Domain of Relation is Image of Inverse Relation

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Theorem

Let $\RR \subseteq S \times T$ be a relation.

Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$.


Then:

$\Dom \RR = \Img {\RR^{-1} }$

That is, the domain of a relation is the image of its inverse.


Proof

By definition:

$\Dom \RR := \set {s \in S: \exists t \in T: \tuple {s, t} \in \RR}$
$\Img {\RR^{-1} } := \set {s \in S: \exists T \in T: \tuple {t, s} \in \RR^{-1} }$


\(\ds x\) \(\in\) \(\ds \Dom \RR\)
\(\ds \leadstoandfrom \ \ \) \(\ds \exists t \in T: \tuple {x, t}\) \(\in\) \(\ds \RR\) Definition of Domain of Relation
\(\ds \leadstoandfrom \ \ \) \(\ds \exists t \in T: \tuple {t, x}\) \(\in\) \(\ds \RR^{-1}\) Definition of Inverse Relation
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\in\) \(\ds \Img {\RR^{-1} }\) Definition of Image of Relation

$\blacksquare$


Also see


Sources