Equivalence of Definitions of Topologically Equivalent Metrics

Theorem
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Definition 1 implies Definition 2
Let $d_1$ and $d_2$ be topologically equivalent by definition 1.

Let $\struct {B, d}$ and $\struct {C, d'}$ be metric spaces:

Let $f: B \to A$ and $g: A \to C$ be mappings.

Then by definition:
 * $f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous
 * $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.

Suppose $U \subseteq A$ is $d_2$-open.

Let $h: A \to A$ be the identity mapping.

That is:
 * $\forall a \in A: \map h a = a$

From Identity Mapping is Continuous, $h$ is $\tuple {d_2, d_2}$-continuous.

Hence because $h$ is $\tuple {d_1, d'}$-continuous $h$ is $\tuple {d_2, d'}$-continuous, $h$ is $\tuple {d_1, d_2}$-continuous.

From Metric Space Continuity by Open Set, it follows that $h^{-1} \sqbrk U$ is $d_1$-open.

But $h^{-1} \sqbrk U = U$, so $U$ is $d_1$-open.

A similar argument, still starting with the same proposition that $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous, shows that if $U$ is $d_1$-open then it is $d_2$-open.

Thus $d_1$ and $d_2$ are topologically equivalent by definition 2.

Definition 2 implies Definition 1
Let $d_1$ and $d_2$ be topologically equivalent by definition 2.

Then by definition:
 * $U \subseteq A$ is $d_1$-open $U \subseteq A$ is $d_2$-open.

From Metric Space Continuity by Open Set, we have that:
 * $f$ is $\tuple {d, d_1}$-continuous for every set $U \subseteq A$ which is open in $\struct {A, d_1}$, $f^{-1} \sqbrk U$ is open in $\struct {B, d}$
 * $f$ is $\tuple {d, d_2}$-continuous for every set $U \subseteq A$ which is open in $\struct {A, d_2}$, $f^{-1} \sqbrk U$ is open in $\struct {B, d}$.

Hence $f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous.

Similarly:
 * $g$ is $\tuple {d_1, d'}$-continuous for every set $U \subseteq C$ which is open in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is open in $\struct {A, d_1}$
 * $g$ is $\tuple {d_2, d'}$-continuous for every set $U \subseteq C$ which is open in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is open in $\struct {A, d_2}$.

Hence $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.

Thus $d_1$ and $d_2$ are topologically equivalent by definition 1.

Note
Note that from the proposition that $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous, we show that $U \subseteq A$ is $d_1$-open  $U \subseteq A$ is $d_2$-open.

From that, we show that both:
 * $f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous
 * $g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.

Hence the proposition that $f$ is $\tuple {d, d_1}$-continuous $f$ is $\tuple {d, d_2}$-continuous is superfluous, as it follows directly from the proposition concerning $g$.