Chebyshev Distance on Real Vector Space is Metric/Proof 1
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Theorem
The Chebyshev distance on $\R^n$:
- $\ds \forall x, y \in \R^n: \map {d_\infty} {x, y}:= \max_{i \mathop = 1}^n {\size {x_i - y_i} }$
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
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Corollary $2.4$