HEIKE MILDENBERGER AND SAHARON SHELAH

Transcription

1 A VERSION OF κ-miller FORCING HEIKE MILDENBERGER AND SAHARON SHELAH Abstract. Let κ be an uncountable cardinal such that 2 <κ = κ or just cf(κ) > ω, 2 2<κ = 2 κ, and ([κ] κ, ) collapses 2 κ to ω. We show under these assumptions the κ-miller forcing with club many splitting nodes collapses 2 κ to ω and adds a κ-cohen real. 1. Introduction Many of the tree forcings on the classical Baire space have various analogues for higher cardinals. Here we are concerned with Miller forcing [4]. For a κ-version of Miller forcing, in addition to superperfectness one usually requires (see, e.g., [2, Section 5.2]) limits of length < κ of splitting nodes be splitting nodes as well and that splitting mean splitting into a club. In this paper we investigate a version of κ-miller forcing where this latter requirement is waived. We show: If cf(κ) > ω, cf(κ) = κ or cf(κ) < 2 cf(κ) κ, 2 2<κ = 2 κ, and there is a κ-mad family of size 2 κ, then this variant of Miller forcing is related to the forcing ([κ] κ, ) and collapses 2 κ to ω. In particular, if ω < κ <κ = κ, then our four premises are fulfilled. Throughout the paper we let κ be an uncountable cardinal. We write for end extension of functions whose domains are ordinals. If dom(t), i are ordinals, we write tˆ i for the concatenation of t with the singleton function {(0, i)}, i.e., tˆ i = t {(dom(t), i)}. We denote forcing orders in the form (P, P ) and let p P q mean that q ist stronger than p. We write λ> κ for the set of functions f : κ for some < λ. The domain of f is also called the length of f. The set of subsets of κ of size κ is denoted by [κ] κ. Definition 1.1. (1) Q 1 κ is the forcing ([κ] κ, ). (2) Q 2 κ is the following version of κ-miller forcing: Conditions are trees T κ> κ that are κ superperfect: for each s T there is s t such that t is a κ-splitting node of T (short t spl(t )). A node t T is called a κ-splitting node if set p (t) = {i < κ : tˆ i T } Date: December 6, Mathematics Subject Classification. Primary 03E05; Secondary 03E04, 03E15. Key words and phrases. Forcing with higher perfect trees. This research, no on the second author s list, was partially supported by European Research Council grant

2 2 HEIKE MILDENBERGER AND SAHARON SHELAH has size κ. We furthermore require that the limit of an increasing in the tree order sequence of length less than κ of κ-splitting nodes is a κ-splitting node if it has length less than κ. For p, q Q 2 κ we write p Q 2 κ q if q p. So subtrees are stronger conditions. (3) For p Q 2 κ and η p we let suc p (η) = {η κ> κ : ( i κ)(η = ηˆ i p)}. (4) Let η p Q 2 κ. We let p η = {ν p : ν η η ν}. (5) For a, b κ we write a κ b if a \ b < κ. Each of the two forcing orders P has a weakest element, denoted by 0 P. Namely, Q 1 κ has as a weakest element 0 Q 1 κ = κ, and Q 2 κ has as a weakest element the full tree κ> κ. We write P ϕ if the weakest condition 0 P forces ϕ. 2. Results about Q 1 κ We will apply the following result for χ = 2 κ. Theorem 2.1. ([5, Theorem 0.5]) (1) Under the assumption of an antichain of size χ in Q 1 κ, Q 1 κ collapses χ to ℵ 0 if ℵ 0 < cf(κ) = κ or if ℵ 0 < cf(κ) < 2 cf(κ) κ. (2) Under the assumption of an antichain of size χ in Q 1 κ, Q 1 κ collapses χ to ℵ 1 in the case of ℵ 0 = cf(κ). Definition 2.2. A family A [κ] κ is called a κ-almost disjoint family if for A B A, A B < κ. A κ-almost disjoint family of size at least κ that is maximal is called a κ-mad family. Observation 2.3. If 2 <κ = κ, there is a κ-mad family A [κ] κ of size 2 κ. Proof. We let f : κ> 2 κ be an injection. We assign to each branch b of κ> 2 a set a b = {f(s) : s b}. Then we complete the resulting family {a b : b branch of κ> 2} to a maximal κ-almost disjoint family. Observation 2.4. If Q 1 κ collapses 2 κ to ω, then there is a κ-mad family A of size 2 κ. Proof. Q 1 κ cannot have the 2 κ -c.c. Hence there is an antichain of size 2 κ. This is a κ-ad family, and we extend it to a κ-mad family. For further use, we indicate the hypothesis for each technical step. Lemma 2.5. Suppose that Q 1 κ collapses 2 κ to ω. Then there is a Q 1 κ-name τ : ℵ 0 2 κ for a surjection, and there is a labelled tree T = (a η, n η, ϱ η ) : η ω> (2 κ ) with the following properties (a) a = κ and for any η ω> (2 κ ), a η [κ] κ. (b) η 1 η 2 implies a η1 a η2.

