RVM, RVC revisited: Clubs and Lusin sets

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1 RVM, RVC revisited: Clubs and Lusin sets Ashutosh Kumar, Saharon Shelah Abstract A cardinal κ is Cohen measurable (RVC) if for some κ-additive ideal I over κ, P(κ)/I is forcing isomorphic to adding λ Cohen reals for some λ. Such cardinals can be obtained by starting with a measurable cardinal κ and adding at least κ Cohen reals. We construct various models of RVC having different properties than this model. Our main results are: (1) κ = 2 ℵ 0 is RVC does not decide S for various stationary S κ. (2) κ is RVC and κ < λ = cf(λ) < 2 ℵ 0 does not decide S for various stationary S λ. (3) κ = 2 ℵ 0 is RVC does not decide the existence of a Lusin set of size κ. We also prove analogues of (1), (2) for real valued measurable cardinals. 1 Introduction A cardinal κ is (atomlessly) real valued measurable (RVM) if for some κ-additive ideal I over κ, P(κ)/I is forcing isomorphic to adding λ random reals for some λ. In [13], Solovay showed that if we start with a measurable cardinal κ and add κ random reals then in the resulting model κ is RVM. The category analogue of RVM was introduced in [5]: A cardinal κ is Cohen measurable (RVC) if for some κ-additive ideal I over κ, P(κ)/I is forcing isomorphic to adding λ Cohen reals for some λ. If we start with a measurable cardinal κ and add κ Cohen reals, then in the resulting model κ is RVC. In [5] it was shown that 2 ℵ 0 is RVC does not decide the statement: For every function f : R R, there is a non meager set X R such that f X i s continuous. We continue this line of investigation by constructing models of RVC with different properties than Prikry s model. In Sections 2 and 3, we mainly deal with the existence of club guessing sequences. In Section 4, we consider the existence of Lusin sets Mathematics Subject Classification: 03E35, 03E55. Key words: Forcing Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J Safra Campus, Givat Ram, Jerusalem 91904, Israel; akumar@math.huji.ac.il; Supported by a Postdoctoral Fellowship at the Einstein Insititute of Mathematics funded by European Research Council grant Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J Safra Campus, Givat Ram, Jerusalem 91904, Israel and Department of Mathematics, Rutgers, The State University of New Jersey, Hill Center-Busch Campus, 110 Frelinghuysen Road, Piscataway, NJ , USA; shelah@math.huji.ac.il; Partially supported by European Research Council grant Paper no

2 1.1 On notation For λ < κ = cf(κ), Sλ κ = {δ < κ : cf(δ) = cf(λ)}. In forcing we use the convention that a larger condition carries more information - So p q means q extends p. For a forcing notion P, G P is the P name for the generic added by P. For a regular infinite cardinal κ, H κ is the family of sets whose transitive closure has size < κ. An uncountable regular cardinal κ is called quasi measurable if there exists a κ-additive ℵ 1 -saturated ideal over κ. For a regular uncountable cardinal κ, N S κ denotes the non-stationary ideal over κ. If S κ is stationary, the non-stationary ideal restricted to S is defined by N S κ S = {X κ : X S N S κ }. For a set X, Leb = Leb X denotes the Lebesgue measure on 2 X ; we may drop X when cle ar from the context. 1.2 Review of diamonds and clubs For a regular uncountable cardinal κ and a stationary S κ of limit ordinals, (a) S is the statement: There exists a sequence A α : α S such that each A α α and for every A κ, the set {α S : A α = A α} is stationary. (b) S is the statement: There exists a sequence A α : α S such that each A α [P(α)] α and for every A κ, the set {α S : A α / A α } is non-stationary. (c) S is the statement: There exists a sequence A α : α S such that each A α is an unbounded subset of α and for every unbounded X κ, the set {α S : A α X} is non empty (equivalently stationary). We need the following facts about diamonds and clubs. Fact 1.1. For