What does this Cantor space in $\left[\frac13,1\right]$ look like? Does it look anything like $\{x\in\Bbb Z[\frac16]\cap[\frac13,1]:\lvert x\rvert_2\geq1\}$ and if so, does it imply some p-adic type completion of $\{x\in\Bbb Z[\frac16]:\lvert x\rvert_2\geq1\}\}$ by $\lvert\cdot\rvert_4$?
I'm looking at the numbers expressible in base $4$ using only the digits drawn from $\{1,3\}$
By the usual bijection to the unit interval it seems obvious it will be in the subset $\left[\frac13,1\right]$ because the smallest number is $.\overline1_4$ and the largest $.\overline3_4$
It can be written $\displaystyle\sum_{n=0}^\infty \frac{a_n}{4^{n+1}}$ with $a_n\in\{1,3\}$
Or it could equally be written:
$\displaystyle\sum_{n=0}^\infty \frac{2a_n+1}{4^{n+1}}$ with $a_n\in\{0,1\}$
I'm struggling to visualise whether the usual removing a centre third structure is preserved or whether it's something degenerate. My instinct says its construction follows the usual diagram of the Cantor set with the centre third removed rather than a half or quarter, but perhaps with the left hand edge tapering in towards the bottom-right at each step.
Or perhaps it's just $\frac23\cdot C+\frac13$
A bit of a random shot in the dark but does this look anything like $\{x\in\Bbb Z[\frac16]\cap[\frac13,1]:\lvert x\rvert_2\geq1\}$ and if so, does it imply some p-adic type completion of $\Bbb Z[\frac16]$ by a "4-adic" valuation $\lvert\cdot\rvert_4$? (please note I'm aware $C$ is an uncountable set including limit points).
Your construction is closely analogous to the usual middle-thirds Cantor set, except at each step you remove the middle half, leaving a fourth at each end.
To wit, if you take your set, scale it by $\frac14$ and add either $\frac 14$ or $\frac34$, you get the two self-similar components of the set.
The inner endpoints of those components are $0.1333\ldots_4=0.2_4=\frac 12$ and $0.3111\ldots_4 = \frac{5}{6}$, so each has length $\frac{1}{6}$ and there is $\frac{2}{6}$ of removed interval between them.