On almost precipitous ideals.

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1 On almost precipitous ideals. Asaf Ferber and Moti Gitik December 20, 2009 Abstract With less than 0 # two generic extensions of L are identified: one in which ℵ 1, and the other ℵ 2, is almost precipitous. This improves the consistency strength upper bound of almost precipitousness obtained in [6], and answers some questions raised there. Also, main results of [5] are generalized- assumptions on precipitousness are replaced by those on -semi precipitousness. As an application it is shown that if δ is a Woodin cardinal and there is an f : ω 1 ω 1 with f = ω 2, then after Col(ℵ 2, δ) there is a normal precipitous ideal over ℵ 1. The existence of a pseudo-precipitous ideal over a successor cardinal is shown to give an inner model with a strong cardinal. 1 Introduction Let us define some basic notions. Definition 1.1 Let κ be a regular uncountable cardinal, τ an ordinal and I a κ-complete ideal over κ. We call I τ-almost precipitous iff every generic ultrapower of I is wellfounded up to the image of τ, where as usual we force with I-positive sets, a generic object G is a V -ultrafilter and (V κ V )/G is called a generic ultrapower. Clearly, any such I is τ-almost precipitous for each τ < κ. Also, note if τ (2 κ ) + and I is τ-almost precipitous, then I is precipitous. Therefore, in fact, if I is τ-almost precipitous for every τ < (2 κ ) +, then I is precipitous. Definition 1.2 Let κ be a regular uncountable cardinal. We call κ almost precipitous iff for each τ < (2 κ ) + there is a τ-almost precipitous ideal over κ. It was shown in [6] that ℵ 1 is almost precipitous if there is an ℵ 1 -Erdős cardinal. The following questions were raised in [6]: 1

2 1. Is an ℵ 1 -Erdős cardinal needed? 2. Can cardinals above ℵ 1 be almost precipitous without there being a measurable cardinal in an inner model? We will answer both questions by constructing two generic extensions of L such that ℵ 1 will be almost precipitous in the first and ℵ 2 in the second. Some of the ideas of Donder and Levinski [1] will be crucial here. Definition 1.3 (Donder- Levinski [1]) Let κ be a cardinal and τ be a limit ordinal of cofinality above κ or τ = On. 1. κ is called τ-semi-precipitous iff there exists a forcing notion P such the following is forced by the weakest condition: There exists an elementary embedding j : V τ M such that (a) crit(j) = κ (b) M is transitive. The restriction to τ s of cofinality above κ allows to apply j to any f : κ V τ, since then f is inside V τ. 2. κ is called < λ- semi-precipitous iff it is τ-semi-precipitous for every limit ordinal τ < λ of cofinality above κ. 3. κ is called semi-precipitous iff it is τ-semi-precipitous for every limit ordinal τ of cofinality above κ. 4. κ is called -semi-precipitous iff it is On-semi-precipitous. Note if κ is a semi-precipitous, then it is not necessarily -semi-precipitous, since by Donder and Levinski [1] semi-precipitous cardinals are compatible with V = L, and -semi-precipitous cardinals imply there exists an inner model with a measurable. 5. Let P and j be as in the item 1 above. We call F = {X κ 0 P κ j (X)} 2

3 a τ-semi-precipitous filter. 1 Note that such an F is a normal filter over κ. Also, F depends on a specific name of j. Note that M in (1) of the definition may be a proper class even if τ is an ordinal. But it is easy to find then i : V τ M with M being a transitive set. Just consider U = {X κ X V and κ j(x)}. Then V κ V τ /U is well founded, since we can embedded it into M using an elementary embedding k defined by k([f] U ) = j(f)(κ). Let M be the transitive collapse of V κ V τ /U and i the corresponding elementary embedding. The paper is organized as follows. In Section 2 we examine basic connections between semi-precipitousness and almost precipitousness are made. The following result that answers a question from [6] is shown: Corollary 2.12 The following are equivalent: 1. Con( there exists an almost precipitous cardinal), 2. Con( there exists an almost precipitous cardinal with normal ideals witnessing its almost precipitousness), 3. Con(there exists < κ ++ -semi-precipitous cardinal κ). In particular the strength of existence of an almost precipitous cardinal is below 0 #. In Section 3, an almost precipitous ideal is constructed over ℵ 2 iterating Namba forcing over L and then applying a variation of a construction of [5]. In Section 4 the following is proved: Theorem 4.1 Assume that 2 κ = κ + and κ carries a λ-semi-precipitous filter for some limit ordinal λ with cof(λ) > κ. Suppose in addition that there is a forcing notion P witnessing the λ-semi-precipitous with corresponding ultrapower of V by generic ultrafilters ill founded. Then 1. if λ < κ ++, then κ is λ-almost precipitous, as witnessed by a normal filter, 2. if λ κ ++, then κ is an almost precipitous, as witnessed by normal filters. 1 M. Foreman [3] 3.37 introduced a similar notion of a pre-precipitous filter. He requires that in a generic extension V Q by a forcing notion Q there is an elementary embedding j : V M with M transitive. The filter defined by picking some Q-name t of element of M, a set Z in V and then setting F = {X Z 0 Q t j (Z)}. So, On-semi-precipitous filters are pre-precipitous. 3

4 The theorem is a kind of interplay between ill foundness and well foundness in which ill foundness helps to produce well foundness. Main results of [5] are generalized here to context of -precipitous ideals: Theorem 4.10 Assume that ℵ 1 is -semi precipitous and 2 ℵ 1 = ℵ 2. Suppose that for some forcing P witnessing this 0 P P i (ℵ 1 ) > (ℵ + 1 ) V. Then ℵ 1 is almost precipitous witnessed by normal filters. Theorem 4.12 Suppose that there is no inner model satisfying ( α o(α) = α ++ ). Assume that ℵ 1 is -semi precipitous and 2 ℵ 1 = ℵ 2. If ℵ 3 is not a limit of measurable cardinals of the core model, then there exists a normal precipitous ideal on ℵ 1. Theorem 4.13 Assume that ℵ 1 is -semi precipitous. Let P be a witnessing this forcing such that 0 P P i (ℵ 1 ) > (ℵ + 1 ) V. Then, after forcing with Col(ℵ 2, P ), there will be a normal precipitous filter on ℵ 1. As a corollary of the last theorem we deduce the following: Corollary 4.17 Suppose that δ is a Woodin cardinal and there is f : ω 1 ω 1 with f ω 2. Then in V Col(ℵ2,δ) there is a normal precipitous ideal over ℵ 1. Section 6 deals with pseudo-precipitous ideals. T. Jech [8] asked how strong is the consistency of there is a pseudo-precipitous ideal on ℵ 1? We show the following: Theorem 6.2 If there is a pseudo-precipitous ideal over a successor cardinal then there is an inner model satisfying ( α o(α) = α ++ ). In particular, an existence of precipitous ideal does not necessary imply an existence of a pseudo-precipitous one. Sections 3,4 rely on the filter construction technique of [5], but otherwise there is not extensive background knowledge required. 2 On semi-precipitous and almost precipitous ideals In this section we consider games that allow to connect semi-precipitousness and almost precipitousness and then deduce some conclusions on the strength of an existence of an almost precipitous filters. Lemma 2.1 Let F be a τ-almost precipitous normal filter over κ for some ordinal τ of cofinality above κ. Then F is τ-semi-precipitous. 4