κ, S as above, S is equivalent to S + κ <κ = κ. Fact 1.2. Let κ be a successor cardinal and S κ be stationary. Assume S. Then for every stationary T S, T holds. Fact 1.3. Let κ, S be as above. Assume S, P is a κ-cc poset and P κ. Then V P S. Furthermore, if V = κ, so does V P. Proof: Using P κ, nice P-names for subsets of κ (union of sets of form W {α} where α < κ and W is an antichain in P) can be thought of as subsets of κ by identifying P κ with κ. If A κ codes such a nice name, then let n(a) denote the corresponding P-name. Fix a S witnessing sequence A α : α S in V. In V [G], define B α : α S by B α = eval G (n(a α )) α if A α codes a nice name for a subset of κ, otherwise B α = 0. Fix a code A for a nice P-name n(a) for a subset of κ. Then for each α < κ, A α also codes a nice name. Moreover, using κ-ccness of P, the set {α < κ : eval G (n(a α)) α = eval G (n(a)) α} contain a club. Since A α = A α on a stationary subset of S, B α = eval G (n(a)) α on a stationary subset as well. Hence B α : α S witnesses S in V [G]. The preservation of κ has a similar proof. Fact 1.4. Suppose κ is regular uncountable and for every stationary S κ, S holds. Let P add κ Cohen reals. Then V P = S for every stationary S κ. 2

3 Proof: Let p S κ is stationary. Put S 1 = {δ < κ : ( p δ p )(p δ δ S)}. So S 1 is stationary. For δ S 1, choose p δ p such that p δ δ S.. By Fodor s lemma, we can find a stationary S 2 S 1 such that {p δ : δ S 2 } forms a delta system with root p p. Note that for every unbounded X S 2, p X S φ. So using S2, we can proceed as in the proof of Fact 1.3. Fact 1.5 (Shelah). Suppose κ = κ ℵ 0 is regular uncountable and S κ is stationary. Suppose that for every S S, stationary in κ, S holds. (1) Let P add κ random reals. Then, in V P, for every stationary S S, S holds. (2) Let Q add κ Cohen reals. Then, in V Q, for every stationary S S, S holds. (3) Suppose ℵ 0 < = cf(), κ < = κ. Let R be a -c.c. forcing of size κ. Then, in V R, for every stationary S S, S holds. Proof: (1) Suppose p S S is stationary in κ. Let S 1 = {δ < κ : δ limit ordinal and ( p δ p )(p δ δ S)}. So S 1 is stationary. For δ S 1, choose p δ p such that p δ δ S. Claim 1.6. There exists p p such that for every p p, {δ S 1 : p, p δ are compatible} is stationary. Proof: If not, let X be a maximal antichain above p such that for every q X, S 1,q = {δ S 1 : q, p δ are compatible} is non-stationary. Since X is countable we can choose δ S 1 \ q X S 1,q. But now p δ is compatible with some q X which is impossible. Claim 1.7. For every stationary S S, there are κ + many stationary subsets of S whose pairwise intersections are bounded below κ. Hence the non-stationary ideal over κ restricted to any stationary subset of S is not κ + -saturated. Proof: Fix a S sequence C δ : δ S. Let {A i : i < κ} be a partition of κ into sets of size κ. Inductively construct {A i : κ < i < κ + } such that each A i [κ] κ and for every i < j < κ +, A i A j < κ. For each i < κ +, let S i = {δ S : C δ A i }. Then {S i : i < κ + } is as required. Claim 1.8. Let Y = {q P : q p }. For q Y, let S 2,q = {δ S 1 : q, p δ are compatible}. There exists a family S 3,q : q Y such that each S 3,q S 2,q is stationary and if q 1 q 2 Y then S 3,q1 S 3,q2 = φ. Proof: Since P = κ ℵ 0 = κ, Y κ. By previous claim, N S κ S is nowhere κ + - saturated. Hence by Theorem 1.2 in [1], we can find a disjoint refinement for any family of size κ. Let Λ be the family of all triplets (p, α, p) where p P, α = α k : k < ω, α k < κ and p = p k : k < ω is a maximal antichain of P above p. Since κ ℵ 0 = κ, we can get a list (p i, α i, p i ) : i < κ in which each member of Λ occurs κ times. For each q Y, let C q δ : δ S 3,q be a S3,q witnessing sequence. We can assume that C δ,q is an unbounded subset of δ of order type cf(δ). Put S 3 = {S 3,q : q Y } and C = C δ : δ S 3 = q Y Cq δ : δ S 3,q. For q Y and δ S 3,q, let D δ = {α i,k : i C δ and ( k < ω)(p i,k G P )}. Let S 4 = {δ S 3 S : D δ is an unbounded subset of δ of order type cf(δ)}. It is enough to show the following. 3

