# 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?

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How do you define $<$ for general sets? –  Asaf Karagila Jan 18 '12 at 18:09
And is one of those supposed to be $\bigcup x\subseteq x$? –  Brian M. Scott Jan 18 '12 at 18:13
$\emptyset$ is an ordinal. –  Michael Greinecker Jan 18 '12 at 18:18
Sorry I made a typo. I just edited it. –  passenger Jan 18 '12 at 18:18
I have one more question.If $x$ is an arbitrary set can hold $x \subset \bigcup x$ ? –  passenger Jan 18 '12 at 19:41
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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, $$\bigcup\alpha=\sup\alpha=\sup\{\beta:\beta<\alpha\}=\begin{cases}\alpha,&\text{if }\alpha\text{ is a limit ordinal or }0\\ \beta,&\text{if }\alpha=\beta+1\;. \end{cases}$$

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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 $\bigcup x = \varnothing\cup\{\varnothing\} = \{\varnothing\}\in x$.

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$\bigcup\alpha=\alpha$ for limit ordinals $\alpha$. $\bigcup(\alpha+1)=\alpha$. –  Brian M. Scott Jan 18 '12 at 18:31
Can you expain a little more the case of equality when x is an ordinal? –  passenger Jan 18 '12 at 18:49
@passenger: It’s not true in general. Indeed, Arturo’s final example shows that $\bigcup 2=1$, and Asaf pointed out that $\bigcup 1=0$. –  Brian M. Scott Jan 18 '12 at 18:52
If i understand correct Arturo said that equality never holds if x is an ordinal. How can I prove this? –  passenger Jan 18 '12 at 19:01
@passenger: $\bigcup\omega=\omega$. –  Asaf Karagila Jan 18 '12 at 19:04
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.