I want to show that
(a) $ \forall \alpha \; \exists \beta$ such that $\beta = \{ dom R \; | \; R \in \alpha \}$
I'm having difficulty proving this axiomatically. I'm not even certain that it can be done. I'm trying to solve this as part of the larger problem of showing that
(b) $\forall \alpha \; dom\bigcup\alpha = \bigcup\{dom R \; | \; R \in \alpha\}$
My approach to solving (a), so far, is the following (supposed to be a fitch-style proof - I couldn't figure out how to do fitch in TeX):
START OF PROOF
(1) Take $A$
(2) Take $s$
(3) Assume $s \in dom \bigcup A$
(4) [Somehow show that $\{dom R \; | \; R \in \alpha\}$ exists for any $\alpha$]
(5) Therefore, there exists $B$ = $\{dom R \; | \; R \in A\}$
(6) From (3), $\exists \psi \;(\in \bigcup A)$
(7) Fix $y$ : $\in \bigcup A$
(8) From (7), $\exists \mu \; (\in \mu \; \wedge \; \mu \in A)$
(9) Fix $R$ : $\in R \; \wedge \; R \in A$
(10) $dom R \in B$ [Remember (5): $B$ = ${dom R \; | \; R \in A}$]
(11) $s \in \bigcup B$ [Because $s \in dom R \; \wedge \; dom R \in B$][This gets me the first side of the bi-conditional]
(12) Assume $s \in \bigcup\{domR \; | \; R \in A\}$
(13) [Show that $s \in dom \bigcup A$][This gets me the second side of the bi-conditional]
(14) From (3-13) : $s \in dom \bigcup A \; \leftrightarrow \; s \in \bigcup \{ dom R \; | \; R \in A\}$(15) $\forall \varphi \; ( \varphi \in dom \bigcup A \; \leftrightarrow \; \varphi \in \bigcup \{ dom R \; | \; R \in A \} )$
$\forall \alpha \forall \varphi \; ( \varphi \in dom \bigcup \alpha \; \leftrightarrow \; \varphi \in \bigcup \{ dom R \; | \; R \in \alpha \} )$
END OF PROOF
My problem is figuring out step (4).
(This is not homework - I'm teaching myself set theory from Enderton's in my spare time, and am a little stumped on this one.)
Thanks!
Max