Pseudometric Defines an Equivalence Relation

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

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.