Understanding a double union of a set
I have an exercise example from Goldrei's Classical Set Theory (For Guided Independent Study) and I am having trouble understanding the axiom of union examples.
Exercise: Given a set F of ordered pairs <a,b>, give a set in ZF in which all the a's and b's lie.
Solution: A typical element <a,b> of F is a set {{a}, {a,b}}. Thus $\cup $F is the set of all {a}'s and {a,b}'s for <a,b> $\in$ F, so that $\cup $($\cup $F) is the set of all a's and b's for which <a,b> $\in$ F. As F is a set, $\cup $($\cup $F) is also a set, by the union axiom.
Question: I am quite confused about the set of $\cup $($\cup $F) and what the elements are. So if I take {{a}, {a,b}} and use the axiom of union I get $\cup $F which results to the set {a,b}. But if I do another union i.e,$\cup $($\cup $F) wouldn't my result just be {a,b} again? I understand elementary union (that we were taught in High school) but since I am learning intro into set theory I'm having trouble grasping the axiom of union and understanding the union of the union of a set.
$\endgroup$3 Answers
$\begingroup$An example may help. Let$$ F = \{(1, 2), (2, 3), (3, 1)\} = \{\{\{1\}, \{1, 2\}\}, \{\{2\}, \{2, 3\}\}, \{\{3\}, \{3, 1\}\}\} $$Then:$$ \begin{align*} \bigcup F &=\{\{1\}, \{1, 2\}, \{2\}, \{2, 3\}, \{3\}, \{3, 1\}\} \end{align*} $$And
$$ \begin{align*} \bigcup\bigcup F &= \bigcup\{\{1\}, \{1, 2\}, \{2\}, \{2, 3\}, \{3\}, \{3, 1\}\} \\ &= \{1, 1, 2, 2, 2, 3, 3, 3, 1\} \\ &= \{1, 2, 3\} \end{align*} $$
$\endgroup$ 3 $\begingroup$We have $$ x\in\bigcup Y\iff \exists y\colon x\in y\in Y$$and$$ x\in\bigcup \bigcup F\iff \exists y\colon x\in y\in \bigcup F\iff \exists y,z\colon x\in y\in z\in F$$Hence if each element of $F$ is a Kuratowski pair, then we obtain the elements of elements elements of these pairs, so indeed what we want.
$\endgroup$ 2 $\begingroup$if I take $\{ \{ a \}, \{ a,b \} \}$ and use the axiom of union I get $\cup F$ which results to the set $\{ a,b \}$.
No; if $(a,b)=\{ \{ a \}, \{ a,b \} \} \in F$ we have that:
$\{ a \}, \{ a,b \} \in \cup F$.
The first union is needed in order to retrieve from the set $F$ of pairs the elements.
Then, we need a second "extraction" in order to retrieve $a,b$, from which $a, b \in \cup (\cup F)$.
Thus:
$\endgroup$ 1$\cup (\cup F) = \{ z \mid \exists w,a,b \ (w= \{ \{ a \}, \{ a,b \}\} \land w \in F \}$.
