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:


 * 1. 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 $$S$$.


 * 2. 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.


 * 3. 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.