Is Every Sequentially Compact, Separable Metric Space Compact?

[Dragonfall]

How do I prove that every metric space that is sequentially compact and separable is compact? I can't seem to use either hypotheses.

 

[StatusX]

A metric space is compact iff it is complete and totally bounded. Also, I don't think the separable condition is needed.

 

[Dragonfall]

That's if you use the sequential definition of compactness. Here 'compactness' alone is defined a la topology; i.e., open covers, etc. Basically, I want to prove that if X is sequentially compact and separable (sequential => separable, but I can assume it without proof, since I proved in some earlier place), then for all open cover there exists a finite subcover for X.

 

[StatusX]

I thought if you couldn't just use that fact, it might at least be a good stepping stone. But actually it's probably easier to use sequential compactness as a stepping stone to get that fact.

If so, I would try assuming there is some open cover that has no finite subcover and use it to construct a sequence with no convergent subsequence. Specifically, construct a sequence of infinitely many disjoint open balls.

 

[Dragonfall]

Even if you construct such a sequence of open balls, how would you construct a sequence of points in X that has no convergent subsequence? X is sequentially compact.

The sequence of points in X must converge to something outside of X, but I can't think of what, since X could be complete.

 

[StatusX]

If X is complete and non-compact, it is not totally bounded. Take the real numbers for example. One open cover of them are the sets (n-1,n+1) as n ranges over the integers. One infinite sequence with no convergent subsequence is n, as n ranges over the integers. The connection won't be as straightforward in general, but you should be able to find one.

 

[Dragonfall]

X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A. Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X. Contradiction.

I don't know where you're getting at with completeness though.

 

[StatusX]

X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A.

I'll agree with your conclusion, but I don't know what "since we can map a dense subset into a countable subcover of A" means.

Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X.

Why not? For example, if X is the set of real numbers and your cover is (n-1,n+1), take x_i=1 for the index corresponding to (-1,1) and x_i=0 for all other terms and you have a convergent sequence. I think you'll need your sets to be disjoint, and then pick an element in each set. I'm not 100% sure though. Sorry, I'm really busy now, but I'll look over this more carefully later tonight. If you need help sooner, maybe someone else has some ideas...

 

[StatusX]

Ok, I've thought about it a bit more. Your idea in post 7 was on the right track, but you need the open sets to be nested. Pick x_i outside of A_i, then you can show that each A_i can only contain finitely many of the x_i, and use this to show there can be no convergent subsequence.