Closure, distributivity and choice

One of the first forcing facts that we learn is that \kappa-closed forcings preserve all sequences of length \kappa. This is usually shown via distributivity, by showing that every \kappa-closed forcing is also \kappa-distributive, and that \kappa-distributivity is equivalent to the forcing not adding any new sequences of length \kappa. I will recall these facts here, and show how they relate to both \textsf{AC}_\kappa and \textsf{DC}_\kappa. Here \textsf{AC}_\kappa is the axiom of \kappa choices, stating that we have choice functions for all sets injecting into \kappa, and \textsf{DC}_\kappa is the axiom of \kappa dependent choices, saying that every pruned tree of height at most \kappa has a branch.

Painting by Julie Bond

Let’s start off by recalling a few definitions of the terms I just mentioned. I’ll be slightly unorthodox here and follow the convention in both Jech (1999) and Schindler (2014) that \kappa-closed and \kappa-distributive are referring to sequences of length \kappa. Others might denote this by \kappa^+-closed and \kappa^+-distributive, respectively.

Definition. Let \kappa be an infinite cardinal and \mathbb P a forcing. Then \mathbb P is…

  • \kappa-closed if every \mathbb P-descending chain \langle p_\alpha\mid\alpha<\kappa\rangle of conditions p_\alpha\in\mathbb P has a lower bound q\in\mathbb P; i.e., q\leq p_\alpha holds for every \alpha<\kappa.
  • \kappa-distributive if the intersection of every \kappa-sized collection of open dense subsets of \mathbb P is again open dense.

The following is the well-known reason for why we care about distributivity.

Lemma 1 (ZF).  Let \kappa be an infinite cardinal and \mathbb P a \kappa-distributive forcing. Then \mathbb P doesn’t add any new \kappa-sequences.

Proof. Let G\subset\mathbb P be V-generic and fix any f\in V[G] such that f:\kappa\to V; say f:\kappa\to x for a set x\in V. Fix p\in g forcing that \dot{f}:\check\kappa\to\check x and define, for each \xi<\kappa,

D_\xi:=\{q\in\mathbb P\mid q\perp p\lor\exists y\in x (q\Vdash\dot{f}(\check\xi)=\check y)\},

all of which are open dense, so \kappa-distributivity of \mathbb P implies that D:=\bigcap_{\xi<\kappa}D_\xi is open dense as well; let q\in G\cap D. Since q\ ||\ p we get for each \xi<\kappa a unique y\in x with q\Vdash\dot{f}(\check\xi)=\check y. We can therefore define, in V, f as the function g:\kappa\to x given as g(\xi)=y\quad\text{iff}\quad q\Vdash\dot{f}(\check\xi)=\check y, so that f\in VQED

In many circumstances we’d simply show that our forcing of choice is \kappa-closed however, when we want to show that it doesn’t add any new \kappa-sequences. But this implication turns out to require some amount of choice, which is the following folklore result:

Proposition (ZF). Let \kappa be an infinite cardinal. Then the following are equivalent:

  1. \textsf{DC}_\kappa
  2. Every \kappa-closed forcing is \kappa-distributive
  3. Every \kappa-closed forcing does not add any new \kappa-sequences.

Proof. Lemma 1 above gives us (2)\Rightarrow (3).

(1)\Rightarrow (2): Let \mathbb P be \kappa-closed and fix a \kappa-sequence \langle D_\alpha\mid\alpha<\kappa\rangle of open dense subsets of \mathbb P. Let p\in\mathbb P and use \textsf{DC}_\kappa to find a \kappa-sequence \langle p_\alpha\mid\alpha<\kappa\rangle with p_\alpha\leq p and p_\alpha\in D_\alpha for every \alpha<\kappa. Since \mathbb P is \kappa-closed we get a q\leq p with q\in\bigcap_{\alpha<\kappa}D_\alpha, showing \kappa-distributivity.

