Compact Sets in Metric Spaces Math 201A, Fall 2016 1 Sequentially compact sets Definition 1. A metric space is sequentially compact if every sequence has a convergent subsequence. Definition 2. A metric space is complete if every Cauchy sequence con- verges. Definition 3. Let > 0. A set fxα 2 X : α 2 Ig is an -net for a metric space X if [ X = B(xα): α2I Definition 4. A metric space is totally bounded if it has a finite -net for every > 0. Theorem 5. A metric space is sequentially compact if and only if it is complete and totally bounded. Proof. Suppose that X is a sequentially compact metric space. Then every Cauchy sequence in X has a convergent subsequence, so, by Lemma 6 below, the Cauchy sequence converges, meaning that X is complete. Next, suppose that X is not totally bounded. Then there exists 0 > 0 such that X has no finite 0-net. Choose a sequence (xn) in X as follows. Pick any x1 2 X. Since X 6= B0 (x1), there exists x2 2 X such that d(x1; x2) ≥ 0. Given fx1; x2; : : : ; xng, there exists xn+1 2 X such that d(xk; xn+1) ≥ 0 for Sn 1 ≤ k ≤ n, since X 6= k=1 B0 (xk). Then d(xm; xn) ≥ 0 for every m 6= n, so (xn) has no Cauchy subsequences and therefore no convergent subsequences, meaning that X is not sequentially compact. It follows that a sequentially compact metric space is complete and totally bounded. 1 Conversely, suppose that a metric space X is totally bounded and com- plete. Let (xn) be a sequence in X. Since X is totally bounded, it has a finite 1-net fam : 1 ≤ m ≤ Mg such that M [ X = B1(am): m=1 At least one ball, say X1 = B1(am), must contain infinitely many terms in the sequence, meaning that xn 2 X1 for infinitely many n 2 N. Choose xn1 2 X1. A subspace of a totally bounded metric space is totally bounded with respect to its subspace metric, so X1 has a finite (1=2)-net, again denoted by fam : 1 ≤ m ≤ Mg, such that M [ X1 = B1=2(am): m=1 At least one ball, say X2 = B1=2(am), contains infinitely many terms in the sequence, so we can choose xn2 2 X2 with n2 > n1. Continuing in this way, given xnj 2 Xj for 1 ≤ j ≤ k and a ball Xk of radius 1=k that contains infinitely many terms in the sequence, we cover Xk by finitely many balls of radius 1=(k + 1), let Xk+1 be one of these balls that contains infinitely many terms in the sequence, and choose xnk+1 2 Xk+1 such that nk+1 > nk. Since xnj 2 Xk for all j ≥ k, it follows that 2 d x ; x < for all i; j ≥ k; ni nj k 1 so the subsequence (xnk )k=1 is a Cauchy sequence. Since X is complete, the subsequence converges, which proves that a complete, totally bounded metric space is sequentially compact. Lemma 6. If a Cauchy sequence has a convergent subsequence, then the Cauchy sequence converges to the limit of the subsequence. 1 Proof. Suppose that (xn)n=1 is a Cauchy sequence in a metric space X with 1 a subsequence (xnk )k=1 that converges to x 2 X. Given > 0, there exists N 2 N such that d(xm; xn) < /2 for all m; n > N, and k 2 N such that nk > N and d(xnk ; x) < /2. Then d(xn; x) ≤ d(xn; xnk ) + d(xnk ; x) < for all n > N, so xn ! x as n ! 1. 2 Definition 7. A metric space has the finite intersection property for closed sets if every decreasing sequence of closed, nonempty sets has nonempty intersection. Theorem 8. A metric space is sequentially compact if and only if it has the finite intersection property for closed sets. Proof. Suppose that X is sequentially compact. Given a decreasing sequence of closed sets Fn, choose xn 2 Fn for each n 2 N. Then (xn) has a convergent subsequence (xnk ) with xnk ! x as k ! 