Standard Bounded Metric is Metric

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Standard_Bounded_Metric_is_Metric&oldid=534058"