Image of Domain of Relation is Image Set
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Theorem
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
The image of the domain of $\RR$ is the image set of $\RR$:
- $\RR \sqbrk {\Dom \RR} = \Img \RR$
where $\Img \RR$ is the image of $\RR$.
Proof
Let $y \in \RR \sqbrk {\Dom \RR}$.
\(\ds y\) | \(\in\) | \(\ds \RR \sqbrk {\Dom \RR}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in \Dom \RR: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | Definition of Image of Subset under Relation | |||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds \Img \RR\) | Definition of Image of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \RR \sqbrk {\Dom \RR}\) | \(\subseteq\) | \(\ds \Img \RR\) | Definition of Subset |
Let $y \in \Img \RR$.
\(\ds y\) | \(\in\) | \(\ds \Img \RR\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in S: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | Definition of Image of Relation | |||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds \RR \sqbrk {\Dom \RR}\) | Definition of Domain of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \Img \RR\) | \(\subseteq\) | \(\ds \RR \sqbrk {\Dom \RR}\) | Definition of Subset |
The result follows by definition of set equality.
$\blacksquare$