Cantor set is boundary of regular open set

Question

Does there exist a regular open set $U$ in $[0,1]$, such that Cantor set is the boundary of $U$?

Answer 1

For $n\in\Bbb N$ let $\Sigma_n$ be the set of sequences $\sigma=\langle\sigma_0,\dots,\sigma_{n-1}\rangle$ of $n$ zeroes and ones. (The unique member of $\Sigma_0$ is the empty sequence $\langle\rangle$.) Let $\Sigma=\bigcup_{n\in\Bbb N}\Sigma_n$. We can index the open intervals removed in the construction of the middle-thirds Cantor set by the elements of $\Sigma$. Specifically, let $I_{\langle\rangle}$ be the open interval $\left(\frac13,\frac23\right)$ removed at stage $0$, the first stage, and let $I_0=\left(\frac19,\frac29\right)$ and $I_1=\left(\frac79,\frac89\right)$ be the open intervals removed at stage $1$. The four intervals removed at stage $2$ are $I_{00}=\left(\frac1{27},\frac2{27}\right)$, $I_{01}=\left(\frac7{27},\frac8{27}\right)$, $I_{10}=\left(\frac{19}{27},\frac{20}{27}\right)$, and $I_{11}=\left(\frac{25}{27},\frac{26}{27}\right)$.

Notice that at stage $n+1$ we remove twice as many open intervals as at stage $n$. Moreover, each removed interval at stage $n$ lies immediately between two of the intervals removed at stage $n+1$. If $\sigma\in\Sigma_n$, so that $I_\sigma$ is one of the open intervals removed at stage $n$, the neighboring intervals removed at stage $n+1$ will be labelled $I_{\sigma^\frown0}$ (to the left of $I_\sigma$) and $I_{\sigma^\frown1}$ (to the right of $I_\sigma$), where $\sigma^\frown i$ is the sequence of length $n+1$ that results from appending $i$ to $\sigma$. (You should check that I followed this scheme in labelling the four intervals removed at stage $2$.)

Let $$U=\bigcup_{\sigma\in\Sigma}I_{\sigma^\frown 0}\quad\text{and}\quad V=\bigcup_{\sigma\in\Sigma}I_{\sigma^\frown 1}\;;$$

$U$ is the union of the removed intervals whose subscripts end in $0$, and $V$ is the union of those whose subscripts end in $1$.

Aside: If you arrange the removed intervals in a full binary tree (like the picture below, but with infinitely many levels), $U$ is the union of the intervals that are left children of their parent intervals, and $V$ is the union of the intervals that are right children of their parent intervals.

If $C$ is the middle-thirds Cantor set itself, show that $\operatorname{cl}U=U\cup C$ and $\operatorname{cl}V=V\cup C$. Conclude that $\operatorname{int}\operatorname{cl}U$ and $\operatorname{int}\operatorname{cl}V$ are regular open sets having $C$ as boundary.