union of ordinal number Which of the following holds: $ \bigcup x \in x ,\quad \bigcup x = x ,\quad  x\in \bigcup x $ ,when:
(a) $x$ is a set, (b) $x$ is an ordinal.
Can someone help me?
Thank you in advance!
 A: HINT for $\bigcup x\in x$ and $x\in\bigcup x$: Is $\bigcup\varnothing\in\varnothing$? Is $\varnothing\in\bigcup\varnothing$?
$x\in\bigcup\varnothing$ iff there is a $y\in\varnothing$ such that $x\in y$, so ... ?
Remember, $\varnothing$ is also the ordinal $0$, so your answer to this question applies to both parts of your problem.
Added: Not much point in a hint now, so I’ll just point out that if $\alpha$ is an ordinal, $$\bigcup\alpha=\sup\alpha=\sup\{\beta:\beta<\alpha\}=\begin{cases}\alpha,&\text{if }\alpha\text{ is a limit ordinal or }0\\
\beta,&\text{if }\alpha=\beta+1\;.
\end{cases}$$
A: Remember that
$$\bigcup x = \{z\mid \exists y(z\in y\land y\in x)\}.$$
Assuming the Axiom of Foundation (Regularity), you can never have $x\in\bigcup x$:
this would require the existence of a $y$ such that $x\in y$ and $y\in x$, which would in turn give you an infinite descending chain of sets ordered by $\in$; Regularity prohibits this. But in the absence of Foundation, that could occur; assuming Aczel's Anti-Foundation Axiom, there is a set $A$ such that $A=\{A\}$, and it is easy to verify that $A\in A=\bigcup A$.
For arbitrary sets, you may or may not have $\bigcup x = x$; specifically, in order for $\bigcup x\subset x$ to hold, $x$ must be transitive and conversely; a set $A$ is transitive if and only if $y\in A$ implies $y\subseteq A$.
Indeed, assume $x$ is transitive: if $z\in \bigcup x$, then there exists $y\in x$ such that $z\in y$; hence, $z\in y\subseteq x$, so $\bigcup x\subseteq x$. Conversely, assume that $\bigcup x \subseteq x$, and let $y\in x$. If $z\in y$, then $z\in \bigcup x\subseteq x$, hence $z\in x$. Thus, $y\subseteq x$. Therefore, $\bigcup x\subseteq x$ if and only if $x$ is transitive. 
Equality is harder, but you should verify that it does in fact hold for ordinals.  As noted, equality need not hold for arbitrary ordinals; in fact, equality holds for an ordinal $\alpha$ if and only if $\alpha$ is a limit ordinal: that is, if for every ordinal $\beta$, $\beta\in\alpha$ implies $\beta\cup\{\beta\}\in\alpha$. 
$\bigcup x\in x$ may or may not hold for arbitrary sets; and never holds for ordinals: assuming you've proven that if $x$ is an ordinal then $\bigcup x = x$, you cannot also have $x=\bigcup x\in x$, since ordinals are, by definition, well-ordered with respect to $\epsilon$. For an example in which $\bigcup x \in x$, take $x = \{\{\varnothing\}, \varnothing\}$. Then $\bigcup x = \varnothing\cup\{\varnothing\} = \{\varnothing\}\in x$. 
A: In general $\bigcup x\notin x$, e.g. $\bigcup\varnothing=\varnothing\notin\varnothing$. This example also shows that often $x\notin\bigcup x$.
Consider also $\bigcup 1=\bigcup\{\varnothing\}=\varnothing$ and so inequality is not necessary either (since ordinals are of course sets, then this is an example for both cases).

The union of an ordinal is an ordinal, therefore $\bigcup x$ is comparable with $x$ so one of the three cases must always hold.
