6
$\begingroup$

It seems that considerable care is taken in the literature to ensure that a countable support (CS) iteration of proper forcings preserves $\aleph_2$. Can you give an example, assuming CH, of a CS iteration of countably closed forcings such that each iterand is forced to have the $\aleph_2$-c.c. but the full iteration does not?

$\endgroup$

2 Answers 2

7
$\begingroup$

Assume CH. We start with a model with a function $F:[\omega_2]^2\to\omega$ such that the following holds: if $\{A_\alpha:\alpha<\omega_2\}$ are pairwise disjoint countable subsets of $\omega_2$ and $i<\omega$ then there are $\alpha<\beta$ such that all values of $F$ between $A_\alpha$ and $A_\beta$ are $>i$. This can be forced wih countable approximations.

Given $P_n$ let $q\in Q_n$ iff $q$ is a countable subset of $\omega_2$ with all values of $F$ on $q$ differing from $n$. $P$ is the full support iteration of the $Q_n$'s.

First it is easy to see that the conditions $p\in P_n$ are dense where $p(i)$ is a real countable set for $i<\omega$. Using this, we can show that $Q_n$ is $\aleph_2$-cc, for all $n$. With the usual trick, this gives a $\gamma_n<\omega_2$ such that if we remove the part of $Q_n$ below $\gamma_n$, then we get a notion of forcing (=no minimal elements).

Clearly, all $Q_n$'s are countably closed.

However, if $p_\alpha$ is the condition with $p_\alpha(n)=\{\alpha\}$, then any two of the $p_\alpha$'s are incompatible.

$\endgroup$
1
  • $\begingroup$ This is interesting, failure after only omega steps. $\endgroup$ Commented Jul 16, 2013 at 18:36
6
$\begingroup$

One of the standard (counter)examples here has to do with uniformization at $\omega_2$, and it is dealt with in Section 3 of the Appendix to Shelah's Proper and Improper forcing:

Let $S=\{\delta<\omega_2:\rm{cf}(\delta)=\omega_1\}$, and for each $\delta\in S$ let $\eta_\delta:\omega_1\rightarrow\delta$ be the increasing enumeration of a closed unbounded subset of $\delta$, and let $\bar{\eta}=\langle \eta_\delta:\delta\in S\rangle.$

Let $\bar{C} =\langle c_\delta:\delta\in S\rangle$ be a collection of functions where $c_\delta$ maps $\omega_1$ to $\{0,1\}$. We say that $\bar{C}$ can be uniformized if there is a function $f:\omega_2\rightarrow\{0,1\}$ such that for every $\delta\in S$ the set $$\{\alpha<\omega_1: c_\delta(\alpha)\neq f(\eta_\delta(\alpha))\}$$ is bounded in $\omega_1$.

Shelah proved that if CH holds, then for any such $\bar{\eta}$ there is a family of colorings $\bar{C}$ as above that cannot be uniformized. Assaf Rinot has a nice write-up of this result on his blog.

Shelah points out in Discussion 3.4A on page 984 of Proper and Improper forcing that for any given $\bar{C}$, there is a very nice $\aleph_1$-closed $\aleph_2$-cc (even $\aleph_2$-centered) forcing that adds a function uniformizing $\bar{C}$.

The relevance to your question is that an iteration of such uniformizing forcings that takes care of all potential $\bar{C}$ for a given $\bar{\eta}$ must collapse $\aleph_2$ as otherwise we contradict the theorem above, and therefore the limit of the iteration certainly cannot have the $\aleph_2$-cc.

The $\aleph_2$-pic (properness isomorphism condition) from Chapter VIII of Proper and Improper Forcing is a strong version of the $\aleph_2$-cc that says (roughly) that our forcing isn't doing something like "uniformizing", and it is one way to ensure that the limit of a countable support iteration possesess the $\aleph_2$-cc.

$\endgroup$
2
  • $\begingroup$ Thanks. I only accepted Peter's answer because it is self contained. $\endgroup$ Commented Jul 16, 2013 at 18:35
  • $\begingroup$ Well, his example is a more drastic and explicit failure! $\endgroup$ Commented Jul 16, 2013 at 20:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .