Open Ball in Cartesian Product under Chebyshev Distance

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

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

Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the Chebyshev distance on $\mathcal A$:
 * $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$

where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.

Let $a = \left({a_1, a_2, \ldots, a_n}\right) \in \mathcal A$.

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

Let $B_\epsilon \left({a; d_\infty}\right)$ be the open $\epsilon$-ball of $a$ in $M = \left({\mathcal A, d_\infty}\right)$.

Then:
 * $\displaystyle B_\epsilon \left({a; d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right)$

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

Let $x = \left({x_1, x_2, \ldots, x_n}\right) \in \mathcal A$.

Then:

And then:

The result follows by definition of set equality.