Chebyshev Distance on Real Vector Space is Metric/Proof 2

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

Proof of $M2$
Let $k \in \closedint 1 n$ such that:

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

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

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

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

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

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

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

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

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