4 Claim 1.9. Suppose q p and q Å [κ]κ. Then for some q q, δ < κ, q (δ S 4 and D δ Å). Proof: Note that since q p, for every δ S 3,q, q, p δ are compatible. Let β j : j < κ be defined by β j = min(å\j). For each j < κ, choose (β j, q j ) such that q j = q j,k : k < ω is a maximal antichain above q, β j = β j,k : k < ω and q j,k β j = β j,k. Let f(j) be the least i [j, κ) such that (q, β j, q j ) = (p i, α i, p i ). Let E κ be a club such that for each δ E, j < δ and k < ω, f(j) < δ and α j,k < δ. Let A = {f(j) : j < κ}. Since C δ : δ S 3,q is a S3,q witnessing sequence, we can choose δ E S 3,q such that C δ A. As p δ, q are compatible, we can choose q p δ, q. Let G be P-generic over V with q G. So δ S = S[G]. Since δ E, D δ δ. As C δ is unbounded in δ and β j j, D δ is also unbounded in δ. Hence δ S 4. Finally note that since C δ A, each member of D δ is of the form α f(j),k where p f(j),k G. As q f(j),k β f(j),k = β j and p f(j),k = q f(j),k, α f(j),k = β f(j),k, we get α f(j),k A. (2) Follows from (3). (3) This has essentially the same proof as (1). In the construction of D δ, we use a modified version of Λ where we consider sequences of length <. The assumptions κ < = κ, R satisfies -c.c. ensure that Λ κ. Fact Let ω σ = cf(σ) < κ = cf(κ). Suppose P satisfies σ + -c.c., S κ is stationary and V P S. Then V S. Proof: Let p Åα : α S is a S witnessing sequence. For each α S, choose B α [P(α)] σ such that the following holds: If for some q p and B α V, q Åα = B α, then B α B α. It is easy to see that, for every B κ, for stationary many α S, B α B α. Let f : κ σ κ be a bijection. Let C κ be a club such that for every α C, f α : α σ α is a bijection. For α C S, let D α = {f[b] : B B α }. Note that for every D σ κ, there are stationary many α C S such that D (σ α) D α. Enumerate D α = {D α,i : i < σ}. Fo r j < σ, let D α,i,j = {β : (j, β) D α,i }. We claim that for some i < σ, D α,i,i : α C S witnesses S. Suppose not and for each i < σ, let D i κ be such that {α C S : D i α D α,i,i } is non-stationary. Put D = {(i, α) : α D i }. But then {α C S : D (σ α) D α } is non-stationary: A contradiction. Fact 1.11 (Kunen). Suppose κ 2 ℵ 0 is quasi measurable with I a witnessing normal ideal. Then, for every I-positive S κ, S holds. Note that the assumption κ 2 ℵ 0 cannot be dropped even for getting κ. For example, if we start with κ measurable and add κ + Cohen reals then in the resulting model κ is quasi measurable (in fact, κ is RVC) but κ fails. Question Suppose κ < 2 ℵ 0 is quasi measurable. Can κ hold? We intend to deal with this in a forthcoming work. 4

5 2 Clubs: When continuum is RVC/RVM Lemma 2.1. It is consistent that κ = 2 ℵ 0 is RVC and for every stationary S κ, S holds. It is also consistent that κ = 2 ℵ 0 is RVM and for every stationary S κ, S holds. Proof: Start with κ measurable and V = L[I] where I is a witnessing normal prime ideal over κ. Then for every stationary S κ, S holds. Now add κ Cohen/random reals and use Fact 1.5. We next show that κ = 2 ℵ 0 is RVC/RVM and S (equivalently, S ) for various stationary S κ is consistent. Suppose λ is inaccessible and GCH holds. Let ω = cf() < λ and S S λ be a stationary set of singular cardinals in λ. Let Q = Q S be the forcing whose conditions are p 2 <λ where p 1 [{1}] S is nowhere reflecting, ordered by end extension. Let S 1 S be the generic added by Q 1. In V [S 1 ], let (A, ρ) = A i, ρ i : i S 1 satisfy: A i is an unbounded subset of i of order type, ρ i : A i 2. Let R = R (A,ρ) V [S 1 ] be the forcing to uniformize (A, ρ). So p R iff p = (β, E, g) where β < λ, E β + 1 is a club and g : β 2 such that if i S 1 E, then ρ i g where means inclusion modulo ρ i restricted to a bounded subset of A i. The orderin g is (β, E, g) (β, E, g ) iff β β, E is an end extension of E and g g. Claim 2.2. Q R is λ-strategically