Theorem: Open Ball in Cartesian Product under Chebyshev Distance

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:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \map {B_\epsilon} {a; d_\infty}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_\infty} {x, a}
      | o = <
      | r = \epsilon
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadsto
      | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, a_i} }
      | o = <
      | r = \epsilon
      | c = {{Defof|Chebyshev Distance}}
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \set {1, 2, \ldots, n}
      | l = \map {d_i} {x_i, a_i}
      | o = <
      | r = \epsilon
      | c = {{Defof|Maximum Element}}
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \set {1, 2, \ldots, n}
      | l = x_i
      | o = \in
      | r = \map {B_\epsilon} {a_i; d_i}
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
      | c = {{Defof|Finite Cartesian Product}}
}}
{{eqn | ll= \leadsto
      | l = \map {B_\epsilon} {a; d_\infty}
      | o = \subseteq
      | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
And then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \set {1, 2, \ldots, n}
      | l = x_i
      | o = \in
      | r = \map {B_\epsilon} {a_i; d_i}
      | c = {{Defof|Finite Cartesian Product}}
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \set {1, 2, \ldots, n}
      | l = \map {d_i} {x_i, a_i}
      | o = <
      | r = \epsilon
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadsto
      | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, a_i} }
      | o = <
      | r = \epsilon
      | c = {{Defof|Maximum Element}}
}}
{{eqn | ll= \leadsto
      | l = \map {d_\infty} {x, a}
      | o = <
      | r = \epsilon
      | c = {{Defof|Chebyshev Distance}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \map {B_\epsilon} {a; d_\infty}
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadsto
      | l = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
      | o = \subseteq
      | r = \map {B_\epsilon} {a; d_\infty}
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
The result follows by definition of set equality.
{{qed}}