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Why does "Totally bounded" need "Complete" in order to imply "Compact"? Shouldn't the definition of totally bounded imply the existence of a convergent subsequence of every sequence?

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    $\begingroup$ Maybe the sequence can converge to a element outside the given set. $\endgroup$ – Tomás Jul 1 '13 at 20:01
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    $\begingroup$ Consider $(0,\,1] \subset \mathbb{R}$. As a subset of a compact set, it is totally bounded. But $\frac1n$ has no limit in that set. $\endgroup$ – Daniel Fischer Jul 1 '13 at 20:02
  • $\begingroup$ It seems (and obviously so) that $\Bbb Q\cap[0,1]$ is the canonical counterexample! $\endgroup$ – Asaf Karagila Jul 1 '13 at 21:18
  • $\begingroup$ Wow. We had a little catastrophe here: four essentially equal answers :-) $\endgroup$ – Mariano Suárez-Álvarez Jul 1 '13 at 21:24
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No, total boundedness of $\langle X,d\rangle$ implies that every sequence in $X$ has a Cauchy subsequence. Completeness of $\langle X,d\rangle$ implies that every Cauchy sequence in $X$ actually has a limit point in $X$ and therefore converges. The two together therefore imply that every sequence in $X$ has a convergent subsequence, i.e., that $X$ is sequentially compact. Finally, there is a theorem that a metric space is sequentially compact if and only if it’s compact, so total boundedness plus completeness imply compactness.

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No, totally bounded does not imply the existence of convergent subsequences. For instance, the space $\mathbb Q \cap [0,1]$, with the usual metric, is totally bounded but is not compact (nor complete of course) as any sequence of rationals converging to an irrational number will show.

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  • $\begingroup$ Why is it obvious that this space is totally bounded? $\endgroup$ – michek Mar 16 '15 at 20:27
  • $\begingroup$ Because you can very easily cover $\mathbb Q\cap [0,1]$ by finitely many intervals of any prescribed length $r>0$. $\endgroup$ – Ittay Weiss Mar 16 '15 at 23:29
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No, it "shouldn't", simply because it doesn't.

For example the set $\mathbb Q\cap[0,1]$ (with its usual metric inherited from $\mathbb R$) is totally bounded and you will surely find sequences in it which do not have convergent subsequences.

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Not at all. Consider $X=\Bbb Q\cap[0,1]$ as a subspace of the real line. This is totally bounded, but not complete/compact.

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