Natural Basis of Product Topology/Lemma 2

Theorem
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
 * $\displaystyle X := \prod_{i \mathop \in I} X_i$

Let $\BB$ be the set of cartesian products of the form $\displaystyle \prod_{i \mathop \in I} U_i$ where:
 * for all $i \in I : U_i \in \tau_i$
 * for all but finitely many indices $i : U_i = X_i$

Then:
 * $\forall B_1, B_2 \in \BB : B_1 \cap B_2 \in \BB$

Proof
Let $B_1, B_2 \in \BB$.

Let $B_1 = \displaystyle \prod_{i \mathop \in I} U_i$ where:
 * for all $i \in I : U_i \in \tau_i$
 * for all but finitely many indices $i : U_i = X_i$

Let $J_1$ be the finite set of indices such that:
 * $J_1 = \set{j \in I : U_i \neq X_i}$

Let $B_2 = \displaystyle \prod_{i \mathop \in I} V_i$ where:
 * for all $i \in I : V_i \in \tau_i$
 * for all but finitely many indices $i : V_i = X_i$

Let $J_2$ be the finite set of indices such that:
 * $J_1 = \set{j \in I : V_i \neq X_i}$

Then:

Now:
 * $\forall i \in I : U_i, V_i \in \tau_i \leadsto U_i \cap V_i \in \tau_i$
 * $\forall i \in I \setminus \paren{J_1 \cup J_2} : U_i, V_i = X_i \leadsto U_i \cap V_i = X_i$

Since $J_1 \cup J_2$ is finite then:
 * for all but finitely many indices $i : U_i \cap V_i = X_i$

Thus $B_1 \cap B_2 \in \BB$