5 Proof. Force with F +. Let i : V N = V κ V/G be the corresponding generic embedding. Set j = i τ. Then j : V τ (V i(τ) ) N. Set M = (V i(τ) ) N. We claim that M is well founded. Suppose otherwise. Then there is a sequence g n n < ω of functions such that 1. g n V 2. g n : κ V τ 3. {α < κ g n+1 (α) g n (α)} G Replace each g n by a function f n : κ τ. Set f n (α) = rank(g n (α)). Clearly, still we have {α < κ f n+1 (α) f n (α)} G. But this means that N is not well-founded below the image of τ. Contradiction. Note that the opposite direction does not necessarily hold. This is because for τ (2 κ ) +, τ-almost precipitousness implies precipitousness and hence a measurable cardinal in an inner model. By Donder and Levinski [1], it is possible to have semi-precipitous cardinals in L. The following is an analog of a game that was used in [6] with connection to almost precipitous ideals. Definition 2.2 (The game G τ (F )) Let F be a normal filter on κ and let τ > κ be an ordinal. The game G τ (F ) is defined as follows: Player I starts by picking a set A 0 in F +. Player II chooses a function f 1 : A 0 τ and either a partition B i i < ξ of A 0 into ξ < κ many pieces or a sequence B α α < κ of disjoint subsets of κ so that α<κ B α A 0. The first player then supposed to respond by picking an ordinal α 2 and a set A 2 F + which is a subset of A 0 and of one of B i s or B α s. At the next stage the second player supplies again a function f 3 : A 2 τ and either a partition B i i < ξ of A 2 into ξ < κ many pieces or a sequence B α α < κ of disjoint subsets of κ so that α<κ B α A 2. 5

6 The first player then responds by picking a set A 4 F + which is a subset of A 2 and of one of the B i s or B α s on which everywhere f 1 is either above f 3 or equal to f 3 or below f 3. In addition he picks an ordinal α 4 such that α 2, α 4 respect the order of f 1 A 4, f 3 A 4, i.e. α 2 < α 4 iff f 1 A 4 < f 3 A 4, α 2 > α 4 iff f 1 A 4 > f 3 A 4 and α 2 = α 4 iff f 1 A 4 = f 3 A 4. Intuitively, α 2n pretends to represent f 2n 1 in a generic ultrapower. The game continues in the same fashion. α 2k 0 < k n should respect the order of f 2k 1 A 2n 0 < k n. Player I wins if the game continues through infinitely many moves. Otherwise Player II wins. Clearly this game is determined. The following lemma is analogous to [6] (Lemma 3). It was proved originally for Cub κ, but the referee of the paper pointed out how to generalize the argument to arbitrary normal filter F. His suggestion is implemented below. Lemma 2.3 Let κ be a regular uncountable cardinal and F a normal filter over κ. Suppose that λ is a κ-erdős cardinal (i.e. λ is the least µ such that µ (κ) <ω 2 ). Then for each ordinal τ,κ < τ < λ, Player I has a winning strategy in the game G τ (F ). Proof. Suppose otherwise. Then Player II has a winning strategy. Let σ be a winning strategy for II. We will first find a set X λ of cardinality κ such that σ does not depend on ordinals picked by Player I from X. In order to get such X let us consider a structure Let X be a set of κ indiscernibles for A. A = H(λ),, λ, κ, P(κ), F, G τ (F ), σ. Pick now a continuous elementary chain M α α < κ of submodels of H(χ) for χ > λ big enough such that for every α < κ the following hold: 1. M α < κ, 6

7 2. σ, X M α, 3. M α κ On, 4. M β β α M α+1. Set κ α = M α κ, for each α < κ. Let E = {α < κ κ α = α}. Then E is a club. For every α < κ, the set H α = {X X F M α } is in F. Hence α<κ H α is in F as well. So H E F. Pick any α H. Then α H β for each β < α = κ α. Now, for every X M α F we have X M β F for some β < α, so X H β and hence α X. Denote M α by M. Then α = M κ and for every X M F we have α X. Let us produce an infinite play in which the second player uses σ. This will give us the desired contradiction. Consider the set S = {f(α) f M, f is a partial function from κ to τ}. Obviously, S has cardinality less than κ. Hence we can fix an order preserving function π : S X. Let one start with A 0 = κ. Consider σ(a 0 ). Clearly, σ(a 0 ) M. It consists of a function f 1 : A 0 τ and, say a sequence B ξ ξ < κ of disjoint subsets of κ so that ξ<κ B ξ A 0. Now, α A 0, hence there is ξ < α such that α B ξ. Then B ξ M, as M α. Hence, A 0 B ξ M and α A 0 B ξ. Let A 2 = A 0 B ξ. Note that A 2 C, for every C F which belongs to M, since α is in both A 2 and C. Hence, by elementarity A 2 F +, so is a legitimate move in the game. Pick α 2 = π(f 1 (α)). Consider now the answer of two which plays according to σ. It does not depend on α 2, hence it is in M. Let it be a function f 3 : A 2 τ and, say a sequence B ξ ξ < κ of disjoint subsets of κ so that ξ<κ B ξ A 2. As above find ξ < α such that α B ξ. Then B ξ M, as M α. Hence, A 2 B ξ M and α A 2 B ξ. Let A 2 = A 2 B ξ. Split it into three sets C <, C =, C > such that C < = {ν A 2 f 3 (ν) < f 1 (ν)}, C = = {ν A 2 f 3 (ν) = f 1 (ν)}, C > = {ν A 2 f 3 (ν) > f 1 (ν)}. 7

8 Clearly, α belongs to only one of them, say to C <. Set then A 4 = C <. Then, clearly, A 4 M, it is in F + and f 3 (α) < f 1 (α). Set α 4 = π(f 3 (α)). Continue further in the same fashion. The next game was introduced by Donder and Levinski in [1]. Definition 2.4 A set R is called κ-plain iff 1. R, 2. R consists of normal filters over κ, 3. for all F R and A F +, F + A R, where F +A is the filter generated by F {A}, i.e. F +A = {X κ Y F X}. Y A Definition 2.5 (The game H R (F, τ)) Let R be κ-plain, F R be a normal filter on κ and let τ > κ be an ordinal. The game H R (F, τ) is defined as follows. Set F 0 = F. Let 1 i < ω. Player I plays at stage i a pair (A i, f i ), where A i κ and f i : κ τ. Player II answers by a pair (F i, γ i ), where F i R and γ i is an ordinal. The rules are as follows: 1. For 0 i < ω, A i+1 (F i ) + 2. For 0 i < ω, F i+1 F i + A i+1 Player II wins iff for all 1 i, k n < ω : (f i < Fn f k ) (γ i < γ k ) Donder and Levinski [1] showed that the existence of a winning strategy for Player II in the game H R (F, τ) for some R, F is equivalent to κ being τ- semi precipitous. The next two lemmas deal with connections between winning strategies for the games G τ (F ) and H R (F, τ). Lemma 2.6 Suppose that Player II has a winning strategy in the game H R (F, τ), for some κ-plain R, a normal filter F R over κ and an ordinal τ. Then Player I has a winning strategy in the game G τ (F ). Proof. Let σ be a winning strategy for Player II in H R (F, τ). We define a winning strategy δ for Player I in the game G τ (F ). Let the first move according to δ be κ. Suppose that Player 8

