Furthermore, f is an identification if f is an identification. Proposition 1.94 Let Y and A be closed subspaces of X, theny/(y A)isa subspace of X/A. (In fact Y/(Y A) is homeomorphic to a subspace of X/A.) Proof Consider the commutative diagram Y i - X p p?? Y/Y A i - X/A Clearly i is a continuous injection. Thus it suffices to show that i : Y/Y A pi[y ]isanopenmap. ThenY/Y A is homeomorphic to the subspace pi[y ]= i[y/y A]ofX/A. Let U Y/Y A be open. Then p 1 [U] isopeniny. 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, eitherp 1 [U] A=Y Aor p 1 [U] A =. In 21

22 the first case, i[u] =p[y] p[[x \ Y ] W ]. Thus, since (X \ Y ) W is open and saturated, i[u] isopeninp[y]. In the second case, i[u] =p[y] p[(x \ A) W ] which is also open in p[y ]sincex\ais saturated and open in X. 1. Let X be a space with a preferred base point, then the reduced suspension SX is the quotient space (X I)/(X {0,1} { } I) wherei=[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 T 1 ) X {0} { } I is closed in X I. Hence by 1.94 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 provided with the quotient topology. The map p id : 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 two closed maps may not be closed.) However we have the following results. Propostion 1.95 Let f : X Y and g : T Z be identifications. Suppose T is compact and Z is regular. Then f g : X T Y Z is an identification. Proposition 1.96 If f : X Y is an identification and Z is locally compact and regular, then f id Z : 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. Letf:A Y be a continuous map. Define an equivalence relation on Y X by : a b if and only if a = b for a, b X, or a, b Y f(a) =f(b) for a, b A a = f(b) for a Y, b A f(a) =b for a A, b Y. Definition 1.97 The quotient space (Y X)/ is called an adjunction space and is denoted by Y 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, thenamapθ f φ: Y fx Z is defined by (θ f φ)(x) = { θ(x) x X φ(x) x Y. 22

23 Note that this is well-defined since if x y then θ(x) =φ(y). Let p : Y X 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 1.98 Let A be closed in X. Then the map i : Y Y f X, defined by i(y) =p(y), is an injective homeomorphism and i[y ]isclosediny f Y. Proof Let B Y be closed. Then p 1 [i[b]] = B f 1 [B] isclosediny X. Hence i[b] isclosediny f X. Therefore i is a closed injective map. In particular, i[y ] is a closed subspace of Y f X. Proposition 1.99 Let A be closed in X. Thenp X\A is an injective homeomorphism onto an open subspace of Y f X. Proof p X\A is obviously continuous and bijective. Let O be open in X \A. Then Ois open and saturated in X. Sincepis an identification, p[o] isopeniny f X. Proposition If X and Y are normal and A X is closed, then Y f X is normal. Proof Let P and Q be disjoint closed sets in Y f X. Then P 1 = P i[y] and Q 1 = Q i[y ] are closed in i[y ]. Since Y is normal, i[y ]isnormalby Therefore there exist neighbourhoods R of P 1 and S of Q 1 such that R, S are open in i[y ] and they have disjoint closures. Note that since i[y ]isclosed in Y f X, R and S have the same closures in i[y ]andiny f X. Let P 2 = p 1 [P R] X and Q 2 = p 1 [Q S] X. ThenP 2 and Q 2 are disjoint and closed in X. Since X is normal, there exist disjoint open neighbourhoods M and N of P 2 and Q 2 respectively. Then D = p[m \ A] R and E = p[n \ A] S are disjoint open neigbourhoods of P and Q respectively. (They are open since, for example p 1 [D] =R (M\A) f 1 [R]=M\(A\f 1 [R]) which is open in Y X.) Proposition The adjunction space construction preserves connectedness, path-connectedness and compactness. When A X is closed, the T 1 property is preserved. Proof Since X and Y are connected (path-connected), their continuous images are also connected (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 compact space Y X, it is compact. Finally, let y Y. Then since Y is closed in Y f X,sois{y}. Similarly if x X \ A, thenp 1 [{x}] is closed in X. Hence {x} is closed in Y f X since p is an identification. Corollary If X and Y are normal Hausdorff and A X is closed, then Y f X is also normal Hausdorff. 23

24 References [1] J. Dugundji, Topology, Prentice-Hall. [2] J.R.Munkres, Topology, Prentice-Hall. 24

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