Union of ordinals

Question

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.

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

score 15 · Accepted Answer · edited Jul 25 '21 at 02:29

15

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 $x\subseteq \bigcup X$. 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 $A\cap\alpha$.

edited Jul 25 '21 at 02:29

gkb0986

131

answered Apr 22 '12 at 15:09

Asaf Karagila

393,674

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
So we need the axiom of regularity, which is why $X$ needs to be a set. But your final paragraph avoids that axiom. Which property of sets is utilized here that prevents $X$ from being a proper class? – zaq Oct 02 '23 at 10:55
@zaq Regularity is not at all needed for this. Full stop. It will simplify some of the proof, yes, but it is not necessary in any sense of the word. The only thing that prevents $X$ from being a proper class is that $\bigcup X$ is a set. – Asaf Karagila Oct 02 '23 at 12:12
@AsafKaragila Thank you. I wonder now what is the reason that we can't say "if $X$ is a class of ordinals, then $\bigcup X$ is an ordinal". Is it because $\bigcup X$ is only defined in general for classes if $X$ is finite? Because a class is actually just something described by a formula and we can only join up finitely many formulas using $\lor$. – zaq Oct 06 '23 at 20:36
1

@zaq: Because an ordinal is a set, and the union over a proper class is not a set. – Asaf Karagila Oct 06 '23 at 21:35

Union of ordinals

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