9 II responds by a function f 1 : κ τ and a partition B 1 of κ to less then κ many subsets or a sequence B 1 = B α α < κ of κ many subsets such that α<κ B α κ. Turn to the strategy σ. Let σ(κ, f 1 ) = (F 1, γ 1 ), for some F 1 F, F 1 R and an ordinal γ 1. Now we let Player I pick A 1 (F 1 ) + such that there is a set B B 1 with A 1 B (he can always choose such an A 1 because F 1 is normal and α<κ B α (F 1 ) + ) and let the respond according to δ be (A 1, γ 1 ). Player II will now choose a function f 2 : A 1 τ and a partition B 2 of A 1 or a sequence B 2 = B α α < κ, α<κ B α A 1. Back in H R (F, τ), we consider the answer according σ of Player II to (A 1, f 2 ), i.e. σ((κ, f 1 ), (A 1, f 2 )) = (F 2, γ 2 ). Choose A 2 (F 2 ) + such that there is a set B B 2 with A 2 B (it is always possible to find such A 2 because F 2 is normal and α<κ B α (F 2 ) + ) on which either f 1 < f 2 or f 1 > f 2 or f 1 = f 2. Let the respond according to δ be (A 2, γ2), where γ2 = γ 2, unless f 1 = f 2 on A 2. If f 1 = f 2 on A 2, then set γ2 = γ 1. Note that in this case the rules of H R (F, τ) allow γ 1 γ 2, but not those of G τ (F ). Continue in a similar fashion. The play will continue for infinitely many moves. Hence Player I will always win by using the strategy δ. Lemma 2.7 Suppose that Player I has a winning strategy in the game G τ (F ), for a normal filter F over κ and an ordinal τ. Then Player II has a winning strategy in the game H R (D, τ) for some κ-plain R and D R. Proof. Let σ be a winning strategy of Player I in G τ (F ). Set J = {X κ X and any of its subsets are never used by σ}, and for every finite play t = t 1,..., t 2n via σ J t = {X κ X and any of its subsets are never used by σ in the continuation of t}. It is not hard to see that such J and J t s are normal ideals over κ. Denote by D and D t the corresponding dual filters. Pick R to be a κ-plain which includes D and all D t s. We define a winning strategy δ for Player II in the game H R (D, τ). Let (A 1, g 1 ) be the first move in H R (D, τ). Then A 1 D +. Hence σ picks a subset of A 1 in a certain play t as a move of Player I in the game G τ (F ). Continue this play, and let Player II respond by a trivial partition of A 1 consisting of A 1 itself and by the function g 1 restricted to A 1. Let 9

10 (B 1, γ 1 ) be the responce of Player I according to σ. Set t 1 = t ({A 1 }, g 1 ). Then B 1 D t1. Now we set the responce of Player II according to δ to be (D t1, γ 1 ). Continue in a similar fashion. Theorem 2.8 Suppose that λ is a κ-erdős cardinal, then κ is τ-semi precipitous for every τ < λ. Proof. By Donder and Levinski [1] the existence of a winning strategy for Player II in the game H R (F, τ) for some R, F is equivalent to κ being τ- semi precipitous. The result follows now by Lemmas 2.3,2.7. Combining the above with Theorem 17 of [6], we obtain the following: Theorem 2.9 Assume that 2 ℵ 1 = ℵ 2 and f = ω 2, for some f : ω 1 ω 1. Let τ < ℵ 3. If there is a τ-semi-precipitous filter over ℵ 1, then there is a normal τ-almost precipitous filter over ℵ 1 as well. Proof. By Donder and Levinski [1], the existence of τ-semi-precipitous filter over ℵ 1 implies that Player II has a winning strategy in the game H R (F, τ) for some ℵ 1 -plain R and a normal filter F R. Then Player I has a winning strategy in the game G τ (F ), by 2.6. Then, as it is easily seen, such a strategy will be a winning strategy in the game G τ (Cub ℵ1 ) as well. Now Theorem 17 of [6] applies. The next result is a small generalization of Donder-Levinski (Theorem 8) and it is well known for precipitous ideals. Theorem 2.10 Let κ be a τ-semi-precipitous cardinal for some τ of cofinality above κ or τ = On, as witnessed by a forcing notion P and F be a correspondent τ-semi-precipitous filter, i.e. 0 P ( there is j : V τ M, crit( j ) = κ, Mis transitive), F = {X κ 0 P κ j (X)}. Suppose that Q is a κ-c.c. forcing notion of cardinality less than τ. Then, for any generic subset G Q of Q, the following hold in V [G Q ]: 1. κ remains a τ-semi-precipitous cardinal in V [G Q ]. 2. There is a forcing notion R V such that 0 R (there is j : (V τ) V [GQ] M, j j), {X κ 0 R κ j (X)} = {X κ Y F Y X}. 10

11 Proof. Clearly (2) implies (1). So let us show (2). It is enough to find some forcing S in V Q that produces j, since then it is possible to absorb all the possibilities for such S inside Col(ω, η) for a large enough η. Let G Q be a generic subset of Q and Y be any F -positive set (in V ). Pick p P which forces κ j (Y ). Suppose that (2) does not hold. Then there is q Q such that: (*) for every generic G Q Q with q Q there is no forcing notion S V [G Q ] such that for some V [G Q ]-generic subset H of S there is G P V [G Q H] generic over V with p G P and with the embedding j = j G P which extends to some j : (V τ ) V [G Q] M. Note that (V τ ) V [G Q] = V τ [G Q ]. It follows by the cardinality assumption we made and since τ is a limit ordinal. Pick a cardinal µ to be above all possible cardinalities of models M that are produced by P, if τ is an ordinal and above all possible cardinalities of 2 j (Q), if τ = On. Let η = µ + 2 P. Consider Col(ω, η). It can be viewed as P T, where T = Col(ω, η)/p. Force with this forcing (over V ). Let G P G T be a generic set with p G P. Then M (given by G P ) is countable in V [G P G T ], so there is G V [G P G T ] which is M-generic subset of j(q) with j(q) G. Set G = {t Q j(t) G }. Recall that Q is κ-c.c. forcing and κ is the critical point of j. Hence G is V -generic subset of Q with q G. Finally we consider the forcing S = Col(ω, η)/g in V [G], i.e. S = {t Col(ω, η) d G (t d G )}, where G is a Col(ω, η)-name of G. It contradicts (*) above. By Donder and Levinski [1], the existence of 0 # implies that the first indiscernible c 0 for L is, in L, τ-semi-precipitous for each τ. They showed [1](Theorem 7) that the property κ is τ-semi-precipitous relativizes down to L. Also it is preserved under κ-c.c. forcings of cardinality κ by Now combine this with 2.9. We obtain the following: Theorem 2.11 Suppose that κ is a < κ ++ -semi-precipitous cardinal in L. Let G be a generic subset of the Levy Collapse Col(ω, < κ). Then for each τ < κ ++, κ carries a τ- almost precipitous normal ideal in L[G]. Proof. In order to apply 2.9, we need to check that there is f : ω 1 ω 1 with f = ω 2. Suppose otherwise. Then by Donder and Koepke [2] (Theorem 5.1) we will have wcc(ω 1 ) 11

