Equivalence Classes are Disjoint

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Theorem

Let $\mathcal R$ be an equivalence relation on a set $S$.


Then all $\mathcal R$-classes are pairwise disjoint:

$\tuple {x, y} \notin \mathcal R \iff \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O$


Proof 1

First we show that:

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


Suppose two $\mathcal R$-classes are not disjoint:

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


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


Therefore, by the Rule of Transposition:

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


Now we show that:

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


Suppose $\tuple {x, y} \in \mathcal R$.

\(\displaystyle \) \(\) \(\displaystyle y \in \eqclass y {\mathcal R}\) Definition of Equivalence Class
\(\displaystyle \) \(\) \(\displaystyle \tuple {x, y} \in \mathcal R\) by hypothesis
\(\displaystyle \) \(\leadsto\) \(\displaystyle y \in \eqclass x {\mathcal R}\) Definition of Equivalence Class
\(\displaystyle \) \(\leadsto\) \(\displaystyle y \in \eqclass x {\mathcal R} \land y \in \eqclass y {\mathcal R}\) Rule of Conjunction
\(\displaystyle \) \(\leadsto\) \(\displaystyle y \in \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R}\) Definition of Set Intersection
\(\displaystyle \) \(\leadsto\) \(\displaystyle \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O\) Definition of Empty Set


Thus we have shown that:

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


Therefore, by the Rule of Transposition:

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


Using the rule of Biconditional Introduction on these results:

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

and the proof is finished.

$\blacksquare$


Proof 2

Suppose that for $x, y \in S$:

$\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O$

Let:

$z \in \eqclass x {\mathcal R}$
$z \in \eqclass y {\mathcal R}$

Then by definition of equivalence class:

$\tuple {x, z} \in \mathcal R$
$\tuple {y, z} \in \mathcal R$


Let $c \in \eqclass x {\mathcal R}$.

That is:

$\tuple {x, c} \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is symmetric so:

$\tuple {z, x} \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is transitive so:

$\tuple {z, x} \in \mathcal R \land \tuple {x, c} \in \mathcal R \implies \tuple {z, c} \in \mathcal R$

and

$\tuple {y, z} \in \mathcal R \land \tuple {z, c} \in \mathcal R \implies \tuple {y, c} \in \mathcal R$

So we have $c \in \eqclass y {\mathcal R}$.

By definition of subset:

$\eqclass x {\mathcal R} \subseteq \eqclass y {\mathcal R}$


Similarly, let $c \in \eqclass y {\mathcal R}$.

That is:

$\tuple {y, c} \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is symmetric so:

$\tuple {z, y} \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is transitive so:

$\tuple {z, y} \in \mathcal R \land \tuple {y, c} \in \mathcal R \implies \tuple {z, c} \in \mathcal R$

and

$\tuple {x, z} \in \mathcal R \land \tuple {z, c} \in \mathcal R \implies \tuple {x, c} \in \mathcal R$

So we have $c \in \eqclass x {\mathcal R}$.

By definition of subset:

$\eqclass y {\mathcal R} \subseteq \eqclass x {\mathcal R}$


That is:

$\eqclass x {\mathcal R} \subseteq \eqclass y {\mathcal R}$

and

$\eqclass y {\mathcal R} \subseteq \eqclass x {\mathcal R}$

By definition of set equality:

$\eqclass x {\mathcal R} = \eqclass y {\mathcal R}$


Thus:

$\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O \implies \eqclass x {\mathcal R} = \eqclass y {\mathcal R}$

and the result follows.

$\blacksquare$


Also see


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