According to an answer here https://mathoverflow.net/questions/25993/sets-with-positive-lebesgue-measure-boundary, if we construct a 'fat' Cantor set by removing the middle open interval of length $\frac{1}{4}$ from $[0,1]$, then removing the $2$ middle length $\frac{1}{4^2}$ open intervals from the remaining intervals, and so on (so that the end result is a Cantor set with measure $\frac{1}{2}$), then the sets:
$$U = \bigcup \{ \text{open intervals removed on odd-numbered steps} \}$$
$$V = \bigcup \{ \text{open intervals removed on even-numbered steps} \}$$
have the fat Cantor set as their common boundary. I fail to see this, can anyone please give me a hint?
Best Answer
This has nothing to do with the Cantor set being fat. Your Cantor set $C$ is the nested intersection $C= \bigcap_{n=0}^{\infty}C_n$ of sets of the form $$C_n=I_{n,1}\cup\dots\cup I_{n,k_n}\text{,}$$
where $I_{n,j}$ are disjoint closed intervals.
Furthermore, each point in the Cantor set is a nested intersection of some of these intervals.
In the construction of the Cantor set, every one of these closed intervals is eventually punctured many times by intervals from $U$ and $V$, and thus each interval $I_{n,j}$ intersects both $U$ and $V$.
Since each point in $C$ is a nested intersection of smaller and smaller intervals that touch both $U$ and $V$, we get $C\subseteq \bar{U}$, $C\subseteq \bar{V}$. Since $C\cap U=\emptyset = C\cap V$, we have $C\subseteq \partial U$ and $C\subseteq \partial V$.
On the other hand, since $U$ and $V$ are open and disjoint from each other, we must have $\partial U\subseteq C$ and $\partial V \subseteq C$.