Characterizing forcings that don't add any dominating reals - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T07:37:08Zhttp://mathoverflow.net/feeds/question/46770http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/46770/characterizing-forcings-that-dont-add-any-dominating-realsCharacterizing forcings that don't add any dominating realsAmit Kumar Gupta2010-11-20T19:54:24Z2010-11-23T07:56:01Z
<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#46780Answer by Stefan Geschke for Characterizing forcings that don't add any dominating realsStefan Geschke2010-11-20T22:05:02Z2010-11-20T22:23:44Z<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#46833Answer by Haim for Characterizing forcings that don't add any dominating realsHaim2010-11-21T16:59:36Z2010-11-21T16:59:36Z<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#47050Answer by Amit Kumar Gupta for Characterizing forcings that don't add any dominating realsAmit Kumar Gupta2010-11-23T04:33:52Z2010-11-23T07:56:01Z<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>$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{n \in \omega} \sum _{n < m < \omega} \sum _{k < f(m)} u _{m,k}$</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>$(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k < f(m))(u _{m,k} \in G)$</p>
</blockquote>
<p>which is to say that there's a real $f$ in the ground model such that:</p>
<blockquote>
<p>$(\forall n)(\exists m > n)(\dot{g}^G (m) < f(m))$</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>$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} \sum _{k < f(m)} u _{m,k}$</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>