Metric over 1 plus Metric forms Metric/Topological Equivalence

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:

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:

The result follows.