Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 1

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 $d_1$ and $d_2$ be Lipschitz equivalent.

Then $d_1$ and $d_2$ are topologically equivalent.

Proof

Consider the identity mapping:

$f: A \to A: \forall x \in A: f \left({x}\right) = x$

Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a Lipschitz equivalence.

The result then follows from Lipschitz Equivalent Metric Spaces are Homeomorphic.

$\blacksquare$