Let \(\theset{X_\alpha \mid \alpha \in A}\) be a family of connected subspaces of a space \(X\) such that there is a point \(p \in X\) which is in each of the \(X_\alpha\).
Show that the union of the \(X_\alpha\) is connected.
Solution
Let \[ Y=\bigcup_{\alpha\in A}X_\alpha. \]
1. Any separation of \(Y\) would separate one of the subspaces \(X_\alpha\).
Suppose \[ Y=U\disjoint V. \] is a separation. Since the common point \(p\) belongs to \(Y\), after interchanging \(U\) and \(V\) if necessary assume \(p\in U\).
Choose \(q\in V\). Since \(q\in Y\), there is some \(\alpha\) with \(q\in X_\alpha\). Then \[ p\in U\cap X_\alpha, \qquad q\in V\cap X_\alpha. \] The sets \(U\cap X_\alpha\) and \(V\cap X_\alpha\) are disjoint, nonempty, open in the subspace \(X_\alpha\), and their union is \(X_\alpha\). Thus they form a separation of \(X_\alpha\).
2. \(Y\) is connected.
If \(Y\) were disconnected, a separation would exist. By 1 it would separate some \(X_\alpha\), contradicting the hypothesis that every \(X_\alpha\) is connected.