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?

  • $\begingroup$ I guess that using properties of the Boolean completion of the forcing poset is not useful for this question, right? $\endgroup$ – Yair Hayut Nov 8 '18 at 13:55
  • $\begingroup$ What do you mean? $\endgroup$ – Asaf Karagila Nov 8 '18 at 14:37
  • $\begingroup$ Are you sure you mean "rank"? Because Namba adds new set of rank $\omega+8$. $\endgroup$ – Not Mike Dec 9 '18 at 8:59
  • $\begingroup$ @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. $\endgroup$ – Asaf Karagila Dec 9 '18 at 9:02
  • $\begingroup$ @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 $V_{\omega+8} \cap L[G] \neq V_{\omega+8} \cap L$ $\endgroup$ – 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.)

  • 1
    $\begingroup$ 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). $\endgroup$ – Asaf Karagila Nov 8 '18 at 14:37
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    $\begingroup$ 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. $\endgroup$ – Will Brian Nov 8 '18 at 14:38
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    $\begingroup$ 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. $\endgroup$ – 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.


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