Every Connected Component is Closed

Last updated Nov 1, 2022

Let $X$ be a Topological Space and let $C \subset X$ be a Connected Component. Then $C$ is Closed.

Consider $x \not\in C$. We know ${x} \cup C$ must be disConnected because $C$ is a Maximal Connected Set. Thus there exists $U_{x}, V_{x} \subset X$ Open that disconnect $C \cup {x}$. Because $C$ is Connected either $C \subset U_{x}$ or $C \subset V_{x}$. Without Loss of Generality suppose $C \subset V_{x}$. Then we must have $x \in U_{x}$ and $U_{x} \cap C = \emptyset$. Thus $U := \bigcup\limits_{x \in X} U_{x}$ is Open and $U^{C} = C$. Thus $C$ is Closed. $\blacksquare$