Characterizing forcings that don't add any dominating reals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:37:08Z http://mathoverflow.net/feeds/question/46770 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46770/characterizing-forcings-that-dont-add-any-dominating-reals Characterizing forcings that don't add any dominating reals Amit Kumar Gupta 2010-11-20T19:54:24Z 2010-11-23T07:56:01Z <p>Regarding reals as functions from $\omega$ to $\omega$, let's say a real $f$ <i>eventually dominates</i> $g$ iff $(\exists n)(\forall m > n)[ f(m) > g(m)]$. Let's say that a (non-trivial separative) forcing poset $P$ <i>doesn't always add a dominating real</i> iff there is a generic extension by $P$ which doesn't contains a real that eventually dominates every real from the ground model. Let's say that $P$ <i>never adds a dominating real</i> iff every generic extension by $P$ doesn't contain any real that eventually dominates all the ground model's reals. I'm interested in combinatorial/order-theoretic conditions which may be necessary or sufficient for either of these notions.</p> <ul> <li><b>$\omega$-closure</b> implies you add no reals, hence you add no dominating reals; Cohen forcing is not $\omega$-closed but it never adds a dominating real</li> <li>one can show that <b>separability</b> implies you never add a dominating real (by separability, I mean containing a countable dense subset); the Cohen forcing that adds uncountably many reals isn't separable but never adds a dominating real</li> <li>Hechler forcing <b>has size at most continuum</b> but always adds a dominating real; the Cohen forcing that adds more than continuum many reals (where "continuum" is "continuum as computed in the ground model" obviously) has size greater than continuum but never adds a dominating real</li> <li>Hechler forcing also has the <b>countable chain condition</b> yet adds a dominating real; the forcing that adds a function $\omega _1 \to \omega _1$ with countable partial functions doesn't have the ccc but it's $\omega$-closed hence adds no new reals and thus never adds any dominating reals.</li> </ul> <p>My question: </p> <p><b>What are some combinatorial/order-theoretic conditions on a poset that are necessary and/or sufficient for the poset to never/not always add a dominating real?</b></p> http://mathoverflow.net/questions/46770/characterizing-forcings-that-dont-add-any-dominating-reals/46780#46780 Answer by Stefan Geschke for Characterizing forcings that don't add any dominating reals Stefan Geschke 2010-11-20T22:05:02Z 2010-11-20T22:23:44Z <p>Let me first say something about not always adding a dominating real (or anything else that can be formalized in the forcing language).</p> <p>Given a forcing notion $P$, the Forcing Theorem (everthing true in the extension is forced by an element of the filter) implies that there is a dense set $D$ of conditions that either force that there is a dominating real in the extension or that there is no dominating real in the extension Let $A$ be a maximal antichain of conditions in $D$. By the density of $D$, $A$ is a maximal antichain in $P$. Hence every generic filter will pick exactly one condition $p\in A$ and we can restrict our attention to the forcing notion that consists of conditions below $p$.</p> <p>In other words, forcing notions that sometimes but not always add a dominating real are characterized by: there is a condition below which the forcing always adds a dominating real and there is a condition below which the forcing does not add a dominating real. </p> <p>So, we are left with characterizing forcing notions that never add dominating reals respectively always add a dominating real.<br> I am sure it is possible (and maybe a good exercise) to write down a characterization of adding or not adding a dominating real, but I am not aware of any crisp, helpful combinatorial characterization of either of these properties. I might be missing something, though.<br> Natural places to check are the Bartoszynski-Judah book Set Theory of the Real Line and Zapletal's Forcing Idealized. </p> <p>Something that is well understood, though, is $\omega^\omega$-boundingness. A forcing notion is $\omega^\omega$-bounding if every function from $\omega$ to $\omega$ in the extension is bounded by a function in the ground model.