Metrics on Space are Lipschitz Equivalent iff Identity Mapping is Lipschitz Equivalence

Theorem
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $I_A$ denote the identity mapping on $A$.

Then:
 * $d_1$ and $d_2$ are Lipschitz equivalent


 * $I_A: M_1 \to M_2$ is a Lipschitz equivalence.
 * $I_A: M_1 \to M_2$ is a Lipschitz equivalence.

Proof
By definition of identity mapping:
 * $\forall x \in A: \map {I_A} x = x$

Sufficient Condition
Let $d_1$ and $d_2$ be Lipschitz equivalent.

Then:

That is, by definition:
 * $I_A: M_1 \to M_2$ is a Lipschitz equivalence.

Necessary Condition
Let $I_A$ be a Lipschitz equivalence.

Then:

That is, by definition:
 * $d_1$ and $d_2$ are Lipschitz equivalent metrics.