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.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces