2

Background Facts

  1. It is known that every number in the Cantor set $\mathcal{C}$ has a ternary expansion, where $a_k \in \{0,2\}$:

$$ x=\sum_{k=1}^{\infty} a_{k} 3^{-k} $$

  1. 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?

J.G.
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Hephaes
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  • I might be wrong about this, but I believe the 2 $k$'s are not the same. – Jia Cheng Sun Mar 21 '21 at 10:31
  • So just understand it as this: if $a_k\neq 0$ for some $k>n$, then $\exists j, x\in (\frac{3j+1}{3^n}, \frac{3j+2}{3^n})$. The $j$ isn't important, you just need to bound the "tail" of $x$ after the $n$-th digit (inclusive) between $\frac{1}{3^n}$ and $\frac{2}{3^n}$. – Jia Cheng Sun Mar 21 '21 at 10:33
  • There are very many Q's and A's about the Cantor set on this site. I answered but you should not be surprised if your Q gets tagged as a duplicate by a moderator. – DanielWainfleet Mar 21 '21 at 12:58

1 Answers1

1

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 $D\subseteq E\subseteq D.$

DanielWainfleet
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  • I expect the reader can verify my assertion in (2) that {$3^nx$}$\in (1/3,2/3)$. – DanielWainfleet Mar 21 '21 at 13:24
  • Thanks for the comprehensive answer. I now supplement the verification:

    $(3k+1)/3^{n+1}<x$ implies $(3^{n+1}k+3^{n})/3^{n+1}<3^{n}x$. And thus $k+\frac{1}{3}<3^{n}x$. We have $0\leq k$ and thus $k+\frac{1}{3}<3^{n}x+k\iff\frac{1}{3}<3^{n}x$.

    We also recognise that $3^{n}x<2/3;\forall n>0\iff3^{n+1}x<2\iff x<\frac{2}{3^{n+1}}$. Since $\frac{k}{3^{n}}>0$ for any $n$, and since $x<(3k+2)/3^{n+1}$, we have $\frac{1}{3}<3^{n}x<\frac{2}{3}$.

     – Hephaes Mar 21 '21 at 15:10
  • @Hephaes. Precisely. Although in Latex you have to keep { and } outside the dollar signs or they disappear, as in your last formula. – DanielWainfleet Mar 22 '21 at 17:43