Pseudometric Defines an Equivalence Relation

Theorem
Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.

For any $x, y \in X$, let $x \sim y$ denote that $d \left({x, y}\right) = 0$.

Then $\sim$ is an equivalence relation, and the equivalence classes consist of sets of elements of $X$ at zero distance from each other.

Proof
Checking in turn each of the criteria for equivalence:

Reflexive
From the definition, we have that $\forall x \in X: d \left({x, x}\right) = 0$.

Symmetric
From the definition, we have that $\forall x, y \in X: d \left({x, y}\right) = d \left({y, x}\right)$.

Transitive
From the definition, we have that $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$.

So if $d \left({x, y}\right) = 0$ and $d \left({y, z}\right) = 0$ it follows directly that $d \left({x, z}\right) = 0$.

Hence the result.