Correspondence Theorem (Set Theory)

This proof is about Correspondence Theorem in the context of Set Theory. For other uses, see Correspondence Theorem.

Theorem

Let $S$ be a set.

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

Let $\mathscr A$ be the set of partitions of $S$ associated with equivalence relations $\RR'$ on $S$ such that:

$\tuple {x, y} \in \RR \iff \tuple {x, y} \in \RR'$

Then there exists a bijection $\phi$ from $\mathscr A$ onto the set of partitions of the quotient set $S / \RR$.

Proof

Denote the equivalence class of an element $x$ of $S$ by $\eqclass x {\RR}$ with respect to the relation $\RR$.

Consider the relation on $S$:

$\phi = \set {\tuple {\eqclass x {\RR'}, \eqclass x \RR}: x \in S}$

We prove that $\phi$ is a bijection.

Let $\eqclass x \RR = \eqclass y \RR$.

Then:

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

Thus:

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

Therefore:

$\eqclass x {\RR'} = \eqclass y {\RR'}$

and so $\phi$ is one-to-many.

Let $\eqclass x {\RR'} = \eqclass y {\RR'}$.

Then:

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

Thus:

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

Therefore $\eqclass x \RR = \eqclass y \RR$

Hence $\phi$ is many-to-one.

As $\phi$ is both one-to-many and many-to-one, it is by definition a one-to-one relation.

From the fact that $\Rng \phi = S / \RR$ we have that $\phi$ is both left-total and right-total.

We have demonstrated that $\phi$ is left-total and many-to-one, and so by definition is a mapping.

We have demonstrated that $\phi$ is right-total, and so by definition is a surjection.

We have demonstrated that $\phi$ is one-to-one, and so by definition is an injection.

Hence by definition $\phi$ is a bijection.

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$: Exercise $18$

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• 1977: Herbert B. Enderton: Elements of Set Theory P60