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4. Compactness 4. Compactness Definition. Let X be a topological space X. A subset K ⊂ X is said to be compact set in X, if it has the finite open cover property: S (f.o.c) Whenever {Di}i∈I is a collection of open sets such that K ⊂ i∈I Di, there exists a finite sub-collection Di1 ,...,Din such that K ⊂ Di1 ∪ · · · ∪ Din . An equivalent description is the finite intersection property: (f.i.p.) If {Fi}i∈I is is a collection of closed sets such that for any finite sub- collection Fi1 ,...,Fin we have K ∩ Fi1 ∩ ...Fin 6= ∅, it follows that \ K ∩ Fi 6= ∅. i∈I A topological space (X, T ) is called compact if X itself is a compact set. Remark 4.1. Suppose (X, T ) is a topological space, and K is a subset of X. Equip K with the induced topology T K . Then it is straightforward from the definition that the following are equivalent: • K is compact, as a subset in (X, T ); • (K, T K ) is a compact space, that is, K is compact as a subset in (K, T K ). The following three results, whose proofs are immediate from the definition, give methods of constructing compact sets. Proposition 4.1. A finite union of compact sets is compact. Proposition 4.2. Suppose (X, T ) is a topological space and K ⊂ X is a compact set. Then for every closed set F ⊂ X, the intersection F ∩ K is again compact. Proposition 4.3. Suppose (X, T ) and (Y, S) are topological spaces, f : X → Y is a continuous map, and K ⊂ X is a compact set. Then f(K) is compact. The following results discuss compactness in Hausdorff spaces. Proposition 4.4. Suppose (X, T ) is toplological Hausdorff space. (i) Any compact set K ⊂ X is closed. (ii) If K is a compact set, then a subset F ⊂ K is compact, if and only if F is closed (in X). Proof. (i) The key step is contained in the following Claim: For every x ∈ X r K, there exists some open set Dx with x ∈ Dx ⊂ X r K. Fix x ∈ X r K. For every y ∈ K, using the Hausdorff property, we can find two open sets Uy and Vy with Uy 3 x, Vy 3 y, and Uy ∩ Vy = ∅. Since we S obviously have K ⊂ y∈K Vy, by compactness, there exist points y1, . , yn ∈ K, such that K ⊂ Vy1 ∪ · · · ∪ Vyn . The claim immediately follows if we then define Dx = Uy1 ∩ · · · ∩ Uyn . 26 §4. Compactness 27 Using the Claim we now see that we can write the complement of K as a union of open sets: [ X r K = Dx, x∈XrK so X r K is open, which means that K is indeed closed. (ii). If F is closed, then F is compact by Proposition 4.2. Conversely, if F is compact, then by (i) F is closed. Corollary 4.1. Let X be a compact Hausdorff space, let Y be a topological space, and let f : X → Y be a continuous map, which is bijective. Then f is a homeomorphism, i.e. the inverse map f −1 : Y → X is continuous. Proof. What we need to prove is the fact that, whenever D ⊂ X is open, it follows that its preimage under f −1, that is, the set f(D) ⊂ Y , is again open. To get this fact we take complements. Consider the set X r D, which is closed in X. Since X is compact, by Proposition 4.2 it follows that X r D is compact. By Proposition 4.3, the set f(X r D) is compact in Y , and finally by Proposition 4.4 it follows that f(X r D) is closed. Now we are done, because the fact that f is bijcetive gives Y r f(D) = f(X r D), and the fact that Y r f(D) is closed means that f(D) is indeed open in Y . Proposition 4.5. Every compact Hausdorff space is normal. Proof. Let X be a compact Hausdorff space. Let A, B ⊂ X be two closed sets with A∩B = ∅. We need to find two open sets U, V ⊂ X, with A ⊂ U, B ⊂ V , and U ∩ V = ∅. We start with the following Particular case: Assume B is a singleton, B = {b}. The proof follows line by line the first part of the proof of part (i) from Proposition 4.4. For every a ∈ A we find open sets Ua and Va, such that Ua 3 a, Va 3 b, and Ua ∩ Va = ∅. Using Proposition 4.4 we know that A is compact, and since we S clearly have A ⊂ a∈A Ua, there exist a1, . , an ∈ A, such that Ua1 ∪· · ·∪Uan ⊃ A. Then we are done by taking U = Ua1 ∪ · · · ∪ Uan and V = Va1 ∩ · · · ∩ Van . Having proven the