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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.

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    $\begingroup$ Just wondering, where did you find this problem? $\endgroup$ – user135971 Mar 21 '14 at 3:30
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    $\begingroup$ I'd like to know where the problem came from too. It stumped me for most of the day and my math professor when I asked him quickly what he thought of it. $\endgroup$ – ruler501 Mar 22 '14 at 4:58
  • $\begingroup$ @user124910 My answer to this question needs to be unaccepted. It is demonstrably wrong $\endgroup$ – ruler501 Mar 27 '14 at 21:04
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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.

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  • $\begingroup$ Looking at this and I'd think you could expand this to have a countable number of disjoint open sets sharing a boundary. Or at the least any finite number you wished. $\endgroup$ – ruler501 Mar 27 '14 at 6:39

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