Assume $V=L$ and let $\kappa$ be a Mahlo cardinal. Let $L[G]$ be the generic extension obatined by Mitchell forcing to make $2^{\aleph_0}=\aleph_2=\kappa.$ It is known that in the extension there are no special $\aleph_2$-Aronszajn trees but there are $\aleph_2$-Aronszajn trees.

Question 1. Is there any $\aleph_2$-Souslin tree in $L[G]?$

I assume the believe is that there are, but I don't know how to prove it. Surprisingly, if, instead of $\aleph_2$, we consider a cardinal $\lambda^+ > \beth_\omega,$ with $\lambda$ regular and apply the Mitchell forcing to get $2^\lambda=\lambda^{++}=\kappa,$ then the results of Assaf Rinot show that there are $\lambda^{++}$-souslin trees in the extension.

My second question is motivated by the work of Laver-Shelah. Assume $V=L$ and $\kappa$ is weakly compact. In order to produce a model of $CH+$there are no $\aleph_2$-Souslin trees, Laver and Shelah, first force with Levy collapse $Col(\aleph_1, < \kappa),$ and over it do an iteration to kill all possible $\aleph_2$-Souslin trees.

Question 2. Are there any $\aleph_2$-Souslin trees just after doing the Levy collapse $Col(\aleph_1, < \kappa)$?


1 Answer 1


About question 1: If $\kappa$ is not weakly compact in $L$, then there is an $\aleph_2$-Suslin tree in $L[G]$.

In $L$ there is a $\kappa$-Suslin tree, $T$, and since Mitchell's forcing is $\kappa$-Knaster, it cannot add an antichain of cardinality $\kappa$ to $T$: If $\dot{\mathcal{A}}$ is a name for unbounded antichain in $T$, we can pick for every $\alpha < \kappa$, a condition $p_\alpha$ in Mitcell's forcing, such that $p_\alpha$ forces that the $\alpha$-th member of $\dot{\mathcal{A}}$ is some specific $t_\alpha\in T$. Let $I\subseteq \kappa$ be an unbounded subset such that for all $\alpha, \beta\in I$, $p_\alpha$ is compatible with $p_\beta$ then $\{t_\alpha \mid \alpha \in I\}$ form an antichain in $L$.

About Question 2: As it is stated in the comments by Assaf, the answer to question 2 is also affirmative.

  • 3
    $\begingroup$ Assaf Rinot also said that they have an explicit example in "A Microscopic approach to Souslin-tree constructions. Part I" as Example 1.19 $\endgroup$
    – Asaf Karagila
    Apr 9, 2016 at 19:28
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    $\begingroup$ Indeed, the proof of Example 1.19 answers Question 2 in the affirmative: assafrinot.com/paper/22 $\endgroup$
    – saf
    Apr 10, 2016 at 5:18
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    $\begingroup$ @saf Thanks Assaf, why not posting your comment as an answer? $\endgroup$ Apr 10, 2016 at 5:36

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