Proposition 4.5.6. Let $\seqi{X}$ be topological spaces and $\seqi{A}$ such that each $A_{i} \subset X_{i}$ is dense, then $\prod_{i \in I}A_{i}$ is dense in $\prod_{i \in I}X_{i}$.
Proof. Let $x \in \prod_{i \in I}X_{i}$ and $U \in \cn(x)$, then there exists $J \subset I$ finite and $\seqf{U_j}$ such that $U_{j} \in \cn(\pi_{j}(x))$ for each $j \in J$, and $x \in \bigcap_{j \in J}\pi_{j}^{-1}(U_{j}) \subset U$. By density of each $A_{j}$, there exists $y \in X$ such that $y_{j} \in U_{j}$ for each $j \in J$. Thus $y \in U$, and $A$ is dense.$\square$