Sequentially Compact Metric Space is Complete

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

Then $M$ is complete.

Proof
Let $M = \left({A, d}\right)$ be compact.

From Sequence of Implications of Metric Space Compactness Properties we have that $M$ is sequentially compact.

Now let $\left \langle {a_k}\right \rangle$ be a Cauchy sequence in $A$.

As $M$ is sequentially compact, $\left \langle {a_k}\right \rangle$ has a subsequence which converges to a point in $M$.

Since the sequence is Cauchy, this implies that the entire sequence has limit $a$.

So $\left({A, d}\right)$ is complete, by definition.