Let \(X\) be a locally path-connected topological space. Prove that: (1) Every open subset \(U \subseteq X\) is locally path-connected. (2) \(X\) is connected if and only if \(X\) is path-connected. (3) Every path component of \(X\) is a connected component of \(X\). (4) Every connected component (and path component) of \(X\) is open in \(X\).
Solution
1. Let \(U\subseteq X\) be open. If \(x\in U\) and \(V\) is a neighborhood of \(x\) in \(U\), then \(V\) contains an open neighborhood \(V'\) of \(x\) in \(X\) with \(V'\subseteq U\). Since \(X\) is locally path-connected, \(V'\) contains a path-connected open neighborhood \(W\) of \(x\). Thus \(U\) is locally path-connected.
2. Every path component \(P\) of \(X\) is open.
2.1. If \(x\in P\), choose a path-connected open neighborhood \(W\) of \(x\).
2.2. Every point of \(W\) can be joined to \(x\) by a path, so \(W\subseteq P\).
2.3. Hence \(P\) is a union of open sets and is open.
3. Path components coincide with connected components.
3.1. Every path-connected set is connected, so each path component \(P\) lies in a connected component \(C\).
3.2. The path components partition \(C\), and each is open in \(C\) by
2.
2.3. If \(C\) contained two or more path components, one path component and the union of the others would form a separation of \(C\). This contradicts connectedness. Hence \(C=P\).
4. Therefore every connected component is also a path component and is open. Components are always closed, so in a locally path-connected space every component is clopen.
5. Finally, path-connectedness always implies connectedness. Conversely, if \(X\) is connected, it has only one connected component; by 3 that component is a path component. Hence \(X\) is path-connected.