# Induction, Cantor Set, Ternary representation

Cantor Set defined by sequence

Let $C$ be a Cantor Set. I'm trying to show that "$x\in C ⇒$ There exists a ternary representation (base $3$) of $x$ such that every term is $0$ or $2$."

Here's my argument following Ayman's advice.

Fix $x\in C$. Let $A=$ {$n\in \mathbb{N}$ | There exists a ternary representation of $x$, with base $3$, such that for all $m≦n$, $m$th decimal is not $1$}.

I used induction and it worked.

Thus, for each $n\in \mathbb{N}$, existence of ternary representation of $x$ such that 'for all $m≦n$, $m$th decimal is not $1$', is guranteed.

Still, i cannot show the existence of ternary representation of $x$ such that every decimal is not $1$.

What am I misunderstanding here?

Let's say $T_n$ is a ternary representation of $x$ such that there is no $1$ till $n$th term. (I'm using AC$_\omega$ here.)

Then $T_n$ and $T_m$ are maybe different.

Likewise, how come existence of $T_1, T_2, ...$ gurantees existence of $T_\mathbb{N}$?

-
Your exposition is not clear. Are you trying to prove that every real number has a ternary presentation with no 1 in it? I hope not --- that can't be done. It's only numbers in the Cantor set that have that kind of representation. –  Gerry Myerson Oct 7 '12 at 23:05
@Gerry I fixed $x$ in Cantor Set without any notion. Edited! –  Katlus Oct 7 '12 at 23:12

You need to prove a little bit more about the representations $T_n$. Specifically, you want to show that if $m<n$, then $T_n$ agrees with $T_m$ on the first $m$ places. Then you can define the $n$-th digit of $T_{\Bbb N}$ to be the $n$-th digit of $T_n$ (which is also the $n$-th digit of $T_m$ for every $m\ge n$).