Basis for Topology on Cartesian Product

Theorem
Let $$T_1 = \left\{{A_1, \vartheta_1}\right\}$$ and $$T_2 = \left\{{A_2, \vartheta_2}\right\}$$ be topological spaces.

Let $$A_1 \times A_2$$ be the Cartesian product of $$A_1$$ and $$A_2$$.

Let $$\vartheta$$ be the product topology for $$A_1 \times A_2$$.

Then $$\vartheta$$ is a topology on $$A_1 \times A_2$$.

Proof
From the definition, $$\vartheta$$ is the topology with basis $$\mathcal{B} = \left\{{U_1 \times U_2: U_1 \in \vartheta_1, U_2 \in \vartheta_2}\right\}$$.

We need to show that conditions B1 and B2 in the definition for basis hold for $$\mathcal{B}$$.


 * B1: Since $$A_1 \in \vartheta_1$$ and $$A_2 \in \vartheta_2$$, $$A_1 \times A_2 \in \mathcal{B}$$.


 * B2: Suppose $$U_1, V_1 \in \vartheta_1$$ and $$U_2, V_2 \in \vartheta_2$$.

From Cartesian Product of Intersections, $$\left({U_1 \times U_2}\right) \cap \left({V_1 \times V_2}\right) = \left({U_1 \cap V_1}\right) \times \left({U_2 \cap V_2}\right)$$.

Since $$\left({U_1 \cap V_1}\right) \in \vartheta_1$$ and $$\left({U_2 \cap V_2}\right) \in \vartheta_2$$, we have $$\left({U_1 \cap V_1}\right) \times \left({U_2 \cap V_2}\right) \in \mathcal{B}$$.

So $$\left({U_1 \times U_2}\right) \cap \left({V_1 \times V_2}\right)$$ is the union of (one) set in $$\mathcal{B}$$.

Hence the result.