- Go to previous talk.
(At Randall’s request, this entry will be more detailed than usual.)
Remark 1. is club in , so any is stationary as a subset of iff it is stationary as a subset of . It follows that proper forcing preserves stationary subsets of .
Remark 2. Proper forcing extensions satisfy the countable covering property with respect to , namely, if is proper, then any countable set of ordinals in is contained in a countable set of ordinals in . We won’t prove this for now, but the argument for Axiom A posets resembles the general case reasonably:
Given a name for a countable set of ordinals in the extension, find an appropriate regular and consider a countable elementary containing , , and any other relevant parameters. One can then produce a sequence such that
- Each is in .
- , where enumerates the dense subsets of in .
Let for all . Then , so is a countable set of ordinals in containing in . A density argument completes the proof.
Woodin calls a poset weakly proper if the countable covering property holds between and . Not every weakly proper forcing is proper, but the countable covering property is a very useful consequence of properness. On the other hand, standard examples of stationary set preserving posets that are not proper, like Prikry forcing (changing the cofinality of a measurable cardinal to without adding bounded subsets of ) or Namba forcing (changing the cofinality of to while preserving are not weakly proper, and account for some of the usefulness of over .
The following is obvious:
Fact. Assume is weakly proper. Then either adds no new -sequences of ordinals, or else it adds a real.
The relation between the reals and the -sequences of ordinals in the presence of strong forcing axioms like is a common theme I am exploring through these talks.
To motivate the main line of results (relating models of forcing axioms to their inner models) I want to discuss some preservation results.
The first is fairly easy.
Fact. is preserved by any proper forcing that does not add subsets of .
For if holds, is such a poset, and is a -name for a proper poset, then and is proper, so .
Trivial as this fact is, removing the assumption of properness of makes matters much more subtle. Closely related to this is the following question:
Open question 1. Assume is a forcing extension of and holds. Does ?
Moore has shown that implies that is a model of choice, so in the situation above, it is a forcing extension of . Even if is an extension of by proper forcing (even if is an extension of by the standard iteration forcing ), I do not see that it is an extension of by proper forcing, so the obvious argument above does not seem to apply.
Following with the theme that is perhaps the most natural strong forcing axiom to study, rather than proving the König-Yoshinobu theorem, I want to prove the following result:
Theorem (Larson). is preserved by -directed closed forcing.
Definition. A poset is -directed closed iff any directed (i.e., for any there is with and ) admits a lower bound in .
Clearly, any such forcing is -closed, which requires to be a decreasing sequence, but being directed closed is more restrictive.
Proof. Assume and let be -directed close. Let be a -name for a stationary set preserving forcing. Then is stationary set preserving as well. Fix also a -name for a sequence of many dense subsets of .
For , let and let be the first-coordinate projection of .
By , there is a filter meeting all the . Let be its first coordinate projection. is a directed subset of , and we can find a directed subset of size at most meeting all the . Since is -directed closed, we can find a lower bound for . Thus,
Remark 3. König and Yoshinobu have shown that for any there is a -closed forcing that destroys . Their argument generalizes the fact that under there are no weak Kurepa trees and in fact every tree of height and size is special; we will review this result later. It also makes essential use of Namba forcing.
Remark 4. The above remark indicates that in a sense the König-Yoshinobu preservation theorem is as strong as possible. There is another sense in which it cannot be strengthened either: Given an ordinal , recall that a poset is strongly -game closed or -strategically closed iff Nonempty has a winning strategy for the game where Empty and Nonempty alternate playing conditions in for stages and Nonempty wins iff there is a lower bound to the sequence produced this way. This is a weaker notion than -closed, and a useful substitute in many instances. However, is not preserved by strongly -game closed forcing. For example, can be added with such a forcing.
Remark 5. Similarly, one can add a -sequence and then destroy it and the whole extension adds no new -sequences of ordinals and preserves , but fails in the intermediate extension, so preservation is a subtler matter and certainly does not depend on the preservation of alone.
Rather than proving the König-Yoshinobu preservation theorem, let me give a qick sketch, emphasizing that the argument is very similar to the proof of Larson’s result given above. Now we consider a -closed poset , a -name for a proper poset, and a -name for a sequence of dense subsets of (the interpretation of) . As in Larson’s proof we can find a directed of size at most meeting the projections of the corresponding dense sets . However, the assumption on does not suffice to pick up a lower bound of . Rather, a new poset (in the extension ) is considered, that adds with countable conditions a decreasing -sequence through . The countable covering property of proper posets is used to see that is -closed. A further use of allows us then to find an -sequence through such that for each there is some with . The closure of now allows us to pick a lower bound for this sequence, from which the proof is concluded as before.
Not exactly a preservation result, but in the same spirit, I want to conclude by mentioning a result of Todorcevic that indicates that considering -sequences of ordinals or at least subsets of is not completely arbitrary in the setting of .
Theorem (Todorcevic). Assume and let be a poset that adds a subset of . Then either adds a reals, or else it collapses .
During the talk I did not finish the proof of this result, which I will continue in the next meeting. The argument is not too complicated, but requires a good understanding of trees on under forcing axioms. I now proceed to the beginning of this analysis.
Definition. Let be a tree of height and size . Then is special (in the restricted sense) iff is a countable union of antichains.
I won’t prove this, but the above is equivalent to the existence of an order preserving embedding of the tree into the real (or even the rational) numbers.
Notice that if is special, then it has no uncountable branches in any outer model, since any uncountable subset of must meet one of the antichains in at least two points. (For the other version, a branch would induce a subset of of order type or , which is impossible.)
Special trees play a key role in Todorcevic’s argument. One begins by showing the following, with which I concluded this lecture:
Theorem. Assume . Then every tree of height and size without uncountable branches is special.
Proof. Let be such a tree. We show that there is a ccc poset that specializes . Begin with , the collection of finite antichains of . We will argue that this is ccc. Now consider the product with finite support of countably many copies of , call it . By Martin’s axiom, this poset is ccc as well. Considering the dense sets for , it follows from Martin’s axiom that there is a partition of into countably many antichains, as required.
To see that is ccc, fix a uniform ultrafilter on and assume otherwise, so there is an uncountable antichain through . Each is a finite antichain through . That they constitute an antichain in means that for any one of the elements of is comparable to one of the elements of . By the -system lemma, we may assume all the have the same size , and are pairwise disjoint. Write .
Set for , . For each , all but countably many countable ordinals are in some , so there are such that . We can fix an uncountable set and numbers such that for all .
Consider , both in . Then there is some in , since in fact this is a set in . But then both and are below so they are in fact comparable. It follows that is linearly ordered, so it generates a branch through . This is a contradiction, and the proof is complete.
Remark 6. One can easily modify the argument above to prove directly (without appealing to ) that is ccc. Hence, the only appeal to comes via arguing that there is already a splitting of into countably many antichains in the ground model.