Standard Bounded Metric is Metric/Topological Equivalence
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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.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 20$