MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Souslin trees on the first inaccessible cardinal it is asked if it is consistent that there are no $\kappa-$Souslin trees at the least inaccessible cardinal $\kappa$.

In this question I would like to ask an apparently simpler question:

Question. Is it consistent that there are no $\kappa-$Souslin trees for some inaccessible not weakly compact cardinal $\kappa$?

Remarks. (A) If $\kappa$ is weakly compact, then the tree property holds at $\kappa$, in particular there are no $\kappa-$Souslin trees,

(B) If $V=L,$ then $\kappa$ is weakly compact iff there are no $\kappa-$Souslin trees.

(C) $\Diamond_\kappa+GCH$ does not imply the existence of a $\kappa-$Souslin tree, for $\kappa$ inaccessible. For example if $\kappa$ is ineffable and $GCH$ holds, then $\Diamond_\kappa+GCH$ holds and there are no $\kappa-$Souslin trees.

share|cite|improve this question
    
What happens if we start with a weakly compact cardinal $\kappa$ in $V$, and then force to kill the Mahloness of $\kappa$ by adding club $C\subset\kappa$ avoiding the regular cardinals below $\kappa$? Does this create a $\kappa$-Suslin tree? – Joel David Hamkins Mar 30 '14 at 13:51
    
I think it does not give the answer. Note that a sufficient condition for having a $\kappa-$Souslin tree for $\kappa$ inaccessible, is square-with-built-in-diamond at $\kappa$, as is implicit in [Sh:624]. A weaker sufficient conditions is, e.g.,: there exists a $\square(\kappa)-$sequence $< C_\alpha: \alpha<\kappa >$, and a stationary set $S$ in $\kappa$, which is disjoint from $acc(C_\alpha)$ for all $\alpha<\kappa$, and $\Diamond(S)$ holds. To force the failure of these principles, we need more than weakly compact cardinals. – Mohammad Golshani Mar 30 '14 at 14:10
    
Does the existence of a $\kappa-$Souslin tree for $\kappa$ inaccessible have any effects on saturated ideals on $\kappa$ or something else? Maybe this way we can reduce the problem to something that we know more about it. – Mohammad Golshani Mar 30 '14 at 14:15
2  
If there is a $\kappa$-closed $\kappa$ saturated ideal on $\kappa$ and $\kappa$ is not measurable then $\kappa$ is inaccessible and there is a $\kappa$-Suslin tree (by splitting the positive sets in a partial binary tree). Actually, in this case every normal $\kappa$-tree contains a Suslin subtree. – Yair Hayut Mar 31 '14 at 17:54
2  
@Mohammad: you don't actually need a $\square (\kappa )$ sequence, all you need is a non reflecting stationary set $S$ with diamond on it. – Yair Hayut May 22 '14 at 17:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.