Exercise 19.1

Proof. We show that $C$ is a basis of the box or product topology using Lemma 13.2. First, it is easy to see that $C$ is a collection of open sets. Consider any $B\in C$ so that $B=\prod <!--\; FUNCTION\; APPLICATION\; -->B_{\alpha }$ where each $B_{\alpha }\in B_{\alpha }$ (for a finitely many $\alpha \in J$ and $B_{\alpha }=X_{\alpha }$ for the rest in the product topology). Since each $B_{\alpha }$ is a basis element of $X_{\alpha }$ (or $X_{\alpha }$ itself), they are open so that $B$ is a basis element of the box or product topology by definition and therefore open. Note that the basis for the product topology is given directly by Theorem 19.1.

Now suppose that $U$ is any open set of the box topology and consider any $x\in U$. Then it follows that there is a basis element ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{U}_{\alpha }$ of the box or product topology containing $x$ where ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{U}_{\alpha }\subset U$. Thus each $U_{\alpha }$ is an open set of $X_{\alpha }$ (or $U_{\alpha }=X_{\alpha }$ for all but finitely many $\alpha \in J$ for the product topology). Also $x\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{U}_{\alpha }$ so that $x={(x_{\alpha })}_{\alpha \in J}$ where each $x_{\alpha }\in U_{\alpha }$. It then follows that there is basis element $B_{\alpha }\in B_{\alpha }$ of $X_{\alpha }$ containing $x_{\alpha }$ where $B_{\alpha }\subset U_{\alpha }$ (for $U_{\alpha }=X_{\alpha }$ we simply set $B_{\alpha }=X_{\alpha }$ as well).

Then clearly $x\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{B}_{\alpha }$ and ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{B}_{\alpha }\in C$. Consider also any $y\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{B}_{\alpha }$ so that $y={(y_{\alpha })}_{\alpha \in J}$ where each $y_{\alpha }\in B_{\alpha }$. Then also each $y_{\alpha }\in U_{\alpha }$ since $B_{\alpha }\subset U_{\alpha }$. This suffices to show that $y\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{U}_{\alpha }\subset U$. Since $y$ was arbitrary this shows that ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{B}_{\alpha }\subset U$. Therefore $C$ is a basis of the box topology by Lemma 13.2. □