# Hilbert Cube is Metric Space

## Theorem

Let $M = \left({I^\omega, d_2}\right)$ be the Hilbert cube.

Then $M$ is a metric space.

## Proof

As defined, $M$ is a subspace of the Hilbert sequence space $\ell^2$.

We have that Hilbert Sequence Space is Metric Space.

The result follows from Subspace of Metric Space is Metric Space.

$\blacksquare$