closed. In fact, it has a dense λ-closed subset. Proof: The completeness player always plays conditions of the form (p, q) where q = (β, E, g) and β dom(p). Definition 2.3. For λ,, S as above, let P = P i, Q j : i λ +, j < λ + be an iteration with < λ support such that the following hold. If j is odd, in V P j, Q j = Q S for some stationary S S λ, ω = cf() < λ. Let S j be a P j+1 name for the generic subset of S added by Q j If j is even, in V P j, Q j = R (A,ρ) where for some j < j, (A, ρ) is a uniformizing candidate on S j Through usual bookkeeping, we ensure that every stationary set of singular cardinals and every uniformizing candidate is taken care of at some stage before κ + Claim 2.4. Let P be as in Definition 2.3. Then P i satisfies λ + -c.c. and is λ-strategically closed for every i λ +. In V P λ +, every stationary set S λ consisting of singular cardinals has a stationary subsets S such that S does not hold. Proof: The main point is that the stationarity of S j V P j+1 is preserved during the later stages of the iteration. For a proof of this, see [8] and more in [9]. Theorem 2.5. Assume GCH. Suppose κ is κ + -supercompact. Then there is a forcing P such that in V P the following hold. 5

6 1. κ is measurable, 2. If S κ is stationary consisting of singular cardinals, then for some stationary S 1 S, S1 does not hold. Proof: Let P α, Q β : α κ + 1, β κ be an Easton support iteration where If α is not inaccessible, Q α is trivial If α is inaccessible, Q α V Pα is the iteration of Definition 2.3, with λ = α. It is easily checked that all regular cardinals are preserved in V P κ+1. GCH holds as well. By Claim 2.4, in V P κ+1, every stationary S κ consisting of singular cardinals has a stationary subset S 1 such that S1 does not hold. To finish, we show that κ remains measurable in V P κ+1. Let j : V M be an elementary embedding witnessing the κ + - supercompactness of κ. So crit(j) = κ, κ ++ < j(κ) < κ +++ and M κ+ M. In M, we can write j(p κ ) = P κ Q κ R (as Q κ M). Let G be P κ -generic over V and let H be Q κ -generic over V [G]. Let R = R[G][H]. Working in V [G][H], we would like to extend j to V [G][H]. We do this in two steps. Claim 2.6. In V [G][H], there exists an R-generic filter K over M[G][H] and an elementary extension j 1 : V [G] M[G][H][K] of j. Proof: As P κ+1 has κ + -c.c., every nice name for a sequence of ordinals of length κ + is in M. Hence (M[G][H]) κ+ V [G][H] M[G][H]. It follows that, in V [G][H], R is κ ++ - strategically closed since M[G][H] = R is κ +++ -strategically closed. So in V [G][H] the completeness player can play for κ ++ moves using this strategy. This works because every initial segment of a play is in M[G][H]. Now in M[G][H], R has j(κ) maximal antichains. In V [G][H], we can list them in order type κ ++ and the use the κ ++ -strategic closure of R to construct a filter K R meeting all of them. Finally since j[g] G H K, we can lift j to j 1 : V [G] M[G][H][K]. Now we would like to extend j 1 to V [G][H]. In M, let j(p κ+1 ) = P κ Q κ R S. Claim 2.7. In V [G][H], there exists an S-generic filter L over M[G][H][K] and an elementary extension j 2 : V [G][H] M[G][H][K][L] of j 1. Proof: Since V [G] = Q κ is κ-strategically closed and has κ + maximal antichains, by elementarity, M[G][H][K] = S is j(κ)-strategically closed and has j(κ + ) maximal antichains. Hence as before, in V [G][H], S is κ ++ -strategically closed so that we can build an S-generic filter over M[G][H][K] above a master condition which we now define as follows. In V [G][H], consider the pointwise image j 1 [H]. Since H can be thought of as a κ + -sequence of subsets of κ, there is a condition q S such that every member of j[h] is below q. So we construct an S-generic filter L over M[G][H][K] with q L. It is now easy to lift j 1 to j 2 : V [G][H] M[G][H][K][L]. Corollary 2.8. It is consistent that κ = 2 ℵ 0 is RVC (resp. RVM) and for every stationary S 1 κ consisting of singular cardinals, there exists stationary S S 1 such that S. Proof: Start with κ measurable, P as above and let Q add κ Cohen reals (resp. random reals). Use Fact

