Lipschitz Equivalence is Equivalence Relation

Theorem
Let $A$ be a set.

Let $\mathcal D$ be the set of all metrics on $A$.

Let $\sim$ be the relation on $\mathcal D$ defined as:
 * $\forall d_1, d_2 \in \mathcal D: d_1 \sim d_2 \iff d_1$ is Lipschitz equivalent to $d_2$

Then $\sim$ is an equivalence relation.

Proof
Let $A$ be a set and let $\mathcal D$ be the set of all metrics on $A$.

In the following, let $d_1, d_2, d_3 \subseteq \mathcal D$ be arbitrary.

Checking in turn each of the criteria for equivalence:

Reflexivity
Let $d_1$ be a metric on $A$.

Then trivially:
 * $\forall x, y \in A: 1 \times d_1 \left({x, y}\right) \le d_1 \left({x, y}\right) \le 1 \times d_1 \left({x, y}\right)$

That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be reflexive.

Symmetry
Let $d_1 \sim d_2$.

That is, let $d_1, d_2$ be Lipschitz equivalent metrics on $A$.

Then by definition:
 * $\forall x, y \in A: h d_1 \left({x, y}\right) \le d_2 \left({x, y}\right) \le k d_1 \left({x, y}\right)$

for some $h, k \in \R_{>0}$.

Then:

and:

That is:
 * $\forall x, y \in A: \dfrac 1 h d_2 \left({x, y}\right) \le d_1 \left({x, y}\right) \le \dfrac 1 k d_2 \left({x, y}\right)$

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.

Transitivity
Let $d_1 \sim d_2$ and $d_2 \sim d_3$.

Then by definition:
 * $\forall x, y \in A: h_1 d_1 \left({x, y}\right) \le d_2 \left({x, y}\right) \le k_1 d_1 \left({x, y}\right)$
 * $\forall x, y \in A: h_2 d_2 \left({x, y}\right) \le d_3 \left({x, y}\right) \le k_2 d_2 \left({x, y}\right)$

for some $h_1, k_1, h_2, k_2 \in \R_{>0}$.

Then:

and:

So:
 * $\forall x, y \in A: h_1 h_2 d_1 \left({x, y}\right) \le d_3 \left({x, y}\right) \le k_1 k_2 d_1 \left({x, y}\right)$

That is, $d_1 \sim d_3$ and so $\sim$ has been shown to be transitive.

$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.