Let $K$ the Cantor set. I have already shown the following result:
Proposition. The Cantor set consists of the numbers $x\in [0,1]$ having ternary development $$\sum_{k=1}^{\infty}\frac{a_k}{3^k},$$
with $a_k=0$ or $a_k=2.$
Note that the previous proposition provides the following characterization $$x\in K\iff x=\sum_{k=1}^\infty\frac{2\epsilon_k}{3^k},\quad\epsilon_k=0\:\text{or}\;\epsilon_k=1\;\text{for all}\;k\in\mathbb{N}, $$ therefore $$K=K^{(1)}\cup K^{(2)}\cup K^{(3)}\tag1,$$
where $$K^{(1)}:=\{x\in K\;|\; \epsilon_k=0\;\text{eventually}\},$$
$$K^{(2)}:=\{x\in K\;|\; \epsilon_k=1\;\text{eventually}\},$$
$$K^{(3)}:=\{x\in K\;|\; \epsilon_k=0\;\text{for infinitely many }\;k\;,\epsilon_k=1\;\text{for infinitely many }\;k\;\}.$$
I would like to show $(1)$.
Clearly, $$K^{(1)}\cup K^{(2)}\cup K^{(3)}\subseteq K.$$We show the other inclusion: let $$x\in K\Rightarrow x=\sum_{k=1}^{\infty}\frac{2\epsilon_k}{3^k},\quad\epsilon_k=0\;\text{or}\; \epsilon_k=1\;\text{for all}\;k\in\mathbb{N}.$$ Then,
\begin{split}\lim_{k\to\infty}\frac{2\epsilon_k}{3^k}=0\Rightarrow& \lim_{k\to\infty}\frac{\epsilon_k}{3^k}=0\\
\Rightarrow& \text{for all}\;\varepsilon > 0\;\text{exists}\; n_0\in\mathbb{N}\;\text{such that}\; \frac{\epsilon_k}{3^k}< \varepsilon\;\text{for all}\; k> n_0.\end{split}
Now, if $\epsilon_k=0$ finally, that is if exists $n\in\mathbb{N}$ such that $\epsilon_k=0$ for $k>n$ then $$\frac{\epsilon_k}{3^k}<\varepsilon\;\text{for all}\; k>n,$$ therefore $x\in K^{(1)}.$
If $\epsilon_k=1$ finally, that is if exists $n\in\mathbb{N}$ such that $\epsilon_k=1$ for $k>n$ then $$\frac{1}{3^k}<\varepsilon\;\text{for all}\; k>n=\log_3\bigg(\frac{1}{\varepsilon}\bigg),$$ therefore $x\in K^{(2)}.$
Question. Why should $x$ belong to $K^{(3)}$?
Thanks!
Best Answer
If $x =\sum \frac {2\epsilon_k} {3^{k}}$ and if it is not true that $\epsilon_k=0$ eventually and it is also not true that $\epsilon_k=1$ eventually then it must be true that $\epsilon_k=0$ for infinitely many $k$ and $\epsilon_k=1$ for infinitely many $k$.