properties of ordinal numbers An ordinal number is a set $\xi$ such that:


*

*$\zeta\in\xi\Rightarrow \zeta\subset\xi$

*$\zeta,\eta\in\xi\Rightarrow \zeta=\eta$ or $\zeta\in\eta$ or $\eta\in\zeta$

*$\emptyset\not=A\subset \xi\Rightarrow \exists\zeta\in A:\zeta\cap A=\emptyset$
I want to prove

1) If $\xi$ is an ordinal number then $\xi\cup\{\xi\}$ is an ordinal number.
2)If $A\not=\emptyset$ is a set of ordinal numbers then $\cup A=\{\zeta:\zeta\in\xi\text{ for some } \xi\in A\}$ is an ordinal number.

I need some help with the third dot in both of them. Thanks in advance!
EDIT:
1) is solved. Any ideas for 2) ?
 A: For 2) I'll make a try. Please let me know if it is correct.
Let $\emptyset\not= B\subset \cup A$. Let $\zeta \in B$. If $\zeta\cap B=\emptyset\checkmark$. If $\zeta\cap B\not=\emptyset$ then $\zeta\cap B\subset \zeta$ which is an ordinal so $\exists \eta\in \zeta\cap B: \eta\cap(\zeta\cap B)=\emptyset$. But $\eta\in\zeta$ so $\eta\cap B=\emptyset$ and $\eta\in B$.
A: 1) I'm assuming the first two points have already been proved.
Assume $\emptyset\ne A\subset\xi\cup\{\xi\}$. Now we can distinguish two cases:


*

*Case 1: $\xi\notin A$. Then clearly $A\subset\xi$ and therefore the claim is true by assumption that $\xi$ is an ordinal.

*Case 2: $\xi\in A$. Then $A\setminus\{\xi\}\subset\xi$, and thus, since $\xi$ is an ordinal, there exists a set $\zeta\in(A\setminus\{\xi\})\subset A$ with $\zeta\cap(A\setminus\{\xi\})=\emptyset$. This means that either $\zeta\cap A=\emptyset$ or $\zeta\cap A=\{\xi\}$.
But if $\zeta\cap A=\{\xi\}$, we obviously have $\xi\in\zeta$. And since $\zeta\in\xi$ and $\xi$ is an ordinal, we have $\zeta\subset\xi$ and thus $\xi\in\xi$, thus $\{\xi\}\subset\xi$. But since by assumtion $\xi$ is an ordinal, this means there is an element of $\{\xi\}$ that is disjoint to $\xi$. But the only element of $\{\xi\}$ is $\xi$. Now if $\xi\ne\emptyset$, we have a contradiction, and thus only the case $\zeta\cap A=\emptyset$ remains.
Thus so far we have that whenever $\xi$ is not empty, $\xi\cup\{\xi\}$ indeed fulfils the third condition. Remains to consider the case $\xi=\emptyset$. Fortunately that is easy to check: $\emptyset\cup\{\emptyset\}=\{\emptyset\}$, of which the only non-empty subset is $\{\emptyset\}$, which in turn has as its only element $\emptyset$. And clearly, $\emptyset\cap\{\emptyset\}=\emptyset$. $\square$
Note: I feel there should be an easier way even without invoking the axiom of foundation (which makes the whole third bullet point pretty trivial).
