Fundamental Theorem on Equivalence Relations

Theorem
Let $\mathcal R \subseteq S \times S$ be an equivalence on a set $S$.

Then the quotient $S / \mathcal R$ of $S$ by $\mathcal R$ forms a partition of $S$.

Proof
To prove that $S / \mathcal R$ is a partition of $S$, we have to prove:


 * $(1): \quad \displaystyle \bigcup {S / \mathcal R} = S$


 * $(2): \quad \eqclass x {\mathcal R} \ne \eqclass y {\mathcal R} \iff \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O$


 * $(3): \quad \forall \eqclass x {\mathcal R} \in S / \mathcal R: \eqclass x {\mathcal R} \ne \O$

Taking each proposition in turn:

Union of Equivalence Classes is Whole Set
The set of $\mathcal R$-classes constitutes the whole of $S$:

Equivalence Class is not Empty
Thus all conditions for $S / \mathcal R$ to be a partition are fulfilled.

Also see

 * Relation Induced by Partition is Equivalence for the converse.