Bounded components of complement of bounded planar domain are simply-connected

Solution 1:

For a subset $B\subset \mathbb{R}^2$ I will use the notation $B^c$ to denote its complement, $\mathbb{R}^2 \setminus B$.

Let me start with a basic observation: If $A$ is a closed subset of $\mathbb{R}^2$, then for each connected component $U$ of $A^c$, the boundary of $U$ is contained in $A$.

Now, back to the question. Let $D$ be an open connected bounded subset of $\mathbb{R}^2$. I will denote $\Omega$ the (unique) unbounded component of $\mathbb{R}^2 \setminus \bar{D}$. Let $E$ be a bounded component of $D^c$. Our goal is to show that $E^c$ contains no bounded components.

It is clear that $\Omega\cap E=\emptyset$.

Let $U$ be a component of $E^c$. Then $U$ and $D$ are open subsets of $E^c$. Since $U$ is a component of $E^c$ and $D$ is connected, the open sets $U$ and $D$ are either disjoint or $D\subset U$.

Case 1: $U\cap D=\emptyset$. Since $D$ is open, we also have that $\bar{U}$ is disjoint from $D$.

By the basic observation, the boundary of $U$ is contained in $E$, hence, $\bar{U}\cap E\ne\emptyset$, thus, $\bar{U}\cup E$ is connected and disjoint from $D$. Thus, $\bar{U}\subset E$ (since $E$ was a component of $D^c$). This is a contradiction since $U\cap E=\emptyset$.

Case 2: $D\subset U$, hence, the boundary of $D$ is contained in the closure of $U$. Hence, $U$ contains the boundary of $\Omega$ and, thus, $\Omega\cap U\ne\emptyset$ and, therefore, $\Omega\cup U$ is connected and is disjoint from $E$. Since $U$ was a component of $E^c$, it follows that $\Omega\subset U$ and, therefore, $U$ is unbounded.

Thus, we conclude that all components of $E^c$ are unbounded. qed

Addendum. The notion of simple connectivity you are using is not equivalent to the usual one even for compact subsets of the plane (the Warsaw circle is a standard example). But your notion is equivalent to simple connectivity in Čech sense. Namely, the following are equivalent for compact connected subsets $K\subset R^2$:

• $K^c$ contains no bounded components.

• $\check{H}^1(K, {\mathbb Z})=0$.

• $\check{\pi}_1(K, x)= \{1\}$ for every $x\in K$, i.e. $K$ is Čech-simply-connected.

Here $\check{H}$ means Čech cohomology and $\check{\pi}_1$ means Čech fundamental group. One proves this equivalence using Alexander duality and Hurewicz theorem.