Background Facts
- It is known that every number in the Cantor set $\mathcal{C}$ has a ternary expansion,
where $a_k \in \{0,2\}$:
$$
x=\sum_{k=1}^{\infty} a_{k} 3^{-k}
$$
- Cantor set has the representation
$$\mathcal{C}=[0,1] \backslash \bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^{n}-1}\left(\frac{3 k+1}{3^{n+1}}, \frac{3 k+2}{3^{n+1}}\right)$$
Problem
What exactly is the relationship between (2.) and (1.)? That is, how should one go about proving that
$$ x \in \mathcal{C} \iff x=\sum_{k=1}^{\infty} a_{k} 3^{-k}\, \,\,\,\,\,\,,a_k \in \{0,2\}$$
Progress
Following the proof which appears in Representation of elements in Cantor set., I have problem understanding why if $a_{k} \neq 0$ for some $k>n,$ then $x \in\left(\frac{3 k+1}{3^{n}}, \frac{3 k+2}{3^{n}}\right)$ for some $n .$
I can see that $x \geq \frac{1}{3^n} + \frac{1}{3^k}$, but how can one prove $x$ indeed lies in the given interval?
Best Answer
Definition. Let $[x]$ be the largest integer not exceeding $x$ and let {$x$}$=x-[x].$
Def'n. Let $D =[0,1]\setminus C,$ according to your representation 2. of the Cantor set.
Def'n. Let $E$ be the set of all $x\in [0,1]$ that $must$ have a digit $1$ in ternary.
(1). Show that $(1/3,2/3)\subset E.$ (Easy. The digit immediately right of the "decimal" point $must$ be $1.$)
(2). If $n>0$ and $0\le k\le 3^n-1$ and if $(3k+1)/3^{n+1}<x<(3k+2)/3^{n+1}$ then $x\in E. $ Because if $x\not\in E$ then {$3^nx$}$\not\in E.$ But {$3^nx$} $\in (1/3,2/3)$ and we have $(1/3,2/3)\subset E$ by (1).
(3). By (1) and (2) we have $D\subseteq E.$
(4). If $x\in E$ let $x=\sum_{n=0}^{\infty}x_n/3^{n+1},$ where each $x_n\in \{0,1,2\}.$ Now let $n^*$ be the least $n$ such that $x_n=1.$ It cannot be that $x_n=0$ for every $n>n^*,$ nor that $x_n=2$ for every $n>n^*,$ otherwise $x$ would have another ternary presentation without the digit $1.$ So $$(\bullet)\quad 0<x-\sum_{n=0}^{n^*}x_n/3^{n+1}<1/3^{n^*+1},\text { and also }x_{n^*}=1$$
Now let $k=3^{n^*}\sum_{n=0}^{n^*-1}x_n/3^{n+1}$ if $n^*>0$, or let $k=0$ if $n^*=0$. By $(\bullet)$ we have $0\le k<3^{n^*}$ and also $0<x-(3k+1)/3^{n^*+1}<1/3^{n^*+1},$ implying $x\in D.$
So $E\subseteq D.$
(5). From (3) and (4) we have $D\subseteq E\subseteq D.$