Let \((X_i)_{i\in I}\) be a family of topological spaces, let \(X \coloneqq \prod_{i\in I} X_i\), and for \(j\in I\) let \(p_j\colon X\to X_j\) be the projection. The box topology on \(X\) is the topology generated by the basis \[ \mathcal B_{\Box} \coloneqq \ts{ \prod_{i\in I} U_i \st U_i \text{ open in } X_i \text{ for every } i }. \] The product topology on \(X\) is the topology generated by the basis \[ \mathcal B_{\prod} \coloneqq \ts{ \prod_{i\in I} U_i \st U_i \text{ open in } X_i \text{ for every } i, \text{ and } U_i = X_i \text{ for all but finitely many } i }. \]
The product topology on \(X\) is the coarsest topology on \(X\) for which every projection \(p_j\) is continuous. It is the topology generated by the subbasis \(\ts{ \inverseof{p_j}(U) \st j\in I,\ U \text{ open in } X_j }\).
Give \(X\) the product topology. For every topological space \(Y\) and every family of continuous maps \(f_i\colon Y\to X_i\), there is a unique continuous map \(f\colon Y\to X\) with \(p_i\circ f = f_i\) for every \(i\in I\). Hence \((X, (p_i)_{i\in I})\) is the product of the \(X_i\) in \(\Top\):
Every element of \(\mathcal B_{\prod}\) lies in \(\mathcal B_{\Box}\), so the box topology is finer than the product topology. The two topologies coincide when \(I\) is finite.
For the box topology, the universal property of the product fails. Let \(X = \prod_{n\geq 1}\RR\) and let \(\Delta\colon\RR\to X\), \(\Delta(t) = (t,t,\ldots)\), whose components are all \(\id_\RR\). Then \(\Delta\) is continuous for the product topology, but not for the box topology: \(\prod_{n\geq1}(-1/n, 1/n)\) is open in the box topology and \(\inverseof{\Delta}\qty{\prod_{n\geq1}(-1/n,1/n)} = \ts{0}\) is not open in \(\RR\).
See [@Mun00].