Let $J_{0,1}:=[0,1]$.
Let's start with the construction of the Cantor set.
Step 1. We remove the central open interval $I_{0,1}=\big(\frac{1}{3},\frac{2}{3}\big)$. We denote with $J_{1,1}:=\big[0,\frac{1}{3}\big]$ and with $J_{1,2}:=\big[\frac{2}{3},1\big]$.
We define $K_1:=J_{1,1}\cup J_{1,2}$.
Step 2. We remove from $J_{1,1}$ and $J_{1,2}$ the central open interval of length $\frac{1}{9}$. Then we define what remains with $K_2:=J_{2,1}\cup J_{2,2}\cup J_{2,3}\cup J_{2,4}$, where $J_{2,1}=\big[0,\frac{1}{9}\big]$, $J_{2,2}=\big[\frac{2}{9},\frac{1}{3}\big]$, $J_{2,3}=\big[\frac{2}{3},\frac{7}{9}\big]$, $J_{2,4}=\big[\frac{8}{9}, 1\big]$.
At the end of step $n$ we will have $2^n$ close interval $J_{n,k}$ for $k=1,\cdots, 2^n$ of length $1/3^n.$
For all $n\in\mathbb{N}$ we define $$K_{n}:=\bigcup_{k=1}^{2^n} J_{n,k}$$
We define the Cantor set $K$ in the follow way $$K:=\bigcap_{n=1}^{+\infty} K_n.$$
I have already shown that: $$\text{If}\quad x\in K\Rightarrow x=\sum_{k=1}^{+\infty}\frac{2\varepsilon_k}{3^k}\quad\text{where}\quad \{\varepsilon_k\}\subseteq\{0,1\}^{\mathbb{N}}$$
Problem We suppose that exists $\overline{n}\in\mathbb{N}$ such that for all $k>\overline{n}$ $\varepsilon_k=0$ then, $$x=\sum_{k=1}^{\overline{n}}\frac{2\varepsilon_k}{3^k}\quad\text{where}\quad \{\varepsilon_k\}\subseteq\{0,1\}^{\mathbb{N}}$$
We can say that $x\in J_{\overline{n},k}$ where $k=1,2,\dots,2^{\overline{n}}$.
My book said that, in this case, $x\in J_{\overline{n},\overline{k}}$, where
$$\overline{k}=\bigg(\sum_{k=1}^{\overline{n}}\varepsilon_k2^{\overline{n}-k}\bigg)+1\tag1$$
Question. Where can I deduce the expression $(1)$?
Thanks!
Best Answer
The tool that unlocks a lot of powerful intuition about this type of constructions is called coding. You can find a very thorough introduction (together with its applications to the Cantor set and other dynamical systems settings) in Chapter 7 of A First Course in Dynamics, by Hasselblatt and Katok.
Here is a summarised application to your problem. We can give each point in the Cantor set an address: if a point $x$ is in the Cantor set, then it's in $K_1$ and, in particular, is in either $J_{1,1}$ or $J_{1,2}$. If it's in $J_{1,1}$, we define the first symbol of the address as $a_1=0$; if it's in $J_{1,2}$, we define it as $a_1=1$.
Now, at the second level, there is again a choice. If $x$ was in $J_{1,1}$, it may be in $J_{2,1}$ or $J_{2,2}$; we associate another symbol of the address, $a_2=0$ or $a_2=1$ respectively. If $x$ was $J_{1,2}$, it may be instead in $J_{2,3}$ or $J_{2,4}$. The symbols we associate here are not $a_2=2$ or $a_2=3$, but again $a_2=0$ or $a_2=1$ respectively. Therefore:
Since we are making a binary choice on each level (whether to look in the left or the right sub-interval), we only need a binary digit for the address.
Repeating this ad infinitum we find that each point of the Cantor set corresponds to a unique infinite address (you can formalise this by proving the Cantor set is totally disconnected, therefore two different points of the set eventually end up in different intervals, see Hasselblatt and Katok). Therefore I can write any point of the Cantor set uniquely as $$x_{a_1 a_2 a_3 \cdots}$$ where $a_i\in\{0,1\}$. For a finite string of symbols $a_1a_2\cdots a_n$, we can define the set $$X_{a_1a_2\cdots a_n}$$ as the set containing all elements of the Cantor set whose addresses begin with $a_1a_2\cdots a_n$. This set is exactly equal to one of the $J_{n,k}$ sets. How to see which one?
Let's try $n=2$ for simplicity, consider $X_{a_1 a_2}$. By checking against the examples above, you can see
We see $k= 1 + a_2 + 2a_1$. With a little induction work, you can show that, in general: $$ X_{a_1a_2\cdots a_n} = J_{n,k}, \qquad k = 1 + \sum_{i=1}^{n} a_i 2^{n-i}. $$
What is nice about the logic above is that it works for any Cantor-type set, as long as we are removing a part of each interval on each iteration. Of course, for the Cantor set (the middle-third Cantor set), we get a little more. The address of a point of the Cantor set corresponds uniquely to its base-3 decimal expansion, which is given by letting $\varepsilon_i=a_i$ and using the formula you have already shown.
Summarising, given a point $$x_{a_1 a_2 a_3 \cdots}$$ in the Cantor set, we see it belongs in a series of sets, $X_{a_1}, X_{a_1 a_2}, \cdots$, and we can identify each of those sets with one of the sub-intervals $J_{n,k}$. In particular, if you take a point $$x_{a_1 a_2 a_3 \cdots a_n 0 0 0 0 0 \cdots},$$ as required by your question, it belongs in $X_{a_1 a_2 a_3 \cdots a_n}$ corresponding exactly to $J_{n, \bar{k}}$ where $\bar{k}$ is given by the formula above.
Do take a look at Hasselblatt and Katok if you get the chance, it's a fantastic book. I hope this helps!