How can I complete this proof the cantor set does not meet $\left(\frac{3s+1}{3^k}, \frac{3s+2}{3^k}\right)$? 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?
 A: 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.
A: $\textbf{Hint:}$
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
$\displaystyle \frac{3s+1}{3^{k+1}}<x<\frac{3s+2}{3^{k+1}}\iff \frac{3s-2\big(3^k\big)+1}{3^k}<3x-2<\frac{3s-2\big(3^k\big)+2}{3^k}$
