Prove that a metrizable space is countably compact iff it is compact.

Question

Prove that a metrizable space is countably compact iff it is compact.

($\Rightarrow$)

I let $\{O_i\}$ be a countable open cover for $(X,T)$ with a finite subcover. Let $\{U_i\}$ be an uncountable open cover (because otherwise it's countable and then trivial). From here I'm stuck. I figure that I can add an uncountable number of additional open balls to the countable set to make in an uncountable set or select a countable subset of the uncountable set, but I'm not sure how I'd do this. To prove this without sequential compactness is what I'm looking for.

Answer 1

I don’t know a direct argument along the lines that you’ve tried to start. I would prove it in steps, as follows.

• Show that if $X$ is countably compact, then $X$ does not have an infinite, closed, discrete subset.
• Show that for each $n\in\Bbb Z^+$ there is a finite $F_n\subseteq X$ such that $\left\{B\left(x,\frac1n\right):x\in F_n\right\}$ covers $X$.
• Show that $\left\{B\left(x,\frac1n\right):n\in\Bbb Z^+\text{ and }x\in F_n\right\}$ is a countable open base for $X$.
• Prove that in fact every open cover of $X$ has a countable subcover (i.e., $X$ is Lindelöf).
• Conclude that $X$ is compact.

I’ll leave it at that for now in case you’d like to try to fill in the details.