Standard Bounded Metric is Metric
Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
- $\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$
Then $\bar d$ is a metric for $A$.
Topological Equivalence
$\bar d$ is topologically equivalent to $d$.
Proof
It is to be demonstrated that $\bar d$ satisfies all the metric space axioms.
Proof of Metric Space Axiom $(\text M 1)$
\(\ds \map {\bar d} {x, x}\) | \(=\) | \(\ds \min \set {1, \map d {x, y} }\) | Definition of Standard Bounded Metric | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {1, 0}\) | as $d$ fulfils Metric Space Axiom $(\text M 1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
So Metric Space Axiom $(\text M 1)$ holds for $\bar d$.
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
Suppose that $\map d {x, y} < 1$ and $\map d {y, z} < 1$.
Then:
\(\ds \map {\bar d} {x, y} + \map {\bar d} {y, z}\) | \(=\) | \(\ds \min \set {1, \map d {x, y} } + \min \set {1, \map d {y, z} }\) | Definition of Standard Bounded Metric | |||||||||||
\(\ds \) | \(=\) | \(\ds \map d {x, y} + \map d {x, y}\) | as both $\map d {x, y} < 1$ and $\map d {x, y} < 1$ | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \map d {x, z}\) | as $d$ fulfils Metric Space Axiom $(\text M 2)$: Triangle Inequality | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \min \set {1, \map d {x, z} }\) |
Now suppose that either $\map d {x, y} > 1$ or $\map d {x, y} > 1$.
Without loss of generality, suppose $\map d {x, y} > 1$.
Then:
\(\ds \map {\bar d} {x, y} + \map {\bar d} {y, z}\) | \(=\) | \(\ds \min \set {1, \map d {x, y} } + \min \set {1, \map d {y, z} }\) | Definition of Standard Bounded Metric | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \min \set {1, \map d {y, z} }\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \min \set {1, \map d {x, z} }\) | as $1 + \epsilon \ge \min {1, \delta}$ for all $\epsilon, \delta \in \R_{>0}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\bar d} {x, z}\) |
The same argument applies for $\map d {y, z} > 1$
So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $\bar d$.
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
\(\ds \map {\bar d} {x, y}\) | \(=\) | \(\ds \min \set {1, \map d {x, y} }\) | Definition of Standard Bounded Metric | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {1, \map d {y, x} }\) | as $d$ fulfils Metric Space Axiom $(\text M 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\bar d} {y, x}\) | Definition of Standard Bounded Metric |
So Metric Space Axiom $(\text M 3)$ holds for $\bar d$.
$\Box$
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 $(\text M 4)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \min \set {1, \map d {x, y} }\) | \(>\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\bar d} {x, y}\) | \(>\) | \(\ds 0\) | Definition of Standard Bounded Metric |
So Metric Space Axiom $(\text M 4)$ holds for $\bar d$.
$\Box$
Thus $\bar d$ 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 $2.6: 6$