12 (the weak Chang Conjecture for ω 1 ). Again by Donder and Koepke [2] (Theorem D), then (ℵ 2 ) L[G] will be almost < (ℵ 1 ) L -Erdös in L. But note that (ℵ 2 ) L[G] = (κ + ) L and in L, 2 κ = κ +. Hence, in L, we must have 2 κ (ω) 2 κ, as a particular case of 2 κ being almost < ℵ 1 -Erdös. But 2 κ (3) 2 κ. Contradiction. Corollary 2.12 The following are equivalent: 1. Con( there exists an almost precipitous cardinal), 2. Con( there exists an almost precipitous cardinal with normal ideals witnessing its almost precipitousness), 3. Con(there exists < κ ++ -semi-precipitous cardinal κ). In particular the strength of existence of an almost precipitous cardinal is below 0 #. 3 An almost precipitous ideal on ω 2 In this section we will construct a model with ℵ 2 being almost precipitous. The initial assumption will be the existence of a Mahlo cardinal κ which carries a (2 κ ) + - semi-precipitous filter F with {ν < κ ν is a regular cardinal } F. Note that in general, if κ is an inaccessible which carries a (2 κ ) + -semi-precipitous filter F, then the set {ν < κ ν is a regular cardinal } need not be in F. It is possible even to have a normal precipitous filter with {τ < κ cof(τ) = ω} inside. Actually this is possible already over the first inaccessible. On the other hand if there there is no inner model with a measurable cardinal (or actually many measurable cardinals) then each (2 κ ) + -semi-precipitous filter F over κ ℵ 3 should concentrate on {ν < κ ν = cof(ν)} If κ = ℵ 2 then a slight variation of the construction below may be used to produce an example of a (2 κ ) + -semi-precipitous filter F over κ which concentrates on {ν < κ cof(ν) = ω}. By Donder and Levinski [1] the initial assumption above is compatible with V = L. If 0 exists, then the first indiscernible will be like this in L. Assume V = L. Let P i, Qj i κ, j < κ be a Revised Countable Support iteration (see [12], Chapter 10, 1) so that for each α < κ, if α is an inaccessible cardinal (in V ), then Qα is Col(ω 1, α) 12

13 which collapses α to ℵ 1 and Qα+1 is the Namba forcing which changes the cofinality of α + (which is now ℵ 2 ) to ω. In all other cases let Qα be the trivial forcing. By [12]( Chapter 11, 4,5,6), the forcing P κ turns κ into ℵ 2, preserves ℵ 1, does not add reals and satisfies the κ -c.c. Let G be a generic subset of P κ. By 2.10, a κ-c.c. forcing of cardinality κ preserves semi-precipitousness of F. Hence F is κ ++ = ℵ 4 -semi-precipitous in L[G]. In addition, {τ < κ cof(τ) = ω 1 } F and {τ < κ cof((τ + ) V ) = ω} F. Now, there is a forcing Q in L[G] so that in L[G] Q we have a generic embedding j : L κ ++[G] M such that M is transitive and κ j(a) for every A F. By elementarity, then M is of the form L λ [G ], for some λ > κ ++, and G j(p κ ) which is L λ -generic. Note that Q κ collapses κ to (ℵ 1 ) M because it was an inaccessible cardinal, and at the very next stage its successor changes the cofinality to ω. That means that there is a function H L κ ++[G] such that j(h)(κ) : ω (ℵ 3 ) L[G] is an increasing and unbounded in (κ + ) L = (ℵ 3 ) L[G] function. Just for each inaccessible in L cardinal α set H(α) : ω (α + ) L to be the generic Namba sequence. We will use such H as a replacement of the corresponding function of [5]. Together with the fact that in the model L[G] we have a filter on ℵ 2 which is ℵ 4 semi precipitous this will allow us to construct a τ- almost precipitous filter on ℵ 2, for every τ < ℵ The construction Work in L[G]. Fix τ < κ ++. Denote by B the complete Boolean algebra RO(Q). Further by we will mean the order of B. For each p B set F p = {X κ p κ j (X)} We will use the following result of Donder and Levinski [1](Theorem 4, Claim 1): Lemma p q F p F q 2. X (F p ) + iff there is a q p, q κ j(x) 3. Let X (F p ) +, then for some q p, F q = F p + X 13

14 Proof. (1) and (2) are trivial. Let us prove (3). Suppose that X (F p ) +. Set q = κ j (X) B p. We claim that F q = F p + X. The inclusion F q F p + X is trivial. Let us show that F p + X F q. Suppose not, then there are Y (F p ) +, Y X and Z F q such that Y Z =. But Y (F p ) +, so we can find s p such that s κ j (Y ). Now, s p and s κ j (X), since Y X. Hence, s q. But then s κ j (Y ), κ j (Z), j (Z Y ) =. Contradiction. Define {A nα α < κ +, n < ω} as in [5]: A nα = {η < κ H(η)(n) = h α (η)}, where h α α < κ + is a sequence of κ + canonical functions from κ to κ (in V = L[G]). Recall that 1 B j (h α )(κ) = α. Note that here H is only cofinal and not onto, as in [5]. The following lemmas were proved in [5] and hold without changes in the present context: Lemma 3.2 For every n < ω there is an ordinal α < κ + such that A nα (F 1B ) +. Lemma 3.3 For every α < κ + and p B there is n < ω and α < β < κ + such that A nβ (F p ) +. Lemma 3.4 Let n < ω and p B. Then the set: {A nα α < κ + and A nα (F p ) + } is a maximal antichain in (F p ) +. The following is an analog of a lemma due Assaf Rinot in [5], 3.5. Lemma 3.5 Let D be a family of κ + dense subsets of B, there exists a sequence p α α < κ + such that for all Z (F 1B ) +,p Q and n < ω if Z n,p = {α < κ + A nα Z (F p ) + } has cardinality κ + then : 14

