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I'm not sure if this is a strong enough condition to imply compactness, since it seems that it must be "totally bounded" and complete. Complete means that X is closed. In R^n bounded and closed set imply compact, but not in general. Any thoughts would be appreaciated

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You use put totally bounded in quotations, but that's an actual term see here. –  JSchlather Apr 5 '11 at 1:30
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you probably need totally bounded –  yoyo Apr 5 '11 at 1:31

2 Answers 2

up vote 7 down vote accepted

No. For example, take $\mathbb{R}$ with the metric $d(x,y)=\min (|x-y|,1)$. Because it's the same as the usual metric for short distances, it's not hard to check that this space is still complete, separable, and not sequentially compact. But it is trivially bounded.

In general, applying this construction to a metric space $X$ yields a bounded space with the same topology, and even the same uniform structure, as $X$. Thus any uniform property which holds under the assumption of boundedness holds without it. In this case, being separable and being sequentially compact are topological properties, and being complete is a uniform property, so "(complete and separable) implies sequentially compact" is a uniform property, and so the boundedness assumption can't be relevant.

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Well, it seems I completely misread and misinterpreted the question. Thanks for clarifying. –  t.b. Apr 5 '11 at 11:24

Since Chris has already answered your question by providing an example, let me give some complementary information.

A metric space $X$ is totally bounded if for every $\varepsilon \gt 0$ there exists a covering of $X$ by finitely many sets of diameter at most $\varepsilon$.

The following are easily seen to be equivalent:

  1. The completion of $X$ is compact.
  2. The space $X$ is totally bounded.
  3. Every sequence in $X$ has a Cauchy subsequence.

The only thing to observe is that $X$ is dense in its completion and that for metric spaces sequential compactness is equivalent to the Heine-Borel property.

Therefore the following are equivalent as well:

  1. The space is compact.
  2. The space is complete and totally bounded.

Many compactness theorems in analysis, such as Heine-Borel or Arzelà-Ascoli (but there are many more involved ones, e.g. the Dunford-Pettis theorem) can be seen as identifying a necessary and sufficient condition for total boundedness in certain circumstances and then adding closed as a supplementary condition:

  • A subset of $\mathbb{R}^{n}$ is totally bounded if and only if it is bounded.
  • If $K$ is a compact metric space, a subset of $C(K)$ is totally bounded if and only if it is bounded and equicontinuous.

These observations are in fact the crux of the proofs of the above-mentioned theorems and as $\mathbb{R}^{n}$, $C(K)$ are both complete, we get the classical formulations as immediate corrolaries.

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