Lipschitz Equivalent Metric Spaces are Homeomorphic

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Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ 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:

$\forall x, y \in A_1: h \map {d_1} {x, y} \le \map {d_2} {\map f x, \map f y} \le k \map {d_1} {x, y}$


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

\(\ds y\) \(\in\) \(\ds \map {B_{h \epsilon} } {\map f x; d_2}\)
\(\ds \leadsto \ \ \) \(\ds \map {d_2} {\map f x, \map f y}\) \(<\) \(\ds h \epsilon\)
\(\ds \leadsto \ \ \) \(\ds \map {d_1} {x, y}\) \(\le\) \(\ds \frac {\map {d_2} {\map f x, \map f y} } h < \epsilon\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\in\) \(\ds \map {B_\epsilon} {x; d_1}\)


and:

\(\ds y\) \(\in\) \(\ds \map {B_{\epsilon / k} } {x; d_1}\)
\(\ds \leadsto \ \ \) \(\ds \map {d_1} {x, y}\) \(<\) \(\ds \frac \epsilon k\)
\(\ds \leadsto \ \ \) \(\ds \map {d_2} {\map f x, \map f y}\) \(\le\) \(\ds k \map {d_1} {x, y} < \epsilon\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\in\) \(\ds \map {B_\epsilon} {\map f x; d_2}\)


Thus:

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

$\Box$


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

Let $x \in U$.

Then:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2} \subseteq U$.

Hence:

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

Thus $U$ is $d_1$-open.


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

Let $x \in U$.

Then:

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

Hence:

$\map {B_{h \epsilon} } {\map f x; d_2} \subseteq U$

Thus $U$ is $d_2$-open.


The result follows by definition of homeomorphic metric spaces.

$\blacksquare$


Also see


Sources