Equivalence of Definitions of Saturated Set Under Equivalence Relation

Theorem
Let $\sim$ be an equivalence relation on a set $S$.

Let $T \subset S$ be a subset.

1 implies 2
Let $T = \overline T$.

By definition of saturation:
 * $T = \ds \bigcup_{t \mathop \in T} \eqclass t {}$

so we can take $U = T$.

1 implies 3
Let $T = \overline T$.

By definition of saturation:
 * $T = q^{-1} \sqbrk {q \sqbrk T}$

so we can take $V = q \sqbrk T$.

2 implies 1
Let $T = \ds \bigcup_{u \mathop \in U} \eqclass u {}$ with $U \subset S$.

Let $s \in S$ and $t \in T$ such that $s \sim t$.

By definition of union:
 * $\exists u \in U : t \in \eqclass u {}$

By definition of equivalence class:
 * $t \sim u$

Because $\sim$ is transitive:
 * $s \sim u$

By definition of equivalence class:
 * $s \in \eqclass u {}$

Thus:
 * $s \in T$

Because $s$ was arbitrary:
 * $\overline T \subset T$

By Set is Contained in Saturation Under Equivalence Relation:
 * $T \subset \overline T$

Thus:
 * $T = \overline T$

3 implies 1
Let $V$ be a subset of the quotient mapping of $S$ by $\sim$:
 * $V \subset S / \sim$

Let $T$ be the preimage of $V$ under $q$:
 * $T = q^{-1} \sqbrk V$

By Quotient Mapping is Surjection and Image of Preimage of Subset under Surjection equals Subset:
 * $q \sqbrk {q^{-1} \sqbrk V} = V$

Thus:

Thus $T$ equals its saturation.