Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A well known theorem by Scott says:

If $\kappa$ is a measurable cardinal and $\mu$ a normal measure on it and $\mu (\lbrace\lambda\in\kappa~|~2^{\lambda}=\lambda^{+}\rbrace)=1$ then $2^{\kappa}=\kappa^{+}$.

So:

If $\text{GCH}$ is false in a measurable cardinal then there is a smaller cardinal which $\text{GCH}$ is false in it too.

Equivalently:

If $\text{GCH}$ is false then its first failure point is not a measurable cardinal.

The role of existence of a normal measure on the cardinal $\kappa$ seems very essential in the proof of Scott's theorem. So perhaps one can prove the following statement:

For any large cardinal type $\text{A}$ smaller than measurables, it is consistent with $\text{ZFC}$ that $\text{GCH}$ be false and its first failure point be a large cardinal of type $\text{A}$.

Question is:

Question: Is the above statement true? For what type of large cardinals less than measurables is there a known result like above statement?

share|improve this question

2 Answers 2

up vote 9 down vote accepted

The statement is not true.

Theorem. If $\kappa$ is strongly unfoldable and the GCH holds below $\kappa$, then it holds at $\kappa$ also.

This is just because the strongly unfoldable cardinals are $\Sigma_2$-reflecting, and the result also holds for any $\Sigma_2$-reflecting cardinal. The strongly unfoldable cardinals are less than measurable in consistency strength, being equiconsistent with an unfoldable cardinal, and in particular, they are consistent with $V=L$.

Another small instance of the GCH overspill phenomenon, by essentially the same argument, occurs when $\kappa$ is $\Sigma_2$-extendible, which means that $V_\kappa\prec_{\Sigma_2}V_\theta$ for some ordinal $\theta$. If the GCH holds up to $\kappa$, then it will also hold up to $\theta$, and so at $\kappa$ itself. Meanwhile, these $\Sigma_2$-extendible cardinals have no strength at all---they provably exist in ZFC, unless you insist that they are also inaccessible, in which case the strength is strictly weaker than the existence of a Mahlo cardinal.

Meanwhile, your statement is indeed true for many of the large cardinals below measurable. Many of the large cardinal properties below measurable have to do with the existence of size $\kappa$ objects, such as embeddings between $\kappa$-models or homogeneous sets for colorings of size $\kappa$. For any such kind of large cardinal property, if $\kappa$ can be made to have this property indestrucitibly after $\text{Add}(\kappa,1)$, then we can make the GCH fail at $\kappa$ simply by forcing with $\text{Add}(\kappa,\kappa^{++})$. The large cardinal property will be preserved since all the size $\kappa$ objects are added by a size $\kappa$-subforcing, which amounts to $\text{Add}(\kappa,1)$. This is how one can show that it is consistent that the GCH fails first at $\kappa$, when $\kappa$ is weakly compact, indescribable, unfoldable, Ramsey, strongly Ramsey and so on. One should simply perform the preparatory forcing to make $\kappa$ indestructible by $\text{Add}(\kappa,1)$, and then appeal to the argument I just gave.

Update. I should not have included the indescribable cardinals on the list, because every $\Pi^2_1$-indescribable cardinal is $(\kappa+1)$-strongly unfoldable and so the failure of the GCH at $\kappa$ will reflect down.

share|improve this answer
    
Dear Prof. Hamkins, your answer is really interesting. Are Ramsey cardinals $\Sigma_2$ reflecting, and if not, then can they be the first cardinal violating the GCH –  Mohammad Golshani Nov 2 '13 at 12:54
    
Interesting and a bit surprising! Thanks Joel. –  Saint Georg Nov 2 '13 at 12:58
1  
@MohammadGolshani, no, Ramsey cardinals are not necessarily $\Sigma_2$-reflecting, and one can make the GCH fail first at them. This is easier to see with the strongly Ramsey cardinals, which can easily be made indestructible by $\text{Add}(\kappa,1)$. –  Joel David Hamkins Nov 2 '13 at 16:12
    
That result about strongly Ramsey cardinals (as well as the definition of these cardinals) is due to Victoria Gitman. –  Joel David Hamkins Nov 2 '13 at 17:56
    
Actually one can prove that $\Sigma_2$-extendible cardinals exist just using the reflection theorem. There is cub-many such cardinals, provided they're not inaccessible (just as you've said). –  Carlo Von Schnitzel Nov 2 '13 at 18:51

As far as I know the best results in this direction are due to Levinski (see his paper "Filters and large cardinals").

The following results are taken from the above paper. First a few definitions:

A premeasure on $\kappa$ is a normal filter $F$ on $\kappa$ such that $P(\kappa)/I$ is $\kappa^+-$distributive, $I$ the dual of $F$. $\kappa$ is premeasurable if it carries a premeasure.

A quasi measure on $\kappa$ is a normal filter $F$ on $\kappa$ such that $P(\kappa)/I$ ha a $\leq \kappa-$closed dense subset. $\kappa$ is quasi measurable if it carries a quasi measure.

Theorem A. Assume GCH+ there exists a measurable cardinal $\kappa$ and let $\lambda \geq \kappa^{++}.$ then there is a generic extension in which $\kappa$ is the first cardinal violating GCH, $2^\kappa \geq \lambda$ and $\kappa$ is premeasurable.

Theorem B. Assume there exists a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}.$ then there is a generic extension in which $\kappa$ is the first cardinal violating GCH, $2^\kappa = \kappa^{++}$ and $\kappa$ is quasi measurable.

share|improve this answer
    
Mohammad, thanks for your useful answer. –  Saint Georg Nov 2 '13 at 12:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.