Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 2

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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

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.

$\blacksquare$


Sources

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