Exercise 19.2

Proof. The basis of the box or product topologies on $\prod <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ is the collection of sets $\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$, where each $V_{\alpha }$ is open in $A_{\alpha }$ and, in the case of the product topology, $V_{\alpha }=A_{\alpha }$ for all but finitely many $\alpha \in J$ (by Theorem 19.1). Denote this basis collection by $C$. By Lemma 16.1, the collection

is a basis of the subspace topology on $\prod <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$, where $B$ is the basis of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$. To prove that $\prod <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ is a subspace of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$, it, therefore, suffices to show that $C=B_A$.

$\text{(⊂)}$ First consider any element $B\in C$ so that $B=\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ for open sets $V_{\alpha }$ in $A_{\alpha }$ (and $V_{\alpha }=A_{\alpha }$ for all but finite many $\alpha \in J$ for the product topology). For each $\alpha \in J$, we then have that $V_{\alpha }=U_{\alpha }\cap A_{\alpha }$ for some open set $U_{\alpha }$ in $X_{\alpha }$ since $A_{\alpha }$ is a subspace of $X_{\alpha }$. Note that this is true even for those $\alpha$ where $V_{\alpha }=A_{\alpha }$ in the product topology since then $V_{\alpha }=A_{\alpha }=X_{\alpha }\cap A_{\alpha }$. In fact, for these $\alpha$ we need to choose $U_{\alpha }=X_{\alpha }$ as will become apparent. We then have the following:

Since $U_{\alpha }=X_{\alpha }$ for all but a finitely many $\alpha \in J$ for the product topology, we have that $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ is a basis element of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$, i.e. $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }\in B$. This shows that $B\in B_A$ so that $C\subset B_A$ since $B$ was arbitrary.

$\text{(⊃)}$ Now suppose that $B\in B_A$ so that $B=B_X\cap \prod <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ for some basis element $B_X\in B$ of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$. We then have that $B_X=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ where each $U_{\alpha }$ is an open set of $X_{\alpha }$ (and $U_{\alpha }=X_{\alpha }$ for all but finitely many $\alpha \in J$ for the product topology). Then let $V_{\alpha }=U_{\alpha }\cap A_{\alpha }$ for each $\alpha \in J$, noting that $V_{\alpha }=X_{\alpha }\cap A_{\alpha }=A_{\alpha }$ when $U_{\alpha }=X_{\alpha }$. Following the above chain of logical equivalences in reverse order then shows that $B=\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ so that $B\in C$ since clearly each $V_{\alpha }$ is open in the subspace topology $A_{\alpha }$. Hence $C\supset B_A$ since $B$ was arbitrary. □