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. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = h d_1 \left({x, y}\right)       | o = \le       | r = d_2 \left({x, y}\right)       | c =  }} {{eqn | ll= \implies       | l = d_1 \left({x, y}\right)       | o = \le       | r = \frac 1 h d_2 \left({x, y}\right)       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = d_2 \left({x, y}\right)       | o = \le       | r = k d_1 \left({x, y}\right)       | c =  }} {{eqn | ll= \implies       | l = \frac 1 k d_2 \left({x, y}\right)       | o = \le       | r = d_1 \left({x, y}\right)       | c =  }} {{end-eqn}} 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. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = h_1 d_1 \left({x, y}\right)       | o = \le       | r = d_2 \left({x, y}\right)       | c =  }} {{eqn | ll= \implies       | l = h_1 h_2 d_1 \left({x, y}\right)       | o = \le       | r = h_2 d_2 \left({x, y}\right)       | c =  }} {{eqn | ll= \implies       | l = h_1 h_2 d_1 \left({x, y}\right)       | o = \le       | r = d_3 \left({x, y}\right)       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = d_2 \left({x, y}\right)       | o = \le       | r = k_1 d_1 \left({x, y}\right)       | c =  }} {{eqn | ll= \implies       | l = k_2 d_2 \left({x, y}\right)       | o = \le       | r = k_1 k_2 d_1 \left({x, y}\right)       | c =  }} {{eqn | ll= \implies       | l = d_3 \left({x, y}\right)       | o = \le       | r = k_1 k_2 d_1 \left({x, y}\right)       | c =  }} {{end-eqn}} 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. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}