# Lipschitz Equivalent Metrics are Topologically Equivalent

## Theorem

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

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

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

## Proof 1

Consider the identity mapping:

$f: A \to A: \forall x \in A: f \left({x}\right) = x$

Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a Lipschitz equivalence.

The result then follows from Lipschitz Equivalent Metric Spaces are Homeomorphic.

$\blacksquare$

## Proof 2

By definition of Lipschitz equivalence:

$\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$

for some $h, k \in \R_{>0}$.

Let $x \in A$.

Let $\epsilon \in \R_{>0}$.

Let $\map {B_{h \epsilon} } {x; d_1}$ denote the open $h \epsilon$-ball with respect to $d_1$ of $x \in A$.

Then:

 $\ds y$ $\in$ $\ds \map {B_{h \epsilon} } {x; d_1}$ $\ds \leadsto \ \$ $\ds \map {d_1} {x, y}$ $<$ $\ds h \epsilon$ $\ds \leadsto \ \$ $\ds \map {d_2} {x, y}$ $\le$ $\ds \frac {\map {d_1} {x, y} } h$ $\ds$ $<$ $\ds \epsilon$ $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds \map {B_\epsilon} {x; d_2}$ $\ds \leadsto \ \$ $\ds \map {B_{h \epsilon} } {x; d_1}$ $\subseteq$ $\ds \map {B_\epsilon} {x; d_2}$

Similarly:

 $\ds y$ $\in$ $\ds \map {B_{\epsilon / k} } {x; d_2}$ $\ds \leadsto \ \$ $\ds \map {d_2} {x, y}$ $<$ $\ds \frac \epsilon k$ $\ds \leadsto \ \$ $\ds \map {d_1} {x, y}$ $\le$ $\ds k \frac {\map {d_2} {x, y} } h$ $\ds$ $<$ $\ds \epsilon$ $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds \map {B_\epsilon} {x; d_1}$ $\ds \leadsto \ \$ $\ds \map {B_{\epsilon / k} } {x; d_2}$ $\subseteq$ $\ds \map {B_\epsilon} {x; d_1}$

Now suppose $U \subseteq A$ is $d_1$-open.

Let $x \in U$.

Then:

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

Thus:

$\map {B_{\epsilon / k} } {x; d_2} \subseteq \map {B_\epsilon} {x; d_1} \subseteq U$

and so $U$ is $d_2$-open.

Mutatis mutandis, if $U \subseteq A$ is $d_2$-open, it follows that $U$ is $d_1$-open.

$\blacksquare$

## Proof 3

By definition of Lipschitz equivalence:

$\exists K_1, K_2 \in \R_{>0}$ such that:

$(1): \quad \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
$(2): \quad \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$
the identity mapping from $M_1$ to $M_2$ is continuous
the identity mapping from $M_2$ to $M_1$ is continuous.

Hence the result by definition of topological equivalence.

$\blacksquare$