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.
The following definitions of the concept of saturated set in the context of Equivalence Relation are equivalent:
Definition 1
$T$ is saturated if and only if it equals its saturation:
- $T = \overline T$
Definition 2
$T$ is saturated if and only if it is a union of equivalence classes:
- $\ds \exists U \subset S : T = \bigcup_{u \mathop \in U} \eqclass u {}$
Definition 3
$T$ is saturated if and only if it is the preimage of some set under the quotient mapping:
- $\exists V \subset S / \sim \; : T = q^{-1} \left[{V}\right]$
Proof
1 implies 2
Let $T = \overline T$.
By definition of saturation:
- $T = \ds \bigcup_{t \mathop \in T} \eqclass t {}$
so we can take $U = T$.
$\blacksquare$
1 implies 3
Let $T = \overline T$.
By definition of saturation:
- $T = q^{-1} \sqbrk {q \sqbrk T}$
so we can take $V = q \sqbrk T$.
$\blacksquare$
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:
- $T = \overline T$
$\blacksquare$
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:
| \(\ds q^{-1} \sqbrk {q \sqbrk T}\) | \(=\) | \(\ds q^{-1} \sqbrk {q \sqbrk {q^{-1} \sqbrk V} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds q^{-1} \sqbrk V\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds T\) |
Thus $T$ equals its saturation.
$\blacksquare$