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 $T_\mathbb{N}$?