15 1. For any p B there exists α Z n,p with p p α. 2. For any D D there exists α Z n,p with p α κ j(a nα Z),p α p and p α D. Proof. Let {S i i < κ + } [κ + ] κ+ be some partition of κ +, {D α α < κ + } an enumeration of D,{q α α < κ + } an enumeration of Q and let be a well ordering of κ + κ + κ + of order type κ +. Now, fix a surjective function ϕ : κ + {(Z, n, p) ((F 1B ) +, ω, Q) Z n,p = κ + }. We would like to define a function ψ : κ + κ + κ + κ + and the sequence p α α < κ +. For that, we now define two sequences of ordinals {L α α < κ + }, {R α α < κ + } and the values of ψ and the sequence on the intervals [L α, R α ] by recursion on α < κ +. For α = 0 we set L 0 = R 0 = 0,ψ(0) = 0 and p 0 = q 0. Now, suppose that {L β, R β β < α} and ψ β<α [L β, R β ] were defined.take i to be the unique index such that α S i.let (Z, n, p) = ϕ(i) and set L α = min(κ + \ β<α [L β, R β ]), R α = min(z n,p \ L α ). Now, for each β [L α, R α ] we set ψ(β) = t,where: t = min (κ + {i} κ + ) \ ψ (Z n,p L α ). If t κ + then we set p β = q t for each β [L α, R α ].Otherwise, t = (i, δ) for some δ < κ + and because A nrα Z F p + and D δ is dense we can find some q D δ, q p, q κ j (A nrα Z) and set p β = q for each β [L α, R α ].This completes the construction. Now, we would like to check that the construction works. Fix Z F 1 + B p Q and n < ω so that Z n,p = κ +.Let i < κ + be such that ϕ(i) = Z n,p and notice that the construction insures that ψ Z n = κ + {i} κ +. (1) Let p B. There exists a t < κ + so that q t p.let α Z n be such that ψ(α) = t, so p α = q t p. (2) Let D D. There exist δ < κ + and α Z n,p such that D δ = D and ψ(α) = (i, δ).then, by the construction we have that p α D δ, p α κ j(a nα Z) and p α p. Define D = {D f f (τ κ ) V }, where D f = {p B γ On p j( ˇf)(κ) = ˇγ} and let p α α < κ + be as in lemma 3.5. We turn now to the construction of filters which will be similar to those of [5]. Start with n = 0. Let α < κ +. Consider three cases: 15

16 Case I: If {ξ < κ + A 0ξ (F 1B ) + } = κ + and p α κ j(a 0α ) then we define q <α> = p α and we associate F q<α> to the sequence < α >. Case II: If I fails but A 0α (F 1B ) + then we define q <α> = κ j(a 0α ) B and we associate F q<α> to the sequence < α >. Case III: If A 0α ˇF 1B (the dual ideal of F 1B ) then q <α> is not defined. Notice that by Lemma 3.2, there exists some α < κ + with A 0α (F 1B ) +, thus {α < κ + q α is defined } is non-empty. Definition 3.6 Set F 0 = {q α α < κ +, q α is defined }, and denote the corresponding dual ideals by I q α and I 0. Clearly, I 0 = {I q α α < κ +, q α is defined }. Also, F 0 F 1B and I 0 ˇF 1B, since each F q α F 1B and I q α ˇF 1B. Note that F 0 is a κ complete, normal and proper filter since it is an intersection of such filters and also I 0 is a κ complete, normal and proper ideal. We now describe the successor step of the construction, i.e., m = n + 1. Let σ : m κ + be a function with q σ defined and α < κ +. There are three cases: Case I: If {ξ < κ + A mξ F q + σ } = κ +, p α q σ and p α κ j(a mα ), then we define q σ α = p α and associate F qσ α to the sequence σ α. Case II: If Case I fails, but A mα (F + q σ ), then let q σ α = κ j(a mα ) B q σ, and associate F qσ α to the sequence σ α. Case III: If A mα I qσ, then q σ α is not defined. This completes the construction. Definition 3.7 Let F n+1 = {F qσ σ : n + 2 κ +, F qσ is defined }, and define the corresponding dual ideals I n+1, I qσ. Notice that each F n and I n is κ-complete, proper and normal, as an intersection of such filters and ideals respectively. Definition 3.8 Let F ω be the closure under ω intersections of n<ω F n. Let I ω = the closure under ω unions of n<ω I n. Lemma 3.9 F F 0... F n... F ω and I I 0... I n... I ω, and I ω is the dual ideal to F ω. 16

17 Lemma 3.10 Let s : m κ + with q s defined; then: 1. {α < κ + q s α is defined } = {ξ < κ + A mξ F + q s }; 2. There exists an extension σ s such that q σ is defined and {ξ < κ + A dom(σ)ξ F + q σ } = κ +. Proof. 1) is clear from the construction above. For 2), let us assume that for every extension σ s such that q σ is defined : {ξ < κ + A dom(σ)ξ F + q σ } κ. That means that Σ = {σ : n κ + n m, σ s and q σ is defined } is of cardinality less than or equal to κ, so ν = σ Σ ran(σ) is less than κ+ and q s will force that j(h)(κ) is bounded by ν, contradiction. From now on the proof that F ω is the desired filter will be almost the same as in [5]( Theorem 2.1)(just some, minor adjustments should be made because here we restricting to filters rather than to sets in [5]. A more detailed argument will be provided further in the proof of 4.1). 4 Constructing almost precipitous ideals from semiprecipitousness Suppose κ is a λ semi-precipitous cardinal for some ordinal λ which is a successor ordinal > κ or a limit one with cof(λ) > κ. Let P be a forcing notion witnessing this. Then, for each generic G P, in V [G] we have an elementary embedding j : V λ M with cp(j) = κ and M transitive. Consider U = {X κ X V, κ j(x)}. Then U is a V normal ultrafilter over κ. Let i U : V V κ V/U be the corresponding elementary embedding. Note that V κ V/U need not be well founded, but it is well founded up to the image of λ. Thus, denote V κ V/U by N. Define k : (V i(λ) ) N M in a standard fashion by setting k([f] U ) = j(f)(κ), 17

18 for each f : κ V λ, f V. Then k will be elementary embedding, and so (V i(λ) ) N is well founded. For every p P set F p = {X κ p κ j (X)}. Clearly, if G is a generic subset of P with p G and U G is the corresponding V -ultrafilter, then F p U G. Note that, if for some p P the filter F p is κ + -saturated, then each U G with p G will be generic over V for the forcing with F p -positive sets. Thus, every maximal antichain in F + p consists of at most κ many sets. Let A ν ν < κ V be such a maximal antichain. Without loss of generality we can assume that min(a ν ) > ν, for each ν < κ. Then there is ν < κ with κ j(a ν ). Hence A ν U G and we are done. It follows that in such a case N which is the ultrapower by U G is fully well founded. See, for example, T. Jech [9], Lemma on page 427. Note that in general if some forcing P produces a well founded N, then κ is -semiprecipitous. Just i and N will witness this. Our aim will be to prove the following: Theorem 4.1 Assume that 2 κ = κ + and κ carries a λ-semi-precipitous filter for some limit ordinal λ with cof(λ) > κ. Suppose in addition that there is a forcing notion P witnessing the λ-semi-precipitous with corresponding N ill founded. Then 1. if λ < κ ++, then κ is λ-almost precipitous, as witnessed by a normal filter, 2. if λ κ ++, then κ is an almost precipitous, as witnessed by normal filters. Proof. The proof will be based on an extension of the method of constructing normal filters of [5] which replaces restrictions to positive sets by restrictions to filters. An additional idea will be to use a witness of the non-well-foundedness in the construction in order to limit it to ω many steps. Let κ, τ, P be as in the statement of the theorem. Preserve the notation that we introduced above. Note also that we use here a poset rather than a Boolean algebra in the previous section. So, 0 P will replace 1 B and the order will be used in the opposite direction. Then 0 P (V i(λ) ) N is well founded and N is ill founded. 18

