# Stationary many subsets of $\kappa^+$ whose order type is a cardinal and whose intersection with $\kappa$ is an inaccessible cardinal

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.

-

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.

-
Jason's article on near supercompactness is available at: sciencedirect.com/science/article/pii/S016800721200142X –  Joel David Hamkins Mar 21 '13 at 1:54
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)... –  Trevor Wilson Mar 21 '13 at 2:02
...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. –  Trevor Wilson Mar 21 '13 at 2:02
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... –  Joel David Hamkins Mar 21 '13 at 2:10
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. –  Trevor Wilson Mar 21 '13 at 2:20