# Is it true that if two disjoint connected open sets in $\mathbb{R}^n$ have the same boundary then the boundary must be connected?

Let us have two disjoint connected open sets $$C_1$$ and $$C_2$$ in $$\mathbb{R}^n$$ such that $$\partial C_1 = \partial C_2$$. My claim is that $$\partial C_1$$ must be connected.

I cannot rigorously prove it but my idea is as follows. Assume that $$\partial C_1$$ is not connected and let $$x$$ and $$y$$ be two points from two different connected components of $$\partial C_1$$. Since open connectedness implies path connectedness in $$\mathbb{R}^n$$ there are paths connecting $$x$$ and $$y$$ in both $$C_1$$ and $$C_2$$. So we can form a closed curve passing through the points $$x$$ and $$y$$ and also through $$C_1$$ and $$C_2$$. I am pretty sure that this curve cannot be shrinked to a point. But I don't know how I can write it mathematically. I appreciate it if someone can help. Thanks.

• If the ideas in this answer are true, then we have a trivial answer in the affirmative when $C_1\cup C_2\cup\partial C_1 = \mathbb{R}^n$. However, it is not always the case that $C_ 1\cup C_2\cup\partial C_1 = \mathbb{R}^n$. The Lakes of Wada shows that there can be three disjoint connected open sets whose closure covers $\mathbb{R}^n$ and all share the same boundary. If we choose $C_1$ and $C_2$ as two of these sets then we can’t use the simple connectedness of $\mathbb{R}^n$ in the same way. Commented Oct 19, 2022 at 21:35
• Thanks for the answer. The link you shared helped me a lot. Commented Oct 22, 2022 at 15:41

Throughout this answer, let $$X=\mathbb{R}^n$$ for simplicity. Note that we can replace $$X$$ (I think) with any simply connected, locally connected metric space.

Lemma 1. If $$A\subseteq X$$ is closed, then $$\partial A = \bigcup_i\partial \Gamma_i$$ where $$\{\Gamma_i\}_i$$ are the connected components of $$A$$.

Proof. Suppose $$x\in \partial A$$, and let $$\Gamma$$ be the connected component of $$A$$ containing $$x$$. Then $$A$$ does not contain a neighborhood (in $$X$$) of $$x$$, so neither does $$\Gamma\subseteq A$$. Thus, $$x\in \partial \Gamma$$.

Now suppose instead that $$x\in \partial \Gamma$$, where $$\Gamma$$ is a connected component of $$A$$. Then $$\Gamma$$ does not contain a neighborhood (in $$X$$) of $$x$$. If $$A$$ contains a neighborhood (in $$X$$) of $$x$$, then it contains a connected neighboorhood $$N$$ of $$x$$, since $$X$$ is locally connected. However, then $$N\subseteq \Gamma$$, which is a contradiction, so $$x\in \partial A$$.

Theorem 1. If $$A$$ is open, connected, and non-empty, then $$\partial A$$ is connected iff $$X-A$$ is connected.

Proof. See this post. Note that the top answer does not make use of OPs extra assumptions. Even boundedness is not necessary to conclude $$k(x)$$ is continuous because $$d(x,U)+d(x,V) = 0$$ implies that $$x\in \overline{U}\cap \overline{V} = U\cap V$$, which contradicts the definition of $$U$$ and $$V$$.

Theorem 2. If $$A,B\subseteq X$$ are open, connected, disjoint, and non-empty, and $$\partial A = \partial B$$, then $$\partial A$$ is connected.

Proof. Since $$B$$ is connected, so is $$\overline{B}$$. Let $$\{\Gamma_i\}_i$$ be the connected components of $$X-A$$, and let $$\Gamma_j$$ be the connected component of $$X-A$$ containing $$\overline{B}$$. Then, $$\partial\Gamma_j\subseteq\partial A = \partial B\subseteq \Gamma_j$$ by lemma 1. Since the $$\Gamma_i$$ are disjoint, $$\partial\Gamma_i = \emptyset$$ for $$i\neq j$$, and thus $$\partial \Gamma_j=\partial A$$ by lemma 1 as well. However, $$\partial \Gamma_i$$ can only be empty if $$\Gamma_i$$ is clopen in $$X$$ (i.e. $$\Gamma_i \in \{X,\emptyset\}$$), which is not possible as $$A$$ and $$\Gamma_i$$ are non-empty by definition. Thus, $$X-A=\Gamma_j$$ is connected, so $$\partial A$$ is connected by theorem 1.