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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...

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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) $\cup\{0,\{7\},\{\{7\}\}=0\cup\{7\}\cup\{\{7\}\}=\{7,\{7\}\}$

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  • $\begingroup$ I was familiar with transitivity because I've studied also other parts after this exercises. Your explanation is very clear! For the 3rd I was almost there, this kind of notation is new for me! Really surprised for $ \cup \omega = \omega$. For confirm, $\{ \omega \}$ are the elements of the natural numbers, and $ \omega $ is THE SET of the natural numbers, right? $\endgroup$ – Alessar Jun 8 at 15:56
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    $\begingroup$ @Alessar: It's not clear to me which thought process led you to describe $\{\omega\}$ as "the elements of the natural numbers", but I can't imagine a way for that description to mean something correct. Rather, $\{\omega\}$ is a set with exactly one element, and that one element is $\omega$. $\endgroup$ – Henning Makholm Jun 8 at 16:19
  • $\begingroup$ Indeed $\omega$ is the set of natural numbers. Further $\{\omega \} $ is a set that has exactly one element which is $\omega$. So it is not correct to say that "$\{\omega \} $ are the elements of the natural numbers". $\endgroup$ – drhab Jun 8 at 16:31
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    $\begingroup$ @Alessar $y\in\{a\}$ means that $y$ is an element of the set $\{a\}$. The notation $\{a\}$ reveals that $\{a\}$ is a set that has exactly one element, which is $a$. So then there is only one choice for $y$: it must be $a$. $\endgroup$ – drhab Jul 5 at 14:44
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    $\begingroup$ Just want to let you all know that I've passed the exam of foundations of mathematics. Thanks again $\endgroup$ – Alessar Jul 11 at 11:09

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