# 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 – Henno Brandsma Feb 9 '18 at 10:52

## 2 Answers

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. – BSplitter Feb 10 '18 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? – BSplitter Feb 8 '18 at 17:52
• @Blake Splitter I have corrected the mistake. – Kavi Rama Murthy Feb 9 '18 at 5:12