Multiple of Metric forms Metric

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Theorem

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

Let $d_1: A^2 \to \R$ be the mapping defined as:

$\forall \tuple {x, y} \in A^2: \map {d_1} {x, y} = k \map d {x, y}$

for some strictly positive $k \in \R_{>0}$.


Then $d_1$ is a metric for $A$.


Proof

It is to be demonstrated that $d_1$ satisfies all the metric space axioms.


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

\(\ds \map {d_1} {x, x}\) \(=\) \(\ds k \map d {x, x}\) Definition of $d_1$
\(\ds \) \(=\) \(\ds k \times 0\) as $d$ fulfils Metric Space Axiom $(\text M 1)$
\(\ds \) \(=\) \(\ds 0\)

So Metric Space Axiom $(\text M 1)$ holds for $d_1$.

$\Box$


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

\(\ds \map {d_1} {x, y} + \map {d_1} {y, z}\) \(=\) \(\ds k \map d {x, y} + k \map d {y, z}\) Definition of $d_1$
\(\ds \) \(=\) \(\ds k \paren {\map d {x, y} + \map d {y, z} }\)
\(\ds \) \(\ge\) \(\ds k \map d {x, z}\) as $d$ fulfils Metric Space Axiom $(\text M 2)$: Triangle Inequality
\(\ds \) \(=\) \(\ds \map {d_1} {x, z}\) Definition of $d_1$

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_1$.

$\Box$


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

\(\ds \map {d_1} {x, y}\) \(=\) \(\ds k \times \map d {x, y}\) Definition of $d_1$
\(\ds \) \(=\) \(\ds k \times \map d {y, x}\) as $d$ fulfils Metric Space Axiom $(\text M 3)$
\(\ds \) \(=\) \(\ds \map {d_1} {y, x}\) Definition of $d_1$

So Metric Space Axiom $(\text M 3)$ holds for $d_1$.

$\Box$


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

\(\ds x\) \(\ne\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \map d {x, y}\) \(>\) \(\ds 0\) as $d$ fulfils Metric Space Axiom $(\text M 4)$
\(\ds \leadsto \ \ \) \(\ds k \map d {x, y}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \map {d_1} {x, y}\) \(>\) \(\ds 0\) Definition of $d_1$

So Metric Space Axiom $(\text M 4)$ holds for $d_1$.

$\Box$


Thus $d_1$ satisfies all the metric space axioms and so is a metric.

$\blacksquare$


Sources

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