Regarding reals as functions from $\omega$ to $\omega$, let's say a real $f$ eventually dominates $g$ iff $(\exists n)(\forall m > n)[ f(m) > g(m)]$. Let's say that a (non-trivial separative) forcing poset $P$ doesn't always add a dominating real 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 $P$ never adds a dominating real 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.

  • $\omega$-closure implies you add no reals, hence you add no dominating reals; Cohen forcing is not $\omega$-closed but it never adds a dominating real
  • one can show that separability 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
  • Hechler forcing has size at most continuum 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
  • Hechler forcing also has the countable chain condition 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.

My question:

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?

  • $\begingroup$ Note that in all nontrivial cases separability just means that you are forcing equivalent to the forcing notion for adding a single Cohen real. $\endgroup$ – Stefan Geschke Nov 20 '10 at 22:34

Stefan's answer pointed me in the right direction, and then talking it over with prof. Leo Harrington we've got an answer:

A complete Boolean algebra $\mathbb{B}$ never adds a dominating real iff for any collection $\{ u _{m,k} : m, k \in \omega \} \subset \mathbb{B}^+$ the following weaker form of weak $(\omega ,\omega )$-distributivity holds:

$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{n \in \omega} \sum _{n < m < \omega} \sum _{k < f(m)} u _{m,k}$

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:

$(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k < f(m))(u _{m,k} \in G)$

which is to say that there's a real $f$ in the ground model such that:

$(\forall n)(\exists m > n)(\dot{g}^G (m) < f(m))$

so $f$ is not dominated by $\dot{g}$, contradiction.

For the forward implication, it should suffice to show it in the case where for each $m$, the set $\{ u _{m,k} : k \in \omega \}$ is an antichain with least upper bound $u$ independent of $m$ (I haven't checked this detail personally). So let $\{ u _{m,k}\}$ be such a collection for which the identity fails. Consider the name:

$\dot{g} = \{ (u _{m,k}, (m,k)) : m, k \in \omega \}$

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.

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:

$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} \sum _{k < f(m)} u _{m,k}$

Forcings that don't add any reals are precisely those that satisfy the following identity:

$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} u _{m,f(m)}$

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".

  • $\begingroup$ @Amit: Nice. Thanks for the update. $\endgroup$ – Andrés E. Caicedo Nov 23 '10 at 6:25
  • $\begingroup$ Andres, no problem! $\endgroup$ – Amit Kumar Gupta Nov 23 '10 at 22:51

Let me first say something about not always adding a dominating real (or anything else that can be formalized in the forcing language).

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 $p$.

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.

So, we are left with characterizing forcing notions that never add dominating reals respectively always add a dominating real.
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.
Natural places to check are the Bartoszynski-Judah book Set Theory of the Real Line and Zapletal's Forcing Idealized.

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.
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.
Examples are random real forcing (forcing with Borel sets of the real line of positive measure) and Sacks forcing.

  • $\begingroup$ Thanks for the pointer in the right direction, I brought it up with prof. Harrington and we figured it out. $\endgroup$ – Amit Kumar Gupta Nov 23 '10 at 3:25

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: http://shelah.logic.at/files/480.pdf

  • $\begingroup$ Am I assuming correctly that non-dominated real means unbounded real (not bounded by a ground model function)? $\endgroup$ – Stefan Geschke Nov 21 '10 at 18:57
  • $\begingroup$ Is the Suslin part known to be required? $\endgroup$ – Justin Palumbo Nov 21 '10 at 19:48
  • 1
    $\begingroup$ @Stefan: Yes, it means "there is no f in V dominating the new real". @Justin: The above result relies quite strongly on the absoluteness properties of Suslin forcing. I'm not aware of a similar result without any definability assumption. $\endgroup$ – Haim Nov 21 '10 at 21:58
  • $\begingroup$ @Justin: In Shelah's result the Souslin assumption is required. In [Definable sets of minimal degree, Yehoshua Bar-Hillel (ed.), Mathematical Logic and Foundations of Set Theory, Amsterdam 1970, 122–128], Jensen used $\lozenge$ to construct a ccc forcing adding (only) a minimal real. Since Cohen reals are far from being minimal, the forcing adds no Cohen real. An overview on this subject is in On Sacks Forcing and the Sacks Property <a href="hausdorff-center.uni-bonn.de/people/geschke/publications"> on my homepage</a>. $\endgroup$ – Stefan Geschke Nov 23 '10 at 8:40

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