0.1. Warm up: Metric and topological spaces 11

A subset K ⊂ X is called sequentially compact if every sequence

from K has a convergent subsequence. In a metric space, compact and

sequentially compact are equivalent.

Lemma 0.11. Let X be a metric space. Then a subset is compact if and

only if it is sequentially compact.

Proof. Suppose X is compact and let xn be a sequence which has no conver-

gent subsequence. Then K = {xn} has no limit points and is hence compact

by Lemma 0.9 (ii). For every n there is a ball Bεn (xn) which contains only

finitely many elements of K. However, finitely many suﬃce to cover K, a

contradiction.

Conversely, suppose X is sequentially compact and let {Oα} be some

open cover which has no finite subcover. For every x ∈ X we can choose

some α(x) such that if Br(x) is the largest ball contained in Oα(x), then

either r ≥ 1 or there is no β with B2r(x) ⊂ Oβ (show that this is possible).

Now choose a sequence xn such that xn ∈

mn

Oα(xm). Note that by

construction the distance d = d(xm,xn) to every successor of xm is either

larger than 1 or the ball B2d(xm) will not fit into any of the Oα.

Now let y be the limit of some convergent subsequence and fix some r ∈

(0, 1) such that Br(y) ⊆ Oα(y). Then this subsequence must eventually be in

Br/5(y), but this is impossible since if d = d(xn1 , xn2 ) is the distance between

two consecutive elements of this subsequence, then B2d(xn1 ) cannot fit into

Oα(y) by construction whereas on the other hand B2d(xn1 ) ⊆ B4r/5(a) ⊆

Oα(y).

In a metric space, a set is called bounded if it is contained inside some

ball. Note that compact sets are always bounded since Cauchy sequences

are bounded (show this!). In

Rn

(or

Cn)

the converse also holds.

Theorem 0.12 (Heine–Borel). In

Rn

(or

Cn)

a set is compact if and only

if it is bounded and closed.

Proof. By Lemma 0.9 (ii) and (iii) it suﬃces to show that a closed interval

in I ⊆ R is compact. Moreover, by Lemma 0.11, it suﬃces to show that

every sequence in I = [a, b] has a convergent subsequence. Let xn be our

sequence and divide I = [a,

a+b

2

] ∪ [

a+b

2

, b]. Then at least one of these two

intervals, call it I1, contains infinitely many elements of our sequence. Let

y1 = xn1 be the first one. Subdivide I1 and pick y2 = xn2 , with n2 n1 as

before. Proceeding like this, we obtain a Cauchy sequence yn (note that by

construction In+1 ⊆ In and hence |yn − ym| ≤

b−a

n

for m ≥ n).

By Lemma 0.11 this is equivalent to