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.

$\blacksquare$