Decomposition of the Cantor set.

Question

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!

Answer 1

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 $k$.

Answer 2

(1) is trivial, and really a statement about sequences of $0$'s and $1$'s: either there are finitely many $0$'s and then the sequence is eventually $1$ (after the last index that has a $0$), or there are finitely many $1$'s and the sequence is eventually $0$ (likewise), or there are infinitely many of both of them. There cannot be finitely many of both of them, as the sequence is an infinite one. So only that last option is ruled out, and the other three mutually exclusive options correspond to being in $K^{(1)}$, $K^{(2)}$ or $K^{(3)}$. No sums etc. are needed. Points are "coded" by sequences.