# Quotient Set Determined by Relation Induced by Partition is That Partition

## Theorem

Let $S$ be a set.

Let $\mathcal P$ be a partition of $S$.

Let $\mathcal R$ be the relation induced by $\mathcal P$.

Then the quotient set $S / \mathcal R$ of $S$ is $\mathcal P$ itself.

## Proof

Let $P \subseteq S$ such that $P \in \mathcal P$.

Let $x \in P$.

Then:

 $\displaystyle y$ $\in$ $\displaystyle \left[\!\left[{x}\right]\!\right]_\mathcal R$ $\displaystyle \iff \ \$ $\displaystyle \left({x, y}\right)$ $\in$ $\displaystyle \mathcal R$ by definition of equivalence class $\displaystyle \iff \ \$ $\displaystyle y$ $\in$ $\displaystyle P$ by definition of relation induced by $\mathcal P$

Therefore:

$P = \left[\!\left[{x}\right]\!\right]_\mathcal R$

and so:

$P \in S / \mathcal R$

and so:

$\mathcal P \subseteq S / \mathcal R$

Now let $x \in S$.

As $\mathcal P$ is a partition:

$\exists P \in \mathcal P: x \in P$

Then by definition of $\mathcal R$:

$\left({x, y}\right) \in \mathcal R \iff y \in \left[\!\left[{x}\right]\!\right]_\mathcal R$
 $\displaystyle y$ $\in$ $\displaystyle P$ $\displaystyle \iff \ \$ $\displaystyle \left({x, y}\right)$ $\in$ $\displaystyle \mathcal R$ by definition of relation induced by $\mathcal P$ $\displaystyle \iff \ \$ $\displaystyle y$ $\in$ $\displaystyle \left[\!\left[{x}\right]\!\right]_\mathcal R$ by definition of equivalence class

Therefore:

$\left[\!\left[{x}\right]\!\right]_\mathcal R = P$

and so:

$\left[\!\left[{x}\right]\!\right]_\mathcal R \in \mathcal P$

That is:

$\mathcal S / \mathcal R \subseteq P$

It follows by definition of set equality that:

$\mathcal S / \mathcal R = P$

Hence the result.

$\blacksquare$