If $x$ has a ternary expansion $\sum \limits_{k=1}^{\infty}\dfrac{c_k}{3^k}$ where each $c_k\in \{0,2\}$ then $x$ belongs to Cantor set.

Proof: Suppose $x$ has a ternary expansion $\sum \limits_{k=1}^{\infty}\dfrac{c_k}{3^k}$ where each $c_k\in \{0,2\}$. We will show $x\in C$ by induction. Clearly $x\in C_0=[0,1]$ since $0\leqslant x \leqslant 1$. Next, if $c_1=0$, then $$x\leqslant (0.0222\dots)_3=(0.1)_3=\frac{1}{3}$$ On the other hand, if $c_1=2$, then $$x\geqslant (0.2000\dots)_3=(0.2)_3=\frac{2}{3},$$and hence, $x\in C_1$. Now we approach the inductive step.

Assume $x\in C_k$ for some $k\in \mathbb{N}$. Then $x$ is in any of the $2^k$ subintervals of $C_k$, each of the length $\frac{1}{3^k}$. WLOG, say $x\in [a,b]$. Since we remove the interior of the middle-thirds, we have two disjoint closed intervals, $$\left[a, a+\frac{1}{3^{k+1}}\right] \quad\text{and} \quad \left[b-\frac{1}{3^{k+1}}, b\right]$$If $c_{k+1}=0$, then $x=\sum \limits_{n=1}^{\infty}\dfrac{c_n}{3^n}\leqslant \sum \limits_{n=1}^{k}\dfrac{c_n}{3^n}+\dfrac{1}{3^{k+1}}$ and $x\in [a,b]$.

Why in this case the right side of inequality is $\leqslant a+\dfrac{1}{3^{k+1}}$?

  • $\begingroup$ Because $\sum\limits_{n=k+2}^\infty\frac2{3^n}=\frac1{3^{k+1}}$. $\endgroup$ – Did Aug 31 '15 at 20:12
  • $\begingroup$ @Did, I know this. I am asking why $\sum \limits_{n=1}^{k}\dfrac{c_n}{3^n}\leqslant a$ $\endgroup$ – ZFR Aug 31 '15 at 20:21
  • 2
    $\begingroup$ Actually, by construction, $\sum\limits_{n=1}^k\frac{c_n}{3^n}=a$... $\endgroup$ – Did Aug 31 '15 at 20:25
  • $\begingroup$ Is it also possible to argue that $x$ is removed in the $k$-th iteration only if $\alpha_k = 1$ for all ternary expansions of $x$? $\endgroup$ – David Nov 5 '15 at 19:37

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