# Example of closed and totally bounded but not complete.

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!

• $(0, 1)$ 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. May 1, 2017 at 10:40
• This is what I was looking for. Thanks! May 1, 2017 at 10:49
• 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 $\mathbb{Q}$ May 1, 2017 at 18:25

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.
• 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.) May 1, 2017 at 10:53