2019-04-15 08:16:54 8 Comments

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?

### Related Questions

#### Sponsored Content

#### 1 Answered Questions

### [SOLVED] What are some good lower bounds on the consistency of the failure of the PCF conjecture?

**2019-02-11 17:11:12****Asaf Karagila****307**View**13**Score**1**Answer- Tags: set-theory lo.logic large-cardinals inner-model-theory pcf-theory

#### 2 Answered Questions

### [SOLVED] The cofinality of the poset $[\kappa]^{<\kappa}$ for a singular cardinal $\kappa$

**2018-10-21 10:18:16****Taras Banakh****206**View**4**Score**2**Answer- Tags: reference-request set-theory order-theory

#### 0 Answered Questions

### Singular compactness for stationary reflection?

**2018-05-19 17:58:09****Otto****95**View**3**Score**0**Answer- Tags: set-theory lo.logic infinite-combinatorics

#### 0 Answered Questions

### Is this Variation of the Continuum Hypothesis Inconsistent with ZFC or ZF?

**2017-10-14 04:31:39****Keith Millar****606**View**18**Score**0**Answer- Tags: lo.logic set-theory continuum-hypothesis

#### 1 Answered Questions

### [SOLVED] Very weak square and good points

**2017-06-24 05:05:56****Jing Zhang****193**View**7**Score**1**Answer- Tags: lo.logic set-theory large-cardinals

#### 2 Answered Questions

### [SOLVED] Slim Kurepa tree at a singular strong limit cardinal of uncountable cofinality

**2014-08-23 16:09:59****Trevor Wilson****335**View**10**Score**2**Answer- Tags: lo.logic set-theory

#### 0 Answered Questions

### Singular Jonsson cardinals

**2015-02-08 05:39:07****Mohammad Golshani****263**View**10**Score**0**Answer- Tags: lo.logic set-theory large-cardinals

#### 2 Answered Questions

### [SOLVED] Prevalent singular cardinals hypothesis

**2013-11-26 09:34:55****Mohammad Golshani****366**View**5**Score**2**Answer- Tags: set-theory forcing cardinal-arithmetic pcf-theory

#### 3 Answered Questions

### [SOLVED] Consistency strength of the failure of Shelah's Strong Hypothesis (SSH)

**2013-11-18 00:04:45****Alberto Levi****488**View**6**Score**3**Answer- Tags: set-theory cardinal-arithmetic pcf-theory

#### 3 Answered Questions

### [SOLVED] Disjoint stationary sets that reflect

**2012-11-03 01:08:40****Ioannis Souldatos****566**View**11**Score**3**Answer- Tags: lo.logic set-theory

## 1 comments

## @Mohammad Golshani 2019-04-15 10:09:07

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.

## @Mohammad Golshani 2019-04-15 10:16:14

Note that you may assume that the sequence does not contain its limit points.

## @Asaf Karagila 2019-04-15 10:16:40

Okay. I sort of expected that to happen. What about the actual question?

## @Mohammad Golshani 2019-04-15 10:17:14

What is the question?

## @Asaf Karagila 2019-04-15 10:17:41

The part in the blockquote, rather than the "side question" below it? Specifically the part where we

dowant to assume that the sequence contain its limits. Maybe evenallof its limits.## @Mohammad Golshani 2019-04-15 10:18:17

Do you require SCH fails at $\kappa$ itself?

## @Asaf Karagila 2019-04-15 10:18:29

Yes. Of course.

## @Asaf Karagila 2019-04-15 12:46:27

Thanks! Why is there in issue with the case of $\theta=\omega_1$?

## @Asaf Karagila 2019-04-15 13:50:02

Also, let me see if I get the point. While probably a bit of an overkill, $o(\kappa)=\kappa^{++}+\theta+1$ gives us a violation at a limit of countable cofinality which is the limit of singulars of cofinality $\theta$ where SCH fails on a club. Right? So $o(\kappa)=\kappa^{++}+\omega_3+3$ would be the reasonable starting point for something of countable cofinality with a cofinal sequence of type $\omega^3$ (or $\omega^\omega$?) of points which are limits of points which are limits of points of failure of cofinality $\omega_3$ on a club?

## @Mohammad Golshani 2019-04-16 04:12:41

About the case $\theta=\omega_1,$ I don't know why, maybe someone expert in inner model theory can say something about it (but maybe it is because then we have a club of order type $\omega_1$ all of whose points have countable cofinality, and for changing cofinality to $\omega$ we really do not need so much strength).

## @Mohammad Golshani 2019-04-16 04:18:36

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 $o(\kappa_{n, m})=\omega_3+1.$

## @Mohammad Golshani 2019-04-16 04:20:06

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 $\omega_3$.

## @Asaf Karagila 2019-04-16 06:31:13

Thanks again, Mohammad!

## @Mohammad Golshani 2019-04-17 04:50:18

You are welcome Asaf. I may say that you can do the first three steps of using Prikry forcing using Radin forcing to get a club of order type $\omega^3$, all of whose points have $o(\lambda)=\omega_3$.