# Some relevant questions about the consistency strength of singularity of $\omega_1$ and $\omega_2$

The following question was asked years ago on MSE, but let me recap it:

Question: Is there anything currently known about the exact consistency strength of "$$\mathsf{ZF}$$ + both $$\omega_1$$ and $$\omega_2$$ are singular?" Or could we find a better bound?

It is well-known that if both $$\omega_1$$ and $$\omega_2$$ are singular, (of course, without choice) then $$0^\sharp$$ exists. It is quite famous that we can make all uncountable cardinals singular from a proper class of strongly compact cardinals. However, if we only consider $$\omega_1$$ and $$\omega_2$$ then Gitik's result is overkill: Apter proved in

Arthur W. Apter. $$\mathsf{AD}$$ and patterns of singular cardinals below $$\Theta$$. J. Symbolic Logic 61 (1996), no. 1, 225–235.

that if we start from $$L(\mathbb{R})\models\mathsf{ZF+AD+DC}$$ then we can find an extension $$N$$ of $$V$$ whose cardinals are the same with that of $$V$$, and both of $$\omega_1$$ and $$\omega_2$$ have cofinality $$\omega$$. Thus we have "$$\mathsf{ZFC}$$ + there are $$\omega$$ many Woodin cardinals" as an upper bound.

(Apter's result for $$\omega_1$$ and $$\omega_2$$ is further generalized in a paper by Apter, Jackson, and Löwe. Also, note that what Apter proved is stronger than I stated: he actually proved that if we work over $$V=L(\mathbb{R})\models\mathsf{ZF+AD+DC}$$ and if $$A$$, $$B$$ are sets partitioning the set of regular cardinals below $$\Theta^{L(\mathbb{R})}$$ then there is an extension $$N$$ of $$V$$ such that cardinals in $$V$$ and those in $$N$$ are the same and cardinals in $$A$$ has cofinality $$\omega$$, but cardinals in $$B$$ are still regular in $$N$$.)

On the other hand, Schindler proved in

Ralf-Dieter Schindler, Successive weakly compact or singular cardinals. J. Symbolic Logic 64 (1999), no. 1, 139–146.

that if $$\kappa$$ is measurable in $$\mathsf{HOD}$$ and inaccessible in $$V\models \mathsf{ZF}$$ and there is $$\delta<\kappa$$ such that $$\delta^+<\kappa$$ and both of $$\delta$$ and $$\delta^+$$ are singular, then there is an inner model of a Woodin cardinal.

It brings possibly related questions that are also curious for me: First, it is easy to see that if we start from $$\mathsf{ZF}$$ with both $$\omega_2$$ and $$\omega_3$$ are singular, then there is a generic extension of $$V$$ that thinks $$\omega_1$$ and $$\omega_2$$ are singular.

Question. How about the opposite direction? So, can we derive the consistency of "$$\mathsf{ZF}$$ + both $$\omega_2$$ and $$\omega_3$$ are singular" from that of "$$\mathsf{ZF}$$ + both $$\omega_1$$ and $$\omega_2$$ are singular"?

Regarding the upper bound, the role of $$\mathsf{AD+DC}$$ in Apter's proof is a normal measure over $$\omega_1$$ and $$\omega_2$$. He then uses a product of Prikry forcings and constructs an inner model of a generic extension. It seems like providing normal measures is the only role of $$\mathsf{AD}$$.

Question. Can we find a model of $$\mathsf{ZF}$$ + "both of $$\omega_1$$ and $$\omega_2$$ have a normal measure" from large cardinal hypotheses weaker than $$\omega$$ many Woodin cardinals?

(The paper by Apter, Jackson, and Löwe pointed out in Theorem 32 that if we start from $$\mathsf{ZFC}$$ two measurable cardinals then we have an extension $$N\models\mathsf{ZF}$$ of $$V$$ that thinks $$\omega_1$$ and $$\omega_3$$ are measurable and $$\omega_2$$ is regular. I have no idea whether it is related to an answer to the above question, and probably not.)

We know little to nothing about the exact strength. Here are the facts, as you already know them.

1. To get $$\omega_1$$ and $$\omega_2$$ both singular we need to have at least a Woodin cardinal.

2. If there are infinitely many Woodin cardinals, then it is consistent that $$\omega_1$$ and $$\omega_2$$ are singular (go to a model of $$\sf AD$$ and collapse $$\omega_1$$ and $$\omega_2$$ or shoot Prikry sequences through their measures to singularise them). Indeed, in this case we have infinitely many consecutive singular cardinals.

The prevailing conjecture, if so, is that successive singular cardinals should have the consistency strength of "roughly a Woodin".1

We also similar things about consecutive measurable cardinals. Essentially it boils down to failure of suitable covering theorems of core models (i.e., the successor of weakly compact cardinals is computed incorrectly, and therefore a Woodin cardinal must hide somewhere in an inner model).

The problem here, which I will use to advertise some of my thoughts on these matters, is that there are two levels here.

1. Consistency strength, which is what we often measure. This requires some sort of canonical inner models, or "slightly less canonical inner models" (à la $$L(\Bbb R)$$ or so) for us to establish these results.

2. Forcing strength, which is what we often can do "by hand". In other words, we want to construct a generic (or symmetric in the choiceless cases) extension in which the statement holds.

These can be very different. For example, "A singular $$\kappa$$ is regular in an inner model", and $$\kappa$$ can always be taken to be $$\aleph_\omega$$ too and the inner model can be assumed to be $$L$$, has the consistency strength of exactly "$$0^\#$$ exists", but if you wanted to do it by forcing the consistency strength jumps to a measurable cardinal.

In order to create a symmetric extension with successive measurable cardinals or successive singular cardinals we need to venture into the zone of "higher compactness". Normally a supercompact will be used, but the proofs can be analysed to some bounded level of sufficient compactness if one insists on doing that. These are of course, only the known bounds, rather than the exact strength of these statement.

It seems to me that we have a phenomenon here of the same flavour as in the case of a singular cardinal that is regular in $$L$$. It seems to me that in order to construct these things "by hand", i.e. to force them, we probably need to rely on much stronger hypotheses.

Now we run into a problem. Since the consistency strength is often analysed by inner models + forcing combined, if you wish to "start from optimal hypothesis and produce a model", this means that the model produced is not produced with forcing, but we hardly have any other method at hand right now for these results.

So, finding the optimal results is hard and remains a mystery. Finding the "forcing strength" is also a mystery and will probably remain open a longer time. One can conjecture that once suitable inner model theory is developed for compact cardinals, we can try the following method "$$V$$ is a symmetric extension in which something happens, force choice in a 'minimally destructive way' and show that some cardinals are generically such and such". However, there is much more work before these are even close to being feasible and working results.

Now. You may argue, and you'd have a significant point there as well, that since $$L(\Bbb R)$$ is a symmetric extension of a model of $$\sf ZFC$$, and perhaps even of $${\rm HOD}^{L(\Bbb R)}$$, it is in fact absolutely the case that consequences of $$\sf AD$$ should be forceable with "roughly the same consistency strength".

Alas, when we look at the results about $${\rm HOD}^{L(\Bbb R)}$$, we see that $$\omega_1^V=\omega_1^{L(\Bbb R)}$$ is in fact the least measurable there, and that the Woodin cardinals actually appear relatively "high". This, in comparison to the forcing based construction where we collapse infinitely many Woodin cardinals to be countable and consider $$L(\Bbb R)$$ of the model containing only bounded reals (i.e. reals added by collapsing of a bounded part below the limit of the Woodins), in which case there are Woodin cardinals hiding well below $$\omega_1$$.

This means that any construction attempting "optimal assumptions", or even "a symmetric extension from reasonably optimal assumptions" is bound to end up as being incredibly complicated. Which, again, puts us fairly far away from being able to understand these optimal results. If you even want to under $$L(\Bbb R)$$ as such a symmetric extension, matters become much worse, since we need to understand the collapsing of intervals of cardinals or singularising and collapsing simultaneously.

And since our main tool for symmetric extensions is homogeneity arguments, it seems that the forcing involved here may or may not have these homogeneity properties. Indeed, the obvious culprit would be some sort of a stationary tower forcing which is notoriously non-homogeneous.

So perhaps the question about "forcing strength" should be about forcing with reasonably behaved partial orders (e.g. sufficiently homogeneous ones). But much, much more needs to be understood before we can even address these questions of optimal strengths.

Footnotes.

1. When I was a student I thought that I saw a paper by a prominent set theorist that proved that, having spent years telling people about that, Yair Hayut eventually suggested we try to read it, having taken some basic inner model theory and stationary tower forcing courses with Menachem (Magidor). It turned out that this was a dream that I had. The result, of course, was that a sharp for a single Woodin cardinal is enough. I asked the alleged author, and they said it would have been a great result, and if I know how to prove it they'd be happy to read or write that paper. Alas, I have no idea how to prove something like that.