Metric over 1 plus Metric forms Metric

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $d_3: A^2 \to \R$ be the mapping defined as:

$\forall \tuple {x, y} \in A^2: \map {d_3} {x, y} = \dfrac {\map d {x, y} } {1 + \map d {x, y} }$


Then $d_3$ is a metric for $A$.


Topological Equivalence

$d_3$ is topologically equivalent to $d$.


Proof

It is to be demonstrated that $d_3$ satisfies all the metric space axioms.


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

\(\ds \map {d_3} {x, x}\) \(=\) \(\ds \dfrac {\map d {x, x} } {1 + \map d {x, x} }\) Definition of $d_3$
\(\ds \) \(=\) \(\ds \dfrac 0 {1 + 0}\) as $d$ fulfils Metric Space Axiom $(\text M 1)$
\(\ds \) \(=\) \(\ds 0\)

So Metric Space Axiom $(\text M 1)$ holds for $d_3$.

$\Box$


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

Note that $\map f x = \dfrac x {1 + x}$ is an increasing function of $x$ for $x = 0$.

In order to simplify the algebra, let $a = \map d {x, y}$, $b = \map d {y, z}$ and $c = \map d {x, z}$.

Then:

\(\ds \map {d_3} {x, y} + \map {d_3} {y, z}\) \(=\) \(\ds \dfrac a {1 + a} + \dfrac b {1 + b}\) Definition of $d_3$
\(\ds \) \(=\) \(\ds \dfrac {a \paren {1 + b} + b \paren {1 + a} } {\paren {1 + a} \paren {1 + b} }\)
\(\ds \) \(=\) \(\ds \dfrac {a + b + 2 a b} {1 + a + b + a b}\)
\(\ds \) \(\ge\) \(\ds \dfrac {a + b + 2 a b} {1 + a + b + 2 a b}\)

But $a + b + 2 a b > c$.

Because $\map f x = \dfrac x {1 + x}$ is an increasing function:

$\dfrac {a + b + 2 a b} {1 + a + b + 2 a b} > \dfrac c {1 + c}$

and it follows by definition of $d_3$ that:

$\map {d_3} {x, y} + \map {d_3} {y, z} \ge \map {d_3} {x, z}$

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_3$.

$\Box$


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

\(\ds \map {d_3} {x, y}\) \(=\) \(\ds \dfrac {\map d {x, y} } {1 + \map d {x, y} }\) Definition of $d_3$
\(\ds \) \(=\) \(\ds \dfrac {\map d {y, x} } {1 + \map d {y, x} }\) as $d$ fulfils Metric Space Axiom $(\text M 3)$
\(\ds \) \(=\) \(\ds \map {d_3} {y, x}\) Definition of $d_3$

So Metric Space Axiom $(\text M 3)$ holds for $d_3$.

$\Box$


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 $(\text M 4)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\map d {y, x} } {1 + \map d {y, x} }\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \map {d_3} {x, y}\) \(>\) \(\ds 0\) Definition of $d_3$

So Metric Space Axiom $(\text M 4)$ holds for $d_3$.

$\Box$


Thus $d_3$ satisfies all the metric space axioms and so is a metric.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Metric_over_1_plus_Metric_forms_Metric&oldid=534057"