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I was trying to demonstrate the result as an exercise, and so have refrained from checking against other sources.

The result in question is that a set $F \subseteq X$, where $(X, d)$ is a complete metric space, is compact iff $F$ is totally bounded and closed. Showing compactness implies closed and totally bounded is simple enough. The part I wanted to be sure I hadn't flubbed is the other direction.

Let $F \subseteq X$ be totally bounded, closed, but assume for contradiction that $F$ is not compact. Then there exists an open cover $\mathcal{U} = \{ U_{i} : i \in I \}$ such that no finite sub-class of $\mathcal{U}$ covers $F$. Since $F$ is totally bounded, there exists a set $\left\{ x_{1}^{(1)}, \ldots, x_{N_{1}}^{(1)} \right\} \subseteq F$ such that $$F \subseteq \bigcup_{j = 1}^{N_1} B \left( x_{j}^{(1)} , 1 \right) \subseteq \bigcup_{j = 1}^{N_1} \overline{B \left( x_{j}^{(1)} , 1 \right)} .$$ Consider the sets $\left\{ \overline{B \left( x_{j}^{(1)} , 1 \right)} \cap F , 1 \leq j \leq N_{1} \right\}$. Then at least one of these sets admits no finite subcover from $\mathcal{U}$ (otherwise $F$ would admit finite subcover), so set $y_{1} \in \left\{ x_{1}^{(1)}, \ldots, x_{N_{1}}^{(1)} \right\}$ such that $\overline{B \left( y_1 , 1 \right)} \cap F = : E_{1}$ admits no finite subcover from $\mathcal{U}$.

Now we construct a sequence $(y_{k})_{k \in \mathbb{N}} \in F^{\mathbb{N}}, (E_{k})_{k \in \mathbb{N}} \in ( \mathcal{P}(F) )^{\mathbb{N}}$ as follows: Suppose we've constructed the elements $y_{1}, \ldots, y_{k}, E_{1}, \ldots, E_{k}$. Then $E_{k}$ by assumption admits no finite sub-cover from $\mathcal{U}$, and moreover is totally bounded, so there exists a family $\left\{ x_{1}^{(k + 1)}, \ldots, x_{N_{k + 1}}^{(k + 1)} \right\} \subseteq E_{k}$ such that $E_{k} \subseteq \bigcup_{j = 1}^{N_{k + 1}} B \left( x_{j}^{(k + 1)} , \frac{1}{k + 1} \right) \subseteq \bigcup_{j = 1}^{N_{k + 1}} \overline{ B \left( x_{j}^{(k + 1)} , \frac{1}{k + 1} \right)}$. Then there exists $y_{k + 1} \in \left\{ x_{1}^{(k + 1)}, \ldots, x_{N_{k + 1}}^{(k + 1)} \right\}$ such that $\overline{B \left( y_{k + 1} , \frac{1}{k + 1} \right)} \cap E_{k} = : E_{k + 1}$ admits no finite sub-cover from $\mathcal{U}$.

Now consider the sequence $(y_{k})_{k \in \mathbb{N}}$. We claim that the sequence is Cauchy. To see this, fix $\epsilon > 0$. If $K \in \mathbb{N}$ such that $1 / K < \epsilon / 2$, and $k_{1}, k_{2} > K$, then $y_{k_{1}} , y_{k_{2}} \in E_{k}$, so $d(y_{k_{1}} , y_{k_{2}} \leq d(y_{k_{1}}, y_{K}) + d(y_{K}, y_{k_{2}}) \leq 2 / K < \epsilon$. Since $X$ is complete, let $y = \lim_{k \to \infty} y_{k}$. Since $F$ is closed and $y_{k} \in F$ for all $k$, we know $y \in F$.

Let $U \in \mathcal{U}$ such that $y \in U$. Then since we're in a metric space, there exists $\delta > 0$ such that $B(y, \delta) \subseteq U$. Let $K \in \mathbb{N}$ sufficiently large that $1 / K < \delta / 2$. Then if $y' \in E_{K}$, then $d(y', y) \leq d(y', y_{K}) + d(y_{K}, y) \leq 2 / K < \delta$, so $E_{K} \subseteq B(y, \delta) \subseteq U \in \mathcal{U}$. Thus $E_{K}$ admits a finite subcover, contradicting the assumption that it not.

Is this right?

AJY
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  • It is simpler to show that a sequence in a closed totally bounded set has a convergent subsequence. – Mariano Suárez-Álvarez Oct 21 '16 at 04:20
  • @bof yes, fixed – AJY Oct 21 '16 at 04:32
  • @MarianoSuárez-Álvarez Could you elaborate? – AJY Oct 21 '16 at 04:32
  • Suppose you have a sequence in your set. Cover the set with finitely many balls of radius 1. The sequence clearly has a subsequence that is entirely contained in one of the balls. Now cover the set with balls of radius 1/2, Finitely many of them. That subsequence has a subsequence which is entirely contained in one of these, and so on. The use a diagonal argument to construct a subsequence of the original sequence which is Cauchy. Since the space is complete, that subsequence converges, and since your set is closed, the limit is in it. – Mariano Suárez-Álvarez Oct 21 '16 at 04:50
  • @MarianoSuárez-Álvarez I see that, but I don't see how to relate that to compactness. – AJY Oct 21 '16 at 12:04
  • You surely know there is some relation between compactness and every sequence containing a convergent subsequence! – Mariano Suárez-Álvarez Oct 21 '16 at 16:27
  • @MarianoSuárez-Álvarez Yes, that compactness implies sequential compactness and vice versa in a metric space. But I was hoping for a more direct proof. – AJY Oct 21 '16 at 18:21
  • @AJY: The argument by way of sequential compactness is probably the simplest: closedness implies completeness, which implies that every Cauchy sequence converges, and total boundedness implies that every sequence has a Cauchy subsequence, so you get sequential compactness and hence compactness. This answer shows one way to avoid explicitly going through sequential compactness. – Brian M. Scott Oct 21 '16 at 22:58
  • Have you already shown that a closed subset of a closed totally bounded set is a totally bounded set? This is needed to justify the assertion that each $E_k$ is totally bounded. – DanielWainfleet Oct 22 '16 at 02:49
  • @user254665 Isn't any subset of a totally bounded set totally bounded? – AJY Oct 22 '16 at 15:24
  • Yes you are quite right – DanielWainfleet Oct 22 '16 at 18:28

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