Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
I have one more question.If $x$ is an arbitrary set can hold $ x \subset \bigcup x$ ? – passenger Jan 18 '12 at 19:41

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, $$\bigcup\alpha=\sup\alpha=\sup\{\beta:\beta<\alpha\}=\begin{cases}\alpha,&\text{if }\alpha\text{ is a limit ordinal or }0\\ \beta,&\text{if }\alpha=\beta+1\;. \end{cases}$$

share|cite|improve this answer

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

share|cite|improve this answer
$\bigcup\alpha=\alpha$ 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.