I am thinking about a proof of the following statement:
"Every open interval of $\mathbb{R}$ contains either infinitely many or no elements in the Cantor set"
and this is what I have thought:
Let $(a,b),\ a,b\in\mathbb{R}, a<b$ be an open interval of $\mathbb{R}$ and define the Cantor set $\mathcal{C}:=[0,1]\setminus\bigcup_{n=1}^{\infty}G_n$, where $G_1=(\frac{1}{3}, \frac{2}{3})$ and $G_n$ for $n>1$ is
the union of the middle-third open intervals in the intervals of $[0, 1]\setminus (\bigcup_{j=1}^{n-1}G_j)$
By the definition of the Cantor set it is clear that if $(a,b)\cap [0,1]=\emptyset$ or $(a,b)$ is or is contained in a middle-third open interval in $[0,1]$ then it contains none of its elements.
Suppose now $(a,b)\cap [0,1]\neq\emptyset$ and $(a,b)$ is not nor is it contained in a middle-third open interval in $[0,1]$:
beginning of hand-wavy argument
then, since as $n$ increases the length of each middle-third open interval decreases (every such interval has length $\frac{1}{3^n}$) there will be, for $n$ large enough, an infinite number of middle-third open intervals in $(a,b)\cap [0,1]$ and since the endpoints of the middle-third open intervals are elements of $\mathcal{C}$ the claim follows.
Now, the last part of this argument is clearly non-rigorous, but I think I have got at least the idea of the proof right so I would appreciate an hint about how to rigorously prove that, in the last part of the proof, $(a,b)$ must contain an infinite number of middle-third intervals, thanks.
EDIT: I have proved the statement by following the advice of the user Joe in the comments below – comment about the proof are welcome
We first prove that the Cantor set contains no isolated points.
We can rewrite the Cantor set as $\mathcal{C}=[0,1]\setminus\bigcup_{n=1}^{\infty}G_n=\bigcap_{n=1}^{\infty}([0,1]\setminus G_n)=\bigcap_{n=1}^{\infty}\mathcal{C}_n$, where $\mathcal{C}_n:=[0,1]\setminus G_n$: then $\mathcal{C}_1=[0,\frac{1}{3}]\cup [\frac{2}{3},1]$, $\mathcal{C}_2=[0,\frac{1}{9}]\cup [\frac{2}{9},\frac{1}{3}]\cup [\frac{2}{3},\frac{7}{9}]\cup [\frac{8}{9},1],\dots$ and note that, being closed, the endpoints of every interval in $\mathcal{C}_n,\ n\geq 1$ belong to $\mathcal{C}_n$ and since they are not affected by the removal of middle-third intervals in the various steps the endpoints of these intervals belong to all the $\mathcal{C}_n$ hence to $\mathcal{C}$.
Let $x\in\mathcal{C}$: then as we said for every $n\geq 1$ there exists $x_n\in\mathcal{C}\cap\mathcal{C}_n=\mathcal{C},\ x_n\neq x$ such that $|x-x_n|\leq\frac{1}{3^n}$, namely one of the two endpoints of $\mathcal{C}_n$, so we can build a sequence $(x_n)_{n=1}^{\infty}$ of elements of $\mathcal{C}$, all distinct from $x$, and converging to $x$; this means that no $x\in\mathcal{C}$ is an isolated point, as desired.
Let now $(a,b),\ a,b\in\mathbb{R}, a<b$ be an open interval of $\mathbb{R}$.
By the definition of the Cantor set it is clear that if $(a,b)\cap [0,1]=\emptyset$ or $(a,b)$ is or is contained in a middle-third open interval in $[0,1]$ (i.e. $(a,b)\subset\bigcup_{n=1}^{\infty}G_n$) then it contains none of its elements.
Suppose now $(a,b)\cap [0,1]\neq\emptyset$ and $(a,b)$ is not nor is it contained in a middle-third open interval in $[0,1]$: then there must be $x\in (a,b)\cap\mathcal{C}$ so there is some $r>0$ such that $(x-r,x+r)\subset (a,b)$. Since $x$ is not an isolated point for every $n>\log_3(\frac{1}{3})+1$ there exists $x_n\in\mathcal{C}$ such that $|x-x_n|<\frac{1}{3^n}<r$ i.e. such that $x_n\in (x-r,x+r)\subset (a,b)$ so $(a,b)$ contains infinitely many elements of the Cantor set, as desired. $\square$
Best Answer
Your hand-wavy argument is definitely on the right track. A couple things to think about that can make it less hand-wavy are: