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.