Direct Image Mapping of Domain is Image Set of Relation
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Theorem
Let $S$ and $T$ be sets.
Let $\powerset S$ and $\powerset T$ be their power sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping of $\RR$:
- $\forall X \in \powerset S: \map {\RR^\to} X = \begin {cases} \set {t \in T: \exists s \in X: \tuple {x, t} \in \RR} & : X \ne \O \\ \O & : X = \O \end {cases}$
Then:
- $\map {\RR^\to} {\Dom \RR} = \Img \RR$
where:
- $\Dom \RR$ is the domain of $\RR$
- $\Img \RR$ is the image set of $\RR$.
Proof
\(\ds y\) | \(\in\) | \(\ds \map {\RR^\to} {\Dom \RR}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists x \in S: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | Definition of Direct Image Mapping of Mapping | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds y\) | \(\in\) | \(\ds \Img \RR\) | Definition of Image Set of Relation |
$\blacksquare$