Positive Multiple of Metric is Metric

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Theorem

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

Let $k \in \R_{>0}$ be a (strictly) positive real number.

Let $d_k: A \times A \to \R$ be the function defined as:

$\forall \tuple {x, y} \in A: \map {d_k} {x, y} = k \cdot \map d {x, y}$


Then $M_k = \struct {A, d_k}$ is a metric space.


Proof

Metric Space Axiom $(\text M 1)$

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

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

$\Box$


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

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

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

$\Box$


Metric Space Axiom $(\text M 3)$

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

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

$\Box$


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 $(\text M 4)$
\(\ds \leadsto \ \ \) \(\ds k \cdot \map d {x, y}\) \(>\) \(\ds 0\) as $k > 0$
\(\ds \leadsto \ \ \) \(\ds \map {d_k} {x, y}\) \(>\) \(\ds 0\) Definition of $d_k$

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

$\blacksquare$


Sources

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