equivalence of compactness and countably compactness

Question

Is there a way to prove that in metric spaces, compactness and countably compactness are equivalent, without using the Bolzano Weierstrass Property?

Answer 1

It’s a bit roundabout, but it can be done.

That compactness implies countable compactness is trivial, so assume that $\langle X,d\rangle$ is countably compact. For $\epsilon>0$ use Zorn’s lemma to get a maximal (with respect to $\subseteq$) $F_\epsilon\subseteq X$ with the property that for all $x,y\in F_\epsilon$, if $x\ne y$, then $d(x,y)\ge\epsilon$. Let $\mathscr{U}_\epsilon=\{B(x,\epsilon):x\in F_\epsilon\}$; maximality of $F_\epsilon$ ensures that $\mathscr{U}_\epsilon$ covers $X$. If $x,y\in F_\epsilon$ and $x\ne y$, then $y\ne B(x,\epsilon)$, so $\mathscr{U}_\epsilon$ is irreducible: it has no proper subcover. It also follows immediately that $F_\epsilon$ is a closed, discrete set in $X$. If $F_\epsilon$ is infinite, let $F_\epsilon'$ be a countably infinite subset of $F_\epsilon$; then $\{B(x,\epsilon:x\in F_\epsilon'\}\cup\{X\setminus F_\epsilon'\}$ is an irreducible countable open cover of $X$, contradicting countable compactness of $X$, so $F_\epsilon$ is finite.

Let $D=\bigcup_{n\in\Bbb N}F_{2^{-n}}$; clearly $D$ is a countable subset of $X$, and it’s easy to check that $D$ is dense in $X$. It’s straightforward to show that $\{B(x,2^{-n}):x\in D\text{ and }n\in\Bbb N\}$ is a countable base for $X$, so $X$ is second countable. A second countable space is trivially Lindelöf, and a countable compact Lindelöf space is trivially compact, so we’re done.

To summarize: countable compactness of the metric space $X$ implies totally boundededness, which implies separability, which implies second countability, which implies that $X$ is Lindelöf and therefore compact.