Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 1
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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
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.
$\blacksquare$