3 (c) n η [lg(η) + 1, ω). A VERSION OF κ-miller FORCING 3 (d) If a [κ] κ then there is some η ω> (2 κ ) such that a a η. (e) If ηˆ β T then a ηˆ β forces n η = ϱ ηˆ β for some ϱ ηˆ β τ nη (2 κ ), such that the ϱ ηˆ β, β 2 κ, are pairwise different. Hence for any η ω> (2 κ ), the family {a ηˆ : < 2 κ } is a κ-ad family in [a η ] κ. Proof. Let τ be a Q 1 κ-name such that Q 1 κ τ : ℵ 0 2 κ is onto. For < 2 κ let AP be the set of objects m satisfying ( ) 1 (1.1) m = (T, ā, n, ϱ) = (T m, ā m, n m, ϱ m ). (1.2) T is a subtree of ( ω> (2 κ ), ) of cardinality + κ and T. (1.3) ā = a η : η T fulfils η ν a ν a η and a = κ and a η [κ] κ. (1.4) n = n η : η T fulfils dom(ϱ ηˆ β ) = n η > lg(η) for any ηˆ β T. (1.5) If ηˆ β T, then a ηˆ β forces a value to τ n η called ϱ ηˆ β and for β γ we have ϱ ηˆ β ϱ ηˆ γ. Hence for any ηˆ β, ηˆ γ T m, β γ implies a ηˆ β a ηˆ γ [κ] <κ. (1.6) For η T m, we let Pos(a η, n η ) = {ϱ nη (2 κ ) : a η Q 1 κ τ n η ϱ}, and require that the latter has cardinality 2 κ. In the next items we state some properties of AP that are derived from ( ) 1. ( ) 2 AP = {AP : < 2 κ } is ordered naturally by AP, which means end extension. ( ) 3 (a) AP is not empty and increasing in. ( ) 4 (b) For infinite, AP is closed under unions of increasing sequences of length < +. Let γ < 2 κ. If m AP γ and η T m and ηˆ T m then there is m AP γ such that m AP m and T m = T m {ηˆ }. Proof: For η T m, U = Pos(a η, n η ) = {ϱ nη (2 κ ) : a η Q 1 κ τ n η ϱ} has size 2 κ, whereas Λ η = {ϱ ηˆ β n η : β 2 κ ηˆ β T m } is of size T m γ + κ. Hence we can choose ϱ U \ Λ η and b [a η ] κ such that b Q 1 κ ϱ = n η. We let ϱ ηˆ = ϱ. Since b τ forces a value of τ n η that is incompatible with the one forced by a ηˆ β for any ηˆ β T m, the set b is κ-almost disjoint from a ηˆ β for any ηˆ β T m. We take b = a m,ηˆ a m,η. Since cf(2 κ ) > ℵ 0 and since {range(ϱ) : ϱ ω> (2 κ ) b Q 1 κ τ n ϱ} = 2 κ,

4 4 HEIKE MILDENBERGER AND SAHARON SHELAH ( ) 5 there is an n such that Pos(b, n) = {ϱ n (2 κ ) : b Q 1 κ τ n ϱ} has cardinality 2 κ. We take the minimal one and let it be n ηˆ. If m AP and a [κ] κ then there is some m m, such that there is η T m with a m,η a. Let U a = {ϱ ω> (2 κ ) : a Q 1 κ ϱ τ }, i.e. U a = {ϱ ω> (2 κ ) : ( b Q 1 κ a)(b Q 1 κ ϱ τ )}. This set has cardinality 2 κ because Q 1 κ τ : ω 2 κ is onto. We take n minimal such that has size 2 κ. We let U a,n = {ϱ n (2 κ ) : ( b Q 1 κ a)(b Q 1 κ ϱ τ )} set + n ( m) = {ϱ η : η T m, lg(ϱ η ) n}. Clearly set + n ( m) T m γ + κ. Thus we can take ϱ a U a,n that is incompatible with every element of set + n ( m). We take some b a [a] κ such that b a Q 1 κ ϱ a τ. The set Λ a = {η T m : b a κ a η } is -linearly ordered by ( ) 1 clauses 1.3 and 1.5 and Λ a. Since b a does not pin down τ, Λ a has a -maximal member η a. Now we take = min{β : η aˆ β T m }. For any η aˆ β T m we have ϱ ηaˆ β and ϱ a are incompatible, and hence a ηaˆ β b a [κ] <κ. Now we choose b 1 a [b a ] κ and ϱ a such that b 1 a Q 1 κ ϱ a τ and lg(ϱ a) n m,ηa > lg(η a ). We let T m = T m {η aˆ }, a ηaˆ = b 1 a, We let n ηaˆ be the minimal n such that Pos(b 1 a, n) 2 κ. So ( ) 5 holds. Now we are ready to construct T as in the statement of the lemma. We do this by recursion on 2 κ. First we enumerate [κ] κ as c : < 2 κ, and we enumerate ω> (2 κ ) as η : < 2 κ such that η η β implies < β. We choose an increasing sequence m by induction on < 2 κ. We start with the tree { }, a = κ, ϱ =, n be minimal such that Pos(κ, n) = 2 κ. In the odd successor steps we take m 2+1 AP m so that a η c for some η T 2+1. This is done according to ( ) 5. In the even successor steps we take m 2+2 AP m 2+1 such that η T 2+2. Since all initial segments of η appeared among the η β, β <, m 2+2 is found according to ( ) 4. In the limit steps we take unions. Then T that is given by the the last three

5 A VERSION OF κ-miller FORCING 5 components of m 2 κ has properties (a) to (e). Since τ = τ [G] is not in V, for any T as in Lemma 2.5 no sequence of first components of a branch, i.e., no a f n : n ω, f ω (2 κ ) V, has a κ-lower bound. 3. Transfer to Q 2 κ In this section we use the tree T from Lemma 2.5 for finding Q 2 κ-names. Definition 3.1. Let µ, λ be cardinals. For ν, ν λ> µ we write ν ν if ν ν and ν ν. Typical pairs (λ, µ) are (ω, 2 κ ) and (κ, κ). An important tool for the analysis of Q 2 κ is the following particular kind of fusion sequence p : < κ <κ in Q 2 κ. Since we do not suppose κ <κ = κ, a fusion sequence can be longer than κ. An important property is that for each ν κ> κ there is at most one < κ <κ such that set p (ν) set p+1 (ν). Lemma 3.2. Let ν : < κ <κ be an injective enumeration of κ <κ such that (3.1) ν ν β < β. Let p, ν, c : < κ <κ be a sequence such that for any λ the following holds: (a) p 0 Q 2 κ. (b1) If = β + 1 < κ <κ and ν β sp(p β ), then (3.2) c β [suc pβ (ν β )] κ and p = p β (ν β, c β ) := {p ν βˆ i β : i c β } {p η β : η ν β ν β η} (b2) If = β + 1 < κ <κ and ν β spl(p β ) then p = p β. (c) p = {p β : β < } for limit κ <κ. Then for any λ κ <κ, p λ Q 2 κ and β < λ, p β Q 2 κ p λ. Proof. We go by induction on λ. The case λ = 0 and the successor steps are obvious. So we assume that λ κ <κ is a limit ordinal and p Q 2 κ for < λ. Since p λ, p λ is not empty, and p λ clearly is a tree. Let t p λ. We show that there is t t that is a splitting node in p λ. We fix the smallest such that ν p0 t is a splitting node in p 0. Then in p 0 there are no splitting nodes in {s : t s ν }. Hence ν spl(p β ) for any β [0, λ]. Now we show that the limit of splitting nodes in p λ is a splitting node. Let ν i : i < γ, γ < κ be an -increasing sequence of splitting nodes of p λ with union ν κ <κ. Then ν is a splitting node of each p, < λ, and also in p λ since set p (ν) : < λ has at most two entries and their intersection has size κ.

