# Characterizing “bounded” distributivity in terms of dense open sets

Prikry forcing is a well-known example of a forcing which adds an $$\omega$$-sequence to some cardinal $$\kappa$$, but does not add bounded subsets to $$\kappa$$.

There are other examples of forcings which add $$\omega$$-sequences, but do not add reals (for example Namba forcing in the presence of $$\sf CH$$).

In these situations the forcing is not $$\sigma$$-distributive, namely there is a countable sequence of dense open sets whose intersection is empty; but the added $$\omega$$-sequences live relatively "far up stairs" in terms of the von Neumann hierarchy.

Question. Is there a well-known characterizations of "$$\Bbb P$$ does not add new sets of rank $$\alpha$$" in terms of dense open sets and the intersection thereof?

• I guess that using properties of the Boolean completion of the forcing poset is not useful for this question, right? – Yair Hayut Nov 8 '18 at 13:55
• What do you mean? – Asaf Karagila Nov 8 '18 at 14:37
• Are you sure you mean "rank"? Because Namba adds new set of rank $\omega+8$. – Not Mike Dec 9 '18 at 8:59
• @NotMike: Yes, I'm pretty sure I mean rank. Also, since reals have rank $\omega$, I don't see how $\omega+8$ contradicts my statement. – Asaf Karagila Dec 9 '18 at 9:02
• @AsafKaragila There's no issue; just trying to clarify what you meant by "far up stairs"; since in the case of something like Namba you might have $L_{\gamma}[G] = L_{\gamma}$ ($\gamma \in (\omega_2)^{L}$) and $V_{\omega+8} \cap L[G] \neq V_{\omega+8} \cap L$ – Not Mike Dec 9 '18 at 9:57

This is just a comment on the question (but too long for a comment box), not an answer.

Let $$\kappa$$ and $$\lambda$$ be cardinals. A complete Boolean algebra $$\mathbb B$$ is called $$(\kappa,\lambda)$$-distributive if $$\prod_{\alpha < \kappa} \ \sum_{\beta < \lambda} u_{\alpha,\,\beta} \,=\, \sum_{f: \kappa \rightarrow \lambda} \ \prod_{\alpha < \kappa} u_{\alpha,\,f(\alpha)}$$ for any $$u_{\alpha,\,\beta}$$ in $$\mathbb B$$. In Jech's set theory book (third edition, Theorem 15.38, p. 246) it is proved that

$$\mathbb B$$ is $$(\kappa,\lambda)$$-distributive if and only if every $$f: \kappa \rightarrow \lambda$$ in the generic extension by $$\mathbb B$$ is already in the ground model.

So for example, $$\mathbb B$$ does not add new reals if and only if it is $$(\aleph_0,2)$$-distributive, and (the Boolean completion of) the Prikry forcing is a good example of an algebra that is $$(\aleph_0,2)$$-distributive but not $$\aleph_0$$-distributive. (It adds new $$\omega$$-sequences, but not new reals.)

Using the highlighted theorem above, we can deduce that

$$\mathbb B$$ adds a set of rank $$\leq\!\alpha$$ if and only if it fails to be $$(|V_\alpha|,2)$$-distributive.

The reason is that a set of rank $$\leq\!\alpha$$ is just a subset of $$V_\alpha$$, and can be identified with a characteristic function $$V_\alpha \rightarrow 2$$. So getting new sets of rank $$\leq\!\alpha$$ is equivalent to getting new functions of this kind.

Of course, I'm not sure this directly answers your question (because $$(\kappa,\lambda)$$-distributivity is not defined in terms of dense open sets). But here is something a little closer: $$\mathbb B$$ is $$(\kappa,\lambda)$$-distributive if and only if every collection of $$\kappa$$ partitions of $$\mathbb B$$, each containing at most $$\lambda$$ members of $$\mathbb B$$, has a common refinement. (This is Lemma 15.37 from Jech.)

• Right. I am aware of these things that you mention. But they are not given in terms of open sets, and ultimately I want to apply them (also) in ZF, where chain conditions are meaningless (so the common refinement argument is somewhat irrelevant, or at least much harder to appeal to). – Asaf Karagila Nov 8 '18 at 14:37
• OK, that's fair enough -- I'll leave this here in case anyone finds it helpful, but you're right that it's not really an answer. – Will Brian Nov 8 '18 at 14:38
• I wonder if you can get a version that makes sense with $\neg\text{AC}$ by using pre-dense sets, instead of antichains. Something like: the intersection of $\kappa$ many dense sets, each having a pre-dense subset of size $\lambda$, is dense. – Joel David Hamkins Jan 10 at 14:11

This is just a translation of @WillBrian's answer into a statement about dense sets.

