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
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) (show that this is possible).
Now choose a sequence xn such that xn
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)
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
the converse also holds.
Theorem 0.12 (Heine–Borel). In
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,
] [
, 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|
for m n).
By Lemma 0.11 this is equivalent to
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