Open Sets of Cartesian Product of Metric Spaces under Chebyshev Distance

Theorem
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.

Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
 * $\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.

For $i \in \set {1, 2, \ldots, n}$, let $U_i$ be open in $M_i$.

Then $\ds \prod_{i \mathop = 1}^n U_i$ is open in $M = \struct {\AA, d_\infty}$.

Proof
A set $U$ is open it is the neighborhood of each of its points.

That is:
 * $\forall a \in U: \exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$

where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$.

Let $I = \set {1, 2, \ldots, n}$.

For all $i \in I$, let $U_i$ be open in $M_i$.

Then:

For all $i \in I$, let $\map d {x_i, a_i} < \delta$.

Then:

Hence $\ds \prod_{i \mathop \in I} U_i$ is open in $M$.