Chebyshev Distance is Metric

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

Then $d_\infty$ is a metric,

Proof of $M1$
So axiom $M1$ holds for $d_\infty$.

Proof of $M2$
Let $k \in \left[{1 \,.\,.\, n}\right]$ such that:

Then by application of axiom $M2$ for metric $d_k$:
 * $d_k \left({x_k, z_k}\right) \le d_k \left({x_k, y_k}\right) + d_k \left({y_k, z_k}\right)$

But by the nature of the $\max$ operation:
 * $\displaystyle d_k \left({x_k, y_k}\right) \le \max_{i \mathop = 1}^n \left\{{d_i \left({x_i, y_i}\right)}\right\}$

and:
 * $\displaystyle d_k \left({y_k, z_k}\right) \le \max_{i \mathop = 1}^n \left\{{d_i \left({y_i, z_i}\right)}\right\}$

Thus:
 * $\displaystyle d_k \left({x_k, y_k}\right) + d_k \left({y_k, z_k}\right) \le \max_{i \mathop = 1}^n \left\{{d_i \left({x_i, y_i}\right)}\right\} + \max_{i \mathop = 1}^n \left\{{d_i \left({y_i, z_i}\right)}\right\}$

Hence:
 * $d_\infty \left({x, z}\right) \le d_\infty \left({x, y}\right) + d_\infty \left({y, z}\right)$

So axiom $M2$ holds for $d_\infty$.

Proof of $M3$
So axiom $M3$ holds for $d_\infty$.

Proof of $M4$
Let $x = \left({x_1, x_2, \ldots, x_n}\right)$ and $y = \left({y_1, y_2, \ldots, y_n}\right)$.

So axiom $M4$ holds for $d_\infty$.