6 6 HEIKE MILDENBERGER AND SAHARON SHELAH Definition 3.3. We assume Q 1 κ collapses 2 κ to ω. Let and T = (a η, n η, ϱ) : η τ ω> (2 κ ) be as in Lemma 2.5. Now let Q T be the set of κ-miller trees p such that for every ν spl(p) there is η p,ν = η ν ω> (2 κ ) such that (3.3) set p (ν) = {ε κ : νˆ ε p} = a ην. By the properties of T, the node η p,ν is unique. Lemma 3.4. Assume that Q 1 κ collapses 2 κ to ω, let T be chosen as in Lemma 2.5, and let Q T be defined from T as above. Then Q T is dense in Q 2 κ. Proof. Let p 0 = T Q 2 κ. Let ν : < κ <κ be an injective enumeration of κ <κ with property (3.1). We now define fusion sequence p, ν, c : κ κ according to the pattern in Lemma 3.2 in order to find p κ <κ T such that p κ <κ Q T. Suppose that p and ν are given. If ν is not in p or is not a splitting node in p, then we let p +1 = p. If ν spl(p ), then according to Lemma 2.5 clause (d) there is η ω> (2 κ ) such that suc p (ν ) a η. We choose such an η of minimal length and call it η(). Then we strengthen p to (3.4) Now we have that p +1 = {p ν : ν = ν ˆ i i a η() } {p η : η ν ν η}. η p+1,ν = η(), c = a η(). For limit ordinals λ κ <κ, we let p λ = {p β : β < λ}. Since the sequence p, ν, c : κ <κ matches the pattern in Lemma 3.2, we have p κ <κ Q 2 κ. By construction, for any < κ <κ for any δ [ + 1, κ <κ ), ν spl(p δ ) implies set p+1 (ν ) = set pδ (ν ) = a η(). Hence the condition p = p κ <κ fulfils Equation (3.3) in its splitting node ν with witness η p,ν = η(). Since all nodes are enumerated, we have p κ <κ Q T. We use only the inclusion set p (ν) a ην from Definition 3.3. Definition 3.5. We assume that Q 1 κ collapses 2 κ to ω and the T is as in Lemma 2.5. For T Q T and a splitting node ν of T we set ϱ T,ν := ϱ ηt,ν ω> (2 κ ). Recall η T,ν is defined in Def. 3.3, and ϱ is a component of T. For p Q T, the relation ν ν p does neither imply η ν η ν nor ϱ ν ϱ ν. However, η ν η ν implies a ην a ην and ϱ ν ϱ ν. Observation 3.6. We assume that Q 1 κ collapses 2 κ to ω. Let p 1, p 2 Q T. If p 1 Q 2 κ p 2 then for ν spl(p 2 ) we have ν spl(p 1 ) and ϱ p1,ν ϱ p2,ν. We introduce dense sets:

7 A VERSION OF κ-miller FORCING 7 Definition 3.7. We assume that Q 1 κ collapses 2 κ to ω. Let n ω. D n = { p Q T : ( ν spl(p))(lg(ϱ p,ν ) > n)}. D n is open dense in Q T and the intersection of the D n is empty. The following technical lemma is the first step of a transformation of a Q 1 κ-name of a surjection from ω onto 2 κ into a Q 2 κ-name of such a surjection. Lemma 3.8. We assume that Q 1 κ collapses 2 κ to ω, cf(κ) > ω and 2 (κ<κ) = 2 κ. Let T : < 2 κ enumerate Q 2 κ such that each Miller tree appears 2 κ times. There