proof that a metric space is compact if and only if it is complete and totally bounded
Proof. Let be a metric space with metric . If is compact, then it is sequentially compact and thus complete. Since is compact, the covering of by all -balls must have a finite subcover, so that is totally bounded.
Now assume that is complete and totally bounded. For metric spaces, compact and sequentially compact are equivalent; we prove that is sequentially compact. Choose a sequence ; we will find a Cauchy subsequence (and hence a convergent subsequence, since is complete).
Cover by finitely many balls of radius (since is totally bounded). At least one of those balls must contain an infinite number of the . Call that ball , and let be the set of integers for which .
Proceeding inductively, it is clear that we can define, for each positive integer , a ball of radius containing an infinite number of the for which ; define to be the set of such .
Each of the is infinite, so we can choose a sequence with for all . Since the are nested, we have that whenever , then . Thus for all , and are both contained in a ball of radius . Hence the sequence is Cauchy.
- 1 J. Munkres, Topology , Prentice Hall, 1975.
|Title||proof that a metric space is compact if and only if it is complete and totally bounded|
|Date of creation||2013-03-22 18:01:06|
|Last modified on||2013-03-22 18:01:06|
|Last modified by||rm50 (10146)|