Let \(X\) and \(X'\) denote a single set in the topologies \(\mathcal{T}\) and \(\mathcal{T}'\), respectively; let \(Y\) and \(Y'\) denote a single set in the topologies \(\mathcal{U}\) and \(\mathcal{U}'\), respectively. Assume these sets are nonempty.
Show that if \(\mathcal{T}' \supset \mathcal{T}\) and \(\mathcal{U}' \supset \mathcal{U}\), then the product topology on \(X' \times Y'\) is finer than the product topology on \(X \times Y\).
Does the converse of (a) hold? Justify your answer.
Solution
A basis for the product topology on \(X\times Y\) consists of \(U\times V\) with \(U\in\mathcal T\) and \(V\in\mathcal U\). Since \(\mathcal T\subseteq\mathcal T'\) and \(\mathcal U\subseteq\mathcal U'\), every such rectangle is open in the product topology on \(X'\times Y'\). Hence the latter topology is finer.
The converse does hold when both underlying sets are nonempty. Suppose the product topology from \((\mathcal T',\mathcal U')\) is finer than that from \((\mathcal T,\mathcal U)\). Let \(U\in\mathcal T\) and choose \(y_0\in Y\). Then \(U\times Y\) is open in the coarser product topology and therefore in the finer one. For each \(x\in U\), choose a basic finer-product neighborhood \[ x\in U'_x,\qquad y_0\in V'_x,\qquad U'_x\times V'_x\subseteq U\times Y. \] Then \(U'_x\subseteq U\), so \[ U=\bigcup_{x\in U}U'_x\in\mathcal T'. \] Thus \(\mathcal T\subseteq\mathcal T'\). The same argument using a fixed \(x_0\in X\) gives \(\mathcal U\subseteq\mathcal U'\).