3
$\begingroup$

The theorem 1.1 of the following paper:

Mohammad Golshani, V, HOD, and the GCH, Journal of Symbolic Logic.

states that:

Theorem: Assume $V\models ZFC+GCH+~\text{There exists a}~(\kappa+4)-\text{strong cardinal}~\kappa$ then there is a generic extension $W$ of $V$ such that:

(1) $\kappa$ remains inaccessible in $W$.

(2) $V_{\kappa}^W=W_\kappa\models ZFC+\forall \lambda, ~2^{\lambda}=\lambda^{+3}$.

(3) $HOD^{W_{\kappa}}\models GCH$.

The theorem shows the consistency of having a model of $ZFC$ (like $W_{\kappa}$) in which $GCH$ fails everywhere (in a finite gap form) but in its $HOD$, $GCH$ holds.

In a remark after the theorem the author says:

Remark: In fact it suffices to have a Mitchell increasing $\kappa^+$-sequence of extenders, each of which is $(\kappa + 3)$-strong. Thus for the conclusion of Theorem 1.1 it suffices to have a cardinal $κ$ such that $o(κ) = \kappa^{+3}+\kappa^+$.

My question is:

Question: Is $o(κ) = \kappa^{+3}+\kappa^+$ necessary for the proof of theorem 1.1. or one could hopefully reduce it to something weaker?

I guess here some sort of core model argument is needed to prove such a result but maybe there are other methods to ensure that the assumption could not be weaker.

$\endgroup$

1 Answer 1

4
$\begingroup$

Thanks for referring to my paper. Here are some notes:

1) In the above paper, we have a fixed gap 3 everywhere, while if someone wants a model in which $GCH$ fails everywhere but its $HOD$ satisfies the $GCH$, then we can have also gap 2. For this we need a measurable with $o(\kappa)=\kappa^{++}+\kappa^+,$ which is weaker than the assumption of the paper. With this assumption the arguments of the above paper work with few changes.

2) By a work of Gitik-Merimovich which is in preparation (see my answer given in Failure of the GCH), to get the total failure of $GCH$ we need large cardinals weaker than what I said above. Now two options may happen:

(A) It might be the case that their forcing is enough homogeneous so that the arguments of the above paper can be applied so that we have $GCH$ in its $HOD$, and then we will have an exact equiconsistency result,

(B) On the other hand it might be the case that their forcing is not enough homogeneous, and then we can not conclude anything from their result.

If I'm forced to make a guess, I'd like to say case (A) happens, so we have an exact equiconsistency result for the result of my paper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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