How can I complete this proof the cantor set does not meet $\left(\frac{3s+1}{3^k}, \frac{3s+2}{3^k}\right)$?

Question

We define the cantor set $K$ as the set of all ternary numbers $0.{a_1a_2a_3\cdots}$ such that $a_i = {0, 2}$ for all $i$, i.e. no digit is allowed to be 1, and the first one is zero.

Here is my proof in progress.

Theorem For all $x \in K$, either $x \leq \frac{3s + 1}{3^k}$ or $x \geq \frac{3s + 2}{3^k}$

Proof We proceed by induction on $k$.

(Base Case) For $k = 1$ we show that $x \in K$ implies $x \leq s + \frac{1}{3}$ or $x \geq s + \frac{2}{3}$ for all natural $s$. By case analysis on the first digit, either $x \leq \frac{1}{3}$ or $x \geq \frac{2}{3}$. Notice $x \leq \frac{1}{3}$ implies $x \leq \frac{1}{3} + s$ for all $s$. For $s \geq 1$, $s + \frac{2}{3} \geq \frac{5}{3} > 1$ and so there is no $x \in K$ such that $x \geq s + \frac{2}{3}$.

(Inductive case) Suppose that $x = 0.a_1a_2\cdots$. If $a_1 = 0$, then $3x \in K$ so by the inductive hypothesis we have two cases. The first case is $3x \leq \frac{3s + 1} {3^k}$ for all $s$, so $x \leq \frac{3s+1}{3^{k + 1}}$ for all $s$. The second case is that $3x \geq \frac{3s + 2}{3^k}$ for all $s$, so $x \geq \frac{3s + 2}{3^{k + 1}}$ for all $s$.

Now here is where I get stuck. If $a_1 = 2$, then $3x - 2 \in K$, so we can apply the inductive hypothesis and similar reasoning to obtain that, for all $s$ either $x \leq \frac{3s + 1}{3^{k + 1}} + \frac{2}{3}$ or $x \geq \frac{3s+2}{3^{k + 1}} + \frac{2}{3}$. The problem is of course that rescaling the interval $[\frac{2}{3}, 1]$ to $[0, 1]$ "shifts" the $s$ value when compared to the intervals excluded in the previous step of the induction. I have an intuition that if $a_1 = 1$ it suffices to consider the case where $\frac{3s + 1}{3^{k + 1}} \geq \frac{2}{3}$, as the other cases are covered by the case where $a_1 = 0$, but I have no idea how to formalise this intuition. So could someone give me hints as to how to complete this proof, or at least how to formalise the aforementioned intuition?

Answer 1

$\textbf{Hint:}$

If $a_1=2$, then $x\ge\frac{2}{3}$, so $x\ge\frac{3s+2}{3^{k+1}}$ whenever

$\frac{3s+2}{3^{k+1}}\le\frac{2}{3}\iff 3s+2\le2\big(3^k\big)\iff 3s+3\le2\big(3^k\big)\iff s\le2\big(3^{k-1}\big)-1$.

Therefore we only need to consider values of $s$ satisfying $2\big(3^{k-1}\big)\le s<3^k$, and

$\displaystyle \frac{3s+1}{3^{k+1}}<x<\frac{3s+2}{3^{k+1}}\iff \frac{3s-2\big(3^k\big)+1}{3^k}<3x-2<\frac{3s-2\big(3^k\big)+2}{3^k}$

Answer 2

Here is a different approach. Consider an interval $I_k = \left(\frac{3s + 1}{3^k}, \frac{3s + 2}{3^k}\right)$. If $x \in [0, 1]$ is in this set, then $s < 3^{k - 1}$. So $s$ can be written in ternary with $k - 1$ digits $s_i$ as follows: $s_1, s_2,\dots, s_{k -1}$. Now, $\frac{3s + 1}{3^k} = \frac{s}{3^{k - 1}} + \frac{1}{3^k}$. Written in ternary, this is: $$0.s_1,s_2,\dots, s_{k -1},1,0,0,\dots$$

Similarly,

$$\frac{3s + 2}{3^k} = \frac{s}{3^{k - 1}} + \frac{2}{3^k} = 0.s_1,s_2,\dots, s_{k -1},1,2,2,\dots$$

Now, any number strictly between* these numbers can only be written as $$0.s_1,s_2,\dots, s_{k -1},1,x_1,x_2,x_3\dots$$ with some $x_i \neq 0$ and some $x_j \neq 2$. Clearly, this representation contains a $1$, and so it cannot be in the Cantor set.

*We do not include the extremes themselves because they are special cases, without unique representations. E.g. $0.s_1,s_2,\dots, s_{k -1},1,0,0,\dots = 0.s_1,s_2,\dots, s_{k -1},0,2,2,\dots$. This is a quirk of place-value number systems.