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For the notion of distributivity of forcings, we have equivalent defintions, one combintorial, the other in terms of what the generic extensions look like. For a partial order $\mathbb{P}$, the following are equivalent:

1) Forcing with $\mathbb{P}$ adds no new functions from ordinals to ordinals with domain $\kappa$.

2) The intersection of $\kappa$ many dense open subsets of $\mathbb{P}$ is dense.

$\kappa$-closure of partial orders is definitely not a forcing invariant, as witnessed just by the fact that an atomless complete boolean algebra is never countably closed. However, we can ask is there a characterization partial orders $\mathbb{P}$ in terms of what happens in their generic extensions that is equivalent to the following: The boolean completion of $\mathbb{P}$ has a $\kappa$-closed dense subset?

EDIT: Changed the question in light of the example mentioned by Dorais.

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  • $\begingroup$ It seems to me that having a $\kappa$-closed dense subset automatically makes you forcing equivalent to a $\kappa$-closed partial order, and in the other direction being forcing equivalent to a $\kappa$-closed partial order means you must have a $\kappa$-closed dense subset; so this stronger property is basically just $\kappa$-closure. Is this correct? $\endgroup$
    – Noah Schweber
    Commented Sep 19, 2012 at 1:45
  • $\begingroup$ Or rather, this stronger property is just "is forcing equivalent to a $\kappa$-closed poset." $\endgroup$
    – Noah Schweber
    Commented Sep 19, 2012 at 4:04
  • $\begingroup$ Noah, yes that's right. But to me this is just a different way of stating a combinatorial property of the forcing. $\endgroup$
    – Monroe Eskew
    Commented Sep 19, 2012 at 5:03
  • $\begingroup$ Sorry, maybe I didn't understand your question. Isn't theorem VII 6.14 in Kunen's book (consequence of lambda-closedness) an answer to your question? Also, how does forcing invariants (in Francois' answer) answer your question? $\endgroup$
    – Eran
    Commented Sep 19, 2012 at 11:25
  • $\begingroup$ Eran, the theorem you mention just shows that closure implies distributivity. What I'm looking for is some property of generic extensions that implies closure. Francois' answer shows that there can be equivalent forcing notions which differ on the property of having a dense countably closed subset, thus the implication I'm looking for cannot be found. However this doesn't really get at the issue I am interested, so I will rephrase my question. $\endgroup$
    – Monroe Eskew
    Commented Sep 19, 2012 at 16:56

2 Answers 2

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Actually, "having a $\kappa$-closed dense subset" is not (necessarily) a forcing invariant. In his paper On the existence of a $\sigma$-closed dense subset [Comment. Math. Univ. Carolin. 51 (2010), 513-517; MR2741884], Jindra Zapletal shows that it is relatively consistent with ZFC that there is a partial order $(P \cup Q,{\leq})$ in which $P$ and $Q$ are both dense, $(P,{\leq})$ is $\omega_1$-closed but $(Q,{\leq})$ has no $\omega_1$-closed dense subset. It is an open problem whether such an example provably exists in ZFC.

This example shows that a nice characterization is unlikely in general. However, Matt Foreman [J. Symbolic Logic 48 (1983), 714-723; MR0716633] and Petr Vojtáš [Comment. Math. Univ. Carolin. 24 (1983), no. 2, 349–369; MR0711272] have found some useful partial results.

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  • $\begingroup$ However, a dense embedding from one seperative partial order to another implies that their respective boolean completions are isomorphic. (right?) For this question I am more interested in combinatorial properties of the boolean completion (because it's the canonical representiative of the forcing-equivalence class) vs. things satisfied by the generic extension. I'm not sure if I can give a rigorous definition of what I'm looking for, but the equivalent formulations of distributivity properties, plus things like weak $(\lambda,\kappa)$-saturation seem to capture the spirit. $\endgroup$
    – Monroe Eskew
    Commented Sep 19, 2012 at 6:07
  • $\begingroup$ The papers by Foreman and Vojtáš have what you want but the results are partial. Their results are also optimal, in some sense, because of Zapletal's example. Zapletal's example shows that it is generally hard to tell which complete Boolean algebras have a $\kappa$-closed dense subset. $\endgroup$
    – François G. Dorais
    Commented Sep 19, 2012 at 7:24
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Jensen showed the boolean completion of $\mathbb P$ has a countably closed dense set if and only if $\mathbb P$ satisfies the following property: For all regular $\theta > | \mathbb P|$, there is a club of countable $M \prec H_\theta$ such that every $(M,\mathbb P)$-generic filter $g$ has a lower bound in $\mathbb P$.

See: MR2840749 Jensen, Ronald . Subcomplete forcing and ℒ-forcing. E-recursion, forcing and C∗-algebras, 83--182, Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap., 27, World Sci. Publ., Hackensack, NJ, 2014.

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