Equivalence of Definitions of Topologically Equivalent Metrics

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 $A$.

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

Let $\left({B, d}\right)$ and $\left({C, d'}\right)$ be metric spaces:

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

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

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

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

That is:
 * $\forall a \in A: h \left({a}\right) = a$

From Identity Mapping is Continuous, $h$ is $\left({d_2, d_2}\right)$-continuous.

Hence because $h$ is $\left({d_1, d'}\right)$-continuous $h$ is $\left({d_2, d'}\right)$-continuous, $h$ is $\left({d_1, d_2}\right)$-continuous.

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

But $h^{-1} \left({U}\right) = U$, so $U$ is $d_1$-open.

A similar argument, still starting with the same proposition that $g$ is $\left({d_1, d'}\right)$-continuous $g$ is $\left({d_2, d'}\right)$-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 $\left({d, d_1}\right)$-continuous for every set $U \subseteq A$ which is open in $\left({A, d_1}\right)$, $f^{-1} \left({U}\right)$ is open in $\left({B, d}\right)$
 * $f$ is $\left({d, d_2}\right)$-continuous for every set $U \subseteq A$ which is open in $\left({A, d_2}\right)$, $f^{-1} \left({U}\right)$ is open in $\left({B, d}\right)$.

Hence $f$ is $\left({d, d_1}\right)$-continuous $f$ is $\left({d, d_2}\right)$-continuous.

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

Hence $g$ is $\left({d_1, d\,'}\right)$-continuous $g$ is $\left({d_2, d\,'}\right)$-continuous.

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

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

From that, we show that both:
 * $f$ is $\left({d, d_1}\right)$-continuous $f$ is $\left({d, d_2}\right)$-continuous;
 * $g$ is $\left({d_1, d'}\right)$-continuous $g$ is $\left({d_2, d'}\right)$-continuous.

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