Show that if \(A\) is a proper subset of a connected space \(X\) and \(B\) is a proper subset of a connected space \(Y\), then \((X\times Y)\setminus(A\times B)\) is connected.
Solution
Since \(A\) and \(B\) are proper, choose \(x_0\in X\setminus A\) and \(y_0\in Y\setminus B\). We use that a union of connected subspaces each of which meets a fixed connected subspace \(S\), together with \(S\), is connected.
1. \((X \times Y) \setminus (A \times B) = \bigl((X \setminus A) \times Y\bigr) \cup \bigl(X \times (Y \setminus B)\bigr)\).
\((x, y) \notin A \times B\) if and only if \(x \notin A\) or \(y \notin B\).
2. \(S = (\{x_0\} \times Y) \cup (X \times \{y_0\})\) is a connected subset of \((X \times Y) \setminus (A \times B)\).
The slices \(\{x_0\} \times Y \cong Y\) and \(X \times \{y_0\} \cong X\) are connected and share the point \((x_0, y_0)\), so their union is connected. It lies in the complement by step 1, since \(x_0\notin A\) and \(y_0\notin B\).
3. Each slice \(\{x\} \times Y\) with \(x \in X \setminus A\) and each slice \(X \times \{y\}\) with \(y \in Y \setminus B\) is a connected subset of the complement meeting \(S\).
1. The slice \(\{x\} \times Y\) contains \((x, y_0) \in S\), and \(X \times \{y\}\) contains \((x_0, y) \in S\).
4. Q.E.D.
By step 1, the complement is the union of the slices in step 3, and it contains \(S\). Each slice is connected and meets the connected set \(S\) of step 2, so the union of \(S\) and all the slices, which is the complement, is connected.