The Proper Forcing Axiom (PFA) implies that every forcing which adds a subset of $\omega_1$ either adds a real or collapses $\omega_2$. Is it consistent that every forcing which adds a subset of $\omega_2$ either adds a subset of $\omega_1$ or collapses $\omega_3$?
Asked
Active
Viewed 418 times
8
-
I think this result is due to Todorcevic, Some Combinatorial Properties of Trees (answering a question of Abraham from his PhD thesis about whether the negation of the first statement in the question is true in ZFC), where he shows that this follows from the conjunction of $2^{\aleph_0}= \aleph_2$ and every $\omega_1$-tree of size $\aleph_1$ is essentially special. This is Theorem 2. The proof seems like it might be helpful here too. – tci Apr 25 '16 at 23:26
-
1For example if we could show that it is consistent that every $\omega_2$-tree of size $\aleph_2$ is $\omega_1$-essentially special and $2^{\aleph_1}=\aleph_3$, it seems like this might give the consistency of the statement you are interested in. – tci Apr 25 '16 at 23:28
-
The paper Two applications of finite side conditions on $\omega_2$ by Neeman also seems to be relevant. See the discussion starting at the bottom of the second page. He considers the problem of adding some sort of specialising function for a single $\omega_2$-tree, and already here there are complications, persistent counterexamples etc. – tci Apr 26 '16 at 01:50
1 Answers
7
Yes, it is consistent. The following theorem is proved by Shelah and myself (our paper is not yet complete):
Theorem. Assume $\kappa$ is weakly compact and $\lambda > \kappa$ is 2-Mahlo. Then there is a generic extension in which $\kappa=\aleph_2, 2^{\aleph_1}=\aleph_3=\lambda$ and any forcing notion which adds new subset to $\aleph_2$, collapses $\aleph_2$ or $\aleph_3.$
Unfortunately the proof is not so easy to state it here.
Note that by a new subset to $\aleph_2$ I mean a subset of $\aleph_2$ which is not in $V$, the ground model, but such that all its initial segments are in $V$.
Remark. It is evident that the above theorem gives the consistency of the requested question. If the forcing does not add subsets to $\aleph_1,$ then it adds new subsets to $\aleph_2$ so the theorem applies.
Added note: The paper is now available at Specializing trees and answer to a question of Williams.
Mohammad Golshani
- 31,966
-
-
-
By a new subset to $\aleph_2$ I mean all its initial segments are in the ground model. – Mohammad Golshani Apr 26 '16 at 03:15
-
Your question follows from the above mentioned theorem. If the forcing adds no new subsets to $\aleph_1,$ then forcing adds a new subset to $\aleph_2$ so the theorem applies. – Mohammad Golshani Apr 26 '16 at 03:19
-
@MohammadGolshani Do you also get this via not having non-special $\omega_2$-trees like Todorcevic? – tci Apr 26 '16 at 03:22
-
-
Judging by some of the things in the Neeman paper I mentioned, that sounds like it will be very interesting. – tci Apr 26 '16 at 03:27
-
Thanks, in fact we were working on a different problem, but later it was realized that our methods work to give the above result. My motivation for the above comes from Foreman's maximality principle. – Mohammad Golshani Apr 26 '16 at 03:29
-
Well, I'm guessing that you do an iteration of $\sigma$-closed forcing after your preparatory forcing, and this is what I would find especially interesting. I guess you have CH in the final model? In any case, I look forward to reading your paper. – tci Apr 26 '16 at 03:37
-
-
In such a model there are no $\aleph_2$-Souslin trees, so at least if $CH$ holds (which is the case in our model), yes we need some large cardinals. – Mohammad Golshani Apr 27 '16 at 03:23