Basis for Topology on Cartesian Product

Theorem
Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be topological spaces.

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

Let $\PP = \set {U_1 \times U_2 : U_1 \in \tau_1, U_2 \in \tau_2}$

Then $\PP$ is a synthetic basis on $A_1 \times A_2$.

Proof
We need to show that conditions $(\text B 1)$ and $(\text B 2)$ in the definition for basis hold for $\PP$.


 * $(\text B 1)$: Since $A_1 \in \tau_1$ and $A_2 \in \tau_2$, $A_1 \times A_2 \in \PP$.


 * $(\text B 2)$: Suppose $U_1, V_1 \in \tau_1$ and $U_2, V_2 \in \tau_2$.

From Cartesian Product of Intersections, $\paren {U_1 \times U_2} \cap \paren {V_1 \times V_2} = \paren {U_1 \cap V_1} \times \paren {U_2 \cap V_2}$.

Since $\paren {U_1 \cap V_1} \in \tau_1$ and $\paren {U_2 \cap V_2} \in \tau_2$, we have $\paren {U_1 \cap V_1} \times \paren {U_2 \cap V_2} \in \PP$.

So $\paren {U_1 \times U_2} \cap \paren {V_1 \times V_2}$ is the union of (one) set in $\PP$.

Hence the result.

Also see

 * Natural Basis of Tychonoff Topology of Finite Product where it is shown that the basis $\PP$ is the natural basis of the Tychonoff topology on a finite Cartesian product