Open Sets Boundary question

I'm really having a hard time with this problem: Find three disjoint, open sets in $\mathbb{R}$ (std. topology) that have the same nonempty boundary. I played around with a few ideas like $\mathbb{Q}$, {$\sqrt{p}+\mathbb{Q}$}, {$\sqrt{q}+\mathbb{Q}$} where $p$ and $q$ are distinct primes, but these sets aren't open. I can find examples that fit two of the conditions, but not all three.

Solution 1:

Let $C$ be the Cantor ternary set on $[0,1]$. Its complement in $[0,1]$ is a countable union of open intervals, which are naturally ranked into "generations", for instance by size. Let $U_i$ be the union of the open sets of generation congruent to$~i$ modulo$~3$, for $i=0,1,2$. Then the opens sets $U_0,U_1,U_2$ are disjoint, and $\partial(U_1)=C$ for $i=0,1,2$.

The essential point of this construction is that although each $U_i$ is missing many of the open intervals of the complement of$~C$, the boundary $\partial(U_i)$ does contain the boundary points of such intervals, because $U_i$ contains intervals arbitrarily close to those points. The same is true for the uncountably many points of $C$ that are not an end point of any interval at all.

One can generalise this construction directly to more than three disjoint open subsets with the same boundary, by choosing appropriate disjoint infinite subsets of the set of generations (which is indexed by the natural numbers). In particular one can find countably many disjoint open subsets all with boundary equal to$~C$. Since any collection of disjoint subsets of$~\Bbb R$ is at most countable, this is about as good (or bad, depending on your point of view) as one can expect it to get.