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I'm looking for references on how to change the successor of a singular cardinal from "more or less" minimal assumptions. If possible, then without adding bounded subsets to the singular either.

In some papers (Gitik's Namba variant paper, for example) I saw footnotes or mentioning that with a Woodin cardinal you can do this using stationary tower forcing. But I couldn't find any references to a written result.

To make this more concrete, let me put the following question.

Suppose that $\aleph_\omega$ is a strong limit cardinal in $V$, and there is a Woodin cardinal. Is it possible to collapse $\aleph_{\omega+1}^V$ without adding bounded subsets to $\aleph_\omega^V$? (I don't mind some structural damage above the new $\aleph_{\omega+1}$, though.)

(I'm saying "a Woodin", but really I mean "approximately a Woodin", in the sense that there's no issue requiring some measurable, or even a second Woodin, above it.)

Any references, or even a sketch of the proof, would be great.

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  • $\begingroup$ Using stationary tower forcing, the example is in Jech page 679: you can collapse succesor of a singular and give it any cofinality you want. Consistency strength is exactly that of a Woodin cardinal $\endgroup$ Oct 12 '14 at 23:25
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Assume $GCH$ holds and there exists a proper class of completely Jonsson cardinals and let $\gamma<\lambda$ be regular cardinals. Let $a=\{ \alpha<\lambda: cf(\alpha) =\gamma\},$ and suppose that $a$ belongs to a $\mathbb{P}_{\infty}$ generic $G$ with associated $j:V\to V[G],$ where $\mathbb{P}_{\infty}$ denote the stationary tower class forcing, using arbitrary stationary sets $a$ as conditions.

Then in $V[G],$ the cofinality of $\lambda$ is changed to $\gamma.$ cardinals below $\lambda$ are preserved and if $2^\delta$ is less $\lambda,$ than then no new subsets of $\delta$ are added.

so for example we can change the cofinality of $\aleph_{\omega+1}$ to say $\aleph_{10},$ without adding bounded subsets of $\aleph_\omega.$

This is known as generalized Namba forcing. A good reference is Larson's book "The Stationary Tower: Notes on a Course by W. Hugh Woodin". Another nice reference is the notes given by Sy Friedman $\mathbb{P}_{max}$ and the stationary tower.

Note that by core model theory, such a weird eeffect cannot be achieved if $ZFC$ is preserved by adding $V$ as an additional predicate, without using more than a Woodin cardinal and probably this would need a supercompact cardinal.

Suppose that $γ < λ < κ$ are regular cardinals below a Woodin cardinal $κ$. Forcing with $\mathbb{P}_{<\kappa}$ below the condition $a= \{α < λ : cof(α) = γ\}$, we get that $j[∪b] ∈ j(b)$, i.e., that in $M, j[λ]$ is an ordinal below $j(λ)$ of cofinality $j(γ).$ This means that the critical point of $j$ is $λ$, and that $cof(λ) = γ$ in $M$.

Furthermore, if $α$ is such that $2^\alpha<\lambda$, then all subsets of $α$ in $M$ are in $V.$ Since $V_\kappa^M= V_\kappa^{V[G]}$ , these facts hold in $V[G]$ also.

See also Larson's notes Six lectures on the stationary tower .

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  • $\begingroup$ Ah, I didn't think to look for "Namba forcing" in Larson's book (because that was the first place to look for anything Stationary Tower related). Now, since my memory is not as good as it never used to be, the existence of a Woodin cardinal implies the existence of a proper class of completely Jonsson cardinals, or just the consistency thereof? $\endgroup$
    – Asaf Karagila
    Oct 12 '14 at 3:51
  • $\begingroup$ (Hm, after the cited paragraph, Larson points out that under additional assumptions on $\kappa$ we can do it without completely Jonsson cardinals, and with $\Bbb P_{<\kappa}$. I suppose that he means that this is the case when $\kappa$ is Woodin, right?) $\endgroup$
    – Asaf Karagila
    Oct 12 '14 at 3:53
  • $\begingroup$ Completely Jonsson cardinals have very low consistency strength. For example if a cardinal is measurable, then it is Completely Jonsson and there are unboundedly Completely Jonsson cardinals below it. $\endgroup$ Oct 12 '14 at 4:03
  • $\begingroup$ Wait, so are you suggesting that if $\kappa$ is measurable we can replace $\Bbb P_\infty$ by $\Bbb P_{<\kappa}$? $\endgroup$
    – Asaf Karagila
    Oct 12 '14 at 4:20
  • $\begingroup$ I think we face some problems. For example if $G$ is $\mathbb{P}_{<\kappa}-$generic, then $V_\kappa$ is not in $Ult(V, G),$ so we may face problems. Another note is that if we want well-foundedness of $Ult(V, G)$ we need more, maybe a Woodin cardinal. I will add more to my answer. $\endgroup$ Oct 12 '14 at 4:30

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