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 topologically equivalent.

Corollary
Let $A$ be a set upon which there are two metrics imposed: $d_1$ and $d_2$.

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

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

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

Then $\exists h, k \in \R: h > 0, k > 0$ 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 $\epsilon$-neighborhood:

... and:

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

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

Let $x \in U$.

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

Hence $N_{\frac \epsilon k} \left({x; d_1}\right) \subseteq U$.

Thus $U$ is $d_1$-open.

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

Let $x \in U$.

Then $\exists \epsilon > 0: N_{\epsilon} \left({x; d_1}\right) \subseteq U$.

Hence $N_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq U$.

Thus $U$ is $d_2$-open.

The result follows by definition of topologically equivalent metric spaces.

Proof of Corollary
If we consider the identity mapping $f: A \to A: \forall x \in A: f \left({x}\right) = x$, we can likewise directly consider $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ as a Lipschitz equivalence.