Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

"Foreman`s maximality principle" states that every non-trivial forcing notion either adds a real or collapses some cardinals. This principle has many consequences including:

1) $GCH$ fails everywhere,

2) there are no inaccessible cardinals,

3) there are no $\kappa-$Souslin trees,

4) Any non-trivial $c.c.c.$ forcing adds a real,

5) Any non-trivial $\kappa^+-$closed forcing notion collapses some cardinals.

Consistency of (1) is proved by Foreman-Woodin, (2) clearly can be consistent and the consistency of (4) is shown in "Forcing with c.c.c forcing notions, Cohen reals and Random reals".

My interest is in the consistency of (5). Let's consider the case $\kappa=\omega.$

Question 1. Is it consistent that any non-trivial $\aleph_1-$closed forcing notion collapses some cardinals?

The above question seems very difficult, and it is not difficult to show that for its consistency we need some very large cardinals. But maybe the following is simpler:

Question 2. Is it consistent that any non-trivial $\aleph_1-$closed forcing notion of size continuum collapses some cardinals? Does its consistency imply the existence of large cardinals?

share|improve this question
    
See related mathoverflow.net/questions/11633/… –  Joel David Hamkins Aug 12 '11 at 14:40
add comment

2 Answers 2

up vote 3 down vote accepted

The answer to the second question is yes, without any large cardinals assumptions.

Claim: if $2^{\aleph_0}$ is singular then every non-trivial $\sigma$-closed forcing of size $2^{\aleph_0}$ collapses the continuum.

The reason is that such forcing must add a new set of ordinals of size $\lambda < 2^{\aleph_0}$, $\tau$ (since the minimal $\lambda$ must be regular). I want to code $2^\omega$ into $\tau$.

If we choose $\lambda$ to be minimal, every initial segment of $\tau$ is in $V$. Now, we can define a function $F\in V$ from all possible initial segments of $\tau$ onto the reals, such that for every $p\in P$ and every $x\in 2^\omega$ there is $q\leq p$ and $\beta < \lambda$ such that $q\Vdash \tau \restriction \beta = \check{a}$ and $F(a)=x$:
Let $\langle (p_i, x_i) | i < 2^{\aleph_0}\rangle$ enumerate $P\times 2^\omega$. For every $\alpha < 2^{\aleph_0}$, we use the $\sigma$-closure of $P$ and the fact that $\tau \notin V$ in order to find $q \leq p_\alpha$ such that $q \neq p_i$ for every $i < \alpha$, $q\Vdash \tau \restriction \beta = \check{a}$ (for some $\beta$) and $F(a)$ is not determined yet, and set $F(a)=x_\alpha$:

Let $p = p_\alpha \in P$. We start by building a tree of $2^{<\omega}$ incompatible conditions $q_s,\, s\in 2^{<\omega}$ such that $q_\emptyset = p$ and for every $s\in 2^{<\omega}$, $q_{s\frown (0)}, q_{s\frown (1)} \leq q_s$, there is $\beta_s < \lambda$ in which $q_{s\frown (i)} \Vdash \tau \restriction \beta_s = a_{s\frown (i)}$ for $i\in\{0,1\}$, $a_{s\frown (0)} \neq a_{s \frown (1)}$. This is possible since $\tau \notin V$ but every initial segment of it is in $V$.

For every $f\in 2^\omega$, let $q_f \in P,\,\forall n\, q_f \leq q_{f\restriction n}$ (by the closure of $P$). Without loss of generality, $\forall f \in 2^\omega\,q_f\Vdash \tau \restriction \beta = \check{a_f}$, and for every $f\neq f^\prime$, $a_f \neq a_{f^\prime}$ (take $\beta = \sup_{s\in 2^{<\omega}} \beta_s$). Since we picked already only $|\alpha |<2^{\aleph_0}$ values, there must be some $f\in 2^\omega$ such that $F(a_f )$ is not determined.

By density arguments, in $V[G]$, $\{F(\tau \restriction \beta) | \beta < \lambda\} = (2^\omega)^V$.

share|improve this answer
    
Thanks, it is really interesting. On the other hand I have proved the consistency of $2^{\aleph_0}$ is arbitrary large regular and there is a non-trivial $\aleph_1-$closed (but not $\aleph_2-$closed) forcing of size continuum which preserves all cardinals. –  Mohammad Golshani Dec 3 '13 at 6:30
    
So we still don't know whether it's possible to have regular continuum while every $\sigma$-closed forcing collapses cardinals. Does your example works for $\lambda = 2^{\aleph_0},\, 2^{<\lambda} > \lambda$? –  Yair Hayut Dec 4 '13 at 8:23
    
Nice question, I should check it. –  Mohammad Golshani Dec 4 '13 at 8:54
add comment

I can't answer your question, but I can give a simpler sounding formulation that might be helpful. Analyze the question in two cases.

Case 1: The continuum hypothesis holds
In this case, the statement is false, because any $<\aleph_1$ -closed forcing of size $\aleph_1$ cannot collapse cardinals. The forcing to add a cohen subset to $\aleph_1$ is a nontrivial example of such a forcing.

Case 2: The continuum hypothesis fails
In this case, it is a theorem that every $<\aleph_1$-closed forcing notion which collapses a cardinal collapses the continuum. (See this question, referenced by Joel in the comments.) But every such forcing is equivalent to the canonical collapse forcing to collapse the continuum to $\aleph_1$. The most general version of this latter theorem that I know of (although the degree of generality makes it hard to follow) can be found in Handbook of Boolean Algebras, Volume 2, Corollary 1.15.

So really, your question boils down to whether every $<\aleph_1$-closed forcing of size continuum is isomorphic to Coll$(\aleph_1, c)$ This sounds strange to me, but I can't prove it's false, and if Foreman is entertaining it, who am I to judge it?

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.