Metric over 1 plus Metric forms Metric/Topological Equivalence

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Theorem

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

Let $d_3: A^2 \to \R$ be the metric on $M$ defined as:

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


$d_3$ is topologically equivalent to $d$.


Proof

That $d_3$ forms a metric on $M$ is demonstrated in Metric over 1 plus Metric forms Metric.


We have that:

$\forall x, y \in A^2: \map {d_3} {x, y} \le \map d {x, y}$

Hence:

$\map {B_\epsilon} {x; d} \subseteq \map {B_\epsilon} {x; d_3}$

where $\map {B_\epsilon} {x; d}$ denotes the open $\epsilon$-ball of $x$ in $\struct {A, d}$.

Hence:

if $U \subseteq A$ is $d_3$-open, the $U$ is $d$-open

where $U$ is a subset of $A$.


Let $U \subseteq A$ be $d$-open.

Let $x \in U$.

Then $\map {B_\epsilon} {x; d} \subseteq U$ for some $\epsilon \in \r_{>0}$.

Let $\delta = \min \set {\dfrac \epsilon 2, \dfrac 1 2}$.

Then:

$\map {d_3} {x, y} < \delta \implies \map {d_3} {x, y} < \dfrac 1 2$

and so:

\(\ds \map d {x, y}\) \(=\) \(\ds \dfrac {\map {d_3} {x, y} } {1 - \map {d_3} {x, y} }\)
\(\ds \) \(\le\) \(\ds \map {d_3} {x, y}\)
\(\ds \) \(<\) \(\ds 2 \map {d_3} {x, y}\)
\(\ds \) \(<\) \(\ds \epsilon\)


Hence:

$\map {B_\epsilon} {x; d_2} = \map {B_\epsilon} {x; d} \subseteq U$

demonstrating that $U$ is $d_2$-open.


Note that $\delta = \dfrac \epsilon {1 + \epsilon}$ is just as good:

$\map f a = \dfrac a {1 + a}$ is strictly increasing for $a \ge 0$

and so:

\(\ds \map {d_3} {x, y}\) \(<\) \(\ds \dfrac \epsilon {1 + \epsilon}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \map d {x, y}\) \(<\) \(\ds \epsilon\)

The result follows.

$\blacksquare$


Sources

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