Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 3

Theorem
Let $M_1 = \left({A, d_1}\right)$ and $M_2 = \left({A, d_2}\right)$ 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:
 * $(1): \quad \forall x, y \in A: d_2 \left({x, y}\right) \le K_1 d_1 \left({x, y}\right)$
 * $(2): \quad \forall x, y \in A: d_1 \left({x, y}\right) \le K_2 d_2 \left({x, y}\right)$

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.