Relation Induced by Partition is Equivalence

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

Let $\mathcal R$ be the relation induced 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 Induced by Partition, we define the relation $\mathcal R$ on $S$ by:


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

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

 * Fundamental Theorem on Equivalence Relations for the converse.