# Show that $A$ is totally bounded if and only if $\overline{A}$ is totally bounded.

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?

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