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.