(3)\Rightarrow (1): Assume \textsf{DC}_\kappa fails, so that there’s a pruned tree T of height \kappa without any branch. As a forcing, T is then trivially \kappa-closed. But if G\subset T is any generic then \bigcup G is a branch through T and therefore V[G] contains a new \kappa-sequence. QED

Okay, so it seems that there’s some nontrivial stuff going on between the distributivity and the closure, and the proposition seems to indicate that we also get the converse of Lemma 1 in ZF. This at least holds if we’re assuming separativity and some choice:

Lemma 2 (\textsf{ZF}+\textsf{AC}_\kappa). Let \kappa be an infinite cardinal and \mathbb P a separative forcing. Then \mathbb P is \kappa-distributive iff it doesn’t add any new \kappa-sequences.

Proof. (\Rightarrow) is just as above, so we show (\Leftarrow): Let \langle D_\alpha\mid\alpha<\kappa\rangle be a sequence of open dense subsets of \mathbb P and fix some p_0\in\mathbb P; we have to show that there’s a q\leq p with q\in\bigcap_{\alpha<\kappa}D_\alpha. Let G\subseteq\mathbb P be a V-generic filter with p_0\in G. In V[G] we may then form the sequence

\langle G\cap D_\alpha\mid\alpha<\kappa\rangle,

all of which are nonempty. Now use \textsf{AC}_\kappa to pick a sequence \langle q_\alpha\mid\alpha<\kappa\rangle of conditions q_\alpha\in G\cap D_\alpha. By assumption this sequence also belongs to V so that we can then define, in V, the set

D:=\{p\leq p_0\mid\exists\alpha<\kappa(p\perp q_\alpha)\lor\forall\alpha<\kappa(p\leq q_\alpha)\}\in V,

which is dense below p_0 by separativity of \mathbb P, so pick q\in G\cap D. Since q\ ||\ q_\alpha for every \alpha<\kappa, as they’re all elements of G, we get that q\leq q_\alpha for every \alpha<\kappa. Since p_0 was arbitrary we get that \bigcap_{\alpha<\kappa}D_\alpha is dense, showing \kappa-distributivity. QED

Note that we need the separativity assumption, as otherwise we could let \kappa=\omega and \mathbb P=(\omega,\ni), which doesn’t add any new sequences (or any sets at all, in fact), but it isn’t \omega-distributive.

If we assume some stronger condition on our forcing we can get the equivalence in ZF, however:

Lemma 3 (ZF). Let \kappa be an infinite cardinal and \mathbb P a separative tree forcing; i.e., where every compatible pair is also comparable. Then \mathbb P is \kappa-distributive iff it doesn’t add any new \kappa-sequences.

Proof. This is exactly like the proof of Lemma 2, but we have to construct the sequence of q_\alpha‘s instead of relying on choice. Again we have, in V[G], the sequence

\langle G\cap D_\alpha\mid\alpha<\kappa\rangle,

all of which are nonempty. Note that every pair of \bigcup_{\alpha<\kappa}G\cap D_\alpha is compatible, being elements of the filter, so they are comparable as well as \mathbb P is a tree forcing. This makes b:=\bigcup_{\alpha<\kappa}G\cap D_\alpha a branch, which is then wellordered by \mathbb P‘s ordering \leq_{\mathbb P}. Define the sequence \langle q_\alpha\mid\alpha<\kappa\rangle\in V[G] as

q_\alpha:=\text{min}_{\leq_{\mathbb P}}(b\cap D_\alpha),

and finish the proof as in Lemma 2. QED

The general question of whether the converse of Lemma 1 holds in ZF seems to be open though, so let’s state it here for convenience:

Open problem? (ZF) Let \kappa be an infinite cardinal and \mathbb P a separative forcing which doesn’t add any new \kappa-sequences. Is \mathbb P then \kappa-distributive?

Update: This might have a positive answer, by @PBL05240969‘s argument in this Twitter thread.