Representation of elements in Cantor set.

Question

Every number in $[0,1]$ has a ternary expansion $$x=\sum_{k=1}^{\infty}a_k3^{-k}$$ where $a_k=0,1,$ or $2$.Note that this decomposition is not unique since, for example, $1/3=\sum_{k=2}^{\infty}2/3^k$.Prove that $x\in \text{Cantor set}$ if and only if $x$ has a representation as above where every $a_k$ is either $0$ or $2$

I have tried some points in Cantor set. for example: $1/3=\sum_{k\geq 2} 2/3^k$ and $1/9=\sum_{k\geq 3} 2/3^k$,but I can't do it for all points in the Cantor set. I think I can't express a general point in that set. thanks very much

What is your definition of "Cantor set"? For me, it is exactly what you're trying to prove. — vadim123, Jun 08 '13 at 17:11
@vadim123 you can find the definition here: http://en.wikipedia.org/wiki/Cantor_set — Laura, Jun 08 '13 at 17:18
You will find the answer to your question on that same page. — vadim123, Jun 08 '13 at 17:19
@vadim123 I would disagree, on that page you do not see a prove but only an illustration. — zesy, Mar 12 '15 at 16:39

score 4 · Accepted Answer · answered Jun 08 '13 at 21:35

Suppose $a_n$ is $1$. We will divide our proof into cases.

Case $1$: If $a_k \ne 0$ for some $k > n$, then $x \in \left( \frac{3k+1}{3^n} , \frac{3k+2}{3^n} \right)$ for some $n$. But this set is taken out of the unit interval during the formation of the Cantor set, so $x$ is not in the Cantor set.

Case $2$: If $a_k = 0$ for all $k > n$, then $x = \sum_{k=1}^\infty b_k 3^{-k}$ with

$$b_k = \begin{cases} a_k & k < n \\ 0 & k = n \\ 2 & k > n \end{cases}$$

This follows because $0 \le x - \sum_{k=1}^N b_k 3^{-k} \le 3^{-N}$ (equality holds if $N \ge n$).

Representation of elements in Cantor set.

1 Answers1

Linked