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Definition: Let $\kappa$ be an $\aleph$ cardinal, we say that $\langle f_\alpha\colon\alpha\to\kappa\mid\alpha<\kappa^+\rangle$ is a ladder if every $f_\alpha$ is injective.

Equivalently this is the range of a choice function from every injection of $\alpha$ into $|\alpha|$ (for $\alpha<\kappa$ we can always assume the identity is taken). Such ladder implies automatically that $\kappa^+$ is regular, since the union of $\kappa$ enumerated sets of size $\kappa$ is at most of size $\kappa$. (Recall that for well ordered sets $\kappa\times\kappa=\kappa$ even without the axiom of choice)

For example, then, if $\omega_1$ is singular then there is no such ladder of countable ordinals, since this would imply that $\aleph_1=\aleph_0^+$ is regular, therefore the existence of ladders for every successor ordinal is not provable in ZF alone.

However if we assume the existence of an inaccessible cardinal we can have the situation where $\aleph_1$ is indeed regular but there is no ladder avail. Indeed this is a necessary requirement since in such situation $\omega_1$ is inaccessible in $L$.

Both the forcing which makes $\aleph_1$ singular and the one which makes it regular without a ladder are essentially the same: collapse a limit cardinal to $\aleph_1$ and take symmetry model which ensures that no ladder exists, while setting $\aleph_1$ to have the cofinality of collapsed cardinal, i.e. singular or regular (if it was inaccessible).

Question I: Can we do this trick by replacing $\aleph_1$ by $\aleph_\alpha$ for any non-limit ordinal? So for example, $\aleph_5$ would be singular or regular without a ladder.

Question II: We do not need an inaccessible cardinal to have $\omega_1$ singular, nor $\omega_2$ singular. However if we want them both to be singular it already implies $0^\#$ exists, and requires Woodin cardinals.

Suppose $\aleph_1$ and $\aleph_2$ are both regular, and neither has a ladder. Can we do that "just" from the existence of two inaccessible cardinals, or would such phenomenon imply that some very large cardinals are playing in the background?

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  • $\begingroup$ I should probably remark that discussing it with several people in one set theory seminar we concluded that finitely many inaccessibles would be enough for finitely many regular successors without ladders; but it was well-pointed that if we want to get it done for all successors, Woodin cardinals join the game to allow changing the successor of singular cardinals. I don't remember whether or not we found a solution for just "infinitely many successors don't have ladders", but I believe it would be relatively simple. $\endgroup$ – Asaf Karagila Jul 25 '12 at 22:13
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Both questions have a position answer. Which, in some sense, indicate that the non-existence of club sequences is somehow a weak property, relatively speaking.

For the first question this is quite trivial and really just requires a straightforward checking of the definitions. The point here being that if we start with a model of $\sf ZFC$ and symmetrically collapse a limit cardinal $\kappa$ to be $\aleph_{\alpha+1}$, then every set of ordinals is given by a bounded collapse, and in particular no ladder system can exist.


As for the second question, If we have two inaccessible cardinals, $\kappa<\lambda$ and we force with the product of the obvious symmetric extensions, $\operatorname{Col}(\omega,<\kappa)\times\operatorname{Col}(\kappa,<\lambda)$. Because $\kappa<\lambda$, this can be seen also as a two-step iteration first collapsing $\lambda$ and then $\kappa$, and we can show that every subset of $\kappa$ was added by the collapse of $\kappa$, and every unbounded subset of $\lambda$ was added by the collapse of $\lambda$, so in either case, after the symmetries worked their magic, the above argument shows again that no ladders exist on $\aleph_1$ or $\aleph_2$.

(Note that if $\kappa$ is a singular cardinal the situation is different, because there is no reasonable way to change the successor of a singular cardinal without using large cardinals. So we may have $\lambda$ singular, but at least $\kappa$ needs to be regular.)

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  • $\begingroup$ I could probably have answered this a couple of years ago... I hope this answer helps others! $\endgroup$ – Asaf Karagila Feb 21 '20 at 20:27

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