# About a proof of “$\bigcup A$ is a limit cardinal”

Assume that if $A$ is a set of cardinals such that $A$ contains no largest element and assume that we have shown that $\bigcup A$ is a cardinal. Now we want to show that $\bigcup A$ is a limit cardinal. By contradiction, we assume that it is a successor cardinal $\kappa^+$ for some cardinal $\kappa$.

The proof in Just/Weese proceeds "Then $A$ must contain an element $\lambda$ such that $\kappa < \lambda$."

But how do we get there?

Question 1: We don't know whether $\kappa \in A$ or not, right?

Question 2: If $\kappa \in A$ and $\bigcup A = \kappa^+$, then how can there be any cardinals between $\kappa$ and $\kappa^+$? (I think there cannot.)

Question 3: Perhaps the reasoning is this? If $A$ does not contain a largest element then for every cardinal $\kappa$ in $\mathbf{Card}$, there is $\lambda \in A$ such that $\kappa < \lambda$?

Thank you for your help.

-

Question 1: No, we do not know if $\kappa \in A$. But since $\kappa < \bigcup A$, this means that $\kappa \in \bigcup A$, and so there must be some $\lambda \in A$ with $\kappa \in \lambda$, or, $\kappa < \lambda$.

Question 2: That is the source of the contradiction!

Question 3: Pretty much, but only restricted to cardinals $\kappa$ less than $\bigcup A$. One can easily produce sets of cardinals which do not have a maximum element, but are yet bounded. The most basic example would be $\omega$ itself, and in general given any limit cardinal $\lambda$ the family $$A = \{ \kappa \in \lambda : \kappa \text{ is a cardinal} \}$$ would be a collection of this type.

Giving a lot of detail to the proof that $\bigcup A$ is a limit cardinal:

If $\bigcup A$ is a successor cardinal, then $\bigcup A = \kappa^+$ for some cardinal $\kappa$. Note that $\kappa < \bigcup A$, and so $\kappa \in \bigcup A$, meaning that there is some cardinal $\lambda \in A$ such that $\kappa \in \lambda$. But as $A$ has no maximal element, there is another cardinal $\mu \in A$ such that $\lambda < \mu$. Since $\mu \in A$ then $\mu \subseteq \bigcup A$, meaning that $\mu \leq \bigcup A$. But look at $$\bigcup A = \kappa^+ \leq \lambda < \mu \leq \bigcup A.$$ Can you see a contradiction?

-
Thank you very much for this clear answer. You have cleared up all my confusion. –  Rudy the Reindeer Dec 10 '12 at 9:17
And yes, I do see the contradiction: $\bigcup A < \bigcup A$. –  Rudy the Reindeer Dec 10 '12 at 9:20

Recall that cardinals (in this context) are just ordinals and that $\bigcup A$ is essentially the supremum of these ordinals.

If $\bigcup A=\kappa^+$ then we are saying that $\sup A=\kappa^+$, so there is some ordinal $\lambda\in A$ such that $\kappa<\lambda$. But we also know that all the members of $A$ are cardinals so $\lambda\geq\kappa^+$. However $A$ does not contain a largest element so there is some $\mu\in A$ such that $\kappa^+\leq\lambda<\mu$, and so $\sup A>\kappa^+$ so $\bigcup A\neq\kappa^+$.

The clearest way to see this, in my opinion, is to transform this into a problem about ordinals.

Let $A'=\{\alpha\in\mathbf{ON}\mid\aleph_\alpha\in A\}$. Show that $A'$ does not have a last element and therefore $\sup A'$ is a limit ordinal, now we have that $\bigcup A$ has to be a limit cardinal.

-
I'm still not too sure. Why can we conclude that there is an ordinal $\lambda$ between $\kappa$ and $\kappa^+$ such that $\lambda \in A$ and $\kappa < \lambda$? –  Rudy the Reindeer Dec 10 '12 at 8:55
Because the supremum of $A$ is $\kappa^+$. It is the least ordinal which is greater or equal than all the members of $A$. But there are so many other ordinals between $\kappa$ and $\kappa^+$ that there has to be some $\lambda$ with this property. –  Asaf Karagila Dec 10 '12 at 8:57
But $A$ is a set of cardinals. How can one have $\lambda \in A$ for an ordinal? That $\lambda$ has to be a cardinal. Which is impossible since by the definition of successor cardinal there are no cardinals between $\kappa$ and $\kappa^+$. –  Rudy the Reindeer Dec 10 '12 at 9:02
Right. We actually must have $\lambda \ge \kappa^+$. –  Rudy the Reindeer Dec 10 '12 at 9:05
@Matt: To the first comment recall that the ordinals in $A$ are special, they are all initial ordinals. So as you said in the second comment, if such ordinal exists then it has to have the property $\lambda\geq\kappa^+$. –  Asaf Karagila Dec 10 '12 at 9:08