One of the first forcing facts that we learn is that -closed forcings preserve all sequences of length . This is usually shown via distributivity, by showing that every -closed forcing is also -distributive, and that -distributivity is equivalent to the forcing not adding any new sequences of length . I will recall these facts here, and show how they relate to both and . Here is the axiom of choices, stating that we have choice functions for all sets injecting into , and is the axiom of dependent choices, saying that every pruned tree of height at most has a branch.
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 -closed and -distributive are referring to sequences of length . Others might denote this by -closed and -distributive, respectively.
Definition. Let be an infinite cardinal and a forcing. Then is…
- -closed if every -descending chain of conditions has a lower bound ; i.e., holds for every .
- -distributive if the intersection of every -sized collection of open dense subsets of is again open dense.
The following is the well-known reason for why we care about distributivity.
Lemma 1 (ZF). Let be an infinite cardinal and a -distributive forcing. Then doesn’t add any new -sequences.
Proof. Let be -generic and fix any such that ; say for a set . Fix forcing that and define, for each ,
all of which are open dense, so -distributivity of implies that is open dense as well; let . Since we get for each a unique with . We can therefore define, in , as the function given as , so that . QED
In many circumstances we’d simply show that our forcing of choice is -closed however, when we want to show that it doesn’t add any new -sequences. But this implication turns out to require some amount of choice, which is the following folklore result:
Proposition (ZF). Let be an infinite cardinal. Then the following are equivalent:
- Every -closed forcing is -distributive
- Every -closed forcing does not add any new -sequences.
Proof. Lemma 1 above gives us .
: Let be -closed and fix a -sequence of open dense subsets of . Let and use to find a -sequence with and for every . Since is -closed we get a with , showing -distributivity.
: Assume fails, so that there’s a pruned tree of height without any branch. As a forcing, is then trivially -closed. But if is any generic then is a branch through and therefore contains a new -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 (). Let be an infinite cardinal and a separative forcing. Then is -distributive iff it doesn’t add any new -sequences.
Proof. is just as above, so we show : Let be a sequence of open dense subsets of and fix some ; we have to show that there’s a with . Let be a -generic filter with . In we may then form the sequence
all of which are nonempty. Now use to pick a sequence of conditions . By assumption this sequence also belongs to so that we can then define, in , the set
which is dense below by separativity of , so pick . Since for every , as they’re all elements of , we get that for every . Since was arbitrary we get that is dense, showing -distributivity. QED
Note that we need the separativity assumption, as otherwise we could let and , which doesn’t add any new sequences (or any sets at all, in fact), but it isn’t -distributive.
If we assume some stronger condition on our forcing we can get the equivalence in ZF, however:
Lemma 3 (ZF). Let be an infinite cardinal and a separative tree forcing; i.e., where every compatible pair is also comparable. Then is -distributive iff it doesn’t add any new -sequences.
Proof. This is exactly like the proof of Lemma 2, but we have to construct the sequence of ‘s instead of relying on choice. Again we have, in , the sequence
all of which are nonempty. Note that every pair of is compatible, being elements of the filter, so they are comparable as well as is a tree forcing. This makes a branch, which is then wellordered by ‘s ordering . Define the sequence as
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 be an infinite cardinal and a separative forcing which doesn’t add any new -sequences. Is then -distributive?