<br> This turns out to be a form of distributivity, so called weak $(\omega,\omega)$-distributivity (this is covered in Jech's Set Theory, The Third Millenium Edition and I would guess in earlier editions as well). Clearly, $(\omega,\omega)$-bounding forcings do not add dominating reals.<br> Examples are random real forcing (forcing with Borel sets of the real line of positive measure) and Sacks forcing.</p> http://mathoverflow.net/questions/46770/characterizing-forcings-that-dont-add-any-dominating-reals/46833#46833 Answer by Haim for Characterizing forcings that don't add any dominating reals Haim 2010-11-21T16:59:36Z 2010-11-21T16:59:36Z <p>There is a somewhat related result by Shelah: Any Suslin ccc forcing which adds a non-dominated real adds a Cohen real. The proof can be found here: <a href="http://shelah.logic.at/files/480.pdf" rel="nofollow">http://shelah.logic.at/files/480.pdf</a></p> http://mathoverflow.net/questions/46770/characterizing-forcings-that-dont-add-any-dominating-reals/47050#47050 Answer by Amit Kumar Gupta for Characterizing forcings that don't add any dominating reals Amit Kumar Gupta 2010-11-23T04:33:52Z 2010-11-23T07:56:01Z <p>Stefan's answer pointed me in the right direction, and then talking it over with prof. Leo Harrington we've got an answer:</p> <p>A complete Boolean algebra $\mathbb{B}$ never adds a dominating real iff for any collection <code>$\{ u _{m,k} : m, k \in \omega \} \subset \mathbb{B}^+$</code> the following weaker form of weak $(\omega ,\omega )$-distributivity holds: </p> <blockquote> <p>$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{n \in \omega} \sum _{n &lt; m &lt; \omega} \sum _{k &lt; f(m)} u _{m,k}$</p> </blockquote> <p>For the reverse implication, suppose "weak weak $(\omega ,\omega )$-distributivity" holds, and for contradiction let $u \neq 0$ be the Boolean value of the sentence "there exists a dominating real," and let $\dot{g}$ be a name witnessing this, i.e. the sentence "$\dot{g}$ is a dominating real" has Boolean value $u$. Define $u _{m,k} = || \dot{g} (m) = k ||$. Now if $G$ is any $\mathbb{B}$-generic filter containing $u$, noting that the left side of the distributivity identity is (at least) $u$, we know that the right side belongs to $G$. It's then not hard to see that:</p> <blockquote> <p>$(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k &lt; f(m))(u _{m,k} \in G)$</p> </blockquote> <p>which is to say that there's a real $f$ in the ground model such that:</p> <blockquote> <p>$(\forall n)(\exists m > n)(\dot{g}^G (m) &lt; f(m))$</p> </blockquote> <p>so $f$ is not dominated by $\dot{g}$, contradiction. </p> <p>For the forward implication, it should suffice to show it in the case where for each $m$, the set <code>$\{ u _{m,k} : k \in \omega \}$</code> is an antichain with least upper bound $u$ independent of $m$ (I haven't checked this detail personally). So let <code>$\{ u _{m,k}\}$</code> be such a collection for which the identity fails. Consider the name:</p> <blockquote> <p><code>$\dot{g} = \{ (u _{m,k}, (m,k)) : m, k \in \omega \}$</code></p> </blockquote> <p>It's not hard to see that the right side of the identity is at most $u$, so assuming the identity fails it's strictly less than $u$, so since $\mathbb{B}$ is separative there's a generic $G$ containing $u$ avoiding the right side of the identity. It's not hard to see from here that $\dot{g}^G$ will dominate all the ground model's reals.</p> <hr/> <p>I should add that if we think of $u _{m,n}$ as saying "$\dot{g}(m) = n$" and replace the Boolean operations with the corresponding quantifiers, then the left side says "$\dot{g}$ is a real," and the right side says "$\dot{g}$ doesn't dominate every real in the ground model." This suggests how we can characterize forcings that don't add any unbounded reals, for example, namely the following identity holds:</p> <blockquote> <p>$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} \sum _{k &lt; f(m)} u _{m,k}$</p> </blockquote> <p>Forcings that don't add any reals are precisely those that satisfy the following identity:</p> <blockquote> <p>$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} u _{m,f(m)}$</p> </blockquote> <p>You can easily generalize this to talking about functions $\kappa \to \lambda$; the above two results so generalized are precisely Theorem 15.38 and Lemma 15.39 in Jech, "Set Theory".</p>