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 $(1)$ and Set Union Preserves Subsets

$\ds $

$\subseteq$

$\ds \bigcup_{x \mathop \in S} S$

from $(2)$ and Set Union Preserves Subsets

$\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

So:

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

By the Rule of Transposition:

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

$\Box$

Now we show that:

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

Suppose:

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

Then:

$\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

So:

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

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:

$\ds 1$

$\RR$

$\ds 3$

$\RR$

$\ds 4$

$\ds 2$

$\RR$

$\ds 6$

$\ds \forall a \in A: \, $

$\ds \size {\eqclass a \RR}$

$=$

$\ds 3$

Then the equivalence classes of $\RR$ are:

$\ds \eqclass 1 \RR$

$=$

$\ds \set {1, 3, 4}$

$\ds \eqclass 2 \RR$

$=$

$\ds \set {2, 5, 6}$

Also see

Relation Induced by Partition is Equivalence for the converse.

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