5
$\begingroup$

Is anything known about the consistency strength of the following statement?

  • $\kappa$ is a Mahlo cardinal and there is a stationary set of $a \in \mathcal{P}_\kappa(\kappa^+)$ such that $a \cap \kappa$ is an inaccessible cardinal and the order type of $a$ is $(a \cap \kappa)^+$?

This statement follows from $\kappa^+$-supercompactness of $\kappa$ and also from subcompactness of $\kappa$. It is a strengthening of the principle $\text{Pr}(\kappa^+)$ that was shown by Burke in "Generic embeddings and the failure of box" to imply $\neg \square_\kappa$, so a lower bound on the consistency strength is the existence of two Mahlo cardinals.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Well, here is a very slight weakening of your $\kappa^+$-supercompactness upper bound, to the assumption merely that $\kappa$ is nearly $\kappa^+$-supercompact. This hypothesis is strictly weaker than $\kappa^+$-supercompactness, but still, under this assumption, the set of such $a$ is stationary as you desire.

Specifically, a cardinal $\kappa$ is nearly $\kappa^+$-supercompact, if for every transitive $M$ of size $\kappa^+$, closed under $\lt\kappa$ sequences, there is an elementary embedding $j:M\to N$ into a transitive set $N$, with critical point $\kappa$, such that $j''\kappa^+\in N$. It follows that $j''\kappa^+$ has your property with respect to $j(\kappa)$ in $N$, and so the set of $a\in P_\kappa(\kappa^+)$ with your property will have measure one with respect to the induced $M$-measure. It follows by the usual normality argument that the set of such $a$ is stationary, as you desire, since any club set $C$ (or the function defining it) can be placed into such an $M$ and so $j''\kappa^+\in j(C)$, meaning that $j$ of the set of $a$ meets $j(C)$, and so there is such an $a$ in $C$.

The nearly $\theta$-supercompact cardinals were introduced by Jason Schanker in his dissertation (written under my supervision), along with the weakly measurable cardinals, and Jason had noted that the near $\theta$-supercompactness hypothesis suffice in many arguments that previously had used full $\theta$-supercompactness. The present case is an additional instance of this phenomenon, where near $\kappa^+$-supercompactness suffices in place of $\kappa^+$-supercompactness.

Meanwhile, I am unsure exactly how the nearly $\kappa^+$-supercompact cardinals relate to the subcompact cardinals, and in any case perhaps a more dramatic weakening is possible.

It may be interesting to note that a recent result proved jointly by Jason Schanker, Brent Cody, Moti Gitik and myself shows that it is relatively consistent with ZFC that the least weakly compact cardinal $\kappa$ is also nearly $\kappa^+$-supercompact (but $2^\kappa\gt\kappa^+$), which might suggest something about the nature of these cardinals.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Jason's article on near supercompactness is available at: sciencedirect.com/science/article/pii/S016800721200142X $\endgroup$
    – Joel David Hamkins
    Mar 21, 2013 at 1:54
  • $\begingroup$ Yes, that is definitely worth mentioning. Ideally, I was hoping for something that implied the stationarity of the set in the question without also implying that $\kappa$ is weakly compact. This is because if we add the additional assumption that $\kappa$ is weakly compact, then $\square(\kappa)$ and $\square_\kappa$ both fail, and the lower bound for this coincides with the current state of the art in inner model theory (I think)... $\endgroup$
    – Trevor Wilson
    Mar 21, 2013 at 2:02
  • $\begingroup$ ...whereas on the other hand if we don't know that $\kappa$ is weakly compact I don't see any way to get more than two Mahlo cardinals out of it, so I thought maybe it could be forced from two Mahlo cardinals somehow (the first being $\kappa$ and the second becoming $\kappa^+$.) I don't know if this is plausible though. $\endgroup$
    – Trevor Wilson
    Mar 21, 2013 at 2:02
  • $\begingroup$ I'll give it some more thought; several ideas I had considered to make it much weaker didn't work out. But perhaps it is possible... $\endgroup$
    – Joel David Hamkins
    Mar 21, 2013 at 2:10
  • $\begingroup$ Thanks. By the way, I should probably share my motivation: I think it is interesting that the simultaneous failure of $\square(\kappa)$ and $\square_\kappa$ is strong whereas the failure of either on its own is weak. I have an application where it's not clear that $\neg \square(\kappa) \And \neg \square_\kappa$ is enough, however, so hopefully I can strengthen $\neg \square(\kappa)$ to "$\kappa$ is weakly compact" and strengthen $\neg \square_\kappa$ slightly to something like in the question, but still maintain the property that they are weaker on their own than they are together. $\endgroup$
    – Trevor Wilson
    Mar 21, 2013 at 2:20

Your Answer

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

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