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In [this] (Bounded components of complement of bounded planar domain are simply-connected) Moishe Kohan says that if $E$ is a closed set and $U$ is a bounded component of the complement of $E$, then the boundary of $U$ is contained in $E$. What is the formal proof of this?

M. Rahmat
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1 Answers1

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Every open set $U$ can be expressed as sum of maximal connected components, say $U_{i\in I}$. Now because for fixed $k\in I$, $U_k$ is separated from $U\setminus U_k $, thus $\overline U\cap (U\U_k)=\emptyset$. It means, that the boundary of any component is contained in the complement of $U$, i.e.: $$\overline {U_k}\setminus U_k\subset U^c$$

Przemek
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  • Sorry I don't follow you. Assume $U$ is a bounded connected component of the complement of $E$. Then $U\setminus U_k=U\setminus U=\emptyset$! And then? – M. Rahmat Nov 02 '20 at 23:16
  • Then $\overline U \setminus U \subset U^c$ as well. That is, if $U$ is itself connected, then by definition its boundary is contained in the complement. But if $E^c=U$, then $U^c=E^{cc}=E$. As for $U$ being bounded, it is used further in Moishe Kohan's answer, but not needed here. – Przemek Nov 02 '20 at 23:32
  • Let me explain more clearly the problem. There are non empty open sets $U$ and $V$ that are disjoint such that $E^c=U\cup V$, and $U$ is bounded. You prove that the boundary of $U$ is included into the complement of $U$(which is always true for any open set $U$). But could you please explain how do you go from there to conclude that the boundary of $U$ is in $E$? – M. Rahmat Nov 03 '20 at 01:05
  • $U$ and $V$ are maximal connected components, thus $U\cup V$ is not only disjoint, but also separated. Thus $\overline U \cap V=\emptyset$. See https://en.wikipedia.org/wiki/Separated_sets#Definitions – Przemek Nov 03 '20 at 01:13
  • OK. But how to conclude that the boundary of $U$ is in $E$? – M. Rahmat Nov 03 '20 at 02:58
  • I'm afraid, I cannot put it simpler. This is direct consequence of boundary definition. Because boundary $\overline U \setminus U$ is disjoint with $U$ and disjoint with $V$, it is subset of $(V\cup U)^c=E$. – Przemek Nov 03 '20 at 08:28