7 3 Clubs: When 2 ℵ 0 > κ and κ is RVC/RVM Lemma 3.1. Suppose κ is RVC or RVM and ω 1 = cf() < κ. Then. Proof: Let I be an RVC (resp. RVM) witnessing ideal over κ. Force with I and form the generic ultrapower M V [G]. Let j : V M be the generic elementary embedding. By [4], G is generic for adding > κ Cohen reals (resp. random reals) over V so fails in V [G] hence also in M as M κ V [G] M. By elementarity, this also holds in V. Lemma 3.2. Suppose κ is measurable and λ = λ ℵ 0 > κ. Let P be the forcing for adding λ Cohen reals (resp. random reals). Then, in V P, κ is RVC (resp. RVM), 2 ℵ 0 = λ and for every uncountable regular < λ, we have. Proof: Use the Cohen/random sequence to outguess every potential club guessing sequence. So we can ask the following Question 3.3. Can we force S for various stationary S where κ < 2 ℵ 0 with κ is RVC (resp. RVM)? together This is addressed in the next two sections. 3.1 Some club preserving forcings On the following forcing, see [2], and for a more general treatment [11]. Definition 3.4. For ω σ κ = cf(κ) < λ, we define Q = Q λ,κ,σ as follows: (A) p Q iff p = (u, f) = (u p, f p ) where u [λ] <κ and f : u 2 <σ (B) For p, q Q, p q iff for some r Q, p pr r (read r is a pure extension of p ) and r apr q (read r is a apure extension of p ), where (C) p pr q iff u p u q and f p = f q u p (D) p apr q iff u p = u q = u and for every α u, f p (α) f q (α) and {α u : f p (α) f q (α)} < σ (E) For p Q, we write Q[p] for the forcing whose conditions are apr -extensions of p ordered under Q (here it is same as apr ) Lemma 3.5. For σ, κ, λ and Q = Q λ,κ,σ as in Definition 3.4, the following hold. (1) If κ <κ = κ, then Q satisfies κ + -c.c. (even κ + -Knaster) (2) Any pr -increasing sequence of length < κ has a least upper bound (3) Q is < σ-directed complete (4) If σ <σ = σ and p Q, then Q[p] satisfies σ + -c.c. and is σ-closed 7

8 (5) Assume σ = σ <σ, σ + < κ. If p Q τ V, then for some q Q, p pr q the set {r Q : q apr r r forces a value to τ} is predense above q (6) Assume σ = σ <σ, σ + < κ. If < κ and p Q τ : V, then for some q Q, p pr q and for every i <, the set {r Q : q apr r r forces a value to τ(i)} is predense above q (7) If σ <σ = σ and κ is inaccessible, then Q preserves all regular cardinals. Proof: Clauses (1)-(4) are easy to check. For (5), suppose p Q τ V. We try to construct (p i, q i ) : i < σ + such that p 0 = p and p i s are pr -increasing continuous for every i < σ +, p i+1 apr q i q i s are pairwise incompatible q i forces a value to τ Suppose we can complete the construction. Let p σ + be the union of p i s for i < σ +. Let r i be the least upper bound of p σ +, q i. Then r i s are pairwise incompatible apr extensions of p σ + which contradicts (4). So the construction cannot be carried out at some stage i < σ +. If i is limit, let q = p i be the union of p j s for j < i. Let r q be any condition deciding τ. If possible suppose r is incompatible with every r j for some j < i (r j is the least upper bound of q, q j ). Then we can set q i = r u q, p i+1 = q r (u r \u q ) contradicting that the construction could not be carried out at stage i. The case when i is a successor is similar. Clause (6) has a similar proof. For clause (7), note that if σ = κ, Q is < κ-closed and has κ + -c.c. so this is clear. So assume σ < κ. Note that all regular car dinal κ + or σ are preserved. Suppose σ < δ κ be regular cardinals and p Q τ : δ. Using clause (6), get q pr p as there. For each i <, choose a maximal antichain in Q[q] deciding τ(i) and let α < δ be the greater than all these values. Then q Q range(τ) α. Lemma 3.6. Let σ = σ <σ κ = κ <κ < λ and Q = Q λ,κ,σ. Let < κ be an infinite regular cardinal. (1) If S S κ+ is stationary in κ + and S holds as witnessed by B = B i : i S, with B i, then V Q = S with the same witness. (2) If λ 1 = cf(λ 1 ) > κ and α < λ 1 = α <κ < λ 1, S S λ 1 is stationary in λ 1 and S holds as witnessed by B = B i : i S, with B i, then V Q = S with the same witness. (3) Assume 2 κ = κ + and let S S κ+ <κ be stationary in κ +. Then S holds in both V and V Q. (4) Suppose 2 κ = κ +, S Sκ κ+ is stationary and S holds. Then in V Q, S holds iff λ = κ +. Moreover, if S holds for every stationary S Sκ κ+ and λ = κ +, then in V Q, S holds for every stationary S Sκ κ+ 8

