Lemma In a metric space , if is a sequence of compact sets such that for and diam. Then , where is a point in .
Proof: First note that cannot contain more than 1 points. If , then . Since diam=0, it forces .
It remains to show is nonempty. Assume it is empty, then . Since metric space is Hausdorff, is closed for all and is an open cover for . Since is compact, there exists finite subcover such that , which contradicts .
Theorem In a metric space (no matter it has countable basis or not), let be a subset in . The following statement are equivalent by assuming axiom of choice:
- is compact;
- is sequentially compact;
- is complete and totally bounded.
I will prove , and then .
Let be a sequence in , we want to show that it has convergent subsequence. Denote by the ball centered at of radius . Then , by compactness . Therefore there must exist a for some that contains infinite elements of . Denote by the intersection of that ball and . Use balls of radius to cover , again there exists a such that contains infinite elements of . Repeating this process we will have a sequence of sets . Pick an element from in each , we will have a cauchy subsequence, still denote by . Since for all , and is compact, then is compact for all . By the above lemma, for some . Therefore we know the cauchy subsequence converges to .
If is sequentially compact, every Cauchy sequence in converges to a point in , which means is complete. To show is also totally bounded, assume it is not, then there exists an such that cannot be covered by finitely many open balls of radius . Pick an , and pick but not in . And then pick that is not in . Since cannot be covered by finitely many balls of radius , we can find a sequence of open balls . Notice that
the sequence will not have convergent subsequence. So must be totally bounded.
If is complete and totally bounded. Given any sequence in , first we cover the set by finite open balls of radius , there must be one ball, say , contains infinite elements of the sequence. Then we cover the set by balls of radius , again there must be one ball, say , such that contains infinite elements of the sequence. Keep doing the process we will have a Cauchy subsequence of , and it converges to some point in since is complete. Hence is sequentially compact.
Assume we have an open cover for , we want to show there exists a finite subcover. Since is totally bounded, it suffices to show there exists an such that for some . This is because can be covered by finitely many , and hence can be covered by finitely many . Now assume for all , there exists an such that is not contained in any . Take for , we then have a sequence such that is not contained in any . By sequential compactness, there exists a convergent subsequence converges to a point . Since is an open cover, for some . Since is open, if we take large such that , , which contradicts our assumption. Therefore there exists an such that for some , and hence there is a finite subcover for .