# Showing that a totally bounded set is relatively compact (closure is compact)

I have been tasked with showing that for a metric space $(X,d)$, a subset $E \subseteq X$ is relatively compact $\iff$ $E$ is totally bounded. I believe I have shown the forward implication $(\Rightarrow)$. However, I'm struggling to show the backward implication.

For clarity, a set $E$ is said to be relatively compact if its closure $\overline{E}$ is compact.

Attempt at a proof:

Assume $E$ is totally bounded. Then by a corollary, we know that every sequence in $E$ has a Cauchy subsequence. Since $\overline{E}$ contains all limits points of $E$, we know that every Cauchy subsequence converges to some point in $\overline{E}$. Then $\overline{E}$ is sequentially compact (every sequence has a convergent subsequence). This implies that $\overline{E}$ is compact, and therefore $E$ is relatively compact.

My issue arises from the fact that I'm not considering what happens to sequences living inside of $\overline{E}\backslash E = \partial E$. I don't know if those converge or have Cauchy subsequences or anything. I only know things about sequences in $E$. Maybe I'm missing something obvious, or perhaps this entire proof is offbase. Any help would be appreciated.

• $(X,d)$ is complete? Otherwise it’s false I think Feb 9, 2018 at 10:52

First of all, you're missing the assumption that $(X,d)$ is a complete metric space. If we don't assume that, the statement is false, as for $X = \mathbb{Q}$ (usual metric) we have that $(0,1) \cap \mathbb{Q}$ is totally bounded and its closure $[0,1] \cap \mathbb{Q}$ is not compact.

Fact: if $E$ is totally bounded then so is $\overline{E}$.

(Proof: suppose that $r>0$ has been given. Then $E$ is covered by finitely many open balls $B(x_i,\frac{r}{2}), i=1,\ldots n$. Then certainly $E \subseteq \cup_{i=1}^n D(x_i, \frac{r}{2})$, where $D(x,s) = \{y \in X: d(y,x) \le s\}$ is a closed ball. The finite union of closed balls is closed, so $\overline{E} \subseteq \cup_{i=1}^n D(x_i, \frac{r}{2}) \subseteq \cup_{i=1}^n B(x_i,r)$. As $r>0$ was arbitary, $\overline{E}$ is totally bounded.)

So we know that $\overline{E}$ is totally bounded and complete (as a closed subset of the complete $(X,d)$). So $\overline{E}$ is compact (every sequence has a Cauchy subsequence, which converges, and so we have sequential compactness).

Relatively compact always implies totally bounded (a sequence in $E$ has a convergent subsequence with limit in $\overline{E}$ by relative compactness, and a convergent subsequence is a Cauchy subsequence, so $E$ is totally bounded), but for the reverse implication we really need completeness of $(X,d)$, as I showed above.

• Thank you! Yes, my professor also later told us that the assigned problem was incorrect because he didn't tell us to assume that the metric space was complete. Feb 10, 2018 at 20:17

Let $\{x_n\}$ be a sequence in $\bar {E}$. For each n choose $y_n$ in E such that $d(x_n,y_n) <1/n$. By your argument applied to the sequence $\{y_n\}$ there is a subsequence ${y_n}_k$ converging to some point $y \in \bar {E}$. Now $d({x_n}_k,y) \leq d({y_n}_k,y)+d({x_n}_k,{y_n}_k) <d({y_n}_k,y)+\frac 1 {n_k} \to 0$ proving that $\{x_n\}$ has a convergent subsequence in $\bar {E}$.

• You say, "$d({x_n}_k,y) \leq d({y_n}_k,y)+d({x_n}_k,{x_n}_k)$." Should the $d({x_n}_k,{x_n}_k)$ be something else? Feb 8, 2018 at 17:52
• @Blake Splitter I have corrected the mistake. Feb 9, 2018 at 5:12