Topological Equivalence is Equivalence Relation

Theorem
Let $A$ be a set.

Let $\mathcal D$ be the set of all metrics on $A$.

Let $\sim$ be the relation on $\mathcal D$ defined as:
 * $\forall d_1, d_2 \in \mathcal D: d_1 \sim d_2 \iff d_1$ is topologically equivalent to $d_2$

Then $\sim$ is an equivalence relation.

Proof
Let $A$ be a set and let $\mathcal D$ be the set of all metrics on $A$.

In the following, let $d_1, d_2, d_3 \subseteq \mathcal D$ be arbitrary.

Checking in turn each of the criteria for equivalence:

Reflexivity
Let $d_1$ be a metric on $A$.

Then trivially:
 * $U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_1$-open.

That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be reflexive.

Symmetry
Let $d_1 \sim d_2$.

That is, let $d_1, d_2$ be topologically equivalent metrics on $A$.

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

It follows from Biconditional is Commutative that:
 * $U \subseteq A$ is $d_2$-open $\iff$ $U \subseteq A$ is $d_1$-open.

That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be symmetric.

Transitivity
Let $d_1 \sim d_2$ and $d_2 \sim d_3$.

Then by definition:
 * $U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.
 * $U \subseteq A$ is $d_2$-open $\iff$ $U \subseteq A$ is $d_3$-open.

Then by Biconditional is Transitive:
 * $U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_3$-open.

That is, $d_1 \sim d_3$ and so $\sim$ has been shown to be transitive.

$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.