Lipschitz Equivalence is Equivalence Relation
Theorem
Let $A$ be a set.
Let $\DD$ be the set of all metrics on $A$.
Let $\sim$ be the relation on $\DD$ defined as:
- $\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is Lipschitz equivalent to $d_2$
Then $\sim$ is an equivalence relation.
Proof
Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for equivalence:
Reflexivity
Let $d_1$ be a metric on $A$.
Then trivially:
- $\forall x, y \in A: 1 \times \map {d_1} {x, y} \le \map {d_1} {x, y} \le 1 \times \map {d_1} {x, y}$
That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be reflexive.
$\Box$
Symmetry
Let $d_1 \sim d_2$.
That is, let $d_1, d_2$ be Lipschitz equivalent metrics on $A$.
Then by definition:
- $\forall x, y \in A: h \map {d_1} {x, y} \le \map {d_2} {x, y} \le k \map {d_1} {x, y}$
for some $h, k \in \R_{>0}$.
Then:
\(\ds h \map {d_1} {x, y}\) | \(\le\) | \(\ds \map {d_2} {x, y}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_1} {x, y}\) | \(\le\) | \(\ds \frac 1 h \map {d_2} {x, y}\) |
and:
\(\ds \map {d_2} {x, y}\) | \(\le\) | \(\ds k \map {d_1} {x, y}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 k \map {d_2} {x, y}\) | \(\le\) | \(\ds \map {d_1} {x, y}\) |
That is:
- $\forall x, y \in A: \dfrac 1 h \map {d_2} {x, y} \le \map {d_1} {x, y} \le \dfrac 1 k \map {d_2} {x, y}$
for some $\dfrac 1 h, \dfrac 1 k \in \R_{>0}$.
That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be symmetric.
Note that in the definition of Lipschitz equivalent metrics, in the expression $d_1 \sim d_2$ it is not explicitly specified which of $d_1$ and $d_2$ goes in the middle of the defining statement.
This result demonstrates that it does not actually matter.
$\Box$
Transitivity
Let $d_1 \sim d_2$ and $d_2 \sim d_3$.
Then by definition:
- $\forall x, y \in A: h_1 \map {d_1} {x, y} \le \map {d_2} {x, y} \le k_1 \map {d_1} {x, y}$
- $\forall x, y \in A: h_2 \map {d_2} {x, y} \le \map {d_3} {x, y} \le k_2 \map {d_2} {x, y}$
for some $h_1, k_1, h_2, k_2 \in \R_{>0}$.
Then:
\(\ds h_1 \map {d_1} {x, y}\) | \(\le\) | \(\ds \map {d_2} {x, y}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds h_1 h_2 \map {d_1} {x, y}\) | \(\le\) | \(\ds h_2 \map {d_2} {x, y}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds h_1 h_2 \map {d_1} {x, y}\) | \(\le\) | \(\ds \map {d_3} {x, y}\) |
and:
\(\ds \map {d_2} {x, y}\) | \(\le\) | \(\ds k_1 \map {d_1} {x, y}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds k_2 \map {d_2} {x, y}\) | \(\le\) | \(\ds k_1 k_2 \map {d_1} {x, y}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_3} {x, y}\) | \(\le\) | \(\ds k_1 k_2 \map {d_1} {x, y}\) |
So:
- $\forall x, y \in A: h_1 h_2 \map {d_1} {x, y} \le \map {d_3} {x, y} \le k_1 k_2 \map {d_1} {x, y}$
That is, $d_1 \sim d_3$ and so $\sim$ has been shown to be transitive.
$\Box$
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Proposal $2.4.3$