Standard Bounded Metric is Metric/Topological Equivalence

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

Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
 * $\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$

$\bar d$ is topologically equivalent to $d$.

Proof
That $\bar d$ forms a metric on $M$ is demonstrated in Standard Bounded Metric is Metric.

We have that:
 * $\forall x, y \in A^2: \map {\bar d} {x, y} \le \map d {x, y}$

Hence:
 * $\map {B_\epsilon} {x; d} \subseteq \map {B_\epsilon} {x; \bar d}$

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

Hence:
 * if $U$ is $\bar d$-open, the $U$ is $d$-open

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

Let $U$ be $d$-open.

Let $x \in U$.

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

Let us take $\epsilon < 1$.

Then:
 * $\map {B_\epsilon} {x; \bar d} = \map {B_\epsilon} {x; d} \subseteq U$

demonstrating that $U$ is $\bar d$-open.

The result follows.