Consistency of $\operatorname{cf}(\omega_1)=\operatorname{cf}(\omega_2)=\omega$

Question

It is known that if $\operatorname{cf}(\omega_1)=\operatorname{cf}(\omega_2)=\omega$ then $0^\#$ exists.

I am familiar with the Feferman-Levy model in which $\omega_1$ is singular, which has the same consistency strength as $ZF$. I have three questions in this matter:

1. Can the Feferman-Levy construction be carried out to turn $\omega_5$ to be singular, while $\omega_1,\ldots,\omega_4$ are still regular; can we force the cofinality of $\omega_5$ to be any of the cardinals smaller than itself by the same method? (Of course $5$ was completely arbitrary, and my question is to generalize this to any $n$)

2. Since having $\operatorname{cf}(\omega_1)=\operatorname{cf}(\omega_2)=\omega$ implies $0^\#$ the consistency strength needs to be relatively strong (i.e. more than weakly-compact, or rather at least $0^\#$...), but how strong is it?

3. I know that it is possible to have all uncountable ordinals singular, and that the proof assumes quite the strong hypothesis (a class of supercompacts strongly compact cardinals, if I recall correct), how much can we deduce from $\operatorname{cf}(\omega_k)=\omega$ for all $k<n$, for some $n$?

(The third question might be less suited to this website, but it fit in here anyway)

Answer 1

The person you want to ask about consistency strength is Ralf Schindler. Years ago ("Successive weakly compact or singular cardinals", Symbolic Logic 64 (1999), no. 1, 139–146) Ralf showed that the existence of two consecutive singular cardinals implies the existence of an inner model with a Woodin cardinal. The paper assumes an additional background assumption that nowadays we now how to remove using recent work of Jensen and Steel, giving the result I have stated.

I expect more can be obtained using the technique known as the "core model induction", but the problem becomes much harder trying to go past one Woodin cardinal (in particular, the core model induction requires something like a bit of choice in the background universe, and we cannot make this assumption here. There are other serious technical obstacles. They appear here and also in the result from the next paragraph).

Ralf's student Daniel Busche looked at this problem in his thesis. In their joint paper "The strength of choiceless patterns of singular and weakly compact cardinals", Ann. Pure Appl. Logic 159 (2009), no. 1-2, 198–248, they show that the assumption that all uncountable cardinals are singular implies determinacy (i.e., $\omega$ Woodin cardinals) in consistency strength.

In fact, I expect this should imply (significantly) more, but you should keep in mind that Arthur Apter proved that from determinacy one can obtain models where all uncountable cardinals below $\Theta$ are singular ($\Theta$ is very large under the assumption of determinacy). His paper is "AD and patterns of singular cardinals below $\Theta$", J. Symbolic Logic 61 (1996), no. 1, 225–235. There he also shows that one can get a model where the only regular cardinals below $\Theta$ are $\omega$ and $\omega_1$.

Let me now make a silly remark on whether a "Feferman-Levy kind of construction" is possible. I assume this means in particular that you want the forcing to be possible over $L$, where $0^\sharp$ does not exist. Suppose, for example, that we want to have $\omega_3$ singular while $\omega_1,\omega_2$ are regular. Then there is a subset $A$ of $\omega_3$ of small order type cofinal in $\omega_3$. Assuming $0^\sharp$ does not exist, it follows from covering that, in $L$, there is a superset $B$ of $A$ (so $B$ can also be taken to be cofinal in $\omega_3$), with $|B|\le|A|+\aleph_1$. This means that true $\omega_3$ is a singular cardinal in $L$. This already would make the construction rather different than for the Feferman-Levy model. I'll see if I can add something of substance later.