Products of Open Sets form Local Basis in Product Space

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 Product 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.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.5$: Products: Proposition $3.5.2$