A space \(X\) is said to be weakly locally connected at \(x\) if for every neighborhood \(U\) of \(x\), there is a connected subspace of \(X\) contained in \(U\) that contains a neighborhood of \(x\). Show that if \(X\) is weakly locally connected at each of its points, then \(X\) is locally connected. [Hint: Show that components of open sets are open.]
Solution
It suffices to show that every component of every open set is open.
Let \(U\subseteq X\) be open, let \(C\) be a component of \(U\), and let \(x\in C\). By weak local connectedness at \(x\), applied to the neighborhood \(U\), there is a connected subspace \(A_x\subseteq U\) that contains a neighborhood \(V_x\) of \(x\). Since \(A_x\) is connected and meets the component \(C\) at \(x\), maximality of components gives \[ A_x\subseteq C. \] Hence \[ x\in V_x\subseteq C. \] So every point of \(C\) is an interior point of \(C\), and \(C\) is open.
Now, given an open neighborhood \(U\) of a point \(x\), the component of \(U\) containing \(x\) is a connected open neighborhood of \(x\) contained in \(U\). Therefore \(X\) is locally connected.