is (p, n, γ ) : < 2 κ such that (a) n < ω, (b) p D n and p T. (c) If β < and n β n then p β p. (d) γ = γ,ν : ν spl(p ). (e) ( ν spl(p ))(a ηp,ν Q 1 κ γ,ν range(ϱ p,ν)). (f) γ,ν 2 κ \ W <,ν with W <,ν = {range(ϱ pβ,ν) : β <, ν spl(p β )}. Proof. Assume that (p β, n β, γ β ) : β < has been defined and we are to define (p, n, γ ). Note that the p β need not be increasing in strength. ( ) 1 The choice of the a η in Lemma 2.5 and the choice Q T and of η pβ,ν for ν spl(p β ), β <, imply that the set W <,ν is well defined and of cardinality + ℵ 0 < 2 κ. Hence we can choose γ,ν 2 κ \ W <,ν. ( ) 2 With the fusion Lemma 3.2 we choose q T, q Q T, such that ( ν spl(q ))(a ηq,ν Q 1 κ γ,ν range(ϱ q,ν). ( ) 3 ( ) 4 For n ω and ν spl(q ) we let U,ν,n = {β < : n β = n, ν spl(p β ) set q (ν) set pβ (ν) = κ}. U,ν = {U,ν,n : n ω}. (a) If n ω and ν spl(q ) then β U,ν ϱ pβ,ν ϱ q,ν. This is seen as follows. We let a = set pβ (ν) set q (ν). Clearly a Q 1 κ ϱ pβ,ν, ϱ q,ν. So either ϱ pβ,ν ϱ q,ν or ϱ pβ,ν ϱ q,ν. τ However, since γ,ν range(ϱ q,ν) \ W <,ν, only ϱ q,ν ϱ pβ,ν is possible. (b) So {ϱ pβ,ν : β U,ν } has at most lg(ϱ q,ν) elements. (c) The assigment β ϱ pβ,ν is is defined between U,ν and {ϱ pβ,ν : β U,ν }. According to properties (e) and (f) in the induction hypothesis, the assigment is injective, and hence U,ν lg(ϱ q,ν).

8 8 HEIKE MILDENBERGER AND SAHARON SHELAH ( ) 5 ( ) 6 ( ) 7 (d) We state for further use that U,ν is finite. We look at the cone above q and show: ( q q )( ν spl(q))( r,ν Q 2 κ q)( c [set q (ν)] κ ) ( r,ν = q(ν, c) ( β U,ν )(r,ν ν p ν β p ν β r,ν) ν ). How do we find r,ν? Given q Q 2 κ q, we enumerate U,ν as β 0,..., β k 1 and choose c 0,..., c k. We start with c 0 = set q (ν). We let r i = q(ν, c i ). So r 0 = q. Given r i, either r ν i let c i+1 = c i, or there is c i+1 [c i ] κ such that r i+1 = q(ν, c i+1 ) fulfils r ν i+1 p ν β i. In the end we take r,ν = q(ν, c k ). p ν β i, in which case we Now we use ( ) 5 iteratively along all ν κ <κ to find a fusion sequence r,ν, ν, c ν : ν < κ <κ with the appropriate instances of ( ) 5. Then we apply the fusion Lemma 3.2 and get an upper bound r of r,ν, ν κ> κ. Note r ν Hence r q and p β iff r ν p ν β ( ν spl(r ))( β U,ν )(r ν and r ν p β iff r ν p β p β r ν ). p ν β. Finally we choose n and p. There are k and ν such that n < ω and ν spl(r ) such that p = r ν fulfils ( β < )(n β k p p β ). Proof of existence. By induction on k ω we try to find ν k, β k : k ω such that (a) ν k spl(r ), (b) ν k ν m for k < m, (c) β k < and n βk k and r ν k p βk. If we succeed, then ν = {ν k : k ω} = ν spl(r ) by Definition 1.1 (2). Here we use that cf(κ) > ω. Hence r ν Q T {D