Proposition 4.7.2. Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\B$ be a filter base on $\prod_{i \in I}X_{i}$, then $\fB$ converges to $x \in \prod_{i \in I}X_{i}$ if and only if $\pi_{i}(\fB)$ converges to $\pi_{i}(x)$ for all $i \in I$.
Proof. $(\Rightarrow)$: Let $i \in I$ and $U \in \cn(\pi_{i}(x))$, then $\pi_{i}^{-1}(U) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset \pi_{i}^{-1}(U)$. In which case, $\pi_{i}(B) \subset U$ and $\pi_{i}(\fB)$ converges to $\pi_{i}(x)$.
$(\Leftarrow)$: Let $U \in \cn(x)$, then there exists $\seqf{i_k}$ and open sets $\seqf{U_k}$ such that $x \in \bigcap_{k = 1}^{n} \pi_{i_k}^{-1}(U_{k}) \subset U$. For each $1 \le k \le n$, since $\pi_{i_k}(\fB)$ converges to $\pi_{i_k}(x)$ for each $1 \le k \le n$, there exists $B_{k} \in \fB$ such that $B_{k} \subset \pi_{i_k}^{-1}(U_{k})$. As $\fB$ is a filter base, there exists $B \in \fB$ with
Therefore $\fB$ converges to $x$.$\square$