Chebyshev Distance is Metric

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

Then $d_\infty$ is a metric.

Proof of $\text M 1$
So axiom $\text M 1$ holds for $d_\infty$.

Proof of $\text M 2$
Let $k \in \closedint 1 n$ such that:

Then by application of axiom $\text M 2$ for metric $d_k$:
 * $\map {d_k} {x_k, z_k} \le \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k}$

But by the nature of the $\max$ operation:
 * $\ds \map {d_k} {x_k, y_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$

and:
 * $\ds \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$

Thus:
 * $\ds \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} } + \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$

Hence:
 * $\map {d_\infty} {x, z} \le \map {d_\infty} {x, y} + \map {d_\infty} {y, z}$

So axiom $\text M 2$ holds for $d_\infty$.

Proof of $\text M 3$
So axiom $\text M 3$ holds for $d_\infty$.

Proof of $\text M 4$
Let $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.

So axiom $\text M 4$ holds for $d_\infty$.