Chebyshev Distance is Metric

Theorem
The maximum metric is a metric.

Proof
We have that:
 * $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{{\left \vert {x_i - y_i} \right \vert}\right\}$

Let $k \in \left[{1 \,.\,.\, n}\right]$ such that:
 * $\displaystyle \left \vert {x_k - z_k} \right \vert = d_\infty \left({x, z}\right) = \max_{i \mathop = 1}^n \left\{{\left \vert {x_i - z_i} \right \vert}\right\}$

Then by the Triangle Inequality:
 * $\left \vert {x_k - z_k} \right \vert \le \left \vert {x_k - y_k} \right \vert + \left \vert {y_k - z_k} \right \vert$

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

and:
 * $\displaystyle \left|{y_k - z_k}\right| \le \max_{i \mathop = 1}^n \left\{{\left|{y_i - z_i}\right|}\right\}$

Thus:
 * $\displaystyle \left|{x_k - y_k}\right| + \left|{y_k - z_k}\right| \le \max_{i \mathop = 1}^n \left\{{\left \vert {x_i - y_i} \right \vert}\right\} + \max_{i \mathop = 1}^n \left\{{\left \vert {x_i - y_i} \right \vert}\right\}$

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