19 Fix a sequence g n n < ω of names of functions witnessing the ill foundedness of N, i.e. 0 P [ g n] > [ g n+1], for every n < ω. Note that, as was observed above, for every p P, the filter F p is not κ + -saturated. Fix some τ < κ ++, τ λ. We should construct a normal τ-almost precipitous filter over κ. For each p P choose a maximal antichain {A pβ β < κ + } in F p +. Let f α α < κ + enumerate all the functions from κ to τ. Fix an enumeration X α α < κ + of F 0 + P. Start now an inductive process of extending F 0P. Let n = 0. Assume for simplicity that there is a function g 0 : κ On V so that 1 P ǧ 0 = g 0. Set p = 0 P. We construct inductively a sequence of ordinals ξ β β < κ + and a sequence of conditions p β β < κ +. Let α < κ +. Case I. There is a ξ < κ + so that ξ ξ β for every β < α, and X α A p ξ F p +. Then let ξ α be the least such ξ. We would like to attach an ordinal to f ξ0α and to decide g 1. Let us pick p P such that 1. p κ j(x α A p ξ α ), 2. for some γ, p j(f ξ α )(κ) = γ, 3. there is a function g 1 : κ On, g 1 V such that p ǧ 1 = g 1. Note that then p ǧ 1 < ǧ 0, since 0 P g 1 < ǧ 0. Set p α = p. Case II. Not Case I. Then set ξ α = 0. We will leave p α undefined in this case. Note that if Case I fails then we have X α β<κ A 0P ξ τ(β) mod F 0P for a surjective τ : κ α. Set F 0 = {F p α α < κ + and p α is defined }, and denote the corresponding dual ideals by I p α and I 0. Clearly, I 0 = {I p α α < κ + and p α is defined }. Also, F 0 F 0P and I 0 ˇF 0P, since 19

20 each F p α F 0P and I p α ˇF 0P. Note that F 0 is a κ complete, normal and proper filter since it is an intersection of such filters and also I 0 is such an ideal. We now describe a successor step of the construction. Let σ : m κ + with p σ defined. Construct by induction a sequence of ordinals ξ σ β β < κ + and a sequence of conditions p σ β β < κ +. Let α < κ + : Case I. There is ξ < κ + so that ξ ξ σ β for every β < α and X α A pσ ξ F + p σ. Then let ξ σ α be the least such ξ. We would like to attach an ordinal to f ξσ α that 1. p p σ, 2. p κ j(x α A 0P ξ σ α ), 3. for some γ, p j(f ξσ α )(κ) = γ, and to decide g m. Let us pick p P such 4. there is a function g m : κ On, g m V such that p ǧ m = g m. Note that then p ǧ m < ǧ m 1, since 0 P g m < g m 1 and p σ ǧ m 1 = g m 1. Set p σ α = p. Case II. Not Case I. Then set ξ σ α = 0. We will leave p σ α undefined in this case. This completes the construction. Set F m = {F pσ α σ : m κ +, α < κ + and p σ α is defined }, and denote the corresponding dual ideals by I pσ α and I m. We will use the following: Definition 4.2 Let F ω be the closure under ω intersections of n<ω F n. Let I ω = the closure under ω unions of n<ω I n. Lemma 4.3 F 0... F n... F ω and I 0... I n... I ω, and I ω is the dual ideal to F ω. Our purpose now will be to show that we cannot continue the construction further beyond ω. Then we will be able to show that F ω is a τ -almost precipitous filter. Lemma 4.4 F + ω = {F pσ σ <ω κ +, p σ is defined}. 20

21 Proof. Let X (F ω ) + and assume that X F pσ for each σ [κ + ] <ω such that p σ is defined. Note that then there are at most κ many σ s such that X F p + σ. It is enough to argue that for every σ [κ + ] <ω such that p σ is defined, the set {ζ < κ + X A pσζ F p + σ } is of cardinality less than or equal κ. Suppose otherwise. Let α < κ + be such that X = X α. Then p σ α is defined according to Case I and X F p α. Contradiction. Set T = {σ [κ + ] <ω p σ is defined, X F p + σ } and for every σ T let B σ = {ζ < κ + X A pσ ζ F + p σ }. Then both sets are of cardinality at most κ. Let σ T. Assume that B σ = κ. The case B σ < κ is treated similar and simpler. Fix ψ σ : κ B σ, for every σ T. Note that X \ β<κ A pσψσ(β) is in the ideal I pσ. Now, let n = 0. Turn the family {A 0P ψ (γ) γ < κ} into a family of disjoint sets as follows: A 0 P ψ (0) := A 0P ψ (0) \ {0} and for each γ < κ let A 0 P ψ (γ) := A 0P ψ (γ) \ ( A 0P ψ (β) (γ + 1)). β<γ Note that β<κ A 0 P ψ (β) = {ν < κ β < ν so that ν A 0 P ψ (β)} and, because ν A 0 P ψ (β) implies that ν > β, we get that the right hand side is equal to {A 0P ψ (γ) γ < κ}. Also note that β<κ A 0 P ψ (β) = β<κ A 0P ψ (β). So {X A 0 P ψ (γ) γ < κ} is still a maximal antichain in F 0 + P mod F 0P. Set R 0 := X \ β<κ A 0 P ψ (β). Then R 0 I 0P. Turn now to n = 1. Let σ T be a sequence of the length 1. below X and X β<κ A 0 P ψ (β) 21

22 We turn the family {A pσ,ψ σ (γ) γ < κ} into a disjoint one {A p σ,ψ σ (γ) γ < κ} exactly as above. Set R σ = (X A p ξ σ ) \ ( A p σ ψ σ (γ) {δ < κ g 1 (δ) < g 0 (δ)}). γ<κ Then R σ I pσ. Define R 1 = {R σ σ T and the length of σ is 1 }. Claim 1 R 1 I 0. Proof. Suppose otherwise. Then R 1 (F 0 ) +. Note that R 1 {X A p ξ σ σ T and the length of σ is 1 } and that the right hand side is a disjoint union. Maximality of the family {X A p ξ σ σ T and the length of σ is 1 } implies that R 1 A p ξ σ F p + σ, for some σ T of the length 1. But R 1 A p ξ σ = R σ and R σ I pσ, contradiction. of the claim. Continue in a similar fashion for each n < ω. We will have R n I n 1. Set R ω := R n. n<ω Then R ω I ω and X \ R ω (F ω ) +. Now, let α X \ R ω. Using disjointness of A s we can find a sequence σ [κ + ] ω such that α n<ω(a p σ n ξ σ n {ν < κ g n+1 (ν) < g n (ν)}). Then g n+1 (α) < g n (α), for every n < ω. Contradiction. Lemma 4.5 The generic ultrapower by F ω is well founded up to the image of τ. Proof. Suppose that h n n < ω is a sequence of (F ω ) + -names of old (in V) functions from κ to τ. Let G (F ω ) + be a generic ultrafilter. Choose X 0 G and a function h 0 : κ τ, h 0 V so that X 0 F + ω ȟ 0 = h 0. Let α 0 < κ + be so that f α0 = h 0. By Lemma 4.4, we can find σ 0 [κ + ] <ω such that p σ0 is defined and X 0 F pσ0. Note that at the next stage of the construction there will be β 0 with A pσ0 α 0 F pσ, and so the value of j(f α0 )(κ) 0 β 0 22

