Prove Theorem 19.2: suppose the topology on each space \(X_\alpha\) is given by a basis \(\mathcal{B}_\alpha\). The collection of all sets of the form \(\prod_{\alpha \in J} B_\alpha\), where \(B_\alpha \in \mathcal{B}_\alpha\) for each \(\alpha\), serves as a basis for the box topology on \(\prod_{\alpha \in J} X_\alpha\); and the collection of all sets of the same form, where \(B_\alpha \in \mathcal{B}_\alpha\) for finitely many indices \(\alpha\) and \(B_\alpha = X_\alpha\) for all the remaining indices, serves as a basis for the product topology.
Solution
For the box topology, let \[ \mathcal B=\left\{\prod_{\alpha\in J}B_\alpha:B_\alpha\in\mathcal B_\alpha\right\}. \] It covers the product because each \(\mathcal B_\alpha\) covers \(X_\alpha\). If \[ x\in\prod B_\alpha\cap\prod C_\alpha, \] then for every \(\alpha\), because \(\mathcal B_\alpha\) is a basis, there is \(D_\alpha\in\mathcal B_\alpha\) such that \[ x_\alpha\in D_\alpha\subseteq B_\alpha\cap C_\alpha. \] Hence \[ x\in\prod D_\alpha\subseteq\left(\prod B_\alpha\right)\cap\left(\prod C_\alpha\right), \] so \(\mathcal B\) is a basis. Each member of \(\mathcal B\) is a product of open sets, hence box-open. Conversely, if \(x\in\prod U_\alpha\) with each \(U_\alpha\) open, choose \(B_\alpha\in\mathcal B_\alpha\) with \(x_\alpha\in B_\alpha\subseteq U_\alpha\); then \(x\in\prod B_\alpha\subseteq\prod U_\alpha\). So every box-open set is a union of members of \(\mathcal B\), and \(\mathcal B\) generates the box topology.
For the product topology, let \(\mathcal B'\) consist of the products \(\prod B_\alpha\) with \(B_\alpha\in\mathcal B_\alpha\) for \(\alpha\) in a finite set \(F\) and \(B_\alpha=X_\alpha\) for \(\alpha\notin F\). It covers the product. If \(x\in\prod B_\alpha\cap\prod C_\alpha\) with finite index sets \(F\) and \(G\), choose \(D_\alpha\in\mathcal B_\alpha\) with \(x_\alpha\in D_\alpha\subseteq B_\alpha\cap C_\alpha\) for \(\alpha\in F\cup G\), and put \(D_\alpha=X_\alpha\) otherwise; then \(\prod D_\alpha\in\mathcal B'\) and \(x\in\prod D_\alpha\subseteq\prod B_\alpha\cap\prod C_\alpha\). So \(\mathcal B'\) is a basis. Its members are basic product-open sets. If \(x\in\prod U_\alpha\) with \(U_\alpha\) open and \(U_\alpha=X_\alpha\) outside a finite set \(F\), choose \(B_\alpha\in\mathcal B_\alpha\) with \(x_\alpha\in B_\alpha\subseteq U_\alpha\) for \(\alpha\in F\) and \(B_\alpha=X_\alpha\) otherwise; then \(x\in\prod B_\alpha\subseteq\prod U_\alpha\). Hence \(\mathcal B'\) generates the product topology.