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

Let $\kappa \ge \aleph_3$ be a regular cardinal that is countably closed ($\alpha^\omega < \kappa$ for every $\alpha < \kappa$.) I'm mostly interested in the case that $\kappa$ is strongly inaccessible. Can there be a homogeneous notion of forcing that makes $\text{cof}(\kappa^{+V}) < \kappa$ without adding any bounded subsets of $\kappa$?

If there is a Woodin cardinal above $\kappa$ then the stationary tower forcing could do this except that it is (probably) not homogeneous.

If there is a forcing notion as desired then I believe the results of the paper "Stacking mice" would give a non-domestic mouse, so some large cardinals would be required to show that such a forcing exists. Can we get one from, e.g. a supercompact cardinal?

share|improve this question
1  
What does "countably closed regular cardinal" mean? –  Noah S Jan 31 '13 at 23:50
1  
Perhaps he means $\delta^\omega\lt\kappa$ whenever $\delta\lt\kappa$. (But he may mean simply that $\kappa^\omega=\kappa$.) –  Joel David Hamkins Feb 1 '13 at 0:03
    
Where can I find a precise definition of "homogeneous forcing"? Thx. –  Carlo Von Schnitzel Feb 12 '13 at 8:49
    
@alephomega When I use that phrase I usually mean "almost homogeneous forcing", which means that for any two conditions $p$ and $q$ in the forcing poset $\mathbb{P}$ there is an automorphism $\pi$ of $\mathbb{P}$ such that $\pi(p)$ is compatible with $q$. The only consequence of this that I am interested in is that the theory of the forcing extension with parameters from the ground model does not depend on the choice of generic filter. –  Trevor Wilson Feb 15 '13 at 2:29
2  

1 Answer 1

up vote 4 down vote accepted

I think that the strongly compact Prikry forcing, for $\kappa^+$ strongly compact cardinal $\kappa$ - that forces $\text{cf }\kappa = \text{cf }(\kappa^+)^V = \omega$ without adding bounded subsets to $\kappa$, is homogeneous, but I couldn't prove it or find a reference.

Instead, I'll show something weaker that still implies that the truth value of statements of the form $\phi(a)$ in $V[G]$, where $a\in V$ doesn't depend on the generic $G$: For every $p,q\in \mathbb{P}$, we will find $p^\prime \leq p,\,q^\prime \leq q$ and an automorphism between $\mathbb{P}\restriction p^\prime = \{r \in \mathbb{P} | r \leq p^\prime\}$ and $\mathbb{P}\restriction q^\prime$.

Let $\mathbb{P}$ be the strongly compact Prikry forcing for changing both $\text{cf }\kappa$ and $\text{cf }\kappa^+$ to $\omega$, without adding bounded subsets to $\kappa$.

Recall that a condition in $\mathbb{P}$ is a tree of finite increasing sequences in $P_\kappa (\kappa^+ )$, with finite trunk and above it every element has $U$-many successors ($U$ is fine $\kappa$-complete ultrafilter over $P_\kappa (\kappa^+)$). For the exact definitions and basic properties see section 1.4 in Gitik's chapter in the Handbook.

Let $t$ be the trunk of $p$ and $s$ the trunk of $q$. By narrowing the trees of $p$ and $q$, if necessary, we may assume that for every $r\in p$ above $t$, $(r\setminus t)\cup s \in q$. For every $a\leq p$ define $\pi (a) = \{(r\setminus t)\cup s | r\in a\}$ - this is the required automorphism.

share|improve this answer
1  
The paper "Consecutive Singular Cardinals and the Continuum Function" by Apter and Cody also contains similar results for supercompact Prikry forcing. –  Mohammad Golshani May 6 at 5:55

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.