# In $\mathbb{R}^n$ how many disjoint open sets can share a boundary

I know that in $\mathbb{R}$ you can have at most $2$ disjoint open sets that share a boundary(I believe my answer to Open Sets Boundary question proves that). My question is is there a way to extend this to $\mathbb{R}^n$ so that we can say there is some function $f(n): \mathbb{N} \to \mathbb{N}$ that tells us how many disjoint open sets can have the same boundary in $\mathbb{R}^n$. I know for $\mathbb{R}^2$ the answer is at least 3(The lakes of wada are an example of 3 disjoint open sets that share a boundary). I also know that $n < m \implies f(n) \leq f(m)$ since if $U$ is open in $\mathbb{R}^n$ then $U\times\mathbb{R}^{m-n}$ is open in $\mathbb{R}^m$ and they'll still share a boundary.

• By the last sentence you for sure mean that if $U$ is open in $\mathbb{R}^n$ then $U × \mathbb{R}^{m - n}$ is open in $\mathbb{R}^m$. – user87690 Mar 23 '14 at 19:08
• Yes, that is what I meant. I edited the question to correct that. – ruler501 Mar 23 '14 at 19:10
• See math.stackexchange.com/questions/122119/… for the plane case. – lhf Mar 23 '14 at 19:29
• So it is a countable infinite in $\mathbb{R}^2$ and higher. Do any of the higher have a larger infinity of sets that could share a boundary? – ruler501 Mar 23 '14 at 19:32
• You can do the Lakes of Wada construction for any finite $n$, and even for a countably infinite family of lakes. – MJD Mar 24 '14 at 13:48

First, for every $n\ge 2$, lake Wada construction gives (infinitely) countably many open subsets of $R^n$ with common frontier. You also cannot have uncountably many disjoint (nonempty) open sets in $R^n$. Indeed every such set will contain a point with rational coordinates and $Q^n$ is countable. Therefore, you cannot have uncountably many open subsets with common frontier.
• Can you just add in that for $n>1$ the answer is countably infinite many because of the Lakes of Wada construction? – ruler501 Mar 24 '14 at 13:51