Quotient Set Determined by Relation Induced by Partition is That Partition

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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:

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

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 $\mathcal R$:

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

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.

$\blacksquare$


Sources