Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 3

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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:

$\exists K_1, K_2 \in \R_{>0}$ 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.

$\blacksquare$


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