Relation Induced by Quotient Set is Equivalence

Theorem

Let $S$ be a set.

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

Let $S / \RR$ be the quotient set of $S$ determined by $\RR$.

Then $\RR' = \RR$.

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

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

By definition of equivalence class, $y \in \eqclass x \RR$ and $x \in \eqclass x \RR$.

By definition of quotient set, $x$ and $y$ both belong to the same element of $S / \RR$.

So, by definition of $\RR'$, it follows that $\tuple {x, y} \in \RR'$.

That is:

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

and so by definition of subset:

$\RR \subseteq \RR'$

Now let $\tuple {x, y} \in \RR'$.

Then $y$ belongs to the same element of $S / \RR$ that $x$ does.

That is:

$y \in \eqclass x \RR$

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

That is:

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

and so by definition of subset:

$\RR' \subseteq \RR$

The result follows by definition of set equality.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.3$