Hilbert Cube is Separable

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

Then $M$ is a separable space.


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.

The result follows by definition of separable space.