Chebyshev Distance on Real Vector Space is Metric/Proof 2

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

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

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

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

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

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

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 $d_\infty$.