Metric Spaces Are Compact Spaces iff They're Countably Compact

# Metric Spaces Are Compact Spaces If and Only If They're Countably Compact

Recall from The Lebesgue Number Lemma page that if $(X, d)$ is a metric space that is also a BW space then for every open cover $\mathcal F$ of $X$ there exists an $\epsilon > 0$ called a Lebesgue number such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$.

We used this very important lemma to prove a very nice result on the Metric Spaces Are Compact Spaces If and Only If They're BW Spaces page. We proved that if $(X, d)$ is a metric space that $X$ is compact if and only if $X$ is a BW space. The first direction of this statement is somewhat trivial as we have already seen that a compact space $X$ is a BW space from the Compact Spaces as BW Spaces page, however, the converse of this result is very useful nevertheless.

We will now look at a nice consequence of these results. We will see that a metric space $X$ is compact if and only if $X$ is countably compact. Therefore, in a metric space $X$, the concept of compactness and countably compactness are in essence the same.

 Theorem 1: Let $X$ be a metric space. Then $X$ is compact if and only if $X$ is countably compact.
• Proof: $\Rightarrow$ Let $X$ be a compact metric space. Then trivially, $X$ is also countably compact (since every open cover $\mathcal F$ of $X$ has a finite subcover $\mathcal F^*$ and a finite number is countable).
• Since $X$ is a BW metric space, we have from the theorem referenced above that then $X$ is a compact metric space. $\blacksquare$