# Sequential compactness

1. Nov 28, 2008

### Ja4Coltrane

Hello.
I'm trying to prove that any sequentially compact metric space is totally bounded (where totally bounded means that for any epsilon > 0, there exists a finite open covering for the space consisting only of balls of radius epsilon.

Does anyone have any advice for proving this? I realize that one thing is that seq compactness => compactness => totally bounded, but I'd like to avoid this if possible...

Thanks!

2. Nov 29, 2008

### adriank

This result is in fact part of the proof that sequentially compact implies compact for a metrizable space X in Munkres' Topology (§28 in the second edition), so one better not use sequentially compact ⇒ compact to prove this!

The proof is one by contradiction that goes roughly like this: Suppose there is an ε > 0 such that X can't be covered by finitely many ε-balls. Let x1 be any point of X, and in general pick xn+1 in X that is not in any ε-ball centered at a previous point (since these ε-balls do not cover X). Argue that the sequence (xn) has no convergent subsequence. (Consider the distance between any pair of points in the sequence.)

Last edited: Nov 29, 2008
3. Nov 29, 2008

### mathwonk

compact is equivalent to complete and totally bounded. totally bounded means every sequence has a cauchy subsequence, which then converges by completeness.

so the previous post no doubt describes how, in the absence of total boundedness, to construct a sequence with no cauchy subsequence.

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