Cantor Space is Complete Metric Space

Theorem

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

Then $T$ is a complete metric space.

Proof

We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space.

We also have Cantor Space is Compact.

The result follows from Compact Metric Space is Complete.

$\blacksquare$