Successive cardinals with the tree property

Transcription

1 UCLA March 3, 2014

2 A classical theorem Theorem (König Infinity Lemma) Every infinite finitely branching tree has an infinite path.

3 Definitions A tree is set T together with an ordering < T which is wellfounded, transitive, irreflexive and such that for all t T the set {x T x < T t} is linearly ordered by < T. The height of an element t is the order-type of the collection of the predecessors of t under < T. That is, the unique ordinal α such that (α, ) ({x T x < T t}, < T ). The α th level of the tree is the collection of nodes of height α. The height of a tree T is the least ordinal β such that there are no nodes of height β. A set b is a cofinal branch through T if b T and (b, < T ) is a linear order whose order-type is the height of the tree.

4 The tree property Theorem (König Infinity Lemma) Every tree of height ω with finite levels has a cofinal branch Let κ be a regular cardinal. Definition A κ-tree is a tree of height κ with levels of size less than κ. Definition A cardinal κ has the tree property if every κ-tree has a cofinal branch. A counterexample to the tree property at κ is called a κ-aronszajn tree.

5 When do Aronszajn trees exist? Theorem (Aronszajn) There is a tree of height ω 1 all of whose levels are countable, which has no cofinal branch. Theorem (Specker) If κ <κ = κ, then there is a κ + -Aronszajn tree. In particular CH implies that there is an ω 2 -Aronszajn tree. Remark The tree constructed is special in the sense that there is a function from T to κ such that f (s) f (t) whenever s < T t.

6 The tree property and large cardinals Definition A uncountable cardinal κ is inaccessible if it is a regular limit cardinal and for all µ < κ, 2 µ < κ. Definition A cardinal κ is weakly compact if κ is uncountable and for all f : [κ] 2 2, there is H κ of size κ such that f is constant on [H] 2. Theorem (Tarski and Keisler) κ is weakly compact if and only if it is inaccessible and has the tree property.

7 What about the tree property at non-inaccessible cardinals? Theorem (Mitchell) The theory ZFC + there is a weakly compact cardinal is consistent if and only if the theory ZFC + ω 2 has the tree property is consistent.

8 Measurable Cardinals Definition A cardinal κ is measurable if there is a transitive class N and an elementary embedding j : V N with critical point κ. Fact κ is measurable implies κ has the tree property. Proof. Let T be a κ-tree and assume that the underlying set of T is κ. Let j witness that κ is measurable. j(t ) is a tree of height j(κ) and j(t ) κ = T. In N choose a point on level κ of j(t ). This point determines a branch through T.

9 Mitchell s Forcing Let λ be a measurable cardinal. We define a poset M as follows: Conditions are pairs (p, f ) such that p P = Add(ω, λ) f is a countable partial function with dom(f ) λ, for all α dom(f ), f (α) is a P α-name for an element of Add(ω 1, 1) V [P α] We set (p 1, f 1 ) (p 2, f 2 ) if and only if p 1 p 2, dom(f 1 ) dom(f 2 ) and for all α dom(f 2 ), p 1 α f 1 (α) f 2 (α).

10 Properties of Mitchell s forcing M is λ-cc M adds a generic for P and P α Add(ω 1, 1) V [P α] for many α. In particular in an extension by M, 2 ω = λ = ω 2. {(p, f ) M p = } ordered as a suborder of M is countably closed. Call this forcing Q. The forcing P Q induces a generic for M in a natural way. In an extension by M α the forcing M/M α is equivalent to a forcing M defined in a similar fashion to M. In particular there are P and Q analogs of P and Q such that P Q generates a generic for M. P is simply a quotient of P over its initial segment.

11 Mitchell continued Claim ω 2 has the tree property in V [M]. Proof. Let j : V M witness that λ is measurable in V. Let T V [M] be an ω 2 Aronszajn tree. Lift the elementary embedding to j : V [M] M[j(M)]. By reflection properties of j, there is an inaccessible α < κ such that T α V [M α] and V [M α] T α is an ω 2 Aronszajn tree. To obtain a contradiction it is enough to show that forcing with Q P over V [M α] cannot add a branch through T α. This follows from standard facts about forcings which do not add branches.

12 Questions Question Is it consistent that all regular cardinals greater than ℵ 1 have the tree property? This question is very hard. So in pursuit of a positive answer we ask: Question What is the largest initial segment of regular cardinals which can have the tree property?

13 Theorem (Abraham) If there is a supercompact cardinal with a weakly compact cardinal above it, then there is a forcing extension in which ℵ 2 and ℵ 3 have the tree property simultaneously. Theorem (Cummings and Foreman) If there are infinitely many supercompact cardinals, then there is a forcing extension in which simultaneously for all n 2, ℵ n has the tree property. Theorem (Neeman) If there are infinitely many supercompact cardinals then there is a forcing extension in which all regular cardinals in the interval [ℵ 2, ℵ ω+1 ] have the tree property.

14 A few notes on the proofs The proofs all have a similar flavor and are quite technical. The original models of Abraham, and Cummings and Foreman are built in a iterative fashion. First establish the tree property at ℵ 2, then at ℵ 3 etc. The key point is to establish the tree property in an indestructible way. There are two main types of indestructibility required.

15 Directed closed forcing The main tool for this is iteration with a prediction mechanism. Theorem (Laver) Assuming there is a supercompact cardinal κ, there is a forcing extension in which the supercompactness of κ is indestructible under κ-directed closed forcing. Going back to Mitchell s construction it is relatively straightforward to fold in this kind of iteration with prediction. Theorem Assuming there is a supercompact cardinal, there is a forcing extension in which ℵ 2 has the tree property and the tree property is indestructible under ℵ 2 -directed closed forcing. We cannot apply this theorem directly in the proof of say Abraham s theorem, but it captures one of the main ideas of the proof.

