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I am looking for an example of a closed, totally bounded and not complete subset of a metric space. I know examples of subsets that are closed and bounded but not complete (e.g. sets with discrete topology). But none of the examples I know is totally bounded.

Further goal is to find an example of closed, totally bounded set which is not compact.

Help appreciated, thanks!

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    $\begingroup$ $(0, 1)$ is closed in $(0, 1)$ (with metric induced by the usual metric on $\mathbb{R}$) and totally bounded, but the Cauchy sequence $1/n$ does not converge. $\endgroup$ – Joppy May 1 '17 at 10:40
  • $\begingroup$ This is what I was looking for. Thanks! $\endgroup$ – ElChorro May 1 '17 at 10:49
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    $\begingroup$ You need to work in an incomplete metric space: a closed and totally bounded subset of a complete metric space is complete, and even compact. Hence the examples in $(0,1)$ or $\mathbb{Q}$ $\endgroup$ – Henno Brandsma May 1 '17 at 18:25
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The set $\mathbf{Q} \cap [0, 1]$ of rationals in $[0, 1]$ with the usual metric/topology is closed (in $\mathbf{Q})$ and totally bounded, but not complete.

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  • $\begingroup$ However, this is still compact, isn't it? My motivation was to find an example that is closed, totally bounded but not compact. This example shows, however how careful one needs to be. $\endgroup$ – ElChorro May 1 '17 at 10:47
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    $\begingroup$ If a space isn't complete, it's certainly not compact. :) Here, pick your favorite irrational $a$ in $(0, 1)$, and consider the open covering $U_{n} = \bigl([0, a - \frac{1}{n}) \cup (a + \frac{1}{n}, 1]\bigr) \cap \mathbf{Q}$. (For a suitable $a$ and a bit of effort, it's possible to express these sets as unions of rational intervals, but you get the idea.) $\endgroup$ – Andrew D. Hwang May 1 '17 at 10:53
  • $\begingroup$ Oh of course. Compact implies complete, need to refresh this proof too. Thanks for clarification! $\endgroup$ – ElChorro May 1 '17 at 10:56

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