ZF Set Theory exercises on axioms, $\cup \{ \omega \}$, $\cup \cup \{ \omega \}$, $\cup \{ 0,\{7\},\{\{7\}\} \}$ I'm new to ZF set theory (second day of study), I'm preparing an exam about it. Here some exercises with my solution, if you could give me some feedback about that!
$1) \cup \{ \omega \} = \{ \{ \omega \} \}$
I know that must be some $b = \cup \{ \omega \}$. But the elements of $b$ are already present here, so I consider the set of union $b \cup \{ \omega \}=\{b, \{ \omega \}\} \rightarrow \{ \{ \omega \} \}$ because it's already inside. Is this legit?
$2) \cup \cup \{ \omega \} =  \{ \{  \{ \{ \omega \} \} \} \} $
Similar to the exercise 1)
$3) \cup \{ 0,\{7\},\{\{7\}\} \} = 0\cup\{7\}\cup\{\{7\}\} = 0\cup\{\{0,1,2,3,4,5,6\}\}\cup\{\{\{0,1,2,3,4,5,6\}\}\}$
Here I know that $0 = \emptyset$ and $7=\{0,1,2,3,4,5,6\}$ but I don't know how to use this info here. I've done some manipulation, I don't know if this is the end and solution...
 A: Let me start with a characterization of $\cup b$ where $b$ is a set:$$x\in\cup b\iff x\in y\in b\text{ for some set }y$$
If $a_i$ is a set for $i=1,2,\dots,n$ then $a_1\cup a_2\cup\cdots\cup a_n$ is a notation of $\cup\{a_1,a_2,\dots,a_n\}$. I use this in 3).

1) $x\in\cup\{a\}\iff\exists y[x\in y\wedge y\in\{a\}]\iff \exists y[x\in y\wedge y=a]\iff x\in a$
This for every $x$ hence showing that $\cup\{a\}=a$. This for every $a$ so also for $a=\omega$.
2) In 1) it is found that $\cup\{\omega\}=\omega$ so that $\cup\cup\{\omega\}=\cup\omega$. We will now prove that $\omega=\cup\omega$
If $n\in\omega$ then $n\in n\cup\{n\}\in\omega$ which implies that $n\in\cup\omega$.
If conversely $n\in\cup\omega$ then $n\in m$ for some $m\in\omega$. From this we are allowed to conclude that $n\in\omega$ because $\omega$ is a transitive set (are you familiar with that? This actually needs a proof on its own.)
Proved is now that: $$\cup\cup\{\omega\}=\cup\omega=\omega$$
3) $\cup\{0,\{7\},\{\{7\}\}=0\cup\{7\}\cup\{\{7\}\}=\{7,\{7\}\}$
