Every Connected Set is contained in a single Component

Last updated Nov 1, 2022

Suppose $X$ is a Topological Space and $S \subset X$ is Connected. Then there exists a Connected Component $C \subset X$ s.t. $S \subset C$. If $S$ is Nonempty, this Connected Component is unique.

If $X = \emptyset$, then $\emptyset$ is a Connected Component, and $S \subset X \subset \emptyset$.

Otherwise recall that Connected Components Partition the Space, so let $\mathcal{C}$ be that Partition of $X$. Since $X$ is Nonempty, $\mathcal{C}$ is Nonempty. If $S = \emptyset$ then any $C \in \mathcal{C}$ will do. Otherwise, suppose $S \neq \emptyset$. Then $\exists x \in S$. Since $\mathcal{C}$ is a Partition, there exists unique $C_{x} \in \mathcal{C}$ so that $x \in C_{x}$. We claim $S \subset C_{x}$. Otherwise if $S \not\subset C_{x}$, because $x \in S \cap C_{x}$, $S \cup C_{x} \supsetneq C_{x}$ is Connected by If Connected Sets share a point, then their Union is Connected and $C_{x}$ is not Maximal $\unicode{x21af}$. So, by contradiction, $S \subset C_{x}$. As noted above $C_{x}$ is the unique element of $\mathcal{C}$ containing $x$. So because $x \in S$, any other $C \in \mathcal{C}$ s.t. $S \subset C$ must also contain $x$, and must therefore be $C_{x}$. $\blacksquare$