> [!definition] > > Let $X$ be a [[Topological Space|topological space]], then $X$ is **locally connected** if it admits a [[Base|base]] consisting of [[Connected|connected]] sets. > [!theorem] > > Let $X$ be topological space, then $X$ is locally connected if and only if for each $U \subset X$ open and each connected component $C \subset U$, $C$ is open in $X$. > [!theorem] > > Let $X$ be a locally connected space and $C \subset X$ be connected, then $C$ is a [[Connected Component|connected component]] if and only if $C$ both [[Open Set|open]] and [[Closed Set|closed]]. > > *Proof*. Suppose that $C$ is connected, open, and closed. Let $C' \supset C$ be a connected component of $X$ containing $C$. If $C' \setminus C \ne \emptyset$, then $C' = C \sqcup (C' \setminus C)$ is a disjoint union of open sets, which is impossible.