Chebyshev Distance on Real Vector Space is Metric/Proof 1

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

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.

$\blacksquare$

Sources

• 1962: Bert Mendelson: Introduction to Topology ... (previous) ... (next): $\S 2.2$: Metric Spaces: Corollary $2.4$