# Basis for Box Topology

## Theorem

Let $\left \langle {\left({S_i, \tau_i}\right)} \right \rangle_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.

Let $S$ be the cartesian product of $\left \langle {S_i} \right \rangle_{i \mathop \in I}$.

That is:

$\displaystyle S := \prod_{i \mathop \in I} S_i$

Define:

$\displaystyle \mathcal B := \left\{{\prod_{i \mathop \in I} U_i: \forall i \in I: U_i \in \tau_i}\right\}$

Then $\mathcal B$ is a synthetic basis on $S$.

## Proof

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

### $(B1)$: Covering

From open set axiom $(O3)$, we have $S_i \in \tau_i$ for all $i \in I$.

Thus $S = \displaystyle \prod_{i \mathop \in I} S_i \in \mathcal B$.

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

$S \subseteq \displaystyle \bigcup \mathcal B$

so $\mathcal B$ is a cover for $S$.

$\Box$

### $(B2)$: Intersections are Unions

Let $A = \displaystyle \prod_{i \mathop \in I} U_i$ and $B = \displaystyle \prod_{i \mathop \in I} V_i$ be in $\mathcal B$.

Then by Cartesian Product of Intersections, we have:

$A \cap B = \displaystyle \prod_{i \mathop \in I} \left({U_i \cap V_i}\right)$

By open set axiom $(O2)$, $U_i \cap V_i \in \tau_i$ for all $i \in I$.

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

$A \cap B = \displaystyle \bigcup \left\{{A \cap B}\right\}$

$\Box$

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

$\blacksquare$