# Do we know the consistency strength of the Singular Cardinal Hypothesis failing on an uncountable cofinality?

Suppose that $$\kappa$$ is a strong limit cardinal. The singular cardinal hypothesis states $$2^\kappa=\kappa^+$$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent with a measurable cardinal $$\kappa$$ satisfying $$o(\kappa)=\kappa^{++}$$.

But this failure is at $$\aleph_\omega$$. Suppose we wanted more.

Suppose that we wanted the failure to happen on a couple isolated points. Well, it's not hard to redo the standard constructions and get just that. But what happens when we have limit points?

Even more, by Silver's theorem if SCH fails at $$\kappa>\operatorname{cf}(\kappa)>\omega$$, then there is a stationary subset of $$\kappa$$ where SCH failed.

What would be the consistency strength when $$\kappa$$ is a singular limit of singular cardinals, and SCH fails cofinally below $$\kappa$$? What if we require $$\kappa$$ to be of uncountable cofinality?

As a side question, what if $$\kappa$$, with uncountable cofinality, does satisfy SCH, but an unbounded subset (which has to be non-stationary, of course) of it does not?

• Could you remind the set-theorist dilettantes what $o(\kappa)$ stands for? Apr 15, 2019 at 10:27
• @Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(\kappa)=1$ means just being a measurable, but $o(\kappa)=2$ means that there is a normal measure which concentrates on those $o(\lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $\kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank Apr 15, 2019 at 11:11

Suppose $$\kappa$$ is a singular cardinal and there are $$cf(\kappa)$$-many measurable cardinals $$\lambda < \kappa$$ with $$o(\lambda)=\lambda^{++}$$ cofinal in $$\kappa.$$ Then you can perform a Prikry type iteration and get the failure of $$SCH$$ at cofinally many singular cardinals below $$\kappa.$$

Now suppose we also want for $$SCH$$ to fail at $$\kappa$$ itself. First let us consider the countable cofinality.

Assume $$\kappa$$ is a measurable cardinal with $$o(\kappa)=\kappa^{++}+1.$$ Then we can get an extension in which $$cf(\kappa)=\omega, 2^\kappa=\kappa^{++}$$ and for cofinally many singular cardinals $$\lambda$$ below $$\kappa$$, we have $$2^{\lambda}=\lambda^{++}$$. I don't know if this assumption is really needed or if it can be reduced.

For uncountable cofinality, say $$\theta$$, a measurable cardinal $$\kappa$$ with $$o(\kappa)=\kappa^{++}+\theta$$ is sufficient. As then you can first find an extension in which $$2^\kappa=\kappa^{++}$$ and such that in the extension, $$o(\kappa)=\theta$$. Then if you perform Magidor forcing for changing cofinality of $$\kappa$$ to $$\theta,$$ you can get a club $$C$$ of singular cardinals below $$\kappa$$ such that for all $$\lambda \in C, 2^\lambda=\lambda^{++}$$.

As far as I know, if we require $$\theta=cf(\kappa)> \omega_1$$, the large cardinal assumption is optimal, but for $$\theta=\omega_1,$$ I think it is open if this assumption is optimal.

• Note that you may assume that the sequence does not contain its limit points. Apr 15, 2019 at 10:16
• Okay. I sort of expected that to happen. What about the actual question? Apr 15, 2019 at 10:16
• What is the question? Apr 15, 2019 at 10:17
• I assume you say something like this: If for example $o(\kappa)=\kappa^+++\omega_3+3$, then first we can find a model with $2^\kappa=\kappa^{++}+o(\kappa)=\omega_3+3$. Then we can make $cf(\kappa)=\omega$ by a cofinal sequence $(\kappa_n: n<\omega)$ of point with $o(\kappa_n)=\omega_3+2.$ then we can add an $\omega$-sequence cofinal in each of these $\kappa_n$'s, say $(\kappa_{n, m}: m<\omega)$ consisting of points of $o(\kappa_{n, m})=\omega_3+1.$ Apr 16, 2019 at 4:18
• Repeat once more, and find the sequence $(\kappa_{n,m,l}: l<\omega)$, cofinal in $\kappa_{n, m}$ with $o(\kappa_{n, m, l})=\omega_3.$ Now apply Magidor's forcing to change the cofinality of each $\kappa_{n, m, l}$ to $\omega_3$. Apr 16, 2019 at 4:20