above particular case, we proceed now with the general case. For every b ∈ B, we use the particular case to find two open sets Ub and Vb, with Ub ⊃ A, Vb 3 b, and Ub ∩ Vb = ∅. Arguing as above, the set B is compact, and S we have B ⊂ b∈B Vb, so there exist b1, . , bn ∈ B, such that Vb1 ∪ · · · ∪ Vbn ⊃ B. Then we are done by taking U = Ub1 ∩ · · · ∩ Ubn and V = Vb1 ∪ · · · ∪ Vbn . Examples 4.1. A. Any closed interval of the form [a, b] is a compact set in R. To prove this, we start with an arbitrary collection D of open subsets of R, with S [a, b] ⊂ D∈D D, and we wish to find D1,...,Dn ∈ D, with D1 ∪ · · · ∪ Dn ⊃ [a, b]. To get this we consider the set M = t ∈ (a, b] : there exist D1,...,Dn ∈ D with D1 ∪ · · · ∪ Dn ⊃ [a, t] , so that all we need is the fact that b ∈ M. We first show that the number s = sup M also belongs to M. On the one hand, there exists some open set D ∈ D with D 3 s, so there exists ε > 0 such that D ⊃ (s − ε, s + ε). On the other hand, there exists t ∈ M with s − ε < t, so there exist D1,...,Dn ⊃ [a, t] ⊃ [a, s − ε]. Since we clearly have D ∪ D1 ∪ · · · ∪ Dn ⊃ [a, s − ε] ∪ (s − ε, s + ε) = [a, s + ε) ⊃ [a, s], 28 CHAPTER I: TOPOLOGY PRELIMINARIES it follows that indeed s belongs to M. Finally, by the same argument as above, it follows that the set M will also contain the intersection [s, s + ε) ∩ [a, b], which clearly forces s = b. B. The above result has a famous generalization - the Bolzano-Weierstrass Theorem - which states that a set X ⊂ R is compact, if and only if (i) X is closed, and (ii) X is bounded, in the sense that there exists some C ≥ 0 such that |x| ≤ C, ∀ x ∈ X. The fact that every compact set X ⊂ R is closed and bounded is clear (use the S∞ finite open cover property with n=1(−n, n) = R ⊃ X). Conversely, if X is closed and bounded, then X is a closed subset of some interval of the form [−C, C], which is compact by A, so X itself is compact. The Bolzano-Weierstarss Theorem has the following important consequence. Proposition 4.6. Let X be a compact space, and let f : X → R be a continu- ous function. Then f attains its maximum and minimum values, in the sense that there exist points x1, x2 ∈ X, such that f(x1) ≤ f(x) ≤ f(x2), ∀ x ∈ X. Proof. Consider the set K = f(X) which is compact in R. Since K is bounded, the quantities t1 = inf K and t2 = sup K are finite. Since K is closed, both t1 and t2 belong to K, so there exist x1, x2 ∈ X such that t1 = f(x1) and t2 = f(x2). The following is a useful technical result, which deals with the notion of uniform convergence. ∞ Theorem 4.1 (Dini). Let X be a compact space, let (fn)n=1 be a sequence of ∞ continuous functions fn : X → R. Assume the sequence (fn)n=1 is monotone, in the sense that either (↑) f1 ≤ f2 ≤ ... , or (↓) f1 ≥ f2 ≥ ... Assume f : X → R is another continuous function, with the property that the ∞ sequence (fn)n=1 converges pointwise to f, that is: lim fn(x) = f(x), ∀ x ∈ X. n→∞ ∞ Then the sequence (fn)n=1 is converges uniformly to f, in the sense that lim max |fn(x) − f(x)| = 0. n→∞ x∈X Proof. Replacing fn with fn − f, we can assume that f = 0, which means that we have limn→∞ fn(x) = 0, ∀ x ∈ X. Replacing (if necessary) fn with −fn, we can also assume that the sequence (fn)n≥1 satisfies (↓). In particular, each fn is non-negative. For each n ≥ 1, let us define sn = maxx∈X fn(x), so that what we need to prove is the fact that limn→∞ sn = 0. Using (↓) it is clear that we have the inequalities s1 ≥ s2 ≥ · · · ≥ 0, so the desired result is equivalent to showing that inf{sn : n ≥ 1} = 0. §4. Compactness 29 We are going to prove this equality by contradiction. Assume there exists some ε > 0, such that sn ≥ ε, ∀ n ≥ 1. For each integer n ≥ 1, let us define the set Fn = {x ∈ X : fn(x) ≥ ε}, −1 so that (use Proposition 4.6) Fn 6= ∅. Since Fn = fn [ε, ∞) , it is clear that Fn is closed, for every n ≥ 1. Using (↓) it is obvious that we have the inclusions F1 ⊃ F2 ⊃ ..., so using the the finite intersection property, it follows that ∞ \ Fn 6= ∅.
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