Cardinal Preserving Elementary Embeddings

Transcription

1 Outline Andrés E. Department of Mathematics California Institute of Technology XIII SLALM, Oaxaca, México, August 2006

2 Outline Outline 1 : Forcing axioms and inner models 2 Inner models M with ω1 M = ω 1 Inner models M with ω2 M = ω 2 3 Elementary embeddings Discontinuities Consistency strength 4 Final Remarks

3 Forcing axioms and inner models The questions that we study here grew out of recent results in the theory of forcing axioms and some interesting conjectures that these results suggest. Our goal here is to try to identify what remains of this structure without the additional assumption of forcing saturation.

4 We seek to study the relation between the universe of sets V and an inner model M under assumptions that imply some form of closeness between M and V. Throughout this talk, M always denotes an inner model. Closeness Assumptions Natural: M computes (some or all) cardinals correctly. Unnatural: M computes cardinals correctly and there is an elementary embedding of M into V (or of V into M).

5 We don t need to know forcing axioms for this talk. It suffices to know that MM (Martin s Maximum) is the strongest forcing axiom of its kind. MM implies PFA (the Proper Forcing Axiom), PFA implies BPFA (a bounded version of PFA), and BPFA implies MA (Martin s Axiom), with none of these implications being reversible. PFA has very high consistency strength.

6 Theorem (C. Veličković) Assume ω M 2 = ω 2 and BPFA holds in M and in V. Then R M. It was a previous result of Veličković that if ω2 M holds in V then R M. Question = ω 2 and MM Does the conclusion of the theorem hold under the assumption that ω M 2 = ω 2 and PFA holds in V?

7 Theorem (Viale) Assume M and V have the same cardinals and reals, PFA holds in V and stationary sets in M are stationary. Then ORD ω M. Question Can we remove the assumption on preservation of stationary sets from the statement of the theorem?

8 Same ω 1 Same ω 2 Inner models that compute cardinals correctly Recall: Theorem (C. Veličković) Assume ω M 2 = ω 2 and BPFA holds in M and in V. Then R M. If ω M 2 = ω 2 and MM holds in V then R M. What can we say about the reals of M under the sole assumption that M and V agree on some cardinals?

9 Same ω 1 Same ω 2 In general, very little can be said when ω1 M = ω 1. For example, if 0 exists, it does not follow that 0 M. In fact, there is in this case (in V ) a set forcing extension of L that computes ω 1 correctly. This observation is folklore: Use the L-indiscernibles (ι α : α ω V 1 ) to guide an inductive construction of a Col(ω,< ω V 1 )-generic over L.

10 Same ω 1 Same ω 2 Inductive construction Only ω many dense sets need to be met to extend a Col(ω,< ι α )-generic G α to a Col(ω,< ι α+1 )-generic G α+1. Since Col(ω,< κ) is κ-cc for κ regular and the L-indiscernibles form a club, it follows that α<κ G α is Col(ω,< ι κ )-generic for κ limit. Question Is there a class forcing extension of L such that no inner model with the same ω 1 is a set forcing extension of L?

11 Same ω 1 Same ω 2 Inner models that compute ω 2 correctly The results in this section are joint work with Martin Zeman. In contrast with the situation for ω 1, if M computes ω 2 correctly and 0 exists, then 0 M. This fact has been known for some time and is probably due to Friedman. Our results extend the situation from 0 to larger mice (like 0 and much more) by means of an induction on the mouse hierarchy, although it is not clear yet where this induction breaks down.

12 Same ω 1 Same ω 2 Question Assume that AD L(R), M V and CAR M = CAR. Does M = AD L(R)? Our methods should show that if x R M then, as long as Kx M is iterable (for example, below Woodin cardinals), any sound x-mouse projecting to ω is in M. This does not help to solve the question, since iterability is not absolute in the presence of Woodin cardinals.

13 Getting 0 in M Same ω 1 Same ω 2 We illustrate our approach in an easy case: If 0 exists, then it belongs to M. Our method is different from Friedman s. See [Zeman] as a reference for this section. Definition We represent 0 as the unique sound mouse N = (J τ,, U) where U is an amenable measure on the largest cardinal of J τ, and 1N = ω.

14 Same ω 1 Same ω 2 We iterate N = 0 by applying the ultrapower construction using U and its images under the corresponding embeddings. Let N α = (J τα,, U α ) be the α-th iterate of N for α ω 1 = ω1 V. Let π α,β, α β ω 1 be the corresponding embeddings. Let (κ α : α ω 1 ) be the corresponding critical sequence.

15 Same ω 1 Same ω 2 It is enough to reconstruct, in M, some map π α,β : N α N β for α < β. If we can, then we can recover U α as U α = {X J τα P(κ α ) : κ α π α,β (X)}. Since ω2 M = ω 2, it is easy to see that τ ω1 < ω2 M. Then, we can find in M some surjection f : ω 1 J τω1. Notice that C = {α < ω 1 : f α = ran(π α,ω1 )} is club.

