Prove Theorem 19.3: let \(A_\alpha\) be a subspace of \(X_\alpha\), for each \(\alpha \in J\). Then \(\prod A_\alpha\) is a subspace of \(\prod X_\alpha\) if both products are given the box topology, or if both products are given the product topology.
Solution
Let \(A=\prod A_\alpha\subseteq X=\prod X_\alpha\).
For the box topology, a basic open set of the subspace \(A\) has the form \[ A\cap\prod U_\alpha=\prod(A_\alpha\cap U_\alpha), \] which is a basic box-open set in the product of the subspaces \(A_\alpha\). Conversely every basic box-open set in \(\prod A_\alpha\) has this form. Thus the topologies agree.
For the product topology the same computation applies, with \(U_\alpha=X_\alpha\) for all but finitely many \(\alpha\). Then \[ A_\alpha\cap U_\alpha=A_\alpha \] for all but finitely many \(\alpha\), exactly the basis condition for the product topology on \(\prod A_\alpha\). Hence the two subspace descriptions agree in the product topology as well.