Cantor Space is Dense-in-itself

Theorem

Let $T = \left({\mathcal C, \tau_d}\right)$ be the Cantor space.

Then $T$ is dense-in-itself.

Proof

Let $U \in \tau_d$ be open in $T$.

Let $p \in U$.

Then $\exists x \in U: \exists \epsilon \in \R: d \left({x, p}\right) < \epsilon$.

Hence the result.

$\blacksquare$