Relation Induced by Partition is Equivalence

Theorem
Let $$\mathbb S$$ be a partition of a set $$S$$.

Let $$\mathcal R$$ be the relation defined by $\mathbb S$.

Then:
 * $$\mathcal R$$ is unique;
 * $$\mathcal R$$ is an equivalence relation on $$S$$.

Hence $$\mathbb S$$ is the quotient set of $$S$$ by $$\mathcal R$$, that is:


 * $$\mathbb S = S / \mathcal R$$

Proof
Let $$\mathbb S$$ be a partition of a $$S$$.

From Relation Defined by Partition, we define the relation $$\mathcal R$$ on $$S$$ by:


 * $$\mathcal R = \left\{{\left({x, y}\right): \left({\exists T \in \mathbb S: x \in T \and y \in T}\right)}\right\}$$

Test for Equivalence
We are to show that $$\mathcal R$$ is an equivalence relation.

Checking in turn each of the critera for equivalence:

Reflexive
$$\mathcal R$$ is reflexive:

$$ $$

Symmetric
$$\mathcal R$$ is symmetric:

$$ $$ $$ $$

Transitive
$$\mathcal R$$ is transitive:

$$ $$ $$

So $$\mathcal R$$ is reflexive, symmetric and transitive, therefore $$\mathcal R$$ is an equivalence relation.

Test for Uniqueness
Now by definition of a partition, we have that:


 * $$\mathbb S$$ partitions $$S \implies \forall x \in S: \exists T \in \mathbb S: x \in T$$

Also:

$$ $$

Thus $$\mathbb S$$ is the family of $$\mathcal R$$-classes constructed above, and no other relation can be constructed in this way.

Also see

 * Quotient Set forms a Partition for the converse.