Cantor Space is Nowhere Dense/Proof 2

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

Then $\mathcal C$ is nowhere dense in $\left[{0 \,.\,.\, 1}\right]$.

Proof
Let $\mathcal C_n$ denote the set $C_{n-1}$ with the middle open intervals of length $\frac{1}{3^{n}}$ removed from every one of the $2^{n-1}$ closed intervals, where $\mathcal C_0 = [0,1]$, and $\mathcal C_\infty = \mathcal C$.

Then the length of every interval in $C_n$ is $\frac{1}{3^n} = 3^{-n}$.

Let $0 \le a < b \le 1$.

Then $(a,b)\subset [0,1]$ is an open interval.

Let $n \in \N$ such that $3^{-n} < b-a$.

Then the length of every interval in $C_n$ is $3^{-n}<b-a$.

Therefore no interval of length $b-a$ exists in $\mathcal C_n$.

Therefore no interval of length $b-a$ exists in $\mathcal C = \mathcal C_\infty \subset \mathcal C_n$.

Since the interval $(a,b)$ was of arbitrary length, there do not exist any open intervals in $\mathcal C$.

Hence the result, by definition of nowhere dense.