Topological Equivalence is Equivalence Relation

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Theorem

Let $A$ be a set.

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

Let $\sim$ be the relation on $\DD$ defined as:

$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is topologically equivalent to $d_2$


Then $\sim$ is an equivalence relation.


Proof

Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.


In the following, let $d_1, d_2, d_3 \subseteq \DD$ 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.

$\Box$


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.

$\Box$


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.

$\Box$


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

Hence by definition it is an equivalence relation.

$\blacksquare$


Sources

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