9 Proof: (1) Follows from (2). (2) Suppose p Q Å is an unbounded subset of λ 1. For each α < λ 1, let p α, γ α be such that (a) p p α (b) γ α < λ 1 and α < β = γ α < γ β (c) p α Q γ α Å By the assumption on λ 1, using delta system lemma, we can find an unbounded W λ 1 such that {u pα : α W } forms a delta system with root u and p α u does not depend on α W. Since {γ α : α W } is an unbounded subset of λ 1, we can find some δ S such that B δ {γ α : α W }. Since B δ, letting q to be the lub of {p α : α B δ }, q Q B δ Å. (3) For some < κ, S = S κ+ S is stationary in κ +. By [10], S holds in V. Now use (1). (4) If λ > κ +, we can diagonalize against any potential S sequence. When λ = κ +, Q satisfies κ + -c.c. and Q κ +. Since 2 κ = κ +, S holds in V so we can use Fact 1.3. Next suppose λ = κ + and S holds for every stationary S Sκ κ+ and let p S Sκ κ+ is stationary. Let S 1 = {α Sκ κ+ : ( p α p)(p α α S)}. By Fodor s lemma, we can find stationary S 2 S 1, such that {p α : α S 2 } forms a delta system with root p p. Note that every unbounded subset of S 2 is forced by p to meet S. Since 2 κ = κ +, S2 holds in V. So we can proceed as in the proof of Fact 1.3. Lemma 3.7. Suppose κ is inaccessible, λ > κ σ = σ <σ, S κ is stationary and S holds as witnessed by B = B i : i S. Let Q = Q λ,κ,σ. Then V Q = S holds with witness B. Proof: This is clear if σ = κ since Q is < κ-closed. So assume σ < κ. Suppose p Å [κ]κ. Construct (p i, u i, q i, α i, F i ) : i < κ such that the following hold 1. For each i < j < κ, α i < α j 2. p 0 = p and p i s are pr -increasing continuous 3. p i apr q i, u i = dom(p i ) 4. every common extension of p i+1 and q i forces α i Å 5. {β u i : p i (β) q i (β)} = F i For i S κ σ, let h(i) be the least j < i such that F i u j. Then h is constant on some some stationary set W. By thinning W further, we can assume that for some F and q : F 2 <σ, we have F i = F and q i F = q for every i W. Now choose δ S such that B δ {α i : i W }. Let j W be larger than sup{i < κ : α i B δ } and q = {q i : i W, i j}. Then q B δ Å. 9

10 3.2 Getting clubs Theorem 3.8. Suppose κ < µ are measurable, µ remains measurable in any < µ-directed closed forcing extension 1 and 2 µ = µ + < λ = λ µ. Let S 0 = S µ µ+. Then for some forcing P, in V P we have (a) No cardinal > µ is collapsed (b) cf(µ) = ℵ 0 (c) 2 ℵ 0 = λ (d) κ is RVC (e) For every regular < µ +, S µ + holds. (f) Moreover, for every stationary S µ + \S 0, S holds. Proof: Remark 3.9. Can we get S for all stationary subsets of µ +? We intend to do this in a subsequent work both for RVC and RVM (see Corollary 3.24 below). Claim For some < µ + -directed closed, µ ++ -c.c. forcing P 1 V, letting V 1 = V P 1, we have V 1 = µ +. Proof: See [6] Chapter 7, H(20). Claim Let P 2 = Q λ,µ,µ V 1. Put V 2 = V P 2 1. Then, in V 2, 2 µ = λ and for every < µ, holds. S µ + Proof: Note that P 2 is < µ-directed closed so κ, µ remain measurable. Also, 2 µ = λ is easily verified. For S µ +, we use Lemma 3.6. Claim In V 2, let P 3 be Prikry forcing w.r.t. some normal ultrafilter D over µ. Then in V 3 = V P 3 2, κ is measurable, µ is a strong limit of countable cofinality, 2 µ = λ and for every regular < µ +, S µ + holds. Proof: We only check the club principle. If ℵ 0 < < µ, then S = (S µ+ ) V 2 = (S µ+ ) V 3. If = ℵ 0, note that S = (S µ+ ) V 2 (S µ+ ) V 3. Hence it is enough to show V 3 = S. In V 2, fix B = B i : i S witnessing S there. Let p P4 Å µ + is unbounded. Construct (p i, γ i ) : i < µ + such that for i < j, γ i < γ j, p p i and p i P4 γ i Å. We can assume that each p i = (s, X i ), s [µ] <ℵ 0, X i D. Let W = {γ i : i < µ + }. Since W is unbounded in µ +, we can choose δ S such that B δ W. Let p = (s, X) where X = {X i : γ i B δ }. Then p P4 B δ Å. Claim In V 3, let P 4 add µ Cohen reals. Then in V 4 = V P 4 3, we have 1 See [3] for examples. 10

