Connectedness of the boundary of a domain

Question

I've been struggling to prove the following lemma: "Let $\Omega\subset\mathbb{R}^{d}$ be open and bounded with a Lipschitz boundary and such that $\mathbb{R}^{d}\setminus\partial\Omega$ has exactly two connected compnents. Then $\partial\Omega$ is connected."

It seems an evident result when you think about it for 5 minutes, but I am struggling with a formal proof (a reference would be really welcome). It also looks like the typical result that you find in some old calculus book, but I haven't been able to find one yet. I have thought a complex analysis approach, but this would only solve the 2-dimensional case, which we know is not equivalent to prove the d-dimensional case and cannot be generalized. So far I have clarified some things to understand the result better:

In all of these cases, the domain $\Omega$ has some different dimensional holes $\Gamma_0$: a $d$-dimsnsional hole in the first picture (however this contradicts the hypothesis of "$\mathbb{R}^{d}\setminus\partial\Omega$ has exactly two connected compnents"), a $d-1$-dimensional slit in the second picture (this contradicts the Lipschitz boundary hypothesis), and a set with positive $d-1$-dimensional Hausdorff measure in the last picture (this also contradicts the Lipschitz boundary hypothesis). It is easy to see these contradictions (with handwaving, sadly), but I cannot seem to prove rigorously the whole lemma (are there any more cases to consider? how do I handle the Lipschitz hypothesis to reach the contradiction in the two latter cases? am I missing something? can it even be trivial?).

So far, my best attempt to the proof is the following:

"There are two possibilities: either $\mathbb{R}^{d}\setminus\Omega$ is connected or $\mathbb{R}^{d}\setminus\Omega$ has at least two disjoint connected components (we will handle the case of $\mathbb{R}^{d}\setminus\Omega$ having exactly two connected components since the cases with more than two connected components are analogous). When $\mathbb{R}^{d}\setminus\Omega$ is connected the result comes straightforward from [Czarnecki-Kulczycki-Lubawski, 2011]. On the other hand, say that $\mathbb{R}^{d}\setminus\Omega=\Gamma_{0}\cup\Gamma_{1}$, where $\Gamma_{0}$ is a connected component which is bounded and $\Gamma_{1}$ is a connected component which is unbounded (the fact that one is bounded and the other is unbounded comes from the fact that $\Omega$ is bounded). Recalling that the $d$-dimensional Hausdorff measure and the $d$-dimensional Lebesgue measure coincide in $\mathbb{R}^{d}$, there are two other possibilities: either $\mathcal{L}^{d}(\Gamma_{0})>0$ or $\mathcal{L}^{d}(\Gamma_{0})=0$. In the first case we have that $\mathbb{R}^{d}\setminus\partial\Omega=int(\Gamma_{0})\cup\Omega\cup\Gamma_{1}$ which are, at least, three disjoint connected components and therefore lead to a contradiction. In the latter case, recalling the set $C=\{x+yn:x\in B_{r}(p)\cap H,-h<y<h\}$ from the definition of a Lipschitz domain, for any $p\in\partial\Gamma_{0}$, when $\mathcal{L}^{d}(\Gamma_{0})=0$ we have that $\partial\Gamma_{0}=\Gamma_{0}$ and hence, that $\Gamma_{0}\cap C=(\partial\Gamma_{0})\cap C$ which is a contradiction with $\Gamma_{0}$ being a Lipschitz domain. Therefore $\Gamma_{0}=\emptyset$ and the result comes straightforward from [Czarnecki-Kulczycki-Lubawski, 2011]."

This looks wrong to me, but I can't say where exactly. Any help will be welcome, thank you.

Answer 1

You do not need to refer to measure. I think you missed p.5 below.

Applying result from the paper:

If $\Omega$ is open bounded subset of $\mathbb{R}^n$ such that $\mathbb{R}^n\setminus \Omega$ is connected, then $\partial\Omega$ is connected.

1. Assume $\Omega$ and $C$ are components of $\mathbb{R}^n\setminus \partial\Omega$ resp.
2. Take $\Omega$ and apply result to it. Need to prove that $\mathbb{R}^n\setminus \Omega$ is connected.
3. $\mathbb{R}^n\setminus \Omega = C \cup \partial\Omega$ is closed and contains $C$ so it contains closure $\overline{C}.$
4. Assuming $x\in (C \cup \partial\Omega)\setminus\overline{C}$ obtain $x\in \partial\Omega\setminus\overline{C}.$ Here is contradiction, because from definition of Lipschitz domain, there are points of $C$ in each neighborhood of $x.$
5. Note the fact: if $X$ is connected open subset (of arbitrary topological space) then closure $X_c = \overline{X}$ is also connected.
   • Assume on the contrary that $X_c$ is not connected, then $X_c = (O_1 \cup O_2)\cap X_c$ where $O_1, O_2$ open subsets and $O_{1,2}\cap X_c \neq \emptyset.$
   • If $O_1\cap X=\emptyset$ then $\mathbb{R}^n\setminus O_1$ is closed set containing $X$ and thus also contains $X_c,$ contradiction with $X_c\cap O_1\neq \emptyset.$ Similarly for $O_2.$
   • Thus $X = (O_1 \cup O_2)\cap X$ and $O_{1,2}\cap X \neq \emptyset$ and $X$ is disconnected, contradiction.
6. As $\partial \Omega$ is locally path-connected and connected then it is path-connected.

UPD 1.

Expansion of p.4.

Introducing notations for Lipschitz domains is rather tedious, refer to, for example, Understanding Lipschitz domain or D.Mitrea "Distributions, partial differential equations and ... "

One needs the following topological property:

For each $x\in \partial \Omega$ there is a map (chart) $\phi:U\to V$ such that

• $\phi$ is homeomorphism of $U$ onto $V,$ both are subsets of $\mathbb{R}^n;$
• $x\in U, \phi(x) = 0;$
• $\phi(U\cap\Omega) = V\cap H^+$ where $H^+=\{x\in\mathbb{R}^n: x_n > 0\}$ is upper half space.
• $\phi(U\cap\partial\Omega) = V\cap \partial H^+$ where $\partial H^+=\{x\in\mathbb{R}^n: x_n = 0\}$ is boundary of upper half space.

This property implies $\phi(U\cap C) = V\cap H^-.$ 0 is in closure $\overline{H^-}$ of lower half space $H^-,$ so $x\in \overline{C}.$