Topological Equivalence is Equivalence Relation
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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Definition $2.4.1$