Metric over 1 plus Metric forms Metric
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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 6$