Products of Open Sets form Local Basis in Product Space

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Theorem

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

Let $\struct{T, \tau} = T_1 \times T_2$ be the product space of $T_1$ and $T_2$.

Let $\tuple{x, y} \in A_1 \times A_2$.

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


Then:

$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple{x, y} \in U_1 \times U_2 \subseteq W$

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


Proof

Let $W \in \tau$ such that $\tuple{x, y} \in W$.

From Natural Basis of Tychonoff Topology of Finite Product, $\tau$ is the topology with basis:

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

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:

$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple{x, y} \in U_1 \times U_2 \subseteq W$

which is what was to be proved.

$\blacksquare$


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