Domain of Relation is Subclass of Union of Union of Relation/Proof

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Theorem

Let $V$ be a basic universe.

Let $\RR \subseteq V \times V$ be a relation.

Let $\Dom \RR$ denote the domain of $\RR$.


Then:

$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$

where $\bigcup \RR$ denotes the union of $\RR$.


Proof

\(\ds x\) \(\in\) \(\ds \Dom \RR\) Definition of Domain of Relation (Class Theory)
\(\ds \leadsto \ \ \) \(\ds \exists x: \, \) \(\ds \tuple {x, y}\) \(\in\) \(\ds \RR\) Definition of Domain of Relation (Class Theory)
\(\ds \leadsto \ \ \) \(\ds \set {\set x, \set {x, y} }\) \(\in\) \(\ds \RR\) Definition of Ordered Pair


But then:

\(\ds \set x\) \(\in\) \(\ds \set {\set x, \set {x, y} }\)
\(\ds \leadsto \ \ \) \(\ds \set x\) \(\in\) \(\ds \bigcup \RR\) Definition of Union of Class
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \map \bigcup {\bigcup \RR}\) Definition of Union of Class
\(\ds \leadsto \ \ \) \(\ds \Dom \RR\) \(\subseteq\) \(\ds \map \bigcup {\bigcup \RR}\) Definition of Subclass

$\blacksquare$


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