Hilbert Cube is Metric Space

Theorem

Let $M = \struct {I^\omega, d_2}$ 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.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $2$