Chebyshev Distance is Metric

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Theorem

Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.

Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:

$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.


Then $d_\infty$ is a metric.


Proof

Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map {d_\infty} {x, x}\) \(=\) \(\ds \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, x_i} }\) Definition of $d_\infty$
\(\ds \) \(=\) \(\ds 0\) as $d_i$ fulfills Metric Space Axiom $(\text M 1)$

So Metric Space Axiom $(\text M 1)$ holds for $d_\infty$.

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

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

\(\ds \map {d_k} {x_k, z_k}\) \(=\) \(\ds \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, z_i} }\)
\(\ds \) \(=\) \(\ds \map {d_\infty} {x, z}\)

Then by application of Metric Space Axiom $(\text M 2)$: Triangle Inequality for metric $d_k$:

$\map {d_k} {x_k, z_k} \le \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k}$

But by the nature of the $\max$ operation:

$\ds \map {d_k} {x_k, y_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$

and:

$\ds \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$

Thus:

$\ds \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} } + \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$

Hence:

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

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_\infty$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

\(\ds \map {d_\infty} {x, y}\) \(=\) \(\ds \max_{i \mathop = 1}^n \map {d_i} {x_i, y_i}\) Definition of $d_\infty$
\(\ds \) \(=\) \(\ds \max_{i \mathop = 1}^n \map {d_i} {y_i, x_i}\) as $d_i$ fulfills Metric Space Axiom $(\text M 3)$
\(\ds \) \(=\) \(\ds \map {d_\infty} {y, x}\) Definition of $d_\infty$

So Metric Space Axiom $(\text M 3)$ holds for $d_\infty$.

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

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

\(\ds x\) \(\ne\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \exists k \in \closedint 1 n: \, \) \(\ds x_k\) \(\ne\) \(\ds y_k\)
\(\ds \leadsto \ \ \) \(\ds \map {d_k} {x_k, y_k}\) \(>\) \(\ds 0\) as $d_k$ fulfills Metric Space Axiom $(\text M 4)$
\(\ds \leadsto \ \ \) \(\ds \max_{i \mathop = 1}^n \map {d_i} {x_i, y_i}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \map {d_\infty} {x, y}\) \(>\) \(\ds 0\) Definition of $d_\infty$

So Metric Space Axiom $(\text M 4)$ holds for $d_\infty$.

$\blacksquare$


Sources

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