k : k < ω} and a ηr,ν determines in Q 1 κ for any k < ω the value of τ k. This is a contradiction. So there is a smallest k such that ν k cannot be defined. We let n = k. We let p be a strengthening of r ν k 1 such that p D n. For finding such a strengthening we again invoke the fusion Lemma 3.2. We show that p p β for β < with n β k. Otherwise, having arrived at r ν k 1 we find some β k, such that n βk k and r ν k 1 is compatible with p βk. Then we can prolong ν k 1 to a splitting node ν k spl(p βk ) spl(r ). By the choice of r the latter implies that p βk. However, now we would have found ν k, β k as required in contradiction to the choice of k. r ν k

9 A VERSION OF κ-miller FORCING 9 Remark 3.9. Conditions (a) to (c) of Lemma 3.8 yield: For any k < ω, {p : n k} is dense in Q 2 κ. Proof. Let k and p be given. There is 0 such that T 0 D 0 and T 0 Q 2 κ p. Then p 0 T 0 and n 0. Then there is 1 > 0 such that T 1 Q 2 κ p 0. Then p 1 T 1 and hence by condition (c), n 1 > n 0 0. We can can repeat the argument k 1 times. Now we drop the component γ from a sequence p, n, γ : < 2 κ given by Lemma 3.8. Then we get a sequence with properties (a), (b), and a weakening (c) with the property stated in the remark. Lemma We assume that Q 1 κ collapses 2 κ to ω, cf(κ) > ω and 2 (2<κ) = 2 κ. Let T : < 2 κ enumerate all Miller trees that such each tree appears 2 κ times. If (p, n ) : < 2 κ are such that (a) n < ω, (b) p D n and p T, (c) if β < and n β = n then p β p, (d) for any k ω, {p : n k} is dense in Q 2 κ. Then there is a Q 2 κ-name τ for a surjection of ω onto 2 κ. Proof. Let G be a Q 2 κ-generic filter over V. We define τ (n), a Q 2 κ-name by (n)[g] τ = if p G and n = n. The name is a name of a function by (c). By (d), the domain of is forced to be τ infinite. For any p Q τ 2 κ we let U p = { : T = p}. U p is of size 2 κ, in particular for 2 κ we have U p = 2 κ. Hence there is f : 2 κ 2 κ such that for any, γ 2 κ and β U pγ with f(β) =. We let τ (n) = f(τ (n)). Next we show Q 2 κ range(τ ) = 2κ. Suppose p Q T and < 2 κ are given. By construction the sequence {p β : β < 2 κ } is dense. Let p p γ. Then there is β U pγ, with f(β) =. However, β U pγ means T β = p γ p β by construction. By the definition of τ, p β τ (n β ) = β, so p β f(τ (n β )) =. So we can sum up: Theorem We assume that Q 1 κ collapses 2 κ to ω and cf(κ) > ω and 2 (κ<κ) = 2 κ. Then the forcing with Q 2 κ collapses 2 κ to ℵ κ-cohen reals and the Levy collapse Another vice of a κ-tree forcing is to add κ-cohen reals. In this section we show that under the above conditions, Q κ 2 adds Cohen reals and is equivalent to the Levy collapse of 2 κ to ℵ 0.

