How to prove that Cantor ternary set has no intervals using ternary expansions?

Question

Found this proof on this site: Cantor's Set Has No Intervals but I don't understand a part of it.

Here is the proof:

Suppose that $x,y\in C$, where

$$x=\sum_{k\ge 1}\frac{a_k}{3^k}\quad\text{and}\quad y=\sum_{k\ge 1}\frac{b_k}{3^k}\;,$$

where all $a_k,b_k\in\{0,2\}$. Let $n\in\Bbb Z^+$ be minimal such that $a_n\ne b_n$, and without loss of generality assume that $a_n=0$ and $b_n=2$. Define $c_k=a_k=b_k$ for $k=1,\dots,n-1$ and $c_k=1$ for $k\ge n$, and let $$z=\sum_{k\ge 1}\frac{c_k}{3^k}\;;$$ then $x<z<y$, and $z\notin C$.

So what makes $x<z<y$ ? I got that that first $n-1$ the terms are equal in $x,y,z$ and at $k=n$ we have $x_n= 0$, $y_n = \dfrac{2}{3^n}$ and $z_n =\dfrac{1}{3^n}$

So,

$\sum_{k\ge 1}^{n}\frac{a_k}{3^k} < \sum_{k\ge 1}^{n}\frac{c_k}{3^k} < \sum_{k\ge 1}^{n}\frac{b_k}{3^k}$

but for the terms whose indices is $k$ where $k=n+1,n+2,...$ what guarantees that still $x<z<y$?

Answer 1

Let $u=\sum_{k=1}^{n-1}\frac{a_k}{3^k}=\sum_{k=1}^{n-1}\frac{b_k}{3^k}=\sum_{k=1}^{n-1}\frac{c_k}{3^k}$. The largest possible value of $x$ occurs when $a_k=2$ for all $k>n$. In that case

$$x=u+\frac{a_n}{3^n}+\sum_{k>n}\frac2{3^k}=u+0+\frac{\frac2{3^{n+1}}}{1-\frac13}=u+\frac1{3^n}\;.$$

The smallest possible value of $y$ occurs when $b_k=0$ for all $k>n$. In that case

$$y=u+\frac{b_n}{3^n}=u+\frac2{3^n}+\sum_{k>n}\frac0{3^n}=u+\frac2{3^n}\;.$$

Finally,

$$z=u+\frac{c_n}{3^n}+\sum_{k>n}\frac{c_k}{3^k}=u+\sum_{k\ge n}\frac1{3^k}=u+\frac{\frac1{3^n}}{1-\frac13}=u+\frac32\cdot\frac1{3^n}\;.$$

Clearly

$$u+\frac1{3^n}<u+\frac32\cdot\frac1{3^n}<u+\frac2{3^n}\;,$$

i.e., $x<z<y$.

Answer 2

It doesn't matter. You have been doing this for a long time, just in base 10 instead of base 3. That is, I'm sure you'll think it's obvious that $$ 0.11<0.114595578<0.12. $$ That's exactly what's happening here: $$ \sum_{k=n+1}^\infty \frac{c_k}{3^n}<\frac2{3^n} $$ no matter the values of $c_k\in\{0,1,2\}$. For a formal proof, $$ \sum_{k=n+1}^\infty \frac{c_k}{3^n}\leq \sum_{k=n+1}^\infty \frac{2}{3^n}=2\frac{\tfrac{1}{3^{n+1}}}{1-\tfrac13}=\frac1{3^n} $$

Answer 3

$z$ is between $x$ and $y$ because the $n$-th term dominates: it is larger than or equal to the sum of all the remaining terms. That is, for instance, $0.012222222222\lt0.02$, and so on for any length tail of $2$'s.

Recall $0.122222222\dots=0.2$.

Just look at the geometric series $\sum(2/3)^n=3$. And remember it's an increasing sequence.

This is, I believe, basic numerical analysis.