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 converges.

Definition 3. Let > 0. A set {xα ∈ X : α ∈ I} is an -net for a metric space X if [ X = B(xα). α∈I 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 ∈ X. Since X 6= B0 (x1), there exists x2 ∈ X such that d(x1, x2) ≥ 0. Given {x1, x2, . . . , xn}, there exists xn+1 ∈ 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 complete. Let (xn) be a sequence in X. Since X is totally bounded, it has a ﬁnite 1-net {am : 1 ≤ m ≤ M} 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 ∈ X1 for infinitely many n ∈ N. Choose xn1 ∈ 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 {am : 1 ≤ m ≤ M}, 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 ∈ X2 with n2 > n1. Continuing in this way, given xnj ∈ 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 ∈ Xk+1 such that nk+1 > nk.

Since xnj ∈ Xk for all j ≥ k, it follows that 2 d x , x < for all i, j ≥ k, ni nj k

∞ 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.

∞ Proof. Suppose that (xn)n=1 is a Cauchy sequence in a metric space X with ∞ a subsequence (xnk )k=1 that converges to x ∈ X. Given > 0, there exists N ∈ N such that d(xm, xn) < /2 for all m, n > N, and k ∈ 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 → ∞.

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 ∈ Fn for each n ∈ N. Then (xn) has a convergent subsequence (xnk ) with xnk → x as k → ∞. Since xnk ∈ Fn for all nk ≥ n T∞ T∞ and Fn is closed, x ∈ Fn for every n ∈ N, so x ∈ 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 = {xk : k > n}.

Then (Fn) is a decreasing sequence of non-empty, closed sets, so there exists

∞ \ x ∈ Fn. n=1

Choose a subsequence (xnk ) of (xn) as follows. For k = 1, there exists xn1 ∈ T1 such that d(xn1 , x) < 1, since x ∈ F1 and T1 is dense in F1. Similarly, since x ∈ Fn1 and Tn1 is dense in Fn1 , there exists xn2 ∈ Tn1 with n2 > n1 such that d(xn2 , x) < 1/2. Continuing in this way (or by induction), given xnk we choose xnk+1 ∈ Tnk , where nk+1 > nk, such that d(xnk+1 , x) < 1/(k + 1).

Then xnk → x as k → ∞, so X is sequentially compact.

Lemma 9 (Lebesgue Covering Lemma). If {Gα ⊂ X : α ∈ I} is an open cover of a sequentially compact metric space X, then there exists δ > 0 such that for every x ∈ X there is some α ∈ I with Bδ(x) ⊂ Gα. Proof. We use proof by contradiction. Suppose that no such δ > 0 exists. Then for every n ∈ N there exists xn ∈ X such that B1/n(xn) is not contained in Gα for any α ∈ I. Since X is sequentially compact, the sequence (xn) has a convergent subsequence (xnk ). Let x = limk→∞ xnk . Then x ∈ Gα for some α ∈ I since {Gα : α ∈ I} covers X, and B(x) ⊂ Gα for some > 0 since Gα is open. Since xnk → x and 1/nk → 0 as k → ∞, there exists k ∈ 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 = {Gα ⊂ X : α ∈ I} such that [ X = Gα. α∈I In that case, we say that C covers X. A subcover of C is a subcollection {Gβ : β ∈ J} 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. S∞ If n=1 Gn = X, then {Gn : n ∈ N} is an open cover of X, so it has a

ﬁnite subcover {Gnk : k = 1, 2,...K} since X is compact. Let

N = max{nk : k = 1, 2,...K}. 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. S∞ It follows that n=1 Gn 6= X and then

∞ ∞ !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

{Gα ⊂ X : α ∈ I}

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

{Bδ(xi): i = 1, 2, . . . , n}

that covers X. Choose αi ∈ I such that Bδ(xi) ⊂ Gαi . Then

{Gαi : i = 1, 2, . . . , n} is a ﬁnite subcover of X, so X is compact.