Cantor Set and Base 3 Decimal Expansions

Question

I'm trying to show that every point in the Cantor Set (obtained by "middle-thirds" removal, starting with $[0,1]$) has a base 3 decimal expansion consisting of only zeros and twos. I think the proof would be by strong induction, but I'm not sure where to start.

Also, is the converse true? That is, is it true that every number of the form $0.a_0a_1a_2...$ (in base 3) with each $a_i$ being $0$ or $2$ is an element of the Cantor Set? I tried constructing a counterexample but haven't had any luck and I'm beginning to think it's true. The question is, how would one prove this?

Thanks in advance!

Answer 1

You can get close to both directions in one blow by proving that the intervals you remove in step $n$ of the construction of the the Cantor set are exactly those numbers whose base-3 expansion has its first 1 at position $n$ after the point.

(A number that has two base-3 expansion will be removed in step $n$ if both expansions have their first 1 at position $n$).

From this you can conclude that (a) any number that has an expansion consisting only of 0 and 2 cannot be removed, and (b) any number that doesn't have such an expansion will be removed in one of the stages.