Chebyshev Distance on Real Vector Space is Metric/Proof 1
Jump to navigation
Jump to search
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 $2.4$