10 10 HEIKE MILDENBERGER AND SAHARON SHELAH Lemma 4.1. If P collapses 2 κ to ℵ 0, cf(κ) > ℵ 0, and 2 2<κ = 2 κ, then Q 2 κ adds a κ-cohen real. Proof. Let G be Q 2 κ-generic over V. Let f : ω 2 <κ be a function in V[G], such that ( η 2 <κ )( kf(k) = η). Such a function exists since 2 <κ 2 κ. Since 2 2<κ = 2 κ, we can enumerate all antichains in C(κ) in 2 κ many steps. In V[G], is countable. We list it as n : n < ω. Now we choose η n C(κ) V by induction on n in V[G]: η 0 =. Given η n we choose k n such that f(k n ) = η n and then we choose η n+1 η n, such that η n+1 I n. Then {η : ( n < ω)(η f(k n ))} is a C(κ)-generic filter over V and it exists in V [G], since it it definable from {f(k n ) : n < ω}. Two forcings P 1, P 2 are said to be equivalent if their regular open algebras RO(P i ) coincide (for a definition of the regular open algebra of a poset, see, e.g., [3, Corollary 14.12]). Some forcings are characterised up to equivalence just by their size and their collapsing behaviour. Definition 4.2. Let B be a Boolean algebra. We write B + = B \ {0}. A subset D B + is called dense if ( b B + )( d D)(d b). The density of a Boolean algebra B is the least size of a dense subset of B. A Boolean algebra B has uniform density if for every a B +, B a has the same density. The density of a forcing order (P, <) is the density of the regular open algebra RO(P). Lemma 4.3. [3, Lemma 26.7]. Let (Q, <) be a notion of forcing such that Q = λ > ℵ 0 and such that Q collapses λ onto ℵ 0, i.e., Then RO(Q) = Levy(ℵ 0, λ). 0 Q Q ˇλ = ℵ 0. Lemma 4.4. If Q 1 κ collapses 2 κ to ℵ 0, then Q 1 κ is equivalent of Levy(ℵ 0, 2 κ ). Proof. Q 1 κ has size 2 κ. Hence Lemma 4.3 yields RO(Q 1 κ) = Levy(ℵ 0, 2 κ ). Definition 4.5. A Boolean algebra is (θ, λ)-nowhere distributive if there are antichains p ε = p ε : < ε of P for ε < θ such that for every p P for some ε < θ { < ε : p p ε } λ. Lemma 4.6. [1, Theorem 1.15] Let θ < λ be regular cardinals. (1) Suppose that P has the following properties (a) to (c). (a) P is a (θ, λ)-nowhere distributive forcing notion, (b) P has density λ, (c) in case θ > ℵ 0, P has a θ-complete subset S. The latter means: ( B [S] <θ )( s S)( b B)(b P s). Then P is equivalent to Levy(θ, λ).

11 A VERSION OF κ-miller FORCING 11 (2) Under (a) and (b) P collapses λ to θ (and may or may not collapse ℵ 0 ). Proposition 4.7. If there is a κ-mad family of size 2 κ the forcing Q 1 κ is (ℵ 0, 2 κ )-nowhere distributive. Proof. Lemma 2.5 gives T such that p n = {a η : η n (2 κ )}, n ω, witnesses (ℵ 0, 2 κ )-nowhere distributivity. By Lemma 4.3 and Theorem 3.11 we get: Proposition 4.8. If Q 1 κ collapses 2 κ to ℵ 0, cf(κ) > ℵ and and 2 (κ<κ) = 2 κ then Q 2 κ is equivalent to Levy(ℵ 0, 2 κ ). References [1] Bohuslav Balcar and Petr Simon. Disjoint refinement. In Handbook of Boolean algebras, Vol. 2, pages North-Holland, Amsterdam, [2] Jörg Brendle, Andrew Brooke-Taylor, Sy-David Friedman, and Diana Carolina Montoya. Cichoń s diagram for uncountable cardinals. Israel J. Math., 225(2): , [3] Thomas Jech. Set Theory. The Third Millenium Edition, revised and expanded. Springer, [4] Arnold Miller. Rational perfect set forcing. In J. Baumgartner, D. A. Martin, and S. Shelah, editors, Axiomatic Set Theory, volume 31 of Contemp. Math., pages American Mathematical Society, [5] Saharon Shelah. Power set modulo small, the singular of uncountable cofinality. Journal of Symbolic Logic, 72: , arxiv:math.lo/ Heike Mildenberger, Albert-Ludwigs-Universität Freiburg, Mathematisches Institut, Abteilung für math. Logik, Ernst Zermelo Straße 1, Freiburg im Breisgau, Germany Saharon Shelah, Institute of Mathematics, The Hebrew University of Jerusalem, Edmond Safra Campus Givat Ram, Jerusalem, Israel