Chapter 2 Topological Spaces This chapter contains a very bare summary of some basic facts from topology. 2.1 Definition of Topology A topology O on a set X is a collection of subsets of X satisfying the following conditions: (T1) ; 2 O (T2) X 2 O (T3) O is closed under finite intersections, i.e. if A; B 2 O then A \ B 2 O (T4) O is closed under unions, i.e. if S ⊂ O then [S 2 O We have used the set-theoretic notation for unions: [S is the union of all the sets in S. Note also that by the usual induction argument, condition (T2) implies that if A1; :::; An are a finite number of open sets then A1 \ · · · \ An is also open. A topological space (X; OX ) is a set X along with a topology OX on it. Usuallly, we just say \X is a topological space." If x 2 X and U is an open set with x 2 U then we say that U is a neighborhood of x. There are two extreme topologies on any set X: • the indiscrete topology f;; Xg 1 2 CHAPTER 2. TOPOLOGICAL SPACES • the discrete topology P(X) consisting of all subsets of X A set A in O is said to be open in the topology O. The complement of an open set is called a closed set. Taking complements of (T1)-(T4) it follows that: (C1) X is closed (C2) ; is closed (C3) the union of a finite number of closed sets is closed (C4) the intersection of any family of closed sets is closed. Let F ⊂ X. The interior F 0 of F is the union of all open sets contained in F (F 0 could be empty, in case the only open set which is a subset of F is the empty set). It is of course an open set and is the largest open set which is a subset of F . The closure F of F is the intersection of all closed sets which contain F as a subset. Thus F is a closed set, the smallest closed set which contains F as a subset. The intersection of any set of topologies on X is clearly also a topology. Let S be any non-empty collection of subsets of X. Consider the collection TS of all topologies which contain S, i.e. for which the sets of S are open. Note that P(X) 2 TS. Then \TS is the smallest topology on X containing all the sets of S. It is called the topology on X generated by S. A topological space X is Hausdorff if any two points have disjoint neigh- borhoods: i.e. for any x; y 2 X with x =6 y, there exist open sets U and V , with x 2 U, y 2 V , and U \ V = ;. 2.2 Continuous maps Let (X; OX ) and (Y; OY ) be topological spaces, and f : X ! Y a mapping. We say that f is continuous if −1 f (OY ) ⊂ OX ; 2.3. COMPACT SETS 3 i.e. if for every open set B ⊂ Y the inverse image f −1(A) is an open subset of X. It is clear that the identity map X ! X is continuous, when X is given any particular topology. The other simple fact is that if f : X ! Y is continuous and g : Y ! Z is continuous then the composite g ◦ f : X ! Z is also continuous. A mapping f : X ! Y between topological spaces is a homeomorphism of f is a bijection and both f and f −1 are continuous. 2.3 Compact Sets In this section X is a topological space. Let A ⊂ X. An open cover of A is a collection S of open sets such that A ⊂ [S It is usually convenient to index the sets of S: i.e we talk of an open cover fUαgα2I of A, for some indexing set I. If S is an open cover of A, then a subcover of S is a subset S0 ⊂ S which also covers A. A set K ⊂ X is compact if every open cover has a finite subcover. Observe that the union of a finite number of compact sets is compact. Lemma 1 A closed subset of a compact set is closed. Proof. Let C be closed and C ⊂ K where K is compact. If S is an open cover of C then we obtain an open cover of K by throwing in the open set Cc along with S. Then S [ fCcg is an open cover of the compact set K and so has a finite subcover S0. The S0 is also automatically a finite cover of C, and remains so if we discard Cc from S0 in case S0 does contain Cc. This yields a finite subset of S which still covers C. QED Lemma 2 In a Hausdorff space X: 4 CHAPTER 2. TOPOLOGICAL SPACES (i) if K is a compact subset of X and y 2 X a point outside K then y and K have disjoint neighborhoods, i.e. there is an open neighborhood Wy of