Countably Compact Metric Space is Compact

Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Then $M$ is countably compact iff $M$ is compact.

Proof 1

We have from Compact Space is Countably Compact that compactness implies countable compactness in all topological spaces regardless of whether they are metric spaces or not.

So all we need to do is demonstrate that if $M$ is countably compact then it is compact.

We have that a countably compact metric space is sequentially compact.

Then we have that a sequentially compact metric space is totally bounded.

A totally bounded metric space is second-countable.

The result follows from the fact that in a second-countable space, compactness is equivalent to countable compactness.

$\blacksquare$

Proof 2

We have from Compact Space is Countably Compact that compactness implies countable compactness in all topological spaces regardless of whether they are metric spaces or not.

So all we need to do is demonstrate that if $M$ is countably compact then it is compact.

We have that a Countably Compact Metric Space is Sequentially Compact.

Then we have that a Sequentially Compact Metric Space is Separable.

For each $n$, a metric space which is countably compact can be covered by finitely many open $\epsilon$-ball neighborhoods : $N_{1/n} \left({x_i}\right)$.

So $\left\{{x_i}\right\}$ is a dense subset of $A$ which is countable.

So if a metric space is countably compact it is by definition second-countable.

The result follows from Second-Countable Space: Compactness Equivalent to Countable Compactness.

$\blacksquare$