# Two complementing consequences of supercompactness

I would like to know if the following two consequences of having a supercompact cardinal are orthogonal:

1) On one hand being supercompact is equivalent to being "A ineffable for all A" (Combinatorial characterization of supercompact cardinals, M. Magidor, Proc. Amer. Math. Soc. 42 (1974), 279-285).

2) On the other hand being supercompact implies having a Laver function (T. Jech, Set Theory, Theorem 20.21).

So from 1) for every "regressive" function on $P_\kappa(A)$ we have a "guessing" subset $B \subset A$. And from 2) we have a "diamond" sequence given by a function f that "guesses" every (small enough) set x in the ultrapower.

I am not sure if there is a connection between the two.

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Compliment = מחמאה; whereas complement = משלים. :-) Your title in Hebrew would read שתי מסקנות מחמיאות מעל-קומפקטיות. Amusing, but I think you didn't mean this title but rather the other one. –  Asaf Karagila Dec 5 '12 at 22:26
Can you say more precisely the particular definition of $A$-ineffable that Magidor uses? There are several different meanings of "regressive" here, and I think you intend a very strong one to get the equivalence. –  Joel David Hamkins Dec 6 '12 at 0:06

One difference is that it is an open question whether these properties are equivalent level-by-level in the supercompactness hierarchy.

Specifically, on the one hand, if $\kappa$ is $\theta$-supercompact, then one gets the ineffability property for all $A\subset P_\kappa\theta$, at the same level of supercompactness. That is, if $\kappa$ is $\theta$-supercompact, witnessed by embedding $j:V\to M$. If $\vec A=\langle A_\sigma\mid \sigma\in A\rangle$, where $A_\sigma\subset \sigma$, then consider $j(\vec A)(j''\theta)$, which is $j''B$ for some $B\subset \theta$. Almost every $\sigma$ has $A_\sigma=B\cap\sigma$, since $j''B=j(\vec A)(j''\theta)=j(B)\cap j''\theta$, and I think that this is the threading property you have in mind.

But meanwhile, on the other hand, although every supercompact cardinal $\kappa$ has a Laver function for $\theta$-supercompactness, Laver's argument is not level-by-level. Specifically, his argument shows only that every $2^{\theta^{\lt\kappa}}$-supercompact cardinal $\kappa$ has a Laver function for $\theta$-supercompactness. One needs a higher level of supercompactness initially than one gets for the Laver function.

In my paper A class of strong diamond principles I call attention to the following open question (question 5):

Question. If $\kappa$ is $\theta$-supercompact, must it have a $\theta$-supercompactness Laver function?

This question is, of course, connected with the number-of-measures problem, since if there is a Laver function for $\theta$-supercompactness, then of course there must be at least $2^\theta$ many such distinct normal fine measures on $P_\kappa\theta$, at least one for guessing each element of $H_{\theta^+}$. It is not known whether than can be exactly one, except in the case $\theta=\kappa$, which reduces to measurability.

Meanwhile, the existence of a $\theta$-supercompactness Laver function, although not yet known to follow from the $\theta$-supercompactness of $\kappa$, does not require any extra consistency strength, since as I show in the paper, if $\kappa$ is $\theta$-supercompact, then fast function forcing with $\kappa$ will add a generic Laver function for $\theta$-supercompactness.

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