Product Topology is Topology

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Theorem

Let $T_1 = \left({A_1, \tau_1}\right)$ and $T_2 = \left({A_2, \tau_2}\right)$ be topological spaces.

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

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


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


Proof

From the definition, $\tau$ is the topology with basis $\mathcal P = \left\{{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}\right\}$.

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

B1: Since $A_1 \in \tau_1$ and $A_2 \in \tau_2$, $A_1 \times A_2 \in \mathcal P$.
B2: Suppose $U_1, V_1 \in \tau_1$ and $U_2, V_2 \in \tau_2$.

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 $\left({U_1 \cap V_1}\right) \in \tau_1$ and $\left({U_2 \cap V_2}\right) \in \tau_2$, we have $\left({U_1 \cap V_1}\right) \times \left({U_2 \cap V_2}\right) \in \mathcal P$.

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

Hence the result.

$\blacksquare$


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