Let \(X\) be a topological space, and \(A,B\subseteq X\) be connected subsets of \(X\). Show that if \(A\cap B\ne\varnothing\) then \(A\cup B\) is a connected subset of \(X\).
Solution
1. Suppose for contradiction that \(A\cup B\) is disconnected.Then there are disjoint nonempty sets \(U,V\), open in the subspace \(A\cup B\), such that\[ A\cup B=U\cup V. \]
This is the definition of disconnectedness of the subspace \(A\cup B\).
2. Since \(A\) is connected, it is contained entirely in \(U\) or entirely in \(V\).
The sets \(A\cap U\) and \(A\cap V\) are disjoint and open in the subspace \(A\), and \[ A=(A\cap U)\cup(A\cap V). \] If both were nonempty they would separate \(A\), contradicting connectedness. Thus one is empty, so \(A\) lies in the other member of the separation.
3. Likewise, \(B\) is contained entirely in \(U\) or entirely in \(V\).
Apply the argument of 2 to the connected subspace \(B\).
4. The sets \(A\) and \(B\) must lie in the same member of the separation.
Choose \(p\in A\cap B\), which exists by hypothesis. If \(A\subseteq U\), then \(p\in U\); since \(p\in B\) and \(B\) lies entirely in one of \(U,V\) by 3, disjointness forces \(B\subseteq U\). The case \(A\subseteq V\) is identical.
5. This contradicts that both \(U\) and \(V\) are nonempty.
By 4, either \(A\cup B\subseteq U\) or \(A\cup B\subseteq V\). Since \(A\cup B=U\cup V\), the other set would then be empty.
6. Therefore \(A\cup B\) is connected.
The assumption of disconnectedness in 1 leads to the contradiction in 5.