11 (1) κ < µ < λ = 2 ℵ 0 (2) κ is RVC (3) For every regular < µ +, S µ + holds Proof: In V 4, since µ is a strong limit of countable cofinality, µ ℵ 0 = 2 µ = λ. Hence V 4 = 2 ℵ 0 = λ. It is clear that κ is RVC in V 4. Also S µ + is preserved by the same witness because P 4 = µ. We next show clause (f). Lemma If (A), then (B) where (A) (a) µ is strongly inaccessible (b) 2 µ = µ + < λ = λ µ and µ + (c) S 0 = S µ+ µ (d) P = Q 0 Q 1 Q 2 where holds (i) Q 0 = Q λ,µ,µ (ii) D is a Q 0 -name such that V Q 0 = D is a normal ultrafilter over µ (iii) Q 1 V Q 0 is Prikry forcing w.r.t. D (iv) Q 2 is the forcing for adding µ Cohen reals (B) V P = S holds for every stationary S µ + \S 0. Proof: Suppose p S µ + \S 0 is stationary. Note that the set of conditions p = (p 0, p 1, p 2 ) P where p 2 is an actual finite partial function from µ to 2 and p 1 = (w, Å) where w is an actual finite subset of µ is dense in P. Let us call such conditions good and assume, WLOG, that p is good. Since P satisfies µ + -c.c., for each α < µ +, we can find a maximal antichain of good conditions p α,i : i u α µ above p deciding α S. Let p α,i = (p 0 α,i, (w α,i, Åα,i), f α,i ). Let u [λ] µ+ be such that for every α < µ +, i u α, p 0 α,i Q u,µ,µ and Åα,i is a Q u,µ,µ-name. Let G be Q u,µ,µ-generic over V with p 0 G. By Fact??, µ still holds in V [G]. Now (Q + λ,µ,µ ) V [G] = Q 0 /G, so we might as well assume that p 0 = φ and for each α < µ +, i u α, p α,i = (φ, (w α,i, A α,i ), f α,i ). Claim There exist j, w, f,, E such that (a) j < µ, w [µ] <ℵ 0, f is finite partial function from µ to 2, < µ is regular infinite, E µ + is a club (b) Letting S 1 = {δ < µ + : cf(δ) =, j u δ, w δ,j = w, f δ,j = f and p δ,j δ S}, and S 2 = {δ S 1 : p δ,j G P }, we have, for every δ E S 1, p δ,j S 2 is stationary. 11

12 Proof: We can easily choose j, w, f, such that S 1 is stationary. Next we can find a club E µ + such that for every δ E S 1, w [A δ,j ] <ℵ 0, the set {α S 1 : w A α,j } is stationary. Let δ S 1 E and suppose p δ,j S 2 is stationary. Hence there is a good q = (q 0, (w, Å), q2 ) p δ,j and a club E µ + (since P satisfies µ + -c.c.) such that q S 2 E = φ. Let δ S 1 E E be such that w A δ,j w. But now q and p δ,j are compatible and p δ,j δ S 2 E - A contradiction. For every w w [µ] <ℵ 0, let S 1,w = {δ S 1 : w\w A δ,j }. Let W = {w : w w [µ] <ℵ 0 and S 1,w is stationary in µ + }. Let E 1 E be a club in µ + such that for every w w [µ] <ℵ 0, if S 1,w is not stationary, then E 1 S 1,w = φ. Claim There exists S 3,w : w W such that (1) S 3,w S 1,w is stationary (2) w w W implies S 3,w S 3,w = φ Proof: Using Theorem 1.2 in [1]. For each w W, choose a w = a δ : δ S 3,w witnessing S3,w. Let a = a δ : δ E 1 S 1. It suffices to show Claim For each δ S 1 E 1, p δ,j a S witnesses S. Proof: Suppose not and let δ 1 S 1 E 1, q p δ1,j, Å be such that q is good and q Å µ+ is unbounded and for every δ S 2, a δ Å. Let (q α, γ α ) : α < µ + be such that for every α < µ +, q q α = (qα, 0 (v α, B α ), qα) 2 is good and q α γ α is the α-th member of Å. Let E 3 = {δ < µ + : δ is limit and α < δ = γ α < δ}. By Fodor s lemma, for some stationary S 3 S µ µ+ E 3, qα 0 : α S forms a delta system with root q 0 and v α = v, qα 2 = q 2 do not depend on α S 3. As q α q p δ1,j and δ 1 S 1 E 1, the set S 1,v is stationary. Hence S 3,v is also stationary so we can find δ 2 E 3 S 3,v such that a δ2 {γ α : α S 3 }. But now recalling the definition of S 1,v, the set {q α : γ α a δ2 } {p δ2,j } has an upper bound. This completes the