Chebyshev Distance on Real Number Plane is not Rotation Invariant

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Theorem

Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.

Let $d_\infty$ denote the Chebyshev distance on $\R^2$.


Then it is not necessarily the case that:

$\forall x, y \in \R^2: \map {d_\infty} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_\infty} {x, y}$


Proof

Proof by Counterexample:

Let $x = \tuple {0, 0}$ and $y = \tuple {1, 1}$ be arbitrary points in $\R^2$.

Then:

\(\ds \map {d_\infty} {x, y}\) \(=\) \(\ds \map {d_\infty} {\tuple {0, 0}, \tuple {1, 1} }\) Definition of $x$ and $y$
\(\ds \) \(=\) \(\ds \max \set {\size {0 - 1}, \size {0 - 1} }\) Definition of Chebyshev Distance on Real Number Plane
\(\ds \) \(=\) \(\ds 1\)


Now let $\alpha = \dfrac \pi 4 = 45 \degrees$.

\(\ds \map {d_\infty} {\map {r_\alpha} x, \map {r_\alpha} y}\) \(=\) \(\ds \map {d_\infty} {\tuple {0, 0}, \tuple {0, \sqrt 2} }\) Definition of Plane Rotation
\(\ds \) \(=\) \(\ds \max \set {\size {0 - 0}, \size {0 - \sqrt 2} }\) Definition of Chebyshev Distance on Real Number Plane
\(\ds \) \(=\) \(\ds \sqrt 2\) simplification
\(\ds \) \(\ne\) \(\ds 1\)

$\blacksquare$


Sources


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