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:


 * $\bigcup {S / \mathcal R} = S$


 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \ne \left[\!\left[{y}\right]\!\right]_\mathcal R \iff \left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing$


 * $\forall \left[\!\left[{x}\right]\!\right]_\mathcal R \in S / \mathcal R: \left[\!\left[{x}\right]\!\right]_\mathcal R \ne \varnothing$

Taking each proposition in turn:

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

Also:

Thus by the definition of Set Equality, $\bigcup {S / \mathcal R} = S$, and so the set of $\mathcal R$-classes constitutes the whole of $S$.

Equivalence Classes are Disjoint
Unequal $\mathcal R$-classes are disjoint.

Suppose $\left[\!\left[{x}\right]\!\right]_\mathcal R \ne \left[\!\left[{y}\right]\!\right]_\mathcal R$.

Then:

So $S / \mathcal R$ is mutually disjoint.

Equivalence Class is Not Empty
No $\mathcal R$-class is empty:

This is immediate, from No Equivalence Class is Null:


 * $\forall \left[\!\left[{x}\right]\!\right]_\mathcal R \subseteq S: \left[\!\left[{x}\right]\!\right]_\mathcal R \ne \varnothing$

Thus all conditions for $S / \mathcal R$ to be a partition are fulfilled.

Also see

 * Equivalence Relation Defined by a Partition for the converse.