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, $T=\displaystyle\bigcup_{t\in T} \left[\!\left[{t}\right]\!\right]$, so we can take $U=T$.

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

By definition of saturation, $T=q^{-1}(q(T))$, so we can take $V=q(T)$.

2 implies 1
Let $T = \displaystyle\bigcup_{u\in U} \left[\!\left[{u}\right]\!\right]$ 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 $T = q^{-1}(V)$ with $V\subset S/\sim$.

By Quotient Mapping is Surjection and Image of Preimage of Subset under Surjection equals Subset, $q(q^{-1}(V)) = V$.

Thus:
 * $q^{-1}(q(T)) = q^{-1}(q(q^{-1}(V))) = q^{-1}(V) = T$

Thus $T$ equals its saturation.