Chebyshev Distance on Real Vector Space is Metric

Theorem

The Chebyshev distance on $\R^n$:

$\displaystyle \forall x, y \in \R^n: d_\infty \left({x, y}\right):= \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$

is a metric.

Proof 1

This is an instance of the Chebyshev distance on the cartesian product of metric spaces $A_1, A_2, \ldots, A_3$.

This is proved in Chebyshev Distance is Metric.

$\blacksquare$

Proof 2

Proof of $M1$

 $\displaystyle \map {d_\infty} {x, x}$ $=$ $\displaystyle \max_{i \mathop = 1}^n \size {x_i - x_i}$ Definition of $d_\infty$ $\displaystyle$ $=$ $\displaystyle 0$

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

$\Box$

Proof of $M2$

Let $k \in \closedint 1 n$ such that:

 $\displaystyle \size {x_k - z_k}$ $=$ $\displaystyle \max_{i \mathop = 1}^n \size {x_i - z_i}$ $\displaystyle$ $=$ $\displaystyle \map {d_\infty} {x, z}$

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 $d_\infty$.

$\Box$

Proof of $M3$

 $\displaystyle \map {d_\infty} {x, y}$ $=$ $\displaystyle \max_{i \mathop = 1}^n \size {x_i - y_i}$ Definition of $d_\infty$ $\displaystyle$ $=$ $\displaystyle \max_{i \mathop = 1}^n \size {y_i - x_i}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle \map {d_\infty} {y, x}$ Definition of $d_\infty$

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

$\Box$

Proof of $M4$

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

 $\displaystyle x$ $\ne$ $\displaystyle y$ $\displaystyle \leadsto \ \$ $\, \displaystyle \exists k \in \closedint 1 n \,$ $\displaystyle x_k$ $\ne$ $\displaystyle y_k$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x_k - y_k}$ $>$ $\displaystyle 0$ Definition of Absolute Value $\displaystyle \leadsto \ \$ $\displaystyle \max_{i \mathop = 1}^n \size {x_i - y_i}$ $>$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \map {d_\infty} {x, y}$ $>$ $\displaystyle 0$ Definition of $d_\infty$

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

$\blacksquare$