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

(Disclaimer: This question was also asked at MSE (http://math.stackexchange.com/questions/71020/can-we-collapse-omega-1-without-adding-a-dominating-real). I'm posting it here because, when I asked it, I was torn between my sense that it was appropriate for MO and my suspicion that this question is much, much easier than I'm making it; and it seems to be attracting no attention at MSE. I'm not a set theorist, so it's hard for me to judge the difficulty of the questions I'm asking; please let me know if this question is too elementary for MO, and I'll delete it.)

The question is precisely that of the title: is there a notion of forcing $\mathbb{P}$ which collapses $\omega_1$ to $\omega$ but does not add a real which dominates every real in the ground model? (Here "real" means "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 $S_f=\lbrace n: \forall m < n, f(m) < f(n)\rbrace$. 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 $\omega$.

On the other hand, looking at the poset of finite increasing functions from $\omega$ to $\omega_1$ (which builds a countable sequence cofinal in $\omega_1$, hence collapsing $\omega_1$ to $\omega$), it's unclear how this would add a dominating real; so perhaps the answer to my question is yes.

share|improve this question
1  
If CH holds in the ground model, then in the forcing extension the ground model reals get collapsed to a countable set, and an easy diagonalization shows that the extension has a dominating real. –  Amit Kumar Gupta Oct 12 '11 at 8:36
add comment

2 Answers 2

up vote 10 down vote accepted

As Amit points out in his comment, if the Continuum Hypothesis holds in the ground model $V$, then the collection of ground model reals becomes countable in the extension $V[G]$ in which $\omega_1$ is collapsed, and therefore there must be a real in $V[G]$ dominating every real in $V$. More generally, consider the dominating number, which is the size of the smallest family of functions such that every function is dominated by a function in the family.

Theorem. The following are equivalent:

  1. There is a forcing extension $V[G]$ collapsing $\omega_1$, but not adding a real dominating every ground model real.
  2. The dominating number of $V$ is at least $\omega_2$.

Proof. Amit's CH observation generalizes to show (the contrapositive of) 1 implies 2, because if $V$ has a dominating family of size $\omega_1$, then this family becomes countable in any $V[G]$ collapsing $\omega_1$, in which case there is a function in $V[G]$ dominating it, and hence dominating any ground model function.

Conversely, suppose that the dominating number is at least $\omega_2$. Now, consider the forcing to collapse $\omega_1$ to $\omega$ by finite conditions. This forcing has size $\omega_1$. Suppose that the forcing adds a function $f={\dot f}_G:\omega\to\omega$ dominating every function in $V$. For any ground model function $h:\omega\to\omega$, there is a condition $p_h$ and a number $n_h$ such that $p_h$ forces that $\dot f$ dominates $h$ beyond $n_h$. Since there are only $\omega_1$ many conditions in the forcing, it must be that some condition $p$ and natural number $n$ works for an unbounded family of functions $h$, since if each such family was bounded in $V$, then we would be able to form a dominating family in $V$ of size $\omega_1$, contrary to hypothesis. So fix $p$ and $n$ such that there is an unbounded family $F$ of functions $h$ such that $p$ forces $\dot f$ dominates $h$ beyond $n$. Now define a function $f^*(m)$ by choosing any condition (the first with respect to some well-ordering) $p_m$ extending $p$, such that $p_m$ decides the value of $\dot f(\check m)$. Since this must be larger than $h(m)$ for any $h\in F$, when $m\gt n$, it follows that $F$ was not unbounded after all, a contradiction. QED

As Todd mentions below, the idea of the proof easily generalizes to larger cardinals. Specifically, no forcing notion of size less than $\frak{d}$ can add a dominating real. So when $\frak{d}$ is very large, you can also collapse $\omega_2$ and larger cardinals without adding a dominating real.

In particular, since it is known to be consistent that the dominating number can be large, the answer to the title question is yes, it is consistent that this can happen.

share|improve this answer
    
Slight generalization: No notion of forcing of size less than $\mathfrak{d}$ can add a dominating real. (But Joel got to it first! ;) ) –  Todd Eisworth Oct 12 '11 at 13:25
    
Hey Joel, I think the first paragraph of your proof should say "1 implies 2" –  Amit Kumar Gupta Oct 12 '11 at 14:33
    
Yes, now fixed. –  Joel David Hamkins Oct 12 '11 at 14:38
    
This is an absolutely fantastic answer! Thanks. –  Noah S Oct 12 '11 at 16:45
1  
(In my comment, trees grow upwards.) –  Noah S Oct 13 '11 at 19:50
show 5 more comments

Hi Noah, I'm adding an answer because this wouldn't fit in a comment. Unless I'm making a silly mistake, the situation is different with respect to adding an escaping real:

Let $P$ be the collection of finite partial functions from $\omega$ to $\omega_1$ as usual, and let $\dot f$ be a $P$-name for the generic surjection from $\omega$ onto $\omega_1$.

Given $g:\omega\rightarrow\omega$ in the ground model, and $n<\omega$, consder the set $D(g, n)$ of conditions $p$ such that for some $m>n$,

  • $m$ is in the range of $p$,
  • the domain of $p$ is an initial segment of $\omega$, and
  • the least $k$ for which $p(k)=m$ is greater than $g(m)$.

This set is dense in $P$ for each $g$ and $n$, and so the real $h$ in the extension defined by setting $h(m)$ equal to the least $k$ such that $\dot f(k)=m$ is not bounded by a ground model real. (So essentially, we are "inverting" the surjection on the initial segment $\omega$ of its range)

Edit: Even simpler, if we define a real $h$ in the extension by setting $h(n)=\dot f(n)$ if $\dot f(n)<\omega$, and $h(n)=0$ otherwise, then $h$ is Cohen over the ground model, hence unbounded.

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.