1 Lecture 7: September 17 Closed sets and compact sets. Last time, we defined compactness in terms of open coverings: a topological space X is compact if, for every open covering U , there are finitely many open sets U1,...,Un U such that X = U1 Un. To prove that a given space is not compact is∈ usually not difficult: we just∪··· have∪ to find one open covering that violates the condition in the definition. To prove that a given space is compact can be quite difficult: we have to verify the condition for all possible open coverings. (This task becomes a little easier when the topology on X is given by a basis, because we only need to consider coverings by basic open subsets.) Since compactness is a strong requirement, compact spaces have a lot of wonderful properties. The following theorem about continuous functions on compact spaces that shows this effect at work. Theorem 7.1. Let f : X Y be a continous function. If X is compact and Y is Hausdorff, then the image→f(X) is closed. Recall from last time that the image of a compact space under a continuous function is again compact; this was an easy consequence of the definitions. The- orem 7.1 requires a little bit more work to prove. Let me first note the following surprising corollary. Corollary 7.2. Let f : X Y be a bijective continous function. If X is compact and Y is Hausdorff, then f→is a homeomorphism. Proof. Let us see how this follows from Theorem 7.1. We have to prove is that 1 the inverse function f − : Y X is continous; this is equivalent to saying that the → 1 preimage of every closed set A X is closed in Y . The preimage of A under f − is exactly f(A), and so we need to⊆ argue that f(A) is closed. As a closed subset of a compact space, A is compact (Proposition 6.13); therefore its image f(A) is again compact by Proposition 6.14.SinceY is Hausdorff, Theorem 7.1 implies that f(A) is closed. Now let us start proving the theorem. By Proposition 6.14, the image f(X)is a compact subspace of the Hausdorffspace Y . The following proposition explains why f(X) must be closed. Proposition 7.3. Let X be a Hausdorfftopological space. If Y X is a compact subspace, then Y is closed in X. ⊆ Proof. We shall argue that X Y is a union of open sets, and hence open. Fix a point x X Y . For any y Y\, we can find disjoint open sets U(y) and V (y)with x U and∈ y\ V ; this is because∈ X is Hausdorff. Therefore ∈ ∈ Y V (y), ⊆ y C ∈ and because Y is compact, there are finitely many points y1,...,yn Y with Y V (y ) V (y ). But then the open set ∈ ⊆ 1 ∪···∪ n U(y ) U(y ) 1 ∩···∩ n is disjoint from V (y1) V (yn), hence contained in X Y ; because it is a neighborhood of x, and∪ because···∪ x X Y was arbitrary, X Y\ must be open. ∈ \ \ 2 Note that this result is false when X is not Hausdorff; a simple example is given by the Sierpi´nski space. More generally, let X be any finite topological space; then all compact subsets of X are closed if and only if the topology on X is discrete if and only if X is Hausdorff. Note. The proof of Proposition 7.3 shows that we can separate points and compact subspaces in a Hausdorffspace by open sets. More precisely, given a compact subspace Y X and a point x Y , there are disjoint open sets U and V with x U and Y⊆ V – in the notation∈ used during the proof, ∈ ⊆ U = U(y ) U(y ) and V = V (y ) V (y ). 1 ∩···∩ n 1 ∪···∪ n If X is both compact and Hausdorff, so that every closed subspace is compact, we can even separate points and arbitrary closed subspaces by open sets. We shall investigate results of this type in more detail later. Examples of compact spaces. Let us now look at a few topological spaces that are compact. The first example is closed intervals in R. You probably already know that closed intervals are “compact” in the analysis sense – every sequence has a convergent subsequence – but we need to do some work to prove that they are also compact in the topology sense. Theorem 7.4. The closed interval [0, 1] is compact. Proof. Let U be an open covering; we have to show that the interval can be covered by finitely many open sets in U . The idea of the proof is very simple: 0 belongs to some open set U U , which contains some interval of the form [0,a); then a 1 ∈ belongs to some open set U2 U , which again contains a maximal interval [a, b); and so on. Of course, the situation∈ could be like in one of Zeno’s paradoxes, with the points a, b, . converging to some limit inside the interval. To avoid this problem, we have to take a slightly less direct approach. Let us introduce the set S = x [0, 1] [0,x] can be covered by finitely open sets in U . ∈ We want to show that 1 S, because this will mean that [0, 1] can be covered ∈ by finitely many sets in U . Using the least upper bound property of R,define s =supS. Now the argument proceeds in three steps: Step 1. Every point t with 0 t<shas to belong to S. Suppose that we had t S. Then every point x S ≤has to satisfy x<t: the reason is that [0,x] can ∈ ∈ be covered by finitely open sets in U , and if t x, then the same open sets would cover [0,t]. This means that t is an upper bound≤ for S, contradicting our choice of s. Consequently, we have [0,s) S. ⊆ Step 2. We show s S.SinceU is an open covering, there is an open set U U with s U.Ifs = 0, this∈ shows that s S.Ifs>0, then U contains an interval∈ of the form∈ (a, s]with0 a<s. By Step∈ 1, a S, and so [0,a]iscoveredbyfinitely ≤ ∈ many open sets in U ;throwinginU, we get a finite open covering of [0,s]. Step 3. We prove that s = 1. Suppose that s<1. We already know that [0,s] is covered by finitely many open sets in U . Their union is a neighborhood of s, and therefore contains an interval of the form [s, b)withs<b 1. Because U is ≤ an open covering, b U for some U U ; but now we have finitely many open sets ∈ ∈ covering [0,b], contradicting our choice of s. In fact, every closed interval of the form [a, b] is compact: this follows either by adapting the above proof, or by noting that any (nontrivial) closed interval in R is 3 homeomorphic to [0, 1]. In the proof, we only used the ordering of R and the least upper bound property; for that reason, the theorem is true is any linearly ordered set with the least upper bound property. The next example is Rn and its subsets. If you are taking analysis, you may already know that a subspace of Rn is compact if and only if it is closed and bounded (in the Euclidean metric). Here bounded means that it is contained in a ball BR(0) of some radius R. I want to explain why this result is also true with our definition of compactness. The following general result about products will be useful for the proof. Proposition 7.5. If X and Y are compact, then X Y is compact. × Proof. Recall that the product topology has a basis consisting of all open sets of the form U V ,withU open in X and V open in Y . Let U be an open covering × of X Y ; as I explained earlier, it suffices to consider the case that U consists of basic× open sets. 1 Step 1. Fix a point x X, and consider the vertical slice p1− (x)= x Y .It is homeomorphic to Y , and∈ therefore compact; one way of seeing this is{ to} consider× the continuous function i: Y X Y , i(y)=(x, y), and to apply Proposition 6.14. → × Since it is covered by the union of all the open sets in U , we can find finitely many 1 open sets U1 V1,...,Un Vn U such that p1− (x) U1 V1 Un Vn. Now define U×(x)=U × U ;thentheentireset∈ U(x⊆) Y×is covered∪···∪ by finitely× 1 ∩···∩ n × many open sets in U . Step 2. The collection of open sets U(x) from Step 1 is an open covering of X. By compactness, there are finitely many points x ,...,x X with 1 m ∈ X = U(x ) U(x ). 1 ∪···∪ m This means that X Y = U(x ) Y U(x ) Y, × 1 × ∪···∪ m × and because each these m sets is covered by finitely many sets in U , the same is true for the product X Y . × More generally, any finite product of compact spaces is again compact; this can easily be proved by induction on the number of factors. We will see later in the semester that arbitrary products of compact spaces are compact (in the product topology); this is an extremely powerful result, but the proof is not that easy.