Let $X_\alpha$ with $\alpha\in A$ be a collection of topological spaces, and let $Z_\alpha\subseteq X_\alpha$ be subsets. Let $$ X=\prod_{\alpha} X_\alpha $$ and $$ Z = \prod_{\alpha} Z_\alpha. $$ In other words, $z\in Z$ means that $z$ is a function $z\colon A\to \cup_\alpha Z_\alpha$ such that $z(\alpha)\in Z_\alpha$ for each $\alpha$ . Thus, $z\in X$ and we have $$ Z\subseteq X $$ as sets.

Proof. Let us denote by $\tau_X$ and $\tau_Z$ the product topologies for $X$ and $Z$ , respectively. Also, let $$ \pi_{X,\alpha}\colon X\to X_\alpha, \quad \pi_{Z,\alpha}\colon Z\to Z_\alpha $$ be the canonical projections defined for $X$ and $Z$ . The subbases for $X$ and $Z$ are given by \begin{eqnarray*} \beta_X &=& \{ \pi_{X,\alpha}^{-1}(U) : \alpha \in A, U\in \tau(X_\alpha) \}, \\ \beta_Z &=& \{ \pi_{Z,\alpha}^{-1}(U) : \alpha \in A, U\in \tau(Z_\alpha) \}, \end{eqnarray*}where $\tau(X_\alpha)$ is the topology of $X_\alpha$ and $\tau(Z_\alpha)$ is the subspace topology of $Z_\alpha\subseteq X_\alpha$ . The claim follows as $$ \beta_Z = \{ B\cap Z : B\in \beta_X \}. $$