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I'm trying to show that if $x \in [0,1]$ has a ternary expansion consisting only of $0$'s and $2$'s, then $x$ is in the Cantor ternary set. The proofs I've seen typically rely on induction. Is it valid to simply argue that $x$ is removed in the $k$-th iteration of the construction of the Cantor set only if the $k$-th digit in all ternary expansions of $x$ is equal to 1?

For example, if $x \in (\frac{1}{3},\frac{2}{3})$, then all ternary expansions of $x$ must begin with 1.

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Yes, if you can show that the set of points removed at $k$th step are those with $1$ in the $k$th digit of (every) ternary expansion, then the conclusion follows.

But I imagine that a demonstration of the above will have an element of induction in it as well, since our knowledge of the position of the middle-third of a line segment depends on the position of the segment itself, which was itself obtained in this process, etc. It's still an inductive proof, reflecting the iterative nature of the construction of the Cantor set.

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