Open Ball in Cartesian Product 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 $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.

Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.

Let $\epsilon \in \R_{>0}$.

Let $\map {B_\epsilon} {a; d_\infty}$ be the open $\epsilon$-ball of $a$ in $M = \struct {\AA, d_\infty}$.

Then:
 * $\ds \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$

Proof
Let $\epsilon \in \R_{>0}$.

Let $x = \tuple {x_1, x_2, \ldots, x_n} \in \AA$.

Then:

And then:

The result follows by definition of set equality.