23 will be decided. Denote this value by γ 0. Set Y 0 = X 0 A pσ0 α 0 {A pσ0 iξ σ0 i i length(σ 0)}. Assume for simplicity that Y 0 is in G (otherwise we could replace X 0 by another positive set using density). Continue below Y 0 and pick X 1 G and a function h 1 : κ τ, h 1 V such that 1. X 1 F + ω ȟ 1 = h 1, 2. {ν < κ h 1 (ν) < h 0 (ν)} X 1 or {ν < κ h 1 (ν) h 0 (ν)} X 1. If {ν < κ h 1 (ν) h 0 (ν)} X 1, then [h 0 ] G [h 1 ] G and so the sequence [h n ] G n < ω is not strictly decreasing. So, suppose that {ν < κ h 1 (ν) < h 0 (ν)} X 1. Let α 1 < κ + be so that f α1 = h 1. By Lemma 4.4, we can find σ 1 [κ + ] <ω such that F pσ1 is defined, σ 1 σ 0 and X 1 F pσ1. Again, note that at the next stage of the construction there will be β 1 with A pσ1 α 1 F pσ, and so the value of j(f α1 )(κ) will be decided. Denote this value by γ 1. Then 1 β 1 we must have γ 1 < γ 0, since p σ 1 β 1 j (h 1) = γ 1, p σ 0 β 0 j (h 0) = γ 0, p σ 1 β 1 p σ 0 β 0, p σ 1 β 1 κ j (X 1) and {ν < κ h 1 (ν) < h 0 (ν)} X 1. Continue the process for every n < ω. There must be k < m < ω such that γ k γ m and Y m G. So the sequence [h n ] G n < ω is not strictly decreasing. Let us deduce now some conclusions concerning the existence of almost precipitous filters. The following answers a question raised in [6]. Corollary 4.6 Assume 0. Let η be a regular cardinal in L. Then there is a generic extension L[G] of L such that 1. L[G] and L have the same cardinals cardinals η, 2. in L[G], η + is an almost precipitous, as witnessed by a normal filter. Proof. Let c > η be a Silver indiscernible for L. By Donder, Levinski [1](Theorem 6) c is a semi-precipitous cardinal in L. Let G Col(η, < c) be L-generic. Then by [1](Theorem 8) 23

24 or by 2.10, c = (η + ) L[G] will be a semi-precipitous cardinal in L[G]. Now the assumptions of 4.1 are satisfied, since it is impossible to have a non-trivial elementary embedding from L to a transitive model N in a generic extension of L. Note that there is no generic extension of L in which the successor of a singular in L cardinal is semi-precipitous (almost precipitous). Just if η is a singular cardinal in L and in a forcing extension L[G] we have η is preserved and η + is a semi-precipitous, then there will be unboundedly many regular L-cardinals below η that change their cofinality in L[G]. This contradicts the Jensen Covering Lemma. Corollary 4.7 Assume that there are class many Ramsey cardinals. Then every uncountable cardinal is almost precipitous, as witnessed by normal filters. Proof. It follows from 2.3 and 4.1. We would like to apply 4.1 in order to characterize almost precipitous cardinals in the model L[U]. The following is likely well known: Theorem 4.8 Assume V = L[U] with U a normal ultrafilter over κ. Then 1. U is the only -semi-precipitous filter. 2. U is the only normal precipitous filter. 3. If W is a precipitous filter, then there are n ω and a partition A k k n such that (a) for every k n, U, W + A k is an ultrafilter over κ isomorphic to a finite power of (b) W = k n (W + A k). Proof. Both (1) and (2) follow immediately from the following general fact about core models: if i : K i(k) is an embedding of a core model formed in its set generic extension, then i is an iterated ultrapower of K by its extenders, see [10] or [13](Theorem 7.4.8). The item (3) requires a little additional argument. Let W be a precipitous filter over some cardinal δ. Force with W + and let i : L[U] L[i(U)] be the corresponding embedding. Then i = i α for some α, where i α is the embedding of the α-th iterated ultrapower by U. 24

25 Claim 2 α < ω, i.e. the iteration is finite. Proof. Suppose otherwise. Then i(κ) κ ω, where κ ω = i ω (κ). Consider η = [id]. Then η is an ordinal in L[i α (U)] the α-th iterated ultrapower. Then there are m < ω, γ 1 <... < γ m η and f : κ m On, f L[U] such that η = i α (f)(κ γ1,..., κ γm ). Pick some n < ω, such that κ n {κ γ1,..., κ γm }. Now, κ n is an ordinal in a generic ultrapower by W, hence for some function g L[U] we must have i(g)(η) = κ n. But then κ n = i(g)(η) = i(g)(i(f)(κ γ1,..., κ γm )) = i(gf)(κ γ1,..., κ γm ). This is impossible since κ n {κ γ1,..., κ γm }. of the claim. Finally back in the ground model, i.e. in L[U], the set S = {ξ X W + X i = i ξ }. By the claim, S ω. Consider a maximal antichain A k k τ of W -positive sets which decide i differentely. Then τ = S and, for every k S, W + A k is an ultrafilter over κ isomorphic to a finite power of U, W = k n (W + A k). Corollary 4.9 Assume V = L[U] with U a normal ultrafilter over κ. Then 1. Every regular uncountable cardinal less than κ is almost precipitous, as witnessed by normal filters and non precipitous. 2. For each τ κ +, κ carries a normal τ-almost precipitous non precipitous filter. Proof. Let η be a regular cardinal less than κ. By 2.8, η is < κ-semi-precipitous. Note that no cardinal less than κ can be -semi precipitous, see for example [10](Lemma 3.47). Hence, η is an almost precipitous, as witnessed by a normal filter, by 4.1. This proves (1). Now, A = {η < κ η is an almost precipitous, as witnessed by a normal filter and a non-precipitous } is in U. Hence, in M κ V/U, for each τ < (κ ++ ) M there is a normal τ-almost precipitous non-precipitous filter F τ over κ. Then F τ remains such also in V, since κ M M. 25

