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I'm working on a problem and I don't think my solution is correct. The problem is, "Prove that if $x\in C$ where $C$ is the Cantor set then $x=\sum_{k=1}^{\infty}a_k3^{-k}$, where each $a_k \in \{0,2\}$". My solution goes as follows, any $x\in[0,1]$ has the expansion $x=\sum_{k=1}^{\infty}a_k3^{-k}$ where $a_k\in\{0,1,2\}$ assume that $x\in C$ and $a_k\neq0,2$. Then $x=\frac{1}{2}\not\in C$. I don't think this is correct, any suggestions? Thanks.

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  • $\begingroup$ When you construct the Cantor set by deleting middle thirds, what do all the points in the first deletion have in common. Hint: Look at their ternary representations. $\endgroup$ – Eric Towers Nov 5 '16 at 21:41
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You can't conclude that $x=\frac{1}{2}$ because all you know is that $a_k=1$ for at least one $k$.

Instead, try showing that the intervals removed at each step of the Cantor set construction correspond exactly to the real numbers in $[0,1]$ having a $1$ in their ternary expansion at that position.

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