Extending Bounded metric spaces to compact spaces

[s.hamid.ef]

Hi
Suppose (X,d) is a bounded metric space. Can we extend (X,d) into (X',d') such that (X',d') is compact and d and d' agree on X?

( The reason for asking the question: To prove a theorem in Euclidean space, I found it convenient to first extend the bounded set in question to a compact one ( its closure, to be exact.) and work with the latter. The proof can be generalized to complete spaces, but to go any further I need the answer to this question.)

Thanks in advance.

 

[micromass]

Your theorem is true if and only if the metric space (X,d) is totally bounded.

Every metric space has a completion. So if your space is totally bounded, then the completion will be complete and totally bounded = compact.

On the other hand, if your space can be isometrically embedded in a compact space, then your space needs to be totally bounded (as each subset of a totally bounded set is totally bounded).


If you don't wish (X,d) to be an isometric to a subset of (X',d') but only homeomorphic to a subset of a compact metric space, then it suffices that (X,d) is separable.

 

[s.hamid.ef]

Thanks so much Micromass! I honestly didn't expect such a clear cut and complete answer to my question. Total boundedness showed up several times here and there but I failed to see it was the key concept , and now it sheds light on a few other things I was struggling with ( such as how uniformly continuous functions act on ( totally!) bounded sets.) Everything is now in it its place!