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 subcollection 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 S Vy with Uy 3 x, Vy 3 y, and Uy ∩ Vy = ∅. Since we 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

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Using the Claim we now see that we can write the complement of K as a union of open sets: [ X rK = 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 = ∅.SUsing Proposition 4.4 we know that A is compact, and since we 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 3Sb, and Ub ∩ Vb = ∅. Arguing as above, the set B is compact, and 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 S this, we start with an arbitrary collection D of open subsets of R, with [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],

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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 continuous 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.

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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},
so that (use Proposition 4.6) Fn 6= ∅. Since Fn = fn−1 [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= ∅. n=1

T∞ But this leads to a contradiction, because if we pick an element x ∈ n=1 Fn , then we will have fn (x) ≥ ε, ∀ n ≥ 1, and then the equality limn→∞ fn (x) = 0 is impossible.  Comment. The reader is strongly encouraged to try to work the two exercises below, before reading Theorems 4.2 and 4.3. Strictly speaking these results do not require the more sofosticated machinery employed in these two theorems. Exercise 1 ♥ . Let X be a compact space. Equip the product space X × R with the product topology (here R is equipped with the standard topology). Prove that a set of the form K = X × [a, b] is compact in X × R. S Hint: Start with an arbitrary collection D of open subsets of X × R, with X × [a, b] ⊂ D∈D D, and prove there exist D1 , . . . , Dn ∈ D, with D1 ∪ · · · ∪ Dn ⊃ X × [a, b]. Argue as above, by considering the set  M = t ∈ (a, b] : there exist D1 , . . . , Dn ∈ D with D1 ∪ · · · ∪ Dn ⊃ X × [a, t] , and prove that b ∈ M , by showing first that sup M ∈ M .

Exercise 2 ♥ . Let n ≥ 1 be an integer, and equip Rn with the product topology, where R is equipped with the standard topology. A. Prove that ”closed boxes” of the form B = [a1 , b1 ] × · · · × [an , bn ] are compact in Rn . B. Consider the coordinate maps πk : Rn → R, k = 1, . . . , n. Declare a set X ⊂ Rn bounded, if all sets πk (X) ⊂ R, k = 1, . . . , n, are bounded. Prove the general Bolzano-Weierstrass Theorem: a set X ⊂ Rn is compact, if and only if X is closed and bounded. C. Prove that a set X ⊂ C is compact, if and only if X is closed and bounded, in the sense that there exists some constant C ≥ 0, such that |ζ| ≤ C, ∀ ζ ∈ X. Hints: A. Use induction and the preceding Exercise. B. Argue exactly as in Example 4.1.B. C. Analyze the two notions of boundedness (the coordinate definition given in B, and the absolute value definition given in C) and show that they are equivalent, under the canonical homeomorphism R2 ' C.

Besides the two equivalent conditions (f.o.c) and (f.i.p.), there are some other useful characterizations of compactness, listed in the following.

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Theorem 4.2. Let (X, T ) be a topological space. The following are equivalent: (i) X is compact. (ii) (Alexander sub-base Theorem) There exists a sub-base S with the finite open cover property: [ (s) For any collection {Si | i ∈ I} ⊂ S with X = Si , there exists a i∈I

finite sub-collection {Si1 , Si2 , . . . , Sin } (for some finite sequence of indices i1 , i2 , . . . , in ∈ I) such that X = Si1 ∪ Si2 ∪ · · · ∪ Sin . (iii) Every ultrafilter in X is convergent. (iv) Every net in X has a convergent subnet. Proof. (i) ⇒ (ii). This is obvious. (In fact any sub-base has the open cover property.) (ii) ⇒ (iii). Let U be an ultrafilter on X. Assume U is not convergent to any point x ∈ X. By Proposition 3.2 it follows that, for each x ∈ X, one can find a set Sx ∈ S with Sx 3 x, but such that Sx 6∈ U. Using property (s), one can find a finite collection of points x1 , . . . , xn ∈ X, such that Sx1 ∪ · · · ∪ Sxn = X.

(1)

Since Sxp 6∈ U, it means that X r Sxp belongs to U, for every p = 1, . . . , n. Then, using (1), we get U 3 (X r Sx1 ) ∩ · · · ∩ (X r Sxn ) = ∅, which is impossible. (iii) ⇒ (iv). Assume condition (iii) holds, and let (xλ )λ∈Λ be some net in X. Denote by Φ the collection of all finite subsets of Λ (including the empty set), and define, for each F ∈ Φ, the sets ΛF = {λ ∈ Λ : λ  µ, ∀ mu ∈ F }, GF = {xλ : λ ∈ ΛF } (use the convention Λ∅ = Λ). Since we obviously have ΛF1 ∩ ΛF2 = ΛF1 ∪F2 , ∀ F1 , F2 ∈ Φ, we get the equalities GF1 ∩ GF2 = GF1 ∪F2 , ∀ F1 , F2 ∈ Φ, which prove the fact that the collection G = {GF }F ∈Φ is a filter in X. Let U be then some ultrafilter with U ⊃ G. By the hypothesis (iii), the ultrafilter U is convergent to some point x ∈ X, which means that U contains the collection Nx of all neighborhoods of x. In particular, we get the fact that (2)

N ∩ GF 6= ∅, ∀ N ∈ Nx , F ∈ Φ.