26 We do not know if (2) remains valid once we replace τ κ + by τ < κ It is unclear to us if it is possible to have a cardinal κ which carries a normal precipitous filter, but κ is not κ + -semi-precipitous via some generic ultrafilter with illfounded ultrapower. Let us turn to the case of -semi precipitous cardinals, which was not covered by Theorem 4.1. Combining constructions of [5] with the present ones (mainly, replacing restrictions to sets by restrictions to filters; required adjustments were explained in the proof of 4.1 and also will be described below in the proof of 4.13) we obtain the following. Theorem 4.10 Assume that ℵ 1 is -semi precipitous and 2 ℵ 1 = ℵ 2. Suppose that for some forcing P witnessing this 0 P P i (ℵ 1 ) > (ℵ + 1 ) V. Then ℵ 1 is almost precipitous witnessed by normal filters. Remark 4.11 Note that, if for some limit λ κ ++ of cofinality bigger than κ there is a λ-semi-precipitous filter with a witnessing forcing producing an ill founded ultrapower, then by 4.1, κ is almost precipitous, as witnessed by normal filters, and without the additional assumptions made in the theorem above. We do not know whether -precipitousness implies always that such λ (i.e. one with ill founded ultrapower) exists. Theorem 4.12 Suppose that there is no inner model satisfying ( α that ℵ 1 is -semi precipitous and 2 ℵ 1 o(α) = α ++ ). Assume = ℵ 2. If ℵ 3 is not a limit of measurable cardinals of the core model, then there exists a normal precipitous ideal on ℵ 1. 2 The referee of the paper answered this question affirmatively. Below is his nice argument. It is enough to find arbitrary large λ and a forcing P witnessing the λ-semi-precipitousness of κ, such that the corresponding normal ultrapower N of V is ill-founded. Then 4.1 can be used to get our witnesses to almost precipitousness, but we must also make sure that the normal filter produced differs from U. Let j : V M Ult(V, U) be the U-ultrapower embedding; note that j(κ ++ ) = κ ++. Fix a limit cardinal λ of cofinality κ +, λ = j(λ). Then κ λ M λ = L λ [j(u)]. Force over M with Col(ω, λ). In M[G], there is an embedding j : M λ j(m λ) = j(m) λ, with crit(j ) = κ, since j M λ : M λ j(m) λ is an embedding of this type. Let D be the normal M λ-ultrafilter derived from j. Then D is in fact an M-ultrafilter and a V -ultrafilter. But N = Ult(M, D) cannot be well-founded, since M satisfies no cardinal less than j(κ) is -semi-precipitous, since this occurs at κ in V. Now if G is in fact V -generic, then Ult(V λ, D) (formed with all functions in κ λ) will be well founded, since κ λ M λ. Therefore also Ult(V λ, D) is well founded. But Ult(V, D) is illfounded, since Ult(M, D) is illfounded. So Col(ω, λ) is a forcing P as required. Apply now the proof of 4.1.Since P forces that the ultrapower N is illfounded, P must force that the derived ultrafilter U is not U. Pick some A κ, A U and p P which forces A U. Fix τ < κ ++, run the construction of 4.1 starting with F p instead of F 0P. This will give A F ω. Then F ω will be τ-almost precipitous, and non-precipitous. 26

27 Theorem 4.13 Assume that ℵ 1 is -semi precipitous. Let P be a witnessing this forcing such that 0 P P i (ℵ 1 ) > (ℵ + 1 ) V. Then, after forcing with Col(ℵ 2, P ), there will be a normal precipitous filter on ℵ 1. Sketch of the proof of Let P be a forcing notion witnessing -semi precipitousness such that 0 P P i (ℵ 1 ) > (ℵ + 1 ) V. Fix a function H such that for some p P p P i (H)(κ) : ω onto (κ + ) V, where here and further κ will stand for ℵ 1. Assume for simplicity that p = 0 P. Let h α α < κ + be a sequence of the canonical functions from κ to κ. For every α < κ + and n < ω set A nα = {ν H(ν)(n) = h α (ν)}. Set F p = {X κ p κ i (X)}, for each p P. Then, the following hold: Lemma 4.14 For every α < κ + and p P there is n < ω so that A nα F + p. Lemma 4.15 Let n < ω and p P. Then the set {A nα α < κ + and A nα F + p } is a maximal antichain in F + p. Denote by Col(ℵ 2, P ) = {t t is a partial function of cardinality at most ℵ 1 from ℵ 2 to P }. Let G Col(ℵ 2, P ) be a generic and C = G. We extend F 0P now as follows. Start with n = 0. If {α A 0α F 0 + P } < κ +, then set F 0 = F 0P. Suppose otherwise. Let α < κ +. If A 0α in the ideal dual to F 0P, then F α will be undefined. 27

28 If A 0α F 0 + P, then we consider F C(α). If A 0α F + C(α), then pick some p α P forcing κ i (A 0α ) and set F α = F p α. If A 0α F + C(α), then pick some p α P, p α C(α) forcing κ i (A 0α ) and set F α = F p α. Set F 0 = {F α α < κ +, F α is defined }. Let now n = 1. Fix some γ < κ + with F γ defined. If {α A 1α F + γ } < κ+, then we do nothing. Suppose that it is not the case. Let α < κ +. We define F γ,α as follows: if A 1α F + γ, then F γ,α will be undefined, if A 1α F + γ, then consider F C(α). If there is no p stronger than both C(α), p γ and forcing κ i (A 1α ), then pick some p γ,α p γ which forces κ i (A 1α ) and set F γ,α = F p γ,α. Otherwise, pick some p γ,α C(α), p γ which forces κ i (A 1α ) and set F γ,α = F p γ,α. Set F 1 = {F γ,α α, γ < κ +, F γ,α is defined }. Continue by induction and define similar filters F s, F n and conditions p s for each n < ω, s [κ + ] <ω. Finally set F ω = the closure under ω intersections of F n. The arguments like those of 4.1 transfer directly to the present context. We refer to [5] which contains more details. Let us prove the following crucial lemma. Lemma 4.16 F ω is a precipitous filter. n<ω Proof. Suppose that g n n < ω is a sequence of F + ω -names of old (in V ) functions from κ On. Let G F ω + be a generic ultrafilter. Pick a set X 0 G and a function g 0 : κ On in V such that X 0 F + ω g 0 = ǧ 0. Pick some t 0 Col(ℵ 2, P ), t 0 C such that t 0, X 0 Col(ℵ2,P ) F + ω g 0 = ǧ 0 28

29 and for some s 0 = ξ 0,..., ξ n [κ + ] <ω t 0 X 0 F s 0, moreover, for each i n, ξ i dom(t 0 ) and t 0 (ξ n ) = p s0. We shrink X 0 to a set X0 = X 0 A iξi. i n Clearly, still t 0 X 0 F s 0. + Claim 3 For each t, Y Col(ℵ 2, P ) F ω t, Y, ρ 0 On and s 0 extending s 0 such that with t, Y t 0, X 0 there are q 0, Z 0 1. q 0 (s 0( s 0 )) p s 0, 2. q 0 Col(ℵ2,P )Ž0 F s 0, 3. p s 0 P i (g 0 )(κ) = ˇρ 0. Proof. Suppose for simplicity that t, Y = t 0, X0. We know that t 0 decides F s0, t 0 (s 0 ( s 0 )) = p(s 0 ) and X 0 F s0. Find s extending s 0 of the smallest possible length such that the set B = {α A s α F s + 0 } has cardinality κ +. Remember that we do not split F s0 before getting to such s. Pick some α B\dom(t 0 ). A s α F s + 0, hence there is some p P, p p s0 which forces κ i (A s α ). Find some p P, p p and ρ 0 such that p P i (g 0 )(κ) = ρ 0. Extend now t 0 to q 0 by adding to it α, p. Let s 0 = s α and Z 0 = X 0 A s α. of the claim. By the genericity we can find q 0, Z 0, s 0 as above with q 0, Z 0 C G. Back in V [C, G], find X 1 Z 0 in G and a function g 1 : κ On in V such that X 1 F + ω g 1 = ǧ 1. Assume [g 1 ] G < [g 0 ] G. Then we can pick X 1 to be a subset of {ν g 1 (ν) < g 0 (ν)}. Proceed as above only replacing X 0 by X 1. This will define q 1, Z 1, s 1 and ρ 1 for g 1 as in the claim. Note 29

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