Hilbert Cube is Separable

Theorem

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.

Then $M$ is a separable space.

Proof

Consider the set $H$ of all points of $M$ which have finitely many rational coordinates and all the rest zero.

Then $H$ forms a countable subset of $A$ which is (everywhere) dense.

This needs considerable tedious hard slog to complete it. In particular: Demonstrate that it is (everywhere) dense. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

The result follows by definition of separable 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: $3$