I'm trying to prove that the Cantor set $\mathcal{C}$ contains all numbers $x \in [0,1]$ with ternary expansion $x = \sum_{k=1}^\infty \frac{a_k}{3^k}$, such that $a_k=0$ or $a_k=2$. I'm going by induction, proving that $x$ belongs to every $\mathcal{C}_k$, where $\mathcal{C} = \bigcap_{k=1}^\infty \mathcal{C}_k$, and $\mathcal{C}_k$ is the usual $k$-th set in the construction of the Cantor set (i.e., a disjoint union of $2^k$ closed intervals, each of length $\frac{1}{3^k}$, etc.).
Here's my work so far:
The base case, $k=1$, was easy: If $a_1=0$, then the geometric sum yields $x \leq \frac{1}{3}$; similarly, if $a_1=2$, then $x \geq \frac{2}{3}$.
For the inductive step, I assume $x \in \mathcal{C}_k$, and try to prove $x \in \mathcal{C}_{k+1}$. First, I assume there is an interval $[a,b]$ of length $\frac{1}{3^k}$ containing $x$, where $[a,b]$ is one of the $2^k$ intervals that make up $\mathcal{C}_k$. I can write $[a,b]$ as
$$ [a,b] = \left[a, a + \frac{1}{3^{k+1}} \right] \cup \left( a + \frac{1}{3^{k+1}}, b - \frac{1}{3^{k+1}} \right) \cup \left[ b- \frac{1}{3^{k+1}}, b \right], $$
and of course I'd like to show, for instance, that $x$ is in the leftmost of these intervals if $a_{k+1} = 0$. The geometric sum gives me the bound $x \leq \sum_{j=1}^k \frac{a_j}{3^j} + \frac{1}{3^{k+1}}$, but I want $x \leq a + \frac{1}{3^{k+1}}$.
I've tried and failed to get past here. I feel like I'm missing something silly, but can anybody help me reach finish off the argument?
Thanks!
You are very close indeed. What you are trying to prove follows from the fact that actually $$a = \sum_{j=1}^k \frac{a_j}{3^j}$$
You may want to strength your statement to include that $C_k$ contains intervals whose LHS endpoints are of that form. Then you could directly use it in your inductive step.