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 $$\exists h, k \in \reals: h > 0, k > 0$$ 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.