This chapter contains a very bare summary of some basic facts from topology.

Transcription

1 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) O (T2) X O (T3) O is closed under finite intersections, i.e. if A, B O then A B O (T4) O is closed under unions, i.e. if S O then S 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 A 1,..., A n are a finite number of open sets then A 1 A n is also open. A topological space (X, O X ) is a set X along with a topology O X on it. Usuallly, we just say X is a topological space. If x X and U is an open set with x U then we say that U is a neighborhood of x. There are two extreme topologies on any set X: the indiscrete topology {, X} 1

2 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 T S of all topologies which contain S, i.e. for which the sets of S are open. Note that P(X) T S. Then T S 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 neighborhoods: i.e. for any x, y X with x y, there exist open sets U and V, with x U, y V, and U V =. 2.2 Continuous maps Let (X, O X ) and (Y, O Y ) be topological spaces, and f : X Y a mapping. We say that f is continuous if f 1 (O Y ) O X,

3 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 {U α } α I of A, for some indexing set I. If S is an open cover of A, then a subcover of S is a subset S 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 C c along with S. Then S {C c } is an open cover of the compact set K and so has a finite subcover S. The S is also automatically a finite cover of C, and remains so if we discard C c from S in case S does contain C c. This yields a finite subset of S which still covers C. QED Lemma 2 In a Hausdorff space X:

4 4 CHAPTER 2. TOPOLOGICAL SPACES (i) if K is a compact subset of X and y X a point outside K then y and K have disjoint neighborhoods, i.e. there is an open neighborhood W y of y and an open set V y K with W y V y =. (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 K, Hausdorffness gives us disjoint open sets U x and F x, with x U x and y F x. The open sets U x, with x running over K, form an open cover of K. By compactness there exist x 1,.., x N K such that U x1,..., U xn cover K, i.e. K V y = U x1 U xn On the other hand we have the open set W y = F y1 F yn which contains y. Observe that V y and W y 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 K c 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 D, there exist disjoint open sets V y and W y with y W y and C V y. The sets W y form an open covering of D which has a finite subcover W y1,..., W ym. Let U = V y1 V ym and V = W y1 W ym. 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 X there is an open set U and a compact set K such that x U K

5 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 neighborhoods, the neighborhood of x having compact closure. Equivalently, if W is an open neigborhood of a point x 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 K c 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 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 C c (X), continuous of compact support, such that 1 K f 1 U Proof. Since X is locally compact Hausdorff and the compact set K is contained in the open set U, there is an open set U 1 with compact closure U 1 and K U 1 U 1 U Applying this argument again to the compact set U 1 lying inside the open set U, we have an open set U 0, with compact closure, such that U 1 U 0 U 0 U Thus we have the open sets U 0 and U 1, each with compact closure, satisfying K U 1 U 1 U 0 U 0 U We shall produce for each rational number r (0, 1) an open set U r, with compact closure, in such a way that U r U s whenever s < r. (2.1) To this end, let r 1, r 2, r 3,... be an enumeration of the rationals in [0, 1], taking r 1 = 0 and r 2 = 1. Suppose n 2 and that we have defined open sets U r1,..., U rn, each having compact closure and satisfying the condition (2.1). We want to define U rn+1 to be an open set with compact closure which will continue to respect the requirement (2.1). So let p be the largest element of {r 1,..., r n } less than r n+1 and q the smallest element larger than r n+1 ; in particular, p < r n+1 < q Since p < q we have U q U p and so there is an open set U rn+1, with compact closure satisfying U q U rn+1 U rn+1 U p

7 2.5. URYSOHN S LEMMA 7 Thus, inductively [or recursively], this allows us to construct the entire family of sets U r, with r running over all the rationals in [0, 1], satisfying (2.1). Now define the function f : X R as follows. On U 1 we set f equal to 1, and outside U 0 we set f equal to 0. [In particular, this ensures that f equals 1 on K and equals 0 outside U.] More generally, let f(x) be the supremum of all the rationals r [0, 1] for which x U r : f(x) = sup{r Q [0, 1] : x U r } and we take the sup of the empty set (i.e. if x / U 0 ) to be 0. Let r be any rational in [0, 1]. If x U r then x also belongs to all lower U q (i.e. for q < r, because then U r U q ), and so f(x) r: if x U r then f(x) r On the other hand, if x / U r then x is certainly not in any higher U q (i.e. for q > r), and so f(x) r: if x / U r then f(x) r In particular, if r < s are rationals in [0, 1] then on the set U r U s, the function f takes values in [r, s]. Now consider any point x 0 X, and any neighborhood (a, b) of f(x 0 ). Assume for the moment that 0 < f(x 0 ) < 1. Let r, s (0, 1) be rationals with a < r < f(x 0 ) < s < b. Since f(x 0 ) > r it follows that x 0 U r. On the other hand, since f(x 0 ) < s it also follows that x 0 / U s. Thus x 0 belongs to the open set U r U s, and on this open neighborhood of x 0 the function f takes values in [r, s] (a, b). This proves that f is continuous at x 0 if f(x 0 ) (0, 1). Now suppose f(x 0 ) = 0. Consider any neighborhood (a, b) of 0. Let s be any rational number with 0 < s < b. Since f(x 0 ) = 0 < s, the point x 0 is outside U s, i.e. x 0 is in the open set U c s. Moreover, on this open set f has values s < b. Thus on the open neighborhood U c s of x 0, the function f takes values in (a, b). So f is continuous at x 0. The argument for the case f(x 0 ) = 1 is similar. QED

8 8 CHAPTER 2. TOPOLOGICAL SPACES 2.6 Product Spaces and Tikhonov s Theorem Consider two topological spaces X a and X b. Then for the Cartesian product X a X b we have two projection maps π a : X a X b X a : (x a, x b ) x a, π b : X a X b X b : (x a, x b ) x b The smallest topology on X a X b which makes both the maps π a and π b continuous is called the product topology on X a X b. Thus if U a is any open subset of X a and U b is any open subset of X b then a (U a ) = U a X b, and π 1 (U b ) = X b U b π 1 are open subsets of X a X b in the product topology, and hence so is their intersection (U a ) π 1 (U b ) = U a U b π 1 a b Indeed, the product topology on X a X b is precisely the collection of all subsets of X a X b which are unions of sets of the form U a U b, with U a open in X a and U b open in X b. More generally, consider an indexing set I and a topological space X α for each α I. The Cartesian product set is the set of all maps X α α I x : I α I X α : α x α for which x α X α for all α I. The element x α X α is the α th coordinate of the element x α I X α. The map π α : α I X α X α : x x α is the α th coordinate of x. The product topology on α I X α is the smallest topology on α I X α for which the projection maps π α are all continuous. A subset of α I X α is open in the product topology if and only if it is the union of sets of the form b π 1 α 1 (U α1 ) π 1 α n (U αn ) with n running over positive integres, α 1,..., α n running over I, and U αi over open sets X αi. runs

9 2.6. PRODUCT SPACES AND TIKHONOV S THEOREM 9 Theorem 2 Suppose X α is a compact Hausdorff space for each α I, where I is any non-empty indecing set. Then the product space α I X α is a compact Hausdorff space.