Can we collapse $\omega_1$ to $\omega$ without adding a dominating real? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:37:21Z http://mathoverflow.net/feeds/question/77891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77891/can-we-collapse-omega-1-to-omega-without-adding-a-dominating-real Can we collapse $\omega_1$ to $\omega$ without adding a dominating real? Noah S 2011-10-12T03:37:20Z 2011-10-12T21:54:10Z <p>(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.)</p> <p>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 $\omega^\omega$.)</p> <p>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 $S_f=\lbrace n: \forall m &lt; n, f(m) &lt; f(n)\rbrace$. 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 $\omega$.</p> <p>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.</p> http://mathoverflow.net/questions/77891/can-we-collapse-omega-1-to-omega-without-adding-a-dominating-real/77922#77922 Answer by Joel David Hamkins for Can we collapse $\omega_1$ to $\omega$ without adding a dominating real? Joel David Hamkins 2011-10-12T13:18:04Z 2011-10-12T14:37:58Z <p>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.</p> <p><b>Theorem.</b> The following are equivalent:</p> <ol> <li>There is a forcing extension $V[G]$ collapsing $\omega_1$, but not adding a real dominating every ground model real.</li> <li>The dominating number of $V$ is at least $\omega_2$. </li> </ol> <p>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. </p> <p>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</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/77891/can-we-collapse-omega-1-to-omega-without-adding-a-dominating-real/77961#77961 Answer by Todd Eisworth for Can we collapse $\omega_1$ to $\omega$ without adding a dominating real? Todd Eisworth 2011-10-12T21:43:34Z 2011-10-12T21:54:10Z <p>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:</p> <p>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 $\omega_1$.</p> <p>Given $g:\omega\rightarrow\omega$ in the ground model, and $n&lt;\omega$, consder the set $D(g, n)$ of conditions $p$ such that for some $m>n$,</p> <ul> <li>$m$ is in the range of $p$,</li> <li>the domain of $p$ is an initial segment of $\omega$, and</li> <li>the least $k$ for which $p(k)=m$ is greater than $g(m)$.</li> </ul> <p>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)</p> <p>Edit: Even simpler, if we define a real $h$ in the extension by setting $h(n)=\dot f(n)$ if $\dot f(n)&lt;\omega$, and $h(n)=0$ otherwise, then $h$ is Cohen over the ground model, hence unbounded.</p>