Show that a metric space $(X,d)$ is totally bounded if and only if the completion of $(X,d)$ is compact.

Question

I am having quite a bit of trouble on the forward direction of this statement. I believe I have gotten a proof for the reverse direction below:

My attempt:

Suppose $(Y,d')$ is the compact completion of $(X,d)$. Then, due to the completion, $Y$ is compact and therefore, complete and totally bounded. Because $X$ is dense in $Y$, it is clear that $X$ is a subset of $Y$. Subsets of totally bounded sets are totally bounded. Thus, $X$ is totally bounded.

Now for the forward direction, my idea is to do the following, but not sure if its correct:

1. Suppose $(X,d)$ is totally bounded and let $(Y,d')$ be the completion of $X$.
2. Because $X$ is totally bounded then there is open cover of epsilon-balls over $X$.
3. Take an arbitrary sequence from $X$ and show that a subsequence of the chosen sequence lies in some ball from the cover.
4. Create a nested sequence of balls with decreasing radii and show that each successive ball has a subsequence of the subsequence of the ball with larger radius before this one. Hence, the sequence we chose has a subsequence which has a subsequence and so on, to where it is converging to the center of the ball.
5. Due to the completion, $X$ is dense in $Y$, so this entire collection of balls must overlap.
6. Hence, all sequences in the completion are sequentially compact.
7. Sequentially compact implies the completion is compact.

Am I on the right track here?

Answer 1

You are over-thinking this. The essential property of completion (other than being complete) is that, if $i:X \to Y$ is the embedding, then for any $y \in Y$ we can find some $x\in Y$ such that $d'(y,i(x))$ is arbitrarily small.

For the forward direction, you just need to find an $\epsilon$-net for $Y$:

Let $\epsilon>0$. Let $x_1,...,x_n \in X$ be a finite ${\epsilon \over 2}$-net for $X$ and let $y_k = i(x_k)$. I claim that the $y_k$ form an $\epsilon$-net for $Y$.

Now suppose $y \in Y$, then there is some $x \in X$ such that $d'(y,i(x)) < { \epsilon \over 2}$. Now choose the $x_k$ such that $d(x,x_k) < {\epsilon \over 2}$ and note that $d'(i(x),i(x_k)) = d(x,x_k) < {\epsilon \over 2}$, and so $d'(y,y_k) < \epsilon$.

For the reverse direction, you just need to find an $\epsilon$-net for $X$:

Let $y_1,...,y_n \in Y$ be an ${\epsilon \over 2}$-net for $Y$. Now choose $x_1,...,x_n \in X$ such that $d'(y_k,i(x_k)) < {\epsilon \over 2}$. I claim that the $x_k$ form an $\epsilon$-net for $X$.

Suppose $x \in X$, then there is some $y_k$ such that $d'(y_k,i(x)) < {\epsilon \over 2}$. Note that $d(x,x_k)=d'(i(x),i(x_k)) < \epsilon$.