Lipschitz Equivalent Metric Spaces are Homeomorphic

Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $M_1$ and $M_2$ be Lipschitz equivalent.

Then $M_1$ and $M_2$ are homeomorphic.

Proof
Let $M_1$ and $M_2$ be Lipschitz equivalent.

Then, by definition, $\exists h, k \in \R_{>0}$ such that:
 * $\forall x, y \in A_1: h \map {d_1} {x, y} \le \map {d_2} {\map f x, \map f y} \le k \map {d_1} {x, y}$

From the definition of open $\epsilon$-ball:

and:

Thus:
 * $\map {B_{h \epsilon} } {\map f x; d_2} \subseteq \map {B_\epsilon} {x; d_1}$
 * $\map {B_{\epsilon / k} } {x; d_1} \subseteq \map {B_\epsilon} {\map f x; d_2}$

Now, suppose $U$ is $d_2$-open.

Let $x \in U$.

Then:
 * $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2} \subseteq U$.

Hence:
 * $\map {B_{\epsilon / k} } {x; d_1} \subseteq U$

Thus $U$ is $d_1$-open.

Similarly, suppose $U$ is $d_1$-open.

Let $x \in U$.

Then:
 * $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$

Hence:
 * $\map {B_{h \epsilon} } {\map f x; d_2} \subseteq U$

Thus $U$ is $d_2$-open.

The result follows by definition of homeomorphic metric spaces.

Also see

 * Lipschitz Equivalent Metrics are Topologically Equivalent