I'm trying to prove the cantor set $C$ is equivalent to the set of all numbers with ternary expansion of $2$'s and $0$'s. That is:
Let $A_0=[0,1]$. $A_n$ is defined to equal $A_{n-1}$ with it's middle third removed. let $C=\bigcap_{n\in\mathbb{N}} A_n$.
Prove $C =\{x=0.a_1a_2a_3…| a_n\in\{0,2\} \text{ for all }n\in\mathbb{N} \} $ where the decimal expansion is in base $3$.
Let $x=0.a_1a_2a_3…$ $\space$ be a ternary expansion of $x$
My strategy is this:
Prove that $a_n=1 \iff x\notin A_n$
(a) Proceed by induction on $n$. For $n=1$ we have:
$a_1 = 1 \iff \frac{1}{3} \le x < \frac{2}{3} \iff x \notin A_1$
Assume it's true for $n$ that is:$\space a_n = 1 \iff x \notin A_n$.
For $n+1$ we have: $a_{n+1}=1 \iff \text{there's some integer $m$ such that:} \frac{1}{3} \le 3^{n}x – m < \frac{2}{3} $
Furthermore $m = a_1a_2a_3…a_{n}$ (in base 3 digit expansion).
Here i got stuck.
Best Answer
Let
$F_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$
$F_2 = [0,\frac{1}{3^2}] \cup [\frac{2}{3^2},\frac{3}{3^2}] \cup [\frac{6}{3^2},\frac{7}{3^2}] \cup [\frac{8}{3^2},1]$
and so on. The Cantor set is then $C=\bigcap^{\infty}_{k=1} F_k$.
Each $x \in C$ can be written in the form $ x = \frac{a_0}{3} + \frac{a_1}{3^2} + \frac{a_2}{3^3} + ...+ \frac{a_n}{3^{n+1}}+ ...$
By Construction, if $x\in F_k$ and $0 \leq j < k$, then $3^jx=a+y$ where $a \in \mathbb{N} \cup \{0\}$ (*) and $y \in F_{k-j}$.
Noting this pattern suppose $x\in C$ therefore $x \in F_k$ for all $k \in \mathbb{N}$. Now assume that for some $m\in \mathbb{N}$ we have $a_m = 1$ Then we have $3^mx=a+\frac{1}{3}+\frac{a_{m+1}}{3^2}+\frac{a_{m+2}}{3^3}+...\ \ \ \Leftrightarrow$ $ \ \ \ \ y=3^mx-a \notin F_1 \Leftrightarrow x \notin F_{m+1} \Leftrightarrow x \notin C$ contradiction.
To see the pattern more clearly, consider $F_3$,
$F_3= [0,\frac{1}{27}] \cup [\frac{2}{27},\frac{3}{27}] \cup [\frac{6}{27},\frac{7}{27}]\cup [\frac{8}{27},\frac{9}{27}]\cup [\frac{18}{27},\frac{19}{27}]\cup [\frac{20}{27},\frac{21}{27}] \cup [\frac{24}{27},\frac{25}{27}] \cup [\frac{26}{27},1]$
Define the "multiplication" of an interval by an scalar, naturally by multiplying the endpoints by that scalar, for example $3F_2= [0,\frac{3}{9}] \cup [\frac{6}{9},1] \cup [\frac{18}{9},\frac{21}{9}] \cup [\frac{24}{9},3]=[0,\frac{1}{3}] \cup [\frac{2}{3},1] \cup [2,\frac{7}{3}] \cup [\frac{8}{3},3]$
This multiplication is important, since it gives us 2 copies of $F_1$, (the 2nd copy is a translation of $F_1$ by an integer, look at (*) )
More generally, $3^jF_k$ gives us $2^j$ copies of $F_{k-j}$
let $x \in F_3$
For $j=0 \ \ \ $, the statement is clear, i.e. $3^0x \in F_{3-0} \Rightarrow x \in F_3$ Trivial !
For $j=1 \ \ \ $Obviously $3^1x \in 3F_3$, i.e. $3^1x$ lies in a translated (by a number $a \in \mathbb{N} \cup \{0\}$) copy of $F_{3-1}=F_2$ Therefore $3x=a+y$ where a is the natural number of translation, and $y\in F_2$
For $j=2 \ \ \ $ Similarly $3^2x \in 3^2F_3 \Rightarrow 3^2x $ lies in a translated copy of $F_{3-2}=F_1$