Connected Components

In the past I learnt connected components in a quite high generality. Special case is of particular importance, the following are the few used implicitly in complex analysis on $\C$.

Fact 1. If $U$ is open in a locally connected topological space $X$, then each of its connected components is open. In particular, if $K$ is closed, $\C\setminus K$ is a countable union of open connected components.

Proof. Suppose $C\subseteq X$ is a connected component, let $x\in C$, then there is a connected neighborhood $V$, $x\in V$. Since $V\cap C\neq \emptyset$, and since $V,C\subseteq X$, it must happen that $V\subseteq C$, as desired.$\qed$

Fact 2. If $D$ is open connected, then $D$ is a component of $\C\setminus \partial D$.

Proof. Let $D\subseteq U\subseteq \C\setminus \partial D$ with $U$ connected, we show that $D=U$. Since $U=D\cup (U\setminus D)$. As $U\cap \partial D=\emptyset$, thus we have $$U=D\cup (U\setminus \ol{D}).$$ Since $U$ is connected, and both $D$ and $U\setminus \ol D$ are open, one of them must be empty, and since $D\neq \emptyset$, we have $U=D$, as desired.$\qed$

Fact 3. If $X$ is locally path connected, then connected subsets are path-connected.

Proof. Let $U\subseteq X$ be connected, fix $x\in U$, then the set $U_x$ of points that can be connected with $x$ by a path in $U$ is both open and closed. Since $x$ and $x$ can be connected by a path in $U$, $U_x\neq \emptyset$, thus $U_x=U$, as desired.$\qed$

Fact 4. If $C$ is a connected component of a locally connected topological subspace $X$ of some $Y$, then $\partial C\subseteq \partial X$. A typical example of $X$ is open subset $\C\setminus K$ of $\C$. This is a particularly important example in potential theory as it pops up quite frequently and naturally when using maximum principle for subharmonic functions (since $\partial C\subseteq \partial (\C\setminus K) =\partial K$).

Proof 1. To show $\ol C\setminus C^\circ \subseteq \ol X\setminus X^\circ$, we pick $x\in \ol C$. We now show that if $x\in X^\circ$, then $x\in C^\circ$. Suppose $x\in X^\circ$, then there is a connected neighborhood $V\subseteq X$ of $x$. Now $V\cap C\neq \emptyset$, thus $V\subseteq C$, so $x\in C^\circ$.

Now we have shown that $\ol C\setminus C^\circ \subseteq \ol C\setminus X^\circ$, thus $\partial C\subseteq \partial X$.$\qed$

Proof 2 (When $C$ is just a component of $\mathbb P\setminus K$ with $K$ compact). This is a very common scenario. Take $x\in \partial C$. For the sake of contradiction, suppose that $x\not\in \partial(\mathbb P \setminus K)=\partial K$, then there is a path connected open $V$, with $x\in V$, such that $V\cap \partial(\mathbb P\setminus K)=\emptyset$. Now $V\cap C\neq \emptyset$ (since $x\in \ol C$) and as $C$ must be open, $V\cup C$ is a connected open set.

We expect $V\cap K=\emptyset$, if not, then there is $v\in V\cap K$, joining this point with a $c\in C\subseteq \mathbb P\setminus K$ using a path in $V\cup C$, we get a point in $$(V\cup C)\cap (\partial K) = C\cap \partial K\subseteq C\cap K=\emptyset,$$ a contradiction. We conclude that $V\subseteq \mathbb P\setminus K$.

Therefore $V\cup C$ is an open connected subset of $\mathbb P\setminus K$, and as $C$ is a component of $\mathbb P\setminus K$, we have $C\cup V=C$, so $V\subseteq C$, a contradiction since $V$ is a neighborhood of $x\in \partial C$.$\qed$