Basis for Box Topology

Theorem
Let $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.

Let $S$ be the cartesian product of $\family {S_i}_{i \mathop \in I}$.

That is:
 * $\ds S := \prod_{i \mathop \in I} S_i$

Define:
 * $\ds \BB := \set {\prod_{i \mathop \in I} U_i: \forall i \in I: U_i \in \tau_i}$

Then $\BB$ is a synthetic basis on $S$.

Proof
Let us check the two conditions for $\BB$ to be a synthetic basis in turn.

$(\text B 1)$: Covering
From open set axiom $(\text O 3)$, we have $S_i \in \tau_i$ for all $i \in I$.

Thus $S = \ds \prod_{i \mathop \in I} S_i \in \BB$.

Hence by Set is Subset of Union: General Result, we have:


 * $S \subseteq \ds \bigcup \BB$

so $\BB$ is a cover for $S$.

$(\text B 2)$: Intersections are Unions
Let $A = \ds \prod_{i \mathop \in I} U_i$ and $B = \ds \prod_{i \mathop \in I} V_i$ be in $\BB$.

Then by Cartesian Product of Intersections, we have:


 * $A \cap B = \ds \prod_{i \mathop \in I} \paren {U_i \cap V_i}$

By open set axiom $(\text O 2)$, $U_i \cap V_i \in \tau_i$ for all $i \in I$.

Hence $A \cap B \in \BB$, and in particular, by Union of Singleton:


 * $A \cap B = \ds \bigcup \set {A \cap B}$

Having verified both conditions, we conclude $\BB$ is a synthetic basis on $S$.

Also see

 * Definition:Box Topology
 * Synthetic Basis and Analytic Basis are Compatible