Cantor Space is Totally Separated

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

Then $\mathcal C$ is totally separated.

Proof
Let $a, b \in \mathcal C$ such that $a < b$.

Then $b - a = \epsilon$.

Consider $n \in \N$ such that $3^{-n} < \epsilon$.

So $\exists r \in \R: a < r < b, r \notin \mathcal C$.

Let $A = \mathcal C \cap \left[{0 .. r}\right)$ and $B = \mathcal C \cap \left({r .. 1}\right]$.

Thus $A \mid B$ is a partition of $\mathcal C$ such that $a \in A, b \in B$.

Such a partition can be found for any two distinct $a, b \in \mathcal C$.

Hence the result by definition of totally separated.