Suppose \(A\), \(B\) are connected subsets of \(\RR^n\) which are not disjoint. Prove that their union \(A \cup B\) is also connected.
Solution
1. Suppose, for contradiction, that \(A\cup B\) is disconnected. Then thereare disjoint nonempty subsets \(U,V\subseteq A\cup B\), open in the subspacetopology, such that\[ A\cup B=U\cup V. \]
This is the definition of disconnectedness.
2. The connected set \(A\) is contained entirely in \(U\) or entirely in \(V\).
We have \[ A=(A\cap U)\cup(A\cap V). \] The two sets on the right are disjoint and open in the subspace topology on \(A\). If both were nonempty, they would separate \(A\), contradicting the connectedness of \(A\). Hence one is empty, so \(A\subseteq U\) or \(A\subseteq V\).
3. The connected set \(B\) is contained entirely in \(U\) or entirely in \(V\).
The same argument as in step 2, applied to \[ B=(B\cap U)\cup(B\cap V), \] uses the connectedness of \(B\).
4. The sets \(A\) and \(B\) must lie in the same member of the separation.
Choose \[ p\in A\cap B, \] which exists because \(A\) and \(B\) are not disjoint. Since \(U\) and \(V\) are disjoint, the point \(p\) belongs to exactly one of them. Steps <1>2–<1>3 then force both \(A\) and \(B\) to lie in that same set.
5. The union \(A\cup B\) is connected.
1. Therefore no separation exists,so \(A\cup B\) is connected.
6. Q.E.D.
Step 5 is the required conclusion.