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We have learned from Joel David Hamkins and Monroe Eskew that:

Answers: Having a measurable cardinal $\delta$ we can force a $\kappa$-Suslin tree for many $\kappa$'s above $\delta$.

But is the opposite true (I am interested in the case when $\delta$ is the least measurable):

Question 1: If we have a set theory with (the least) measurable cardinal $\delta$ can we have a set theory with $\delta$ (still the least measurable) but without any $\kappa$-Suslin trees for $\kappa>\delta$?

Actually I hadn't expected Souslin trees above $\delta$... Since I seem to be unable to remove another assumption from my constructions let me ask:

Question 2: Does the validity of statements in Answer and Question 1 above change if we assume that $\delta$ (the least measurable) is strongly compact?

Edit: I edited the text to indicate that, to me, the most interesting case is when $\delta$ is the least measurable cardinal. Inserted pieces are in parentheses.

Why am I interested in this? Let $\mathbb{Z}^\kappa=A\oplus B$ be a direct sum decomposition of the product of the group of integers. By old results, if no measurable cardinal exists below $\kappa$ then both $A$ and $B$ are themselves isomorphic to some products of integers. For larger $\kappa$'s little is known except that $A$ or $B$ has to be of this form. I think that the problem of whether both $A$ and $B$ have to be products depends on the existence of Suslin trees above $\delta$.

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  • $\begingroup$ Do you want SH to hold in a single cardinal above the measurable $\delta$ or in every (regular) cardinal above $\delta$? $\endgroup$
    – Yair Hayut
    Commented Apr 1, 2015 at 17:58
  • $\begingroup$ @Yair Hayut - in every cardinal above $\delta$. I edited the question, hopefully it is more clear now. $\endgroup$
    – Adam Przeździecki
    Commented Apr 1, 2015 at 18:48
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    $\begingroup$ I don't know if the situation in Question 1 is consistent, but if it is, it would require much larger cardinals than just a measurable. For example, in order to get SH at successor to strong limit singular you must either violate SCH or have $\neg \square_\mu$, where $\mu$ is the strong limit, by a result of Ben-David and Shelah. So you need at least a cardinal $\kappa$ with $o(\kappa) = \kappa^{++}$, and probably much more. $\endgroup$
    – Yair Hayut
    Commented Apr 1, 2015 at 20:49

1 Answer 1

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By Laver-Shelah: If $\kappa=\lambda^+$ is successor of a regular uncountable cardinal and there is a weakly compact cardinal above it, then there is a $\lambda-$directed closed forcing notion, forcing $\kappa$-Souslin hypothesis. From this question 1 follows immediately.

Also if we start with supercompct $\delta$, and consider $\kappa>\delta^+$ successor of a regular cardinal, assume there is a weakly compact cardinal above $\kappa$ and if we assume $\delta$ is Laver indestructible, then we can force $\kappa$-Souslin hypothesis, preserving the supercompactness of $\delta.$

If we require $\kappa$-Souslin hypothesis holds at all regular cardinals above $\delta,$ then the question is open and its consistency requires large cardinals much stronger than measurability (see Yair Hayut's comment).

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  • $\begingroup$ making sure: the forcings you describe kill Suslin trees only at one $\kappa$ at a time, not all the $\kappa$-Suslin trees between $\delta$ and the weakly compact above it? $\endgroup$
    – Adam Przeździecki
    Commented Apr 1, 2015 at 18:57
  • $\begingroup$ It just forces the tree property at $\kappa,$ and in fact it make the weakly compact cardinal in the extension to become $\kappa.$ $\endgroup$
    – Mohammad Golshani
    Commented Apr 2, 2015 at 1:20

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