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.