Let \(X\) be locally path connected. Show that every connected open set in \(X\) is path connected.
Solution
Let \(U\subseteq X\) be open and connected. Since \(X\) is locally path connected, the subspace \(U\) is locally path connected: if \(x\in U\), choose a path-connected open neighborhood \(V\) of \(x\) contained in \(U\).
In a locally path-connected space, path components are open. Indeed, if \(P\) is a path component and \(x\in P\), choose a path-connected open neighborhood \(V\) of \(x\); then every point of \(V\) is path connected to \(x\), so \(V\subseteq P\). Thus each path component of \(U\) is open in \(U\).
Distinct path components are disjoint, and their union is \(U\). If there were more than one, one path component and the union of all the others would give a separation of connected \(U\). Hence \(U\) has only one path component and is path connected.