1. Since xnk 2 Fn for all nk ≥ n T1 T1 and Fn is closed, x 2 Fn for every n 2 N, so x 2 n=1 Fn, and n=1 Fn 6= ;. Conversely, suppose that X has the finite intersection property. Let (xn) be a sequence in X and define Fn = Tn;Tn = fxk : k > ng: Then (Fn) is a decreasing sequence of non-empty, closed sets, so there exists 1 \ x 2 Fn: n=1 Choose a subsequence (xnk ) of (xn) as follows. For k = 1, there exists xn1 2 T1 such that d(xn1 ; x) < 1, since x 2 F1 and T1 is dense in F1. Similarly, since x 2 Fn1 and Tn1 is dense in Fn1 , there exists xn2 2 Tn1 with n2 > n1 such that d(xn2 ; x) < 1=2. Continuing in this way (or by induction), given xnk we choose xnk+1 2 Tnk , where nk+1 > nk, such that d(xnk+1 ; x) < 1=(k + 1). Then xnk ! x as k ! 1, so X is sequentially compact. Lemma 9 (Lebesgue Covering Lemma). If fGα ⊂ X : α 2 Ig is an open cover of a sequentially compact metric space X, then there exists δ > 0 such that for every x 2 X there is some α 2 I with Bδ(x) ⊂ Gα. Proof. We use proof by contradiction. Suppose that no such δ > 0 exists. Then for every n 2 N there exists xn 2 X such that B1=n(xn) is not contained in Gα for any α 2 I. Since X is sequentially compact, the sequence (xn) has a convergent subsequence (xnk ). Let x = limk!1 xnk . Then x 2 Gα for some α 2 I since fGα : α 2 Ig covers X, and B(x) ⊂ Gα for some > 0 since Gα is open. Since xnk ! x and 1=nk ! 0 as k ! 1, there exists k 2 N such that d(x; xnk ) < /2 and 1=nk < /2. It follows that B1=nk (xnk ) ⊂ B(x), so B1=nk (xnk ) is contained in Gα, contradicting the choice of xnk . 3 2 Compact sets Definition 10. An open cover of a metric (or topological) space X is a collection of open sets C = fGα ⊂ X : α 2 Ig such that [ X = Gα: α2I In that case, we say that C covers X. A subcover of C is a subcollection fGβ : β 2 Jg of sets in C, where J ⊂ I, that covers X. The subcover is finite if J is finite. Definition 11. A metric (or topological) space is compact if every open cover of the space has a finite subcover. Theorem 12. A metric space is compact if and only if it is sequentially compact. Proof. Suppose that X is compact. Let (Fn) be a decreasing sequence of c closed nonempty subsets of X, and let Gn = Fn. S1 If n=1 Gn = X, then fGn : n 2 Ng is an open cover of X, so it has a finite subcover fGnk : k = 1; 2;:::Kg since X is compact. Let N = maxfnk : k = 1; 2;:::Kg: SN Then n=1 Gn = X, so N N !c \ [ FN = Fn = Gn = ;; n=1 n=1 contrary to our assumption that every Fn is nonempty. S1 It follows that n=1 Gn 6= X and then 1 1 !c \ [ Fn = Gn 6= ;; n=1 n=1 meaning that X has the finite intersection property for closed sets, so X is sequentially compact. Conversely, suppose that X is sequentially compact. Let fGα ⊂ X : α 2 Ig 4 be an open cover of X. By Lemma 9, there exists δ > 0 such that every ball Bδ(x) is contained in some Gα. Since X is sequentially compact, it is totally bounded, so there exists a finite collection of balls of radius δ fBδ(xi): i = 1; 2; : : : ; ng that covers X. Choose αi 2 I such that Bδ(xi) ⊂ Gαi . Then fGαi : i = 1; 2; : : : ; ng is a finite subcover of X, so X is compact. 5.