# 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...

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

• 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? – Alessar Jun 8 at 15:56
• @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 $\omega$. – Henning Makholm Jun 8 at 16:19
• 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". – drhab Jun 8 at 16:31
• @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 $a$. – drhab Jul 5 at 14:44
• Just want to let you all know that I've passed the exam of foundations of mathematics. Thanks again – Alessar Jul 11 at 11:09