> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is locally path-connected if it has a [[Base|base]] consisting of [[Path-Connected|path-connected]] sets. > [!theorem] > > Let $X$ be a locally path-connected topological space, then > 1. The [[Connected Component|connected components]] of $X$ are open. > 2. The path components of $X$ are equal to its components. > 3. $X$ is connected if and only if it is path-connected. > 4. Every [[Open Set|open]] subset of $X$ is locally path-connected. > > ### Connected Components are Open > > Let $E \subset X$ be a connected component, then $E$ is open. > > *Proof*. Let $x \in E$, then since $X$ is locally path-connected, there exists a connected neighbourhood $U \in \cn(x)$. As $E$ is a connected component, $U \subset E$, so $x \in U^o \subset E^o \subset E$. Therefore $E^o = E$ and $E$ is open. > > > ### Path Components Coincide with Components > > Let $x \in X$, $P$ be a path-connected component containing $x$, then the connected component containing $P$ is equal to $P$. > > *Proof*. Let $C \supset P$ be the connected component containing $P$. Suppose that $C$ is not path-connected. Let $\seqi{P}$ be the family of all path-connected components in $C$. Let $P' = \bigcup_{P_i \ne P}P_i$, then $P$ and $P'$ are disjoint open sets with $C = P \cup P'$, which contradicts the fact that $C$ is connected. > > ### Inheritance > > Let $U \subset X$ be [[Open Set|open]], then $U$ is also locally path-connected. > > *Proof*. Let $x \in U$ and $V \in \cn^o(x)$ be open in the relative topology, then there exists $\widehat{V}$ such that $V = \widehat{V} \cap U$, making $V$ open in the original topology as well. Therefore there exists a connected neighbourhood of $x$ contained in $V$.