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

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 topologically equivalent.

Then $M_1$ and $M_2$ are homeomorphic.

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


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

Set $B = C = A$.

The result follows by definition of homeomorphic metric spaces.