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:
 * $T = \displaystyle \bigcup_{t \mathop \in T} \left[\!\left[{t}\right]\!\right]$

so we can take $U = T$.

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

By definition of saturation:
 * $T = q^{-1} \left[{q \left[{T}\right]}\right]$

so we can take $V = q \left[{T}\right]$.

2 implies 1
Let $T = \displaystyle\bigcup_{u \mathop \in U} \left[\!\left[{u}\right]\!\right]$ with $U \subset S$.

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

By definition of union:
 * $\exists u \in U : t \in \left[\!\left[{u}\right]\!\right]$

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

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

By definition of equivalence class:
 * $s \in \left[\!\left[{u}\right]\!\right]$

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 $q$:
 * $T = q^{-1} \left[{V}\right]$

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

Thus:

Thus $T$ equals its saturation.