We are now in position to define a subnet of (xλ )λ∈Λ , which will prove to be convergent to x. Consider the set Σ = {(N, λ) ∈ Nx × Λ : xλ ∈ N }, equipped with the ordering  (N1 , λ1 )  (N2 , λ2 ) ⇐⇒

N1 ⊂ N2 λ1  λ2

Let us remark that Σ is a directed set. Indeed, if we start with two elements σ1 = (N1 , λ1 ) and σ2 = (N2 , λ2 ) in Σ, then using (2) with N = N1 ∩ N2 , and F = {λ1 , λ2 }, we get the existence of some λ ∈ Λ with λ  λ1 , λ  λ2 , and such

§4. Compactness

31

that xλ ∈ N . This means that the pair σ = (N, λ) belongs to Σ, and it also satisfies σ  σ1 , σ  σ2 . Consider also the map φ : Σ 3 (N, λ) 7−→ λ ∈ Λ. Again using (2), it follows that φ is a directed map. Define then the net (yσ )σ∈Σ by yσ = xφ(σ) , ∀ σ ∈ Σ. By construction, (yσ )σ∈Σ is a subnet of (xλ )λ∈Λ . We show now that (yσ )σ∈Σ is convergent to x. Start with some arbitrary neighborhood N of x. Use (2) with F = ∅, to find some λN ∈ Λ, such that xλN ∈ N . Put σN = (N, λN ). If σ = (V, λ) ∈ Σ is such that σ  σN , then in particular we have xλ ∈ M ⊂ N , i.e. yσ ∈ N . In other words we have yσ ∈ N, ∀ σ  σN , thus proving that (yσ )σ∈Σ is indeed convergent to x. (iv) ⇒ (i). Assume (iv), and let us prove that X has the finite intersection property (f.i.p.). Let {Fi }i∈I is is a collection of closed sets such that for any finite sub-collection Fi1 , . . . , Fin we have Fi1 ∩ . . . Fin 6= ∅, and let us prove that T non-empty subsets of I, and i∈I Fi 6= ∅. Denote by Ω the collection of all finite T define, for each J ∈ Ω, the non-empty closed set F = J i∈J Fi . With this notation, T what we have to prove is the fact that J∈Ω FJ 6= ∅. The advantage here is the fact that Ω is directed (by inclusion). Since FJ 6= ∅, for each J ∈ Ω, we can choose an element xJ ∈ FJ . This way (use the Axiom of Choice) we can construct a net (xJ )J∈Ω . Using (iv) there exists a subnet (yσ )σ∈Σ ⊂ (xJ )J∈Ω which is convergent φ

to some point x ∈ X. In order to finish the proof, it suffices to show that x ∈ FJ , ∀ J ∈ Ω. As above, denote by Nx the collection of all neighborhoods of x. Since each FJ is closed, all we need to prove is the fact that N ∩ FJ 6= ∅, ∀ N ∈ Nx , J ∈ Ω. Fix N ∈ Nx and J ∈ Ω. On the one hand, since φ : Σ → Ω is a directed map, there exists σ1 ∈ Σ, such that φ(σ)  J, ∀ σ  σ1 . On the other hand, since (yσ )σ∈Σ is convergent to x, there exists some σ2 ∈ Σ, such that yσ ∈ N , ∀ σ  σ2 . If we choose σ ∈ Σ such that σ  σ1 and σ  σ2 , then on the one hand we have yσ = xφ(σ) ∈ Fφ(σ) ⊂ FJ (here we use the fact that, for J, J1 ∈ Ω, one has J1  J ⇒ FJ1 ⊂ FJ ), and on the other hand we also have yσ ∈ N . This means precisely that yσ ∈ N ∩ FJ .  An interesting application of the above result is the following: Theorem 4.3 (Tihonov). Suppose one Q has a familiy (Xi , Ti )i∈I of compact topological spaces. Then the product space Xi is compact in the product topology. i∈I

Proof. We are going to use the ultrafilter characterization (iii) from the preQ ceding Theorem. Let U be an ultrafilter on X = Xi . Denote by πi : X → Xi , i∈I

i ∈ I the coordinate maps. Since each Xi is compact, it follows that, for every i ∈ I, the ultrafilter πi∗ (U) (in Xi ) is convergent to some point xi ∈ Xi . If we form the element x = (xi )i∈I ∈ X, this means that πi∗ (U) is convergent to πi (x), for every i ∈ I. Then, by the ultrafilter characterization of the product topology (see section 3) it follows that U is convergent to x. 

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Comment. Another interesting application of Theorem 4.2 is the following construction. Suppose (X, T ) is a compact Hausdorff space, and (xi )i∈I ⊂ X is an arbitray family of elements. (Here I is an arbitrary set.) Suppose U is an ultrafilter on I. If we regard the family (xi )i∈I simply as a function f : I → X, then we can construct the ultrafilter f∗ (U) on X. More explicitly  f∗ (U) = W ⊂ X : the set {i ∈ I : xi ∈ W } belongs to U . Since X is compact Hausdorff, the ultrafilter f∗ (U) is convergent to some unique point x ∈ X. This point is denoted by lim xi . U