Lipschitz Equivalence yields Lipschitz Equivalent Metric
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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 $f: M_1 \to M_2$ be a Lipschitz equivalence.
Let $d_3: A^2 \to \R$ be a mapping defined as:
- $\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$
Then $d_3$ yields a metric on $M_1$ which is Lipschitz equivalent to $d_1$.
Proof
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Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics