About a proof of "$\bigcup A$ is a limit cardinal"

Question

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.

Answer 1

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?

Answer 2

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 $\bigcup A\neq\kappa^+$.

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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.