Let $A\subseteq X$. Show that $A$ is totally bounded if and only if $\overline{A}$ is totally bounded. If $A$ is totally bounded subset of a complete metric space, then show that $\overline{A}$ is compact.

$(\Rightarrow)$Let $A$ be totally bounded.

Then $\forall \epsilon >0\; \exists n(\epsilon/2)\in N$ such that $\exists x_1\ldots x_n \in A(\subseteq \overline{A})$ such that $A\subseteq \cup_{i=1}^{n}B_{\epsilon/2}(x_i) \rightarrow \overline{A}\subseteq \overline{\cup_{i=1}^{n}B_{\epsilon/2}(x_i)} = \cup_{i=1}^{n}\overline{B_{\epsilon/2}(x_i)}\subseteq \cup_{i=1}^{n}B_{\epsilon}(x_i)$

So we proved that $\forall \epsilon>0\; \exists n(\epsilon)>0$ such that $\exists x_1 \ldots x_n \in \overline{A}$ such that

$\overline{A}\subseteq \cup_{i=1}^{n}B_{\epsilon}(n)$ so that $\overline{A}$ is totally bounded.

(Please check above inclusions if I have written them correctly).

$(\Leftarrow)$Let $\overline{A}$ be totally bounded. Then $A$ becomes bounded since it is a subset of a totally bounded set.

For the other part, $A$ is totally bounded so $\overline{A}$ is totally bounded.

What is left is to show $\overline{A}$ is complete.

But $\overline{A}$ is a closed subset of complete metric space, hence complete. So $\overline{A}$ becomes compact.

is my proof correct?

  • $\begingroup$ @KaboMurphy because a subset of a metric space is compact iff it's complete and totally bounded $\endgroup$ – Henno Brandsma Oct 27 '19 at 11:47
  • $\begingroup$ Sorry. I did not read the second part of the question when I made the comment. $\endgroup$ – Kavi Rama Murthy Oct 27 '19 at 11:48
  • $\begingroup$ Your proof looks fine. $\endgroup$ – Kavi Rama Murthy Oct 27 '19 at 11:50

Yes, your proof is correct, provided you've already shown that $\overline{B_{\delta}(x)} \subseteq B_{\delta'}(x)$ for $\delta' > \delta$ e.g. or some such inclusion (you now seem to assume the inclusion without proof).

The reverse implication is indeed that trivial and the remark on completeness of closed subsets finishes the proof.


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