Lemma 15.2.2. Let $\bracs{(X_i, \cm_i)}_{i \in I}$, let
\[\ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j}\]
then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_{i}$.
Proof.
For any $j \in I$, $\emptyset = \pi_{j}^{-1}(\emptyset) \in \ce$.
Let
\[\bigcap_{j \in J}\pi_{j}^{-1}(E_{j}), \bigcap_{j \in J'}\pi_{j}^{-1}(F_{j}) \in \ce\]Assume without loss of generality that $J = J'$, then
\[\bigcap_{j \in J}\pi_{j}^{-1}(E_{j}) \cap \bigcap_{j \in J'}\pi_{j}^{-1}(F_{j}) = \bigcap_{j \in J}\pi_{j}^{-1}(E_{j} \cap F_{j}) \in \ce\]For any $j \in I$, $\prod_{i \in I}X_{i} = \pi_{j}^{-1}(X_{j}) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_{j}^{-1}(E_{j}) \in \ce$, then
\[\braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^{c} = \bigcup_{j \in J}\pi_{j}^{-1}(E_{j}^{c}) = \bigsqcup_{\emptyset \subsetneq K \subset J}\underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}\]
$\square$