Lipschitz Equivalent Metric Spaces are Homeomorphic

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ 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:
 * $\forall x, y \in A_1: h d_1 \left({x, y}\right) \le d_2 \left({f \left({x}\right), f \left({y}\right)}\right) \le k d_1 \left({x, y}\right)$

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

... and:

Thus:
 * $B_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq B_\epsilon \left({x; d_1}\right)$
 * $B_{\epsilon / k} \left({x; d_1}\right) \subseteq B_\epsilon \left({f \left({x}\right); d_2}\right)$

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

Let $x \in U$.

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

Hence:
 * $B_{\epsilon / k} \left({x; d_1}\right) \subseteq U$

Thus $U$ is $d_1$-open.

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

Let $x \in U$.

Then:
 * $\exists \epsilon \in \R_{>0}: B_\epsilon \left({x; d_1}\right) \subseteq U$

Hence:
 * $B_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq U$

Thus $U$ is $d_2$-open.

The result follows by definition of homeomorphic metric spaces.