Let $E$ be a Polish space. A set $A\subset E$ is totally bounded if and only if $A$ is relatively compact.
First we suppose $A$ is totally bounded. Since $\overline A$ is closed and $E$ is complete, $\overline A$ is complete. Let $x$ be the limit of a sequence $\{x_n\}$ in $A$ and $\varepsilon>0$. Choose $y_1,\ldots, y_n\in A$ so that $A\subset \bigcup_{i=1}^n B(y_i,\varepsilon/2)$. Choose $N$ such that $d(x_n,x)<\varepsilon/2$ and let $j$ be such that $x_N\in B(y_j,\varepsilon/2)$. Then $$d(x,y_j)<d(x,x_n)+d(x_n,y_j)<\varepsilon,$$ so that $x\in B(x,y_j)$. It follows that $\overline A$ is totally bounded, and thus compact.
I am a bit unsure of this since I did not use the separability of $E$. I also do not see how to show the converse. Any tips would be appreciated.
Best Answer
Separability is not required. The equivalence is true in any complete metric space. For the converse note that $\overset {-} A$ can be covered by a finite number of balls of radius $\epsilon$ (because it is compact and it is covered by all balls of radius $\epsilon)$. Hence $\overset {-} A$ is totally bounded. Any subset of a totally bounded set is also totally bounded, so $A$ is totally bounded.
Proof of the fact that if $A \subset B$ and $B$ is totally bounded the $A$ is totally bounded: cover $B$ by balls $B(b_i,\epsilon /2), 1\leq i \leq n$. Without loss of generality assume that each of these balls has non-empty intersection with $A$. Let $a_i \in A\cap B(b_i,\epsilon /2)$. You can now verify that $A$ is covered by the balls $B(a_i,\epsilon)$.