A topological space \(X\) is said to be locally connected if the connected components of each point form a base of neighborhoods of \(X\). Prove that in a locally connected space the connected components of \(X\) are open in \(X\).
Solution
The source’s phrase “the connected components of each point form a base of neighborhoods” does not specify a neighborhood base. We use the standard definition: \(X\) is locally connected if every point has a neighborhood base consisting of connected open sets.
1. Let \(C\) be a connected component of \(X\) and let \(x\in C\).There is a connected open neighborhood \(U_x\) of \(x\).
By local connectedness, the point \(x\) has a neighborhood base of connected open sets. In particular, taking the neighborhood \(X\) itself gives a connected open neighborhood \[ x\in U_x\subseteq X. \]
2. The neighborhood \(U_x\) from 1 satisfies\[ U_x\subseteq C. \]
Both \(C\) and \(U_x\) are connected, and they meet at \(x\). The union of two connected subsets with nonempty intersection is connected, so \[ C\cup U_x \] is connected. Since \(C\) is a connected component, it is maximal among connected subsets of \(X\). Because \[ C\subseteq C\cup U_x, \] maximality forces \[ C\cup U_x=C. \] Hence \(U_x\subseteq C\).
3. The component \(C\) is open in \(X\).
By <1>1–<1>2, for every \(x\in C\) there is an open set \(U_x\) satisfying \[ x\in U_x\subseteq C. \] Therefore \[ C = \bigcup_{x \in C} U_x. \] An arbitrary union of open sets is open, so \(C\) is open.
4. Hence every connected component of a locally connected space is open.
The component \(C\) in 1 was arbitrary, and 3 proves it is open.