Square of Metric does not necessarily form Metric

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $d_4: A^2 \to \R$ be the mapping defined as:

$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \paren {\map d {x, y} }^2$


Then $d_4$ may or may not be a metric for $A$.


Proof

Let $d$ be the standard discrete metric on $M$.

Then:

$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \map d {x, y}$

and indeed in this case $d_4$ is a metric for $A$.

$\Box$


Let $M = \struct {\R, d}$ be the real number line with the usual (Euclidean) metric.

Let $x = 1$, $y = \dfrac 1 2$ and $z = 0$.

We have:

\(\ds \map {d_4} {x, y}\) \(=\) \(\ds \size {1 - \dfrac 1 2}^2\)
\(\ds \) \(=\) \(\ds \dfrac 1 4\)
\(\ds \map {d_4} {y, z}\) \(=\) \(\ds \size {\dfrac 1 2 - 0}^2\)
\(\ds \) \(=\) \(\ds \dfrac 1 4\)
\(\ds \map {d_4} {x, z}\) \(=\) \(\ds \size {1 - 0}^2\)
\(\ds \) \(=\) \(\ds 1\)

So we have:

$\map {d_4} {x, y} + \map {d_4} {y, z} < \map {d_4} {x, z}$

and it is seen that $d_4$ does not satisfy Metric Space Axiom $(\text M 2)$: Triangle Inequality.

$\blacksquare$


Sources

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