Fundamental Theorem on Equivalence Relations

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Theorem

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


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


Proof

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

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


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 $\mathcal R \subset S \times S$ be a relation on $S$ defined as:

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


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

\(\ds \eqclass 1 {\mathcal R}\) \(=\) \(\ds \set {1, 2, 3}\)
\(\ds \eqclass 4 {\mathcal R}\) \(=\) \(\ds \set {4, 5}\)
\(\ds \eqclass 6 {\mathcal R}\) \(=\) \(\ds \set 6\)


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:

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


Also see


Sources