y and an open set Vy ⊃ K with Wy \ Vy = ;. (ii) every compact subset of X is closed. (iii) any two disjoint compact subsets of X have disjoint open neighborhoods, i.e. if C and D are compact subsets of X then there exist open sets U ⊃ C and V ⊃ D with U \ V = ;. The finiteness argument used in the proof here is typical. Proof. Let y be a point outside the compact set K. For each x 2 K, Hausdorffness gives us disjoint open sets Ux and Fx, with x 2 Ux and y 2 Fx. The open sets Ux, with x running over K, form an open cover of K. By compactness there exist x1; ::; xN 2 K such that Ux1 ; :::; UxN cover K, i.e. K ⊂ Vy = Ux1 [ · · · [ UxN On the other hand we have the open set Wy = Fy1 \ · · · \ FyN which contains y. Observe that Vy and Wy are disjoint. This proves (i). From (i) we see that each point outside the compact set K has an open neighborhood contained entirely inside the complement K c, which implies Kc is open, i.e. K is closed. This proves (ii). The finiteness argument used to prove (i) also proves (iii). By (i), for any y 2 D, there exist disjoint open sets Vy and Wy with y 2 Wy and C ⊂ Vy. The sets Wy form an open covering of D which has a finite subcover Wy1 ; :::; Wym . Let U = Vy1 \ · · · \ Vym and V = Wy1 [ · · · [ Wym . Then U and V are open sets, they are disjoint, and C ⊂ U and D ⊂ V . QED 2.4 Locally Compact Hausdorff Spaces Let X be a topological space. We say that X is locally compact if for each point x 2 X there is an open set U and a compact set K such that x 2 U ⊂ K 2.4. LOCALLY COMPACT HAUSDORFF SPACES 5 If X is Hausdorff, then K would be closed and so for a Hausdorff space, being locally compact means that each point has a neighborhood U whose closure U is compact. The following separation result is useful: Lemma 3 Suppose X is a locally compact Hausdorff space. If C is a closed subset of X and x a point outside C then x and C have disjoint open neigh- borhoods, the neighborhood of x having compact closure. Equivalently, if W is an open neigborhood of a point x 2 X then there is an open neighborhood V of x such that the closure V is compact and V ⊂ W . Proof We prove the second formulation. Because X is locally compact Hausdorff, there is an open neighborhood U of x and compact K with U ⊂ K. Since W c and K are closed, W c \ K is closed. Being closed and a subset of the compact set K, it is also compact. So there is an open neighborhood F of x with closure F lying in the complement of W c \ K. Let V = F \ U. Then V is an open neighborhood of x, and V ⊂ U ⊂ K So V is compact. Moreover, since V is a subset of F the set V lies in the complement of W c \ K, i.e. V ⊂ W [ Kc But we have already seen that V is a subset of K, so V ⊂ W which is what we wanted. QED The following stronger separation result follows easily from the preceding result: Lemma 4 Suppose X is a locally compact Hausdorff space. If C is a closed subset of X and D a compact set disjoint from C then C and D have disjoint open neighborhoods, the neighborhood of D being also compact. Equivalently, if K is a compact set and U and open set with K ⊂ U then there is an open set V , with compact closure V , such that K ⊂ V ⊂ V ⊂ U 6 CHAPTER 2. TOPOLOGICAL SPACES 2.5 Urysohn's Lemma The main theorem of use for us is Urysohn's Lemma: Theorem 1 Suppose X is a locally compact Hausdorff space. Let K be a compact subset of X and U an open subset of X with K ⊂ U. Then there is a function f 2 Cc(X), continuous of compact support, such that 1K ≤ f ≤ 1U Proof. Since X is locally compact Hausdorff and the compact set K is contained in the open set U, there is an open set U1 with compact closure U 1 and K ⊂ U1 ⊂ U 1 ⊂ U Applying this argument again to the compact set U 1 lying inside the open set U, we have an open set U0, with compact closure, such that U 1 ⊂ U0 ⊂ U 0 ⊂ U Thus we have the open sets U0 and U1, each with compact closure, satisfying K ⊂ U1 ⊂ U 1 ⊂ U0 ⊂ U 0 ⊂ U We shall produce for each rational number r 2 (0; 1) an open set Ur, with compact closure, in such a way that U r ⊂ Us whenever s < r.