Cantor Space is Nowhere Dense/Proof 1

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

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

Proof
From Cantor Set Closed in Real Number Space, $\mathcal C$ is closed.

So from Closed Set Equals its Closure:
 * $\mathcal C^- = \mathcal C$

where $\mathcal C^-$ denotes the closure of $\mathcal C$.

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

Then $I = \left({a \,.\,.\, b}\right)$ is an open interval of $\left[{0 \,.\,.\, 1}\right]$.

Let $\epsilon = b - a$.

Clearly $\epsilon > 0$.

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

So there exists an open interval of $\left[{0 \,.\,.\, 1}\right]$ which has been deleted from $\left[{0 \,.\,.\, 1}\right]$ during the process of creating $\mathcal C$.

Thus no open interval of $\left[{0 \,.\,.\, 1}\right]$ is disjoint from all the open intervals deleted from $\left[{0 \,.\,.\, 1}\right]$.

So any open interval of $\left[{0 \,.\,.\, 1}\right]$ can not be a subset of $\mathcal C = \mathcal C^-$.

Hence the result, by definition of nowhere dense.