Relation Induced by Partition is Equivalence

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

Let $\RR$ be the relation induced by $\mathbb S$.

Then:
 * $(1): \quad \RR$ is unique
 * $(2): \quad \RR$ is an equivalence relation on $S$.

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


 * $\mathbb S = S / \RR$

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

From Relation Induced by Partition, we define the relation $\RR$ on $S$ by:


 * $\RR = \set {\tuple {x, y}: \paren {\exists T \in \mathbb S: x \in T \land y \in T} }$

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

Checking in turn each of the critera for equivalence:

Reflexive
$\RR$ is reflexive:

Symmetric
$\RR$ is symmetric:

Transitive
$\RR$ is transitive:

So $\RR$ is reflexive, symmetric and transitive.

Hence, by definition, $\RR$ 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 set of $\RR$-classes constructed above, and no other relation can be constructed in this way.

Also see

 * Fundamental Theorem on Equivalence Relations for the converse.