Why is every limit cardinal a supremum of regular succesor cardinals?
Of course, working within ZFC, why does this hold? I stumbled upon this statement in Kunen and I don't see exactly why it holds, especially the regular part of the claim.
Thank you very much in advance for any hints/ideas!
All infinite successor cardinals are regular.
To see that $\delta^+$ is regular, observe first that the union of $\delta$ many sets of size $\delta$ has size $\delta$. This is a generalization of the familiar fact that the countable union of countable sets is countable. One can prove this with the Cantor pairing function, which provides a well-ordering of $\kappa\times\kappa$ in order type $\kappa$ for any infinite cardinal $\kappa$. We just place pairs $(\alpha,\beta)$ in order first by their supremum and then by the lexical order.
Given this claim, it now follows that successor cardinals $\delta^+$ are regular, because the supremum of $\delta$ many ordinals below $\delta^+$ will still have size $\delta$, and hence be bounded below $\delta^+$.
Every limit cardinal is of course the supremum of the successor cardinals below it, since if $\kappa$ is a limit cardinal and $\delta<\kappa$, then also $\delta^+<\kappa$, since otherwise $\kappa$ wouldn't be a limit cardinal.
It follows that every uncountable limit cardinal is the limit of regular successor cardinals.
(And as User14111 notes in the comments, one must indeed say "uncountable limit cardinal", since $\aleph_0$ is a limit cardinal that is not the limit of regular successor cardinals.)
Note. The argument takes place in ZFC, and it uses the axiom of choice, because it is known to be consistent with ZF that every uncountable cardinal is singular, including every infinite successor cardinal.