Chebyshev Distance on Real Vector Space is Metric/Proof 2

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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

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

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

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:

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

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:

$\ds \size {x_k - y_k} \le \max_{i \mathop = 1}^n \size {x_i - y_i}$

and:

$\ds \size {y_k - z_k} \le \max_{i \mathop = 1}^n \size {y_i - z_i}$

Thus:

$\ds \size {x_k - y_k} + \size {y_k - z_k} \le \max_{i \mathop = 1}^n \size {x_i - y_i} + \max_{i \mathop = 1}^n \size {y_i - z_i}$

Hence:

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

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

$\Box$


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

\(\ds \map {d_\infty} {x, y}\) \(=\) \(\ds \max_{i \mathop = 1}^n \size {x_i - y_i}\) Definition of $d_\infty$
\(\ds \) \(=\) \(\ds \max_{i \mathop = 1}^n \size {y_i - x_i}\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \map {d_\infty} {y, x}\) 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 $y = \tuple {y_1, y_2, \ldots, y_n}$.

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

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

$\blacksquare$


Sources

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