Saturation Under Equivalence Relation in Terms of Graph
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Theorem
Let $\RR \subset S \times S$ be an equivalence relation on a set $S$.
Let $\pr_1, \pr_2 : S \times S \to S$ denote the projections.
Let $T\subset S$ be a subset.
Let $\overline T$ denote its saturation.
Then the following hold:
- $\overline T = \map {\pr_1} {\RR \cap \map {\pr_2^{-1} } T}$
- $\overline T = \map {\pr_2} {\RR \cap \map {\pr_1^{-1} } T}$
Proof
Let $s \in S$.
We have:
\(\ds \) | \(\) | \(\ds s \in \map {\pr_1} {\RR \cap \map {\pr_2^{-1} } T}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds \exists t \in S: \tuple {s, t} \in \RR \cap \map {\pr_2^{-1} } T\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds \exists t \in S : \tuple {s, t} \in \RR \cap (S \times T)\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds \exists t \in T : \tuple {s, t} \in \RR\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(\) | \(\ds s \in \overline T\) | Definition of Saturation Under Equivalence Relation |
A similar reasoning proves the second identity.
$\blacksquare$