For each $$\delta \in \mathsf{On}$$, fix $$\dot{R}_\delta \in V^{\mathbb{P}}$$ of minimal rank, such that $$1 \Vdash_\mathbb{P} \dot{R}_\delta = \{ x : \mathsf{rank}(x) < \check{\delta}\}$$. Then, because of how the rank function works, we always have that the least $$\delta \in \mathsf{On}$$, such that $$1 \Vdash_{\mathbb{P}} \dot{R}_\delta \neq \check{V}_\delta$$ is a successor; so the thing to do is characterize when $$1 \Vdash_{\mathbb{P}} \dot{R}_{\gamma + 1} = \check{V}_{\gamma + 1}$$. To this end,

Theorem: For every $$\delta \in \mathsf{On}$$ such that $$1 \Vdash_{\mathbb{P}} \dot{R}_\delta = \check{V}_\delta$$, the following are equivalent,

1. $$1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$$,

2. For every $$p \in \mathbb{P}$$ and every $$\{ \mathcal{B}_x: x \in V_{\delta} \} \subset \mathcal{P}(\mathbb{P})$$, there is some $$p_0 \le p$$, such that, for any $$x \in V_{\delta}$$, either $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$ is dense below $$p_0$$.

Proof: Fix $$\delta \in \mathsf{On}$$, and assume that $$1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$$. Then, for every $$p\in \mathbb{P}$$ and $$\dot{g} \in V^{\mathbb{P}}$$ such that $$p \Vdash_{\mathbb{P}} \dot{g}:\check{V}_\delta \rightarrow \check{2}$$, the set $$\{ q \le p : (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} ) \}$$ is dense. So let $$\{ \mathcal{B}_x : x \in V_\delta\} \subset \mathcal{P}(\mathbb{P})$$ be arbitrary, and define $$\dot{g},\sigma_x, \tau_x \in V^{\mathbb{P}}$$ ($$x\in V_\delta$$) as follows $$\dot{g} = \{ \langle\sigma_x, 1\rangle: x\in V_\delta \}$$, $$\sigma_x = \mathsf{pair}_\mathbb{P}(\check{x}, \tau_x)$$, and $$\tau_x = \{ \langle \emptyset, p \rangle: p \in \mathcal{B}_x \}$$. Then, $$1 \Vdash_{\mathbb{P}} \dot{g}: \check{V}_\delta \rightarrow \check{2}$$ and so by assumption the set $$\{ q \in \mathbb{P}: (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} )\}$$ is dense open, hence given $$p \in \mathbb{P}$$ we can find $$p_0 \le p$$ and $$f:V_\delta \rightarrow 2$$ such that, for every $$x \in V_\delta$$, $$p_0 \Vdash \dot{g}(\check{x}) = \check{f}(\check{x})$$ or equivalently, $$p_0 \Vdash_{\mathbb{P}} "\tau_x = \check{f}(\check{x})" \iff$$ $$f(x) = 1$$ and $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$$ is dense below $$p_0$$, or $$f(x)=0$$ and $$\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$$ is dense below $$p_0$$.

For the converse, assume (2), then fix $$p\in\mathbb{P}$$ and $$\dot{a}\in V^{\mathbb{P}}$$ such that $$p \Vdash_\mathbb{P} \dot{a} \in \dot{R}_{\delta+1}$$. Then, for $$x\in V_\delta$$ defining $$\mathcal{B}_x = \{ q \in \mathbb{P}: q\Vdash_{\mathbb{P}} \check{x} \in \dot{a} \}$$, we can find $$p_0 \le p$$ (by 2) such that for every $$x$$, either $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$ is dense below $$p_0$$. As such, letting $$A = \{ x \in V_\delta : \{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\} \text{ is dense below } p_0 \}$$ we have $$p_0 \Vdash \dot{a} = \check{A} \in \check{V}_{\delta+1}$$. $$\square$$

Remark: If in (2), the requirement "either $$A^x_1=\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$$, or $$A^x_0=\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$$ is dense below $$p_0$$", is changed to "$$p_0$$ is an element of $$A^x_0$$ or $$A^x_1$$", then the new statement is equivalent to $$\mathbb{P}$$ being $$\vert V_\delta \vert$$-baire.