16 A recipe for indestructibility Let κ be supercompact and F be a Laver function for κ. Let P be a poset such that there are posets P α for α < κ such that in a given ultrapower P is the class of α P α and there are projections π α : P P α. Let G be P generic. In V [G] define a diagonal Easton support iteration as follows. The definition is by induction on α < κ. Suppose we have defined U α for some α. We allow α into the domain of conditions when Pα U α F (α) is α-directed closed. In this case we allow our partial functions to return P α U α-names for elements of F (α).

17 Recipe continued The ordering is given by u u if and only if dom(u) dom(u ) and there is p G such that for all α dom(u ), (π α (p), u α) u(α) u (α). Remark To obtain a model where the tree property at ℵ 2 is indestructible under ℵ 2 -directed closed forcing, we use this construction with M in place of P and M α in place of P α.

18 More indestructibility Also important in the Abraham, and Cummings and Foreman results are the ideas need to prove the following theorem: Theorem The tree property at ℵ 2 in Mitchell s model is indestructible under adding an arbitrary number of Cohen subsets of ω 1 with a poset from the ground model. The proof of this fact goes by repeating the proof of Mitchell s theorem in the presence of this extra forcing.

19 Preservation lemmas The method of proof requires us to force further to see that we get branches through all trees. We then argue that this further forcing could not have added these branches. Lemma (Silver) If R is χ + -closed and 2 χ η, then forcing with R cannot add a branch through an η-tree. Lemma (U) If P is χ + -cc, R is χ + -closed and 2 χ η, then forcing with R over V P cannot add a cofinal branch through an η-tree.

20 Preservation Lemmas continued Lemma (Kunen and Tall) If P is τ-knaster, then P cannot add a branch through a branchless tree of height τ. Lemma (U) Suppose that P P is τ-cc. Forcing with P cannot add a cofinal branch through a tree of height τ.

21 How do you add ℵ ω+1? The original proof of the consistency of the tree property at ℵ ω+1 is due to Magidor and Shelah and it uses a little more than a huge cardinal. This upperbound was improved by Sinapova to only ω supercompact cardinals using a Prikry forcing construction. Neeman was able to obtain the same result as Sinapova, but only using a product of Levy Collapses. The key point in all of these theorems is the selection of the cardinal that will become ω 1. Neeman s theorem getting the consistency of the tree property at every regular cardinal in the interval [ℵ 2, ℵ ω+1 ] proceeds by trying to combine ideas from the Cummings and Foreman constrution with ideas from the proof with the product of Levy Collapses. Ideas of indestructibility play a role in the proof, but their presence is less obvious because the forcing has become more complex.

22 Scales Scales are a notion from Shelah s PCF theory. The general setting is a singular cardinal ν with an increasing and cofinal sequence ν i i < cf(ν) of regular cardinals less than ν. Given members f, g i ν i we say that f < g if and only if there is a j < cf(ν) such that for all i j, f (i) < g(i). A sequence of functions f α α < ν + is a scale of length ν + in i ν i if it is increasing and cofinal in i ν i under the ordering <. A point γ < ν + with cf(γ) > cf(ν) is good for a scale f of length ν + if there are A γ cofinal and j < cf(ν) such that for all i j the sequence f α (i) α A is strictly increasing.

23 Scales continued A scale f is good if there is a club C ν + such that all γ in C of cofinality greater than cf(ν) are good for f. A scale f is bad if it is not good. We say there is a bad scale of length ν + if there is a bad scale of length ν + in some product ν i. Relating scales with other principles we have: Theorem (Shelah) For a singular ν, ν implies ν + I [ν + ] implies All scales of length ν + are good. Moreover we have Theorem (Jensen) There is a special ν + -Aronszajn tree if and only if ν holds.

24 Theorems from supercompactness It turns out that all of the weak square principles above fail in the presence of supercompactness. Theorem Let κ be a supercompact cardinal and ν > κ be singular with cf(ν) < κ, every scale of length ν + is bad. Let f α α < ν + be a scale in the product i ν i. Let j : V M witness that κ is ν + -supercompact. In particular crit(j) = κ and ν+ M M.

25 A proof continued Let γ = sup j ν + and note that γ < j(ν + ). It is not hard to show that the function, i sup j ν i is a so called exact upper bound for j(f ) α α < γ. It follows that γ is bad for j( f ). Standard reflection arguments show that there is a stationary set S ν + of bad points for f. Moreover S concentrates on cofinalities µ + where µ is a singular cardinal of cofinality cf(ν).

26 Bringing results down to ℵ ω Theorem Assuming there is a supercompact cardinal it is relatively consistent that there is a bad scale of length ℵ ω+1, in particular ℵ ω fails. Ideas of the proof: Fix a scale of length κ +ω+1 in the product κ +n. By our previous argument the scale has stationarily many bad points of some cofinality µ + < κ. Force with Coll(ω, µ) Coll(µ ++, < κ). Show that the scale is still a scale. Show that the stationary set of bad points is preserved. Show that it is still a set of bad points, but now concentrating on cofinality ω 1.

27 A parting theorem Let κ n n < ω be an increasing sequence of supercompact cardinals and let R be the Cummings-Foreman iteration for getting the tree property at each ℵ n. Theorem (U) There are a µ < κ 0 and a generic object d G for Coll(ω, µ) (R) V [Coll(ω,µ)] such that in V [d G], there are a bad scale of length ℵ ω+1 and a non-reflecting stationary subset of ℵ ω+1.