Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question is exactly that in the title: is there a forcing which collapses $\omega_1$ to $\omega$ but does not add a dominating real ("real" here meaning "element of $\omega^\omega$")?

It seems like the answer should be "no," and I've attempted to prove this myself. The problem is that the easiest way to do so would be to define a dominating function in terms of an arbitrary surjection $f: \omega\twoheadrightarrow\omega_1;$ however, there seems to be no clear way to do this. My first thought was to look at the set $S_f=\lbrace n: \forall m(m<n\implies f(m)<f(n) \rbrace$. This is certainly a real, but there is no reason it should be dominating, let alone not present in the original model already; in fact, we can alter the usual collapsing poset to demand that $S_f$ be precisely the evens, or precisely the powers of 17, or in fact any infinite co-infinite subset of $\omega$.

Hence, my question. Additionally, I have a meta-question: I am not a set theorist, so when I think about set theory I have a hard time judging which of the questions that occur to me are trivial and which are actually hard. In particular, I have placed this question here (as opposed to at MO) because I strongly suspect it is not actually very hard at all. My meta-question is, "Would this question have been appropriate at MO?"

share|improve this question
    
It's fine to ask this question here. It's also nice to see someone else using the [foring] tag, for a change! –  Asaf Karagila Oct 9 '11 at 5:31
    
The surjection $f$ is not arbitrary. Generic seems like the correct term. –  Asaf Karagila Oct 10 '11 at 5:30
    
The surjection $f$ is indeed not arbitrary, but I stand by my statement that the easiest way to settle this would be to show how one could define a dominating function in terms of an arbitrary surjection. (This would show the more general result that any outer model $V'$ of $V$ in which $\omega_1^V$ is countable, there is some real $f$ which dominates every $g\in V$.) –  user13568 Oct 10 '11 at 5:51
    
Regarding your meta-question, my opinion is that this question is fine at MO, and any non-trivial question about forcing is fine there. –  JDH Oct 12 '11 at 14:19

1 Answer 1

up vote 1 down vote accepted

In my answer over at the MathOverflow version of your question, I showed that indeed it is possible over some models of set theory to collapse $\omega_1$ or even much larger cardinals to $\omega$, without adding a dominating real. The necessary hypothesis for this to occur is that the dominating number of the ground model be large. In particular, under CH is it impossible, but it is known to be consistent with the failure of CH that the dominating number is very large. Over these models, the ordinary collapse forcing will have size less than the dominating number, and no such forcing can add a dominating real, for the reasons I give in the other answer.

share|improve this answer

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.