Lecture 4: Compact Metric Spaces

18.S190: Introduction to Metric Spaces Lecturer: Paige Bright IAP 2023 4 January 19, 2023 Compact Metric Spaces Last time, we showed that a set in Rn is sequentially compact if and only if it is topologically compact, by showing sequentially compact ⇐⇒ closed and bounded Heine-Borel ⇐⇒ topologically compact. However, by the previous remark, we don’t have Heine-Borel for arbitrary metric spaces. Which begs the question: is sequentially compact the same as topologically compact in metric spaces? The answer is yes. To prove this, we first show a handful of preliminary results. Lemma 1 (Lebesgue Number Lemma) Let (X, d) be a sequentially compact metric space and {Ui }i∈I be an open cover of X. Then, there exists an r > 0 such that for each x ∈ X, Br (x) ⊆ Ui for some i ∈ I. Proof: Before proving this, try to visualize the result! We prove this lemma through contradiction. Assume that for some r > 0 there exists an x ∈ X (possibly depending on r ) such that for each i ∈ I, Br (x) ̸⊆ Ui . Consider the sequence {xn }n in X such that B1/n (xn ) ̸⊆ Ui for all i ∈ I. Given that X is sequentially compact, {xn } has a convergent subsequence {xnk }k . Let xnk → x ∈ X. Given that {Ui } is an open cover of X, there exists a Ui0 such that x ∈ Ui0 . Given Ui0 is open, it also follows that there exists an r0 such that Br0 (x) ⊆ Ui0 . Hence, choose N large enough such that d(x, xN ) < r0 2 and 1 N < r0 2. Then, if y ∈ B1/N (xN ), then d(x, y ) ≤ d(x, xN ) + d(xN , y ) < r0 . Therefore, y ∈ Br0 (x) ⊆ Ui0 . Hence, B1/N (xN ) ⊆ Br0 (x) ⊆ Ui0 which is a contradiction. We call this r the Lebesgue number of the open cover of X, which is useful in other applications. Definition 2 A metric space X it totally bounded if, for every ϵ > 0, there exists x1 , x2 , . . . , xk ∈ X with k finite such that {Bϵ (xi ) | 1 ≤ i ≤ k} is an open cover of X. 1 Lemma 3 A metric space X being sequentially compact implies that X is totally bounded. Proof: Assume that X is sequentially compact and not totally bounded. Therefore, there exists an ϵ > 0 such that X cannot be covered by a collection of open sets of only finitely many ϵ-balls. Hence, let x1 ∈ X, x2 ∈ X \ Bϵ (x1 ), then x3 ∈ X \ Bϵ (x1 ) \ Bϵ (x2 ) and so on. We know that there exists such xi by the previous statement. Hence, for all i ̸= j, d(xi , xj ) ≥ ϵ. Therefore, {xn }n has no convergent subsequence as if there was a convergent subsequence it would be Cauchy, and the previous line shows that no subsequence of {xn } will be Cauchy. This is a contradiction to X being sequentially compact. Theorem 4 A metric space X is (topologically) compact if and only if X is sequentially compact. Proof: We first show that sequentially compact implies topologically compact, applying the two lemmas developed thus far. Let X be sequentially compact and let {Ui }i∈I be an open cover of X. By the Lebesgue number lemma, there exists an r > 0 such that for each x ∈ X, Br (x) ⊂ Ui for some i ∈ I. Furthermore, by Lemma 5, X is totally bounded. Hence, there exists y1 , . . . , yk ∈ X such that X ⊂ Br (y1 ) ∪ · · · ∪ Br (yk ). However, for each i ∈ I, we have Br (yi ) ⊂ Uj(i) for some j(i ) ∈ I. (This notation just means for each i , there exists a j ∈ I which depends on i such that Br (yi ) ⊆ Uj .) Thus, {Uj(1) , . . . , Uj(k) } is a finite subcover for X. Therefore, every open cover of X has a finite subcover, and thus sequentially compact implies topologically compact. We now prove the other direction. Assume for the sake of contradiction there there exists a sequence {xn }n in X with no convergent subsequence. Notice that no term in the sequence can appear infinitely many times, as otherwise there would be a trivial subsequence of {xn }. Hence, we assume without loss of generality that xi ̸= xj if i ̸= j. Furthermore, notice then that for every n there exists an ϵn > 0 such that Bϵn (xn ) contains no other terms in the sequence. If this wasn’t the case, then there would again be a convergent subsequence of {xn }n . Therefore, for each i , there exists an open ball Ui centered at xi such that xj ∈ / Ui for all i ̸= j. Additionally, consider U0 = X \ {xn | n ∈ N}. U0 is open, as U0c = {xn | n ∈ N} is closed (it contains all of it’s limit points). Hence, U0 ∪ {Un | n ∈ N} is an open cover of X. However, this open cover has no finite subcover as any finite collection of the cover will fail to include infinitely many terms from the sequence {xn }n . This is a contradiction, and thus topologically compact implies sequentially compact. Remark 5. Notice that we technically could’ve used this proof in the previous lecture, but the Heine-Borel Theorem is so vastly important that I decided to do that proof before today’s lecture. We will now start to look at some illuminating applications of compact sets to reach an even more powerful theorem. Recall 6 Let X, Y be metric spaces and f : X → Y be a continuous function. Then, for all U open in Y , f −1 (U) is open in X. 2 Theorem 7 Let X, Y be metric spaces and f : X → Y be continuous. Given K ⋐ X, f (K) ⊂ Y is compact. Proof: Let {Ui }i∈I be an open cover of f (K). Then, define