Lemma
If $\mathfrak{X}=\{X_i:i\in I\}$ is a collection of topological spaces and if $\mathfrak{B}=\{\mathcal{B_i}: i\in I\}$ is a collection of basic neighbourhood system for $\pi_i(x)$ for any $i\in I$ and for $x\in\prod_{i\in I}X_i$ then the collection $\mathcal{B}=\{B\subseteq\prod_{i\in I}X_i:\pi_i[B]\in\mathcal{B}_i,\forall i\in I\}$ is a basic neighbourhood system for $x$.
Proof. Since the proiections are open then if $V$ is a neighbourhood of $x\in\prod_{i\in I}X_i$ then
$\pi_i[V]$ is a neighbourhood of $\pi_i(x)$ for each $i\in I$ and so there exist $B_i\in\mathcal{B}_i$ such that $\pi_i(x)\in B_i\subseteq\pi_i[V]$ so that $x\in\bigcap_{i\in I}\pi^{-1}_i[B_i]\subseteq\bigcap_{i\in I}\pi^{-1}_i\big[\pi_i[V]\big]=V$ but $\bigcap_{i\in I}\pi^{-1}_i[B_i]\in\mathcal{B}$ because $\pi_j\big[\bigcap_{i\in I}\pi^{-1}_i[B_i]\big]=B_j$ for any $j\in J$.
So I ask if the statement of the lemma is true and if not I ask to take a counterexample. Furthermore if the proof is correct: in particular I suspect that the equilities $\bigcap_{i\in I}\pi^{-1}_i\big[\pi_i[V]\big]=V$ and $\pi_j\big[\bigcap_{i\in I}\pi^{-1}_i[B_i]\big]=B_j$ are false so I ask to prove them. So could someone help me, please?
Best Answer
The sets in $\mathcal{B}$ are not even neighbourhoods in the product topology, so they certainly do not form a neighbourhood system at all, if $I$ is infinite.
A better choice (that does work)
$$\mathcal{B}(x) = \{\bigcap_{i \in F} \pi^{-1}[B_i]: F \subseteq I \text{ finite and } \forall i \in F: B_i \in \mathcal{B}_i\}$$
Using this we can e.g. show that a countable product of first countable spaces is first countable.