Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic

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 topologically equivalent.

Then $M_1$ and $M_2$ are homeomorphic.

Proof
By definition, $d_1$ and $d_2$ are topologically equivalent on $A$ iff
 * For all metric spaces $\left({B, d}\right)$ and $\left({C, d'}\right)$:
 * For all mappings $f: B \to A$ and $g: A \to C$:


 * $(1): \quad f$ is $\left({d, d_1}\right)$-continuous iff $f$ is $\left({d, d_2}\right)$-continuous
 * $(2): \quad g$ is $\left({d_1, d'}\right)$-continuous iff $g$ is $\left({d_2, d'}\right)$-continuous.

Set $B = C = A$.

The result follows by definition of homeomorphic metric spaces.