Characterization of Relative Compactness in Complete Metric Spaces

Question

Let $(M, d)$ be a complete metric space and $A\subseteq M$. I want to show that $A$ is relatively compact (i.e., $\overline A$ is compact) iff there exists no infinite subset $S\subseteq A$ such that $\inf_{x,y\in S, x\neq y}d(x,y)>0$.

This is apparently "well-known", but I am not sure about my solution:

"$\Rightarrow$": Assume that there is an infinite $S\subseteq A$ such that $\inf_{x,y\in S, x\neq y}d(x,y)>0$. Pick a sequence $(s_n)_{n\in\mathbb N}$ in $S$. Since $\inf_{n,m\in \mathbb N, s_n\neq s_m}d(s_n,s_m)>0$ by assumption, we found a sequence in $A$ with no convergent subsequence.* Hence $A$ cannot be relative compact. (correct?)

"$\Leftarrow$": Suppose that there exists no infinite subset $S\subseteq A$ such that $\inf_{x,y\in S, x\neq y}d(x,y)>0$. That is, for each infinite $S\subseteq A$ it holds that $\inf_{x,y\in S,x\neq y}d(x,y) = 0$. Now take any sequence $(a_n)_{n\in\mathbb N}$ in $A$, and put $S := \{a_n : n\in\mathbb N\}$. By assumption, $\inf_{n,m\in\mathbb N,a_n\neq a_m}d(a_n,a_m) = 0$, so there exists a convergent subsequence of $(a_n)_{n\in\mathbb N}$.** That is, $A$ is relatively compact.

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`* Assume there exists a convergent subsequence $(s_{\phi(n)})_{n\in\mathbb N}$. Then for each $\epsilon>0$ there is an $N\in\mathbb N$ such that $d(s_{\phi(n)},s_{\phi(m)}) < \epsilon = 0 + \epsilon$. But this means $\inf_{n,m\in\mathbb N,s_{\phi(n)}\neq s_{\phi(m)}} d(s_{\phi(n)}, s_{\phi(m)}) = 0$. A contradiction.

`** Assume that $\inf_{n,m\in\mathbb N,a_n\neq a_m}d(a_n,a_m) = 0$ and $(a_n)_{n\in\mathbb N}$ has no convergent subsequence. $\leadsto$ stuck: I think I have to invoke completeness here?

Answer 1

I would give a very different proof using the following standard characterization of relative compactness: A subset of a complete metric space is relatively compact if and only if it is totally bounded.

Definition. A subset $S$ of a metric space $(M,d)$ is said to be $r$-separated if every pair of distinct elements $x, y\in S$ satisfies $d(x,y)>r$. A subset $S$ is said to be separated if it is $r$-separated for some $r>0$.

Note that the condition that you are imposing on subsets $S\subset A$ exactly means that $S$ is separated. You are trying to prove:

Proposition. A subset $A$ of a complete metric space $(M,d)$ is relatively compact if and only if it contains no infinite separated subsets.

Proof. 1. Suppose that $\bar{A}$ is compact and contains a separated subset $S$. Then (since $S$ is separated) $S$ is closed in $M$ and, moreover, has no accumulation points in $M$. In particular, $S$ is compact (as a closed subset of a compact subset $\bar{A}$). But a compact infinite subset of a metric space has to have at least one accumulation point, see here. A contradiction.

2. Suppose that $A\subset M$ does not contain an infinite separated subset. I will prove (using the Axiom of Choice) that $A$ is totally bounded, hence, relatively compact. Fix $r>0$. Consider the poset $E_r\subset 2^A$ consisting of all $r$-separated subsets of $A$, with the order given by inclusion. Then, by Zorn's Lemma, $E_r$ contains a maximal element $S$. By the assumption, $S$ has to be finite (since $A$ contains no infinite separated subsets). In view of maximality of $S$, $$ A\subset \bigcup_{x\in S} B(x, 2r). $$
   Since $S$ is finite and $r>0$ is arbitrary, it follows that $A$ is totally bounded. qed