Vi = {f −1 (Ui )}i∈I , which is open as f is continuous. Therefore, {f −1 (Ui )}i∈I is an open cover of K. Hence, there exists a finite subcover {Vi1 , . . . Vik } of K as K is compact. Thus, {Ui1 , . . . Uik } = {f (Vi1 ), . . . , f (Vik )} is a finite subcover of f (K). Therefore, f (K) is compact. Corollary 8 Let X be a metric space and K ⋐ X. Then, given a continuous function f : X → R, f obtains a maximum and minimum finite value on K. Proof: The proof follows from the previous theorem, a problem on PSET 2. Corollary 9 Sometimes in particular we want to study bounded continuous functions, and the previous corollary gives us a nice property. Given a compact metric space X, every continuous function on f is bounded. Proof: Follows immediately. Theorem 10 (Cantor’s Intersection Theorem) If K1 ⊃ K2 ⊃ K3 ⊃ . . . is a decreasing sequence of nonempty sequentially compact subsets of Rn , then T i≥1 Ki is non-empty. Proof: Choose a sequence {an }n such that an ∈ Kn for each n. We know that there exists such an an as each Kn is nonempty. Then, {an }n is a sequence in K1 , and thus there exists a convergent subsequence {ank }k such that ank → a ∈ K1 . Furthermore, {an }∞ n=2 is a sequence in K2 , and thus contains a a convergent subsequence. Therefore, T a ∈ K2 . Continuing this process, we get that a ∈ Ki for all i . Thus, a ∈ i≥1 Ki . Definition 11 (Finite Intersection Property) A collection of closed sets {Ci }i has the finite intersection property if every finite subcollection has a nonempty intersection. Given Lemma 3 and the Cantor Intersection Theorem, it is clear that there are some relations between compact sets, nonempty intersections of sets, and totally bounded sets. We hence show the following theorem. Theorem 12 Given a metric space (X, d), the following are equivalent. (1) X is compact. (2) X is sequentially compact. (3) X is Cauchy complete and totally bounded. (4) Every collection of closed subsets of X with the finite intersection property has a non-empty intersection. 3 We have shown (1) ⇐⇒ (2), and thus we show (1) ⇐⇒ (4) and (2) ⇐⇒ (3) to finish the proof. Proof: (1) =⇒ (4): Assume for the sake of contradiction that there exists a collection of closed subsets {Ci }i∈I T with the finite intersection property such that i∈I Ci = ∅. Given Ci is closed in X for all i , Ui = Cic is open in X for each i . Then, !c [ Ui = i∈I [ Cic \ = i∈I = ∅c = X. Ci i∈I Hence, the Ui cover X. Given X is compact, there exists a finite subcover {Ui1 , . . . , Uik } of X. Thus, X= k [ Uin = n=1 Therefore, Tk n=1 k \ !c Uicn k \ = n=1 !c Cin . n=1 Cin = ∅ which is a contradiction with the finite intersection property. (4) =⇒ (1): Suppose that {Ui }i∈I is an open cover of X, and let Ci = Uic for each i ∈ I. Assume for the sake of contradiction that no finite subset of the Ui covers X. We show that Ci has the finite intersection property. Assume for the sake of contradiction that {Cn1 , . . . , Cnk } satisfies Cn1 ∩ · · · ∩ Cnk = ∅. Then, k [ Uni = i=1 k \ !c Unci = i=1 k \ !c Cik = ∅c = X. i=1 This is a contradiction with the assumption that no subset of the Ui covers X. Hence, {Ci }i∈I satisfies the finite T S intersection property. Therefore, {Ci }i∈I has non-empty intersection; i.e. i∈I Ci ̸= ∅. Then, i∈I Ui ̸= X, which is a contradiction to the Ui being an open cover for X. Thus, every open cover of X has a finite open subcover. (2) =⇒ (3): We have already shown that X being sequentially compact implies totally bounded, and hence we only need show that sequentially compact implies Cauchy complete. Let {xn } be a Cauchy sequence in X. Given {xn } is a sequence in X, there exists a convergent subsequence {xnk } in X such that xnk → x ∈ X. Let ϵ > 0, and choose N such that d(xi , xj ) < ϵ/2 whenever i , j ≥ N. Next, choose nk > N such that d(xnk , x) < ϵ/2. Then, d(x, xN ) ≤ d(x, xnk ) + d(xnk , xN ) < ϵ. Thus, xn → x ∈ X as n → ∞. Therefore, every Cauchy sequence in X converges to a point in X. Hence, X is Cauchy complete. (3) =⇒ (2): This part of the proof is quite difficult. Consider a sequence {xn }n in X. Given X is totally bounded, (n) (n) 1 1 for every n ∈ N, there exists a finite set of points {y1(n) , . . . , yr(n) (n) } such that X ⊂ B (y1 ) ∪ · · · ∪ B (yr (n) ). Define n n Sn = {y1(n) , . . . , yr(n) (n) }. We want to find a convergent subsequence of {xn }n . We do so by construction. Given S1 is finite, there exists a (1) (1) yn(1) ∈ S1 such that B1 (yn(1) ) contains infinitely many points from {xn }n . Choose z1 from this ball. Then, given S2 is (2) (1) (2) finite, there is a yn(2) such that infinitely many points from {xn }n are in B1 (yn(1) ) ∩ B1/2 (yn(2) ). Choose z2 from this Tk (k) set. Continue this procedure for each k > 1, selecting a zk from i=1 B 1 (yn(k) ). Then, we show {zn }n is Cauchy. Let k ϵ > 0. Then, there exists an N ∈ N such that 1 N < ϵ. Hence, for all n, m ≥ N, d(zn , zm ) < 1 < ϵ. N Therefore, by the Cauchy completeness of X, {zn } converges to a point in X. Remark 13. Where do we use the fact that each ball has infinitely many points? We do in fact use this property in 4 the proof. Try to figure out how! 5