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.
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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: $3$