Complement of Middle-Third Cantor set is Dense in $[0,1]$

Question

Does anyone have a simple proof to show that the complement of the Cantor Set ($C$) is dense in $[0,1]$?

Let $x,y \in [0,1]$. Suppose $x<y$.

Suppose $y\in[0,1]-C$

Since $[0,1]-C$ is open $\exists \delta$ s.t $ (y-\delta,y+\delta) \subseteq [0,1]-C $

Hence, we can choose $\delta_1$ small enough i.e $\delta_1 \le min\{\frac{y-x}{2},\delta\}$

Therefore, $\exists z\in (y-\delta_1,y+\delta_1)$ s.t $x<z<y$

However, I'm stuck with the case when $y\in C$.

Any hints or solutions?

Edit:

I might have another solution.

Since $x<y$. By Density of Rationals in Reals: $\exists q\in Q$ $ (x<q<y)$

If $q\in [0,1]-C $ then we are done.

Suppose $q\in C$. Hence $q$ is a limit-point of $[0,1]-C$.

Let $\delta=min\{\frac{y-q}{2},\frac{q-x}{2}\}$

Therefore $\exists z\in(q-\delta,q+\delta) \land z\in [0,1]-C$ s.t $(x<z<y)$

Answer 1

The only thing you've used about $C$ thus far is that it is closed, so one would expect the case with $y\in C$ to involve the construction of $C$. Indeed it does. In fact, by looking at the construction of $C$, we can avoid arguing by cases.

Recall that $C$ is the intersection of a nested sequence of sets $A_n$, where each $A_n$ is a disjoint union of closed intervals of length $1/3^n$. Suppose $x,y\in[0,1]$ such that $x<y$. Find $n\in\mathbb{N}$ such that $y-x>1/3^n$. Thus no closed interval of length $1/3^n$ contains $(x,y)$, so $(x,y)\not\subseteq A_n$. Hence $(x,y)\not\subseteq C$, so that there is some $r\in(x,y)$ such that $r\not\in C$. Therefore the complement of $C$ is dense in $[0,1]$.

Answer 2

It suffices to show that the Cantor set is nowhere dense in $[0, 1]$ since a closed subset of a metric space is nowhere dense $\iff$ its complement is dense in the space.

Suppose the the contrary, that the Cantor set was not nowhere dense. Let $C^{o}$ and $\overline{C}$ denote the interior and closure of Cantor set $C$ respectively. Then since $\overline{C} = C$ and by our assumption,

$\Rightarrow C^{o} \neq \emptyset$

$\Rightarrow \exists \, x\in C$ and $\varepsilon >0$ such that $(x- \varepsilon, x+\varepsilon) \subset C$ contradicting the fact that $C$ does not contain an interval.

Therefore since $C$ is a closed, nowhere dense subset of $[0, 1]$ it follows that its complement is dense in $[0,1].$