Lipschitz Equivalent Metric Spaces are Homeomorphic

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

Then $$\exists h, k \in \reals: h > 0, k > 0$$ such that $$\forall x, y \in A: h d_1 \left({x, y}\right) \le d_2 \left({x, y}\right) \le k d_1 \left({x, y}\right)$$.

From the definition of $\epsilon$-neighborhood:

$$ $$ $$ $$

... and:

$$ $$ $$ $$

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

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

Let $$x \in U$$.

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

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

Thus $$U$$ is $d_2$-open.

The result follows by definition of topologically equivalent metrics.