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Let $X$ denote a metric space.

Supposing every subset of $X$ that is closed and totally bounded is also compact, is $X$ necessarily complete?

What I've got so far. Assume every $X$ subset of $X$ that is closed and totally bounded is also compact, and consider a Cauchy sequence $a : \omega \rightarrow X.$ Then the image of $a$ (call it $A$) is totally bounded. (However, $A$ is not necessarily closed; now what?)

goblin GONE
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    Consider the closure of the image. – Najib Idrissi Mar 28 '14 at 15:22
  • @nik, thanks. So okay, consider the closure of $A$, call it $B$. Then $B$ is both closed and totally bounded; hence compact. Thus $I=\bigcap_{n=0}^\infty (\mathrm{cl} \circ \mathrm{im})(a \restriction [n,\infty))$ is non-empty... Any ideas how to finish the proof from here? – goblin GONE Mar 28 '14 at 15:35

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As you wrote, consider the image of the cauchy sequence, $A$, which is totally bounded by the cauchy property. So it's closure, $B$, is totally bounded and closed, so compact by hypothesis.

To finish the proof just use sequential compactness. You know a subsequence converges, but it is a basic fact (easy to prove) that if a subsequence of a cauchy sequence converges then the cauchy sequence converges.

Seth
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