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 |
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 8$ Relations: Exercise $8.1$