70 CHAPTER 3. METRIC SPACES

Definition 3.5.6 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 3.5.7 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⊆∪n

k=1B(xk,ε) in which case, you havefound an ε net or this does not happen in which case, you can pick xn+1 /∈ ∪n

k=1B(xk,ε).The process must terminate since otherwise, the sequence would need to have a convergentsubsequence which is not possible because every pair of terms is farther apart than ε . SeeLemma 3.2.4. Thus for every ε > 0, there is an ε net. Thus the metric space is totallybounded. Let Nε denote an ε net. Let D = ∪∞

k=1N1/2k . Then this is a countable dense set. Itis countable because it is the countable union of finite sets and it is dense because given apoint, 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.

Theorem 3.5.8 For (X ,d) a metric space, the following are equivalent.

1. (X ,d) is compact.

2. (X ,d) is sequentially compact.

3. (X ,d) is complete and totally bounded.

Proof: By Theorem 3.5.5, the first two conditions are equivalent.2.=⇒ 3. If (X ,d) is sequentially compact, then by Lemma 3.5.7, it is totally bounded.

If {xn} is a Cauchy sequence, then there is a subsequence which converges to x ∈ X byassumption. However, from Theorem 3.2.2 this requires the original Cauchy sequence toconverge.

3.=⇒ 1. Since (X ,d) is totally bounded, there must be a countable dense subset of X .Just take the union of 1/2k nets for each k ∈ N. Thus (X ,d) is completely separable byTheorem 3.4.6 has the Lindeloff property. Hence, if X is not compact, there is a countableset of open sets {Ui}∞

i=1 which covers X but no finite subset does. Consider the nonemptyclosed sets Fn and pick xn ∈ Fn where

X \∪ni=1Ui ≡ X ∩ (∪n

i=1Ui)C ≡ Fn

Let{

xkm}Mk

m=1 be a 1/2k net for X . We have for some m,B(xk

mk,1/2k

)contains xn for in-

finitely many values of n because there are only finitely many balls and infinitely manyindices. Then out of the finitely many

{xk+1

m}

where B(xk+1

m ,1/2k+1)

has nonempty in-

tersection with B(xk

mk,1/2k

), pick one xk+1

mk+1such that B

(xk+1

mk+1,1/2k+1

)contains xn for

infinitely many n. Then obviously{

xkmk

}∞

k=1is a Cauchy sequence because

d(

xkmk,xk+1

mk+1

)≤ 1

2k +1

2k+1 ≤1

2k−1