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

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.