Suppose \(A, B\subseteq \RR^n\) are connected and not disjoint. Prove that \(A\union B\) is also connected.
Solution
Use that \(X\) is connected iff \(\Hom_{\Top}(X, S^0) = \ts{c_{-1}, c_1}\), i.e. every continuous map from \(X\to \ts{-1, 1}\) is a constant map \(x \mapsvia{c_{-1}} -1\) or \(x \mapsvia{c_1} 1\). Let \(f: A\union B \to S^0\) be arbitrary, and let \(f_1 \da \ro{f}{A}\) and \(f_2 \da \ro{f}{B}\). By connectedness of \(A\), \(f_1\) is a constant map, as is \(f_2\). On the intersection, for \(x\in A \intersect B \neq \emptyset\), we have \(f_1(x) = f_2(x)\) since \(x\in A\) and \(x\in B\). So \(f_1\) and \(f_2\) are constant functions that must map to the same constant, so \(f\) is constant and this \(A\union B\) is connected.