Products of Open Sets form Local Basis in Product Space

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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 $\left({T, \tau}\right) = T_1 \times T_2$ be the topological product of $T_1$ and $T_2$.

Let $\left({x, y}\right) \in A_1 \times A_2$.

Let $W \in \tau$ be an open set of $T$ such that $\left({x, y}\right) \in W$.


Then:

$\exists U_1 \in \tau_1, U_2 \in \tau_2: \left({x, y}\right) \in U_1 \times U_2 \subseteq W$

That is, products of open sets from $T_1$ and $T_2$ form a local basis at $\left({x, y}\right)$.


Proof

Let $W \in \tau$ such that $\left({x, y}\right) \in W$.

From the definition of the product topology, $\tau$ is the topology with basis:

$\mathcal B = \left\{{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}\right\}$

Thus $W$ is the union of sets of the form $U_1 \times U_2$ where $U_1 \in \tau_1$ and $U_2 \in \tau_2$.

That is:

$\exists U_1 \in \tau_1, U_2 \in \tau_2: \left({x, y}\right) \in U_1 \times U_2 \subseteq W$

which is what was to be proved.

$\blacksquare$


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