# Hilbert Cube is Separable

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## 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: $3$