Does the set given by $X\cdot\langle2\rangle\cdot\langle3\rangle$ exactly cover $\Bbb Z\left[\frac16\right]\cap(0,\infty)$? Is there a similar method using a Cantor set?
Where $X=\{x\in\Bbb N:x\equiv\pm1\pmod 6\}$ and $\langle p\rangle=\{p^i:i\in\Bbb Z\}$
I think it does because if we exclude from the positive integers the multiples of $2$ and $3$, we get the numbers equivalent to $\pm1\pmod 6$ then reintroducing the factors $2^i\cdot3^i$ restores $\Bbb N>0$ and also all the dyadic and ternary denominators.
Is there a similar cover using the middle two-thirds Cantor set $C_{2/3}$ in place of $X$? The middle two-thirds Cantor set is the set of strings in base $6$ constructed out of 5s and 0s, so ends in $\pm1$. So I imagine there is some similar way of constructing $\Bbb Z[\frac16]$ using $C_{2/3}$, $\langle2\rangle$ and $\langle3\rangle$.
Obviously a direct replacement doesn't work but is there a relatively simple way of swapping the Cantor set in, such as e.g. a sum of two cantor sets?
Yes $X.\langle2\rangle.\langle3\rangle$ covers $\mathbb{Z}[\frac{1}{6}] \cap (0, \infty)$. You can show this bit pretty easily by seeing that any element $x \in \mathbb{Z}[\frac{1}{6}] \cap (0, \infty)$ will be of the form $\frac{z}{6^n} = \frac{2^a3^bz'}{6^n} = 2^{a'}3^{b'}z'$ (where $z'$ and 6 are coprime), which cleary are contained in $X.\langle2\rangle.\langle3\rangle$.