A space \((X,\mathcal{T})\) is called locally path-connected if, for every \(x \in X\), every neighborhood of \(x\) contains a path-connected neighborhood of \(x\). Show that a connected, locally path-connected space is path-connected.
Solution
If \(X=\varnothing\), then \(X\) is path-connected vacuously. Assume henceforth that \(X\ne\varnothing\).
Fix \(x_0\in X\) and let \[ P=\{x\in X:\text{there is a path in }X\text{ from }x_0\text{ to }x\} \] be the path component of \(x_0\).
1. The set \(P\) is open in \(X\).
Let \(x\in P\). Apply local path-connectedness to the neighborhood \(X\) of \(x\). There is a path-connected neighborhood \(N\) of \(x\) with \[ x\in N\subseteq X. \] Because \(N\) is a neighborhood of \(x\), there is an open set \(U\) such that \[ x\in U\subseteq N. \]
For every \(y\in N\), path-connectedness of \(N\) gives a path in \(N\) from \(x\) to \(y\). Since \(x\in P\), there is a path in \(X\) from \(x_0\) to \(x\). Concatenating these paths gives a path from \(x_0\) to \(y\), so \[ N\subseteq P. \] Consequently \[ x\in U\subseteq P. \] Thus every point of \(P\) has an open neighborhood contained in \(P\), and \(P\) is open.
2. Every path component of \(X\) is open.
The proof of 1 uses only that the chosen point lies in the path component under consideration. Applying the same argument with a basepoint in any path component shows that every path component is open.
3. The set \(P\) is closed in \(X\).
Path components partition \(X\). By 2, every path component other than \(P\) is open, so \[ X\setminus P \] is a union of open sets and hence is open. Therefore \(P\) is closed.
4. We have \(P=X\), and hence \(X\) is path-connected.
The set \(P\) is nonempty because \(x_0\in P\) via the constant path. By 1 and 3 it is both open and closed. Since \(X\) is connected and \(P\ne\varnothing\), connectedness forces \[ P=X. \] Thus every point of \(X\) can be joined to \(x_0\) by a path. Given any \(x,y\in X\), concatenate a reversed path from \(x\) to \(x_0\) with a path from \(x_0\) to \(y\). This gives a path from \(x\) to \(y\), so \(X\) is path-connected.