46 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

it has no Cauchy subsequence. Therefore, by Lemma 2.3.2, it would have no convergentsubsequence. This would contradict K is sequentially compact. Now for B(xi,δ ) , pickUi ∈ C and {U1, ...,Um} covers K.⇒ Now suppose K is compact. If it is not sequentially compact, then there exists a

sequence {xn} which has no convergent subsequence to a point of K. In particular, nopoint of this sequence is repeated infinitely often. The set of points ∪n {xn} has no limitpoint in K. If it did, you would have a subsequence converging to this point since everyball containing this point would contain infinitely many points of ∪n {xn}. Now considerthe sets Hn ≡ ∪k≥n {xk}∪H ′ where H ′ denotes all limit points of ∪n {xn} in X which isthe same as the limit points of ∪k≥n {xk}. Therefore, each Hn is closed thanks to Theorem2.2.7. Now let Un ≡HC

n . This is an increasing sequence of open sets whose union containsK thanks to the fact that there is no constant subsequence. However, none of these opensets covers K because Un is missing xn, violating the definition of compactness.⇒Another proof of the second part of the above is as follows. Suppose K is not se-

quentially compact. Then there is {xn} such that no x ∈ K is the limit of a convergentsubsequence. Hence if x ∈ K, there is rx > 0 such that B(x,rx) contains xn for only finitelymany n. Otherwise, B

(x, 1

k

)would contain xn for infinitely many n and there would exist{

xnk

}∞

k=1 with nk < nk+1 for all k and xnk ∈ B(x, 1

k

)so this subsequence would converge to

x. By compactness, there are finitely many of these balls B(x,rx) which cover K. Now thisis a contradiction because one of these balls must now contain xn for infinitely many n. ■

Definition 2.5.6 Let X be a metric space. Then a finite set of points {x1, · · · ,xn} iscalled an ε net if X ⊆ ∪n

k=1B(xk,ε) . If, for every ε > 0 a metric space has an ε net, thenwe say that the metric space is totally bounded.

Lemma 2.5.7 If a metric space (K,d) is sequentially compact, then it is separable andtotally bounded.

Proof: Pick x1 ∈K. If B(x1,ε)⊇K, then stop. Otherwise, pick x2 /∈B(x1,ε) . Continuethis way. If {x1, · · · ,xn} have been chosen, either

K ⊆ ∪nk=1B(xk,ε)

in which case, you have found an ε net or this does not happen in which case, you can pickxn+1 /∈ ∪n

k=1B(xk,ε). The process must terminate since otherwise, the sequence wouldneed to have a convergent subsequence which is not possible because every pair of termsis farther apart than ε . See Lemma 2.3.2. Thus for every ε > 0, there is an ε net. Thus themetric space is totally bounded. Let Nε denote an ε net. Let D = ∪∞

k=1N1/2k . Then this is acountable dense set. It is countable because it is the countable union of finite sets and it isdense because given a point, there is a point of D within 1/2k of it. ■

Also recall that a complete metric space is one for which every Cauchy sequence con-verges to a point in the metric space.

The following is the main theorem which relates these concepts. Note that if (X ,d) is ametric space, then so is (S,d) whenever S⊆ X . You simply use the metric on S.

Theorem 2.5.8 For (X ,d) a metric space, the following are equivalent.

1. (X ,d) is compact.

2. (X ,d) is sequentially compact.