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:
- $(1): \quad \forall x, y \in A: \map {d_2} {x, y} \le K_1 \map {d_1} {x, y}$
- $(2): \quad \forall x, y \in A: \map {d_1} {x, y} \le K_2 \map {d_2} {x, y}$
By Identity Mapping between Metrics separated by Scale Factor is Continuous:
- 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.
$\blacksquare$