Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

From Jech (pg. 20): If $X$ is a nonempty set of Ordinals, than $\bigcup X$ is an ordinal.

This should be easy enough to prove but I don't see how; also, I guess this is not the case if $X$ is a proper class, so the difference must show in the proof and I don't see where it comes in.

share|cite|improve this question
What is Jech's definition of "ordinal"? – Chris Eagle Apr 22 '12 at 14:46
Transitive set that is strictly well-ordered by inclusion. – Gadi A Apr 22 '12 at 14:48
This depends on whose definition of ordinal and proper class you take. Under Monk's definition in 'Introduction to Set Theory', the collection of all ordinals IS an ordinal and a proper class. – James Auld Apr 22 '12 at 14:50
Actually, Jech simply says "a set is an ordinal if...", so he doesn't say that the class of all ordinals is not an ordinal. still, my basic question is why the union of a set of ordinals is an ordinal. – Gadi A Apr 22 '12 at 14:54
Gadi, "inclusion" is ambiguous in Hebrew but in English it almost always mean $\subseteq$. Jech defines an ordinal to be a transitive set which is well-ordered by $\in$, which is membership. – Asaf Karagila Apr 22 '12 at 15:20
up vote 5 down vote accepted

Suppose that $X$ is a set of ordinals. First note that by the axiom of union we have that $\bigcup X$ is a set as well.

First we want to show that $\bigcup X$ is transitive, that is $x\in\bigcup X$ then $x\subseteq X$. Suppose that $x\in\bigcup X$, then there is some $\alpha\in X$ such that $x\in\alpha$. Since $\alpha$ is an ordinal it is a transitive set and therefore $x\subseteq\alpha$, and since $\alpha\subseteq\bigcup X$ we have $x\subseteq\bigcup X$ as wanted.

Now we want to show that it is linearly ordered by $\in$, so take $x,y\in\bigcup X$. Then there are $\alpha,\beta\in X$ such that $x\in\alpha$ and $y\in\beta$. By Lemma 2.11 (p. 19) we know that $\alpha\in\beta$ or $\beta\in\alpha$ or $\alpha=\beta$. Either way we may assume without loss of generality that $x,y\in\alpha$. Since $\alpha$ is linearly ordered by $\in$ we have that $x\in y$ or $y\in x$ or $x=y$ as wanted.

Lastly, to show that every non-empty set has a least element we can resort to the axiom of regularity which says that $\in$ is well-founded. To see that directly we can also argue as following:

Let $A\subseteq\bigcup X$ be non-empty. There is some $\alpha\in X$ such that $A\cap\alpha\neq\varnothing$. Since $\alpha$ is well-ordered we have that $\alpha\cap A$ has a minimal element, call it $x$. Suppose that $x$ was not minimal in $A$, then there is some $y\in A$ such that $y\in x$, however $x\in\alpha$ therefore $x\subseteq\alpha$, so $y\in\alpha$ and therefore $y\in A\cap\alpha$ which contradicts the fact that $x$ was the minimal element of $A\cap\alpha$.

share|cite|improve this answer
The axiom of regularity is the main thing I was missing. Thanks! – Gadi A Apr 22 '12 at 15:34
(Is that axiom an absolute must here?) – Gadi A Apr 22 '12 at 15:34
@Gadi: No, not needed. I have added the needed argument. – Asaf Karagila Apr 22 '12 at 15:44

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.