16 Same ω 1 Same ω 2 Let α < β in C. Since f α M and J τα is its transitive collapse (and similarly for β), it follows that π α,β is the collapse of the inclusion map f α f β. So π α,β (and therefore 0 ) is in M.

17 Discontinuities Consistency strength Cardinal preserving elementary embeddings Recall: Theorem (Viale) Assume M and V have the same cardinals and reals, PFA holds in V and stationary sets in M are stationary. Then ORD ω M. Question ( ) Can we remove the assumption on preservation of stationary sets from the statement of the theorem? (We cannot remove both this assumption and the agreement on cardinals).

18 Discontinuities Consistency strength Viale has shown that if PFA holds and M computes the same cardinals and reals, but ORD ω M, then the first κ such that κ ω M is in M a singular cardinal of cofinality ω. More is known. For example, Question Does CAR M = CAR imply that (for all κ in M of cofinality ω) there is in M a sequence (S α : α < κ) of disjoint subsets of Sω κ+ that are stationary in V? (Yes for κ = ω, or if M = HOD, or if we only require a sequence of length ω) If so, then question ( ) has a positive answer.

19 Discontinuities Consistency strength Here is one way in which question ( ) could fail: Assume that there is an elementary embedding. Under mild assumptions, these embeddings can be obtained using the stationary tower forcing, or by means of elementary hulls. (But they cannot be created by set forcing.)

20 Discontinuities Consistency strength Foreman has shown that for any such embedding, ORD ω M (see [Welch-Vickers]). Thus, if M has the same cardinals as V, and PFA holds in V, then this solves the question negatively. Thus, either PFA rules out these embeddings, or ( ) can have a negative answer. Most likely, this indicates that such embeddings cannot exist. Here we point out that they would be highly discontinuous. Theorem Suppose that is elementary, and CAR M = CAR. Let κ = cp(j). Then for all λ > κ, j(λ) > λ. In particular, if j λ λ, then cf M (λ) κ.

21 Discontinuities Consistency strength It follows from this result that the critical point of such an embedding is Π 1 -indescribable in a very strong sense. For example: Corollary If there is a cardinal preserving then there is a proper class of weakly inaccessible cardinals. Proof. Let κ = cp(j). Any weakly inaccessible cardinal λ in V is also weakly inaccessible in M and therefore j(λ) is (another) weakly inaccessible cardinal. κ is weakly inaccessible in M, so there are (in V, thus in M) weakly inaccessible cardinals above κ. If there are only set many of them, their supremum would be a fixed point of j.

22 Discontinuities Consistency strength It is expected that there are no cardinal preserving embeddings. Since such an M computes incorrectly many cofinalities, covering fails very badly for M, and it should be no surprise that such an embedding would require considerable consistency strength. For example: Theorem Assume that there is a cardinal preserving embedding. Then there are inner models with strong cardinals.

23 Proof. Discontinuities Consistency strength Assume otherwise. Then K exists, it is rigid (i.e., there are no nontrivial elementary embeddings π : K K ) and any universal proper class mouse W is an iterate of K, so in particular there is an elementary π : K W. Universality of a class W = L[E] is a technical assumption, but below strong cardinals, (by the covering lemma [Mitchell]) if (µ + ) W = µ + for cofinally many cardinals µ, then W is universal.

24 Discontinuities Consistency strength K M is iterable. It follows from the previous paragraph that K M is universal by the covering lemma and the fact that CAR M = CAR. It follows that there is an embedding π : K K M, so j π : K K is nontrivial. Contradiction. At the moment, I do not see how to extend this result to obtain the existence of inner models with Woodin cardinals.

25 How about cardinal preserving embeddings j : V M? Fact If there is a cardinal preserving j : V M, then there is a cardinal preserving j : V N such that ORD ω N. It follows that these embeddings (in the presence of forcing axioms) also solve question ( ) negatively.

26 Definition A cardinal κ is µ-measurable iff there is an elementary embedding π : V W with critical point κ such that the measure {X P(κ) : κ π(x)} belongs to W. Theorem If there is a cardinal preserving elementary embedding j : V M then there is an inner model with a µ-measurable cardinal.

27 This follows from the covering lemma. However, no embedding j : V M obtained from (set sized) extenders can be cardinal preserving, so any such j should have much larger consistency strength. Finally, in the presence of a proper class of strongly compact cardinals, embeddings j : V V[G] can be obtained by a variation of Woodin s stationary tower forcing that exhibit (local) discontinuity properties, although they are not cardinal preserving.

28 W. Mitchell The Covering Lemma To appear in the Handbook of Set Theory, forthcoming. P. Welch and J. Vickers On Elementary Embeddings of an Inner Model to the Universe The Journal for Symbolic Logic, 66 (3), (2001),

29 M. Zeman. Inner Models and Large Cardinals. de Gruyter series in logic and its applications, vol 5, 2001.