Completely Metrizable Topology is Metrizable Topology

Theorem

Let $\struct {S, \tau}$ be a completely metrizable topological space.

Then, $\struct {S, \tau}$ is a metrizable topological space.

Proof

Since $\struct {S, \tau}$ is completely metrizable, $\tau$ is induced by a (complete) metric $d$.

In particular, $\struct {S, d}$ is a metric space.

As $\struct {S, \tau}$ is homeomorphic to itself by the identity mapping, it is metrizable.

$\blacksquare$