Cantor Space is Nowhere Dense

Theorem
Let $\left({\mathcal C, \tau_d}\right)$ be the Cantor set considered as a topological subspace of the real number space $\R$ under the Euclidean topology $\tau_d$.

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

Proof 1
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.

Proof 2
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.