# Fundamental Theorem on Equivalence Relations

## Theorem

Let $\RR \subseteq S \times S$ be an equivalence on a set $S$.

Then the quotient set $S / \RR$ of $S$ by $\RR$ forms a partition of $S$.

## Proof

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

$(1): \quad \ds \bigcup {S / \RR} = S$
$(2): \quad \eqclass x \RR \ne \eqclass y \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$
$(3): \quad \forall \eqclass x \RR \in S / \RR: \eqclass x \RR \ne \O$

Taking each proposition in turn:

### Union of Equivalence Classes is Whole Set

The set of $\RR$-classes constitutes the whole of $S$:

We have that:

 $\ds \forall x \in S: \,$ $\ds x$ $\in$ $\ds \eqclass x \RR$ Definition of Equivalence Class $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \set {y \in S: x \mathrel \RR y}$ Definition of Equivalence Class

and:

 $\ds \eqclass x \RR$ $=$ $\ds \set {y: x \mathrel \RR y}$ Definition of Equivalence Class $\text {(2)}: \quad$ $\ds$ $\subseteq$ $\ds S$ Definition of Subset

Then:

 $\ds S$ $=$ $\ds \bigcup_{x \mathop \in S} \set x$ Definition of Union of Set of Sets $\ds$ $\subseteq$ $\ds \bigcup_{x \mathop \in S} \eqclass x \RR$ from $(1)$ $\ds$ $\subseteq$ $\ds \bigcup_{x \mathop \in S} S$ from $(2)$ $\ds$ $=$ $\ds S$

$\Box$

### Equivalence Classes are Disjoint

First we show that:

$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$

Suppose two $\RR$-classes are not disjoint:

 $\ds \eqclass x \RR \cap \eqclass y \RR \ne \O$ $\leadsto$ $\ds \exists z: z \in \eqclass x \RR \cap \eqclass y \RR$ Definition of Empty Set $\ds$ $\leadsto$ $\ds \exists z: z \in \eqclass x \RR \land z \in \eqclass y \RR$ Definition of Set Intersection $\ds$ $\leadsto$ $\ds \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {y, z} \in \RR}$ Definition of Equivalence Class $\ds$ $\leadsto$ $\ds \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {z, y} \in \RR}$ Definition of Symmetric Relation: $\RR$ is symmetric $\ds$ $\leadsto$ $\ds \tuple {x, y} \in \RR$ Definition of Transitive Relation: $\RR$ is transitive

Thus we have shown that $\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \tuple {x, y} \in \RR$.

Therefore, by the Rule of Transposition:

$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$

Now we show that:

$\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$

Suppose $\tuple {x, y} \in \RR$.

 $\ds$  $\ds y \in \eqclass y \RR$ Definition of Equivalence Class $\ds$  $\ds \tuple {x, y} \in \RR$ by hypothesis $\ds$ $\leadsto$ $\ds y \in \eqclass x \RR$ Definition of Equivalence Class $\ds$ $\leadsto$ $\ds y \in \eqclass x \RR \land y \in \eqclass y \RR$ Rule of Conjunction $\ds$ $\leadsto$ $\ds y \in \eqclass x \RR \cap \eqclass y \RR$ Definition of Set Intersection $\ds$ $\leadsto$ $\ds \eqclass x \RR \cap \eqclass y \RR \ne \O$ Definition of Empty Set

Thus we have shown that:

$\tuple {x, y} \in \RR \implies \eqclass x \RR \cap \eqclass y \RR \ne \O$

Therefore, by the Rule of Transposition:

$\eqclass x \RR \cap \eqclass y \RR = \O \implies \paren {x, y} \notin \RR$

Using the rule of Biconditional Introduction on these results:

$\eqclass x \RR \cap \eqclass y \RR = \O \iff \paren {x, y} \notin \RR$

and the proof is complete.

$\Box$

### Equivalence Class is not Empty

 $\ds \forall \eqclass x \RR \subseteq S: \exists x \in S: \,$ $\ds x$ $\in$ $\ds \eqclass x \RR$ Definition of Equivalence Class $\ds \leadsto \ \$ $\ds \eqclass x \RR$ $\ne$ $\ds \O$ Definition of Empty Set

$\Box$

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

$\blacksquare$

## Examples

### Arbitrary Equivalence on Set of $6$ Elements: $1$

Let $S = \set {1, 2, 3, 4, 5, 6}$.

Let $\RR \subset S \times S$ be a relation on $S$ defined as:

$\RR = \set {\tuple {1, 1}, \tuple {1, 2}, \tuple {1, 3}, \tuple {2, 1}, \tuple {2, 2}, \tuple {2, 3}, \tuple {3, 1}, \tuple {3, 2}, \tuple {3, 3}, \tuple {4, 4}, \tuple {4, 5}, \tuple {5, 4}, \tuple {5, 5}, \tuple {6, 6} }$

Then $\RR$ is an equivalence relation which partitions $S$ into:

 $\ds \eqclass 1 \RR$ $=$ $\ds \set {1, 2, 3}$ $\ds \eqclass 4 \RR$ $=$ $\ds \set {4, 5}$ $\ds \eqclass 6 \RR$ $=$ $\ds \set 6$

### Arbitrary Equivalence on Set of $6$ Elements: $2$

Let $S = \set {1, 2, 3, 4, 5, 6}$.

Let $\RR \subset S \times S$ be an equivalence relation on $S$ with the properties:

 $\ds 1$ $\RR$ $\ds 3$ $\ds 3$ $\RR$ $\ds 4$ $\ds 2$ $\RR$ $\ds 6$ $\ds \forall a \in A: \,$ $\ds \size {\eqclass a \RR}$ $=$ $\ds 3$

Then the equivalence classes of $\RR$ are:

 $\ds \eqclass 1 \RR$ $=$ $\ds \set {1, 3, 4}$ $\ds \eqclass 2 \RR$ $=$ $\ds \set {2, 5, 6}$