proof of Theorem 3.8. What happens when we add µ random reals to V 3? In this case, we get a weak version of S which we now describe. Definition For ℵ 0 < κ = cf(κ) and S κ stationary, the principle inf S says the following: There exists A δ : δ S such that each A δ is an unbounded subset of δ of order type cf(δ) and for every A [κ] κ, for every ε > 0, for stationary many (equivalently, some) δ S, for every α A δ, lim inf n { Aδ A [α, β] n } : α < β < δ n = A δ [α, β] 1 ε Remark Suppose κ < λ = λ ℵ 0, κ is measurable and P adds λ random reals. Then, in V P, for every < λ = 2 ℵ 0, inf does not hold. 12

13 Definition For ℵ 0 < κ = cf(κ), S κ stationary, ℵ 0, σ < κ, the principle sp(,σ) S says the following: There exists A δ : δ S such that each A δ is an unbounded subset of δ of order type cf(δ) and for every A [κ] κ, f α : α A, where each f α : u α σ with u α [κ] <, there exists stationary many (equivalently, some) δ S such that A δ A and f α : α A δ forms a delta system - i.e., for some u, for every α β A δ, u α u β = u and f α u = f β u. Claim Let κ, σ,, S be as in Definition Suppose for every α < κ, α < < κ. Then, sp(,σ) S is equivalent to S via the same witness. Proof: Easily follows from: Whenever f α : α A is as in Definition 3.20, for some B [A] κ, f α : α A forms a delta system. Claim Suppose µ is measurable with a normal ultrafilter D and S S µ+ is a stationary subset of µ + where = cf() < µ. Assume S. Let P be Prikry forcing w.r.t. D. Then, for every, σ < µ, V P = sp(,σ) S. Proof: Note that V P = µ ℵ 0 µ + so it is not enough to just check S. Using Claim 3.21, fix a sp(,σ) S witnessing sequence A δ : δ S in V. We ll show that sp(,σ) S holds in V P with the same witness. So suppose p forces Å [µ+ ] µ+, f α : α Å, ů α [µ + ] <, f α : ů α σ. Let A 1 = {α < µ + : ( p α p)(p α α Å)}. For each α A 1, let p α = (w α, B α ), ξ α < be such that p α p, w α [µ] <ℵ 0, B α D and p α forces α Å and order type of u α is ξ α. Choose A 2 [A 1 ] µ+, w and ξ such that for each α A 2, w α = w, ξ α = ξ. For α A 2, let γ α,i : i < ξ list ů α in increasing order. Since P does not add bounded subsets of µ, we can also assume that each p α forces a value ζ α,i to f α ( γ α,i ). Let X = A 2 ξ. Let E be the equivalence relation on X define by ((α 1, i 1 ), (α 2, i 2 )) E iff for some B D, (w, B) γ α1,i 1 = γ α2,i 2. Let g : X µ + be such that for all x, y X, (x, y) E iff g(x) = g(y). Define f α : α A 2 by dom(f α) = {g(α, i) : i < ξ }, f α(g(α, i)) = ζ α,i. Choose δ S such that A δ A 2 and f α : α A δ forms a delta system. Put B = {B α : α A δ }, q = (w, B). Then q A δ Å. Using the Prikry property, it is easily verified that q f α : α A δ forms a delta system. Claim Suppose 2 ℵ 0 < µ, = cf() < µ, S S µ+ is a stationary subset of µ +, {S i : i < µ} is a partition of S into stationary sets and A = A δ : δ S is such that for each i < µ, A S i witnesses sp(ℵ 1,ℵ 0 ) S i with A as a witness.. Let P add µ random reals. Then V P = inf S holds Proof: Suppose p Å [µ+ ] µ+ and 0 < ε < 1. Let A 1 = {α < µ + : ( p α p)(p α α Å)}. For α A 1, choose a compact p α such that p p α α Å. By thinning down A 1, we can also assume that for some clopen set q, Leb(q p α ) < (1 ε/8)leb(p α ). Let γ α,i : i < ξ α where γ α,i < µ are increasing with i and ω ξ α < ω 1 be a list containing supports of p α, p, q. Using 2 ℵ 0 < µ, choose A 2 [A 1 ] µ+, ξ, K 2 ξ compact such that for every α A 2, ξ α = ξ and p α = {x 2 µ : x α K where x α 2 ξ satisfies x α (i) = x(γ α,i )}. Let i = {i n : n < ω}. For α A 2, define f α = {(γ α,in, n) : n < ω}. Choose δ S such that A δ A 2 and f α : α A δ forms a delta system so for some v, for every α, β A δ, α β, {γ α,i : i < ξ } {γ β,i : i < ξ } = v and f α v = f β v. Let q be the constant projection 13

14 of p α to v. Since both p, q are supported in v, p q, q q and for all α A δ, q p α hence Leb(p α ) > (1 ε/8)(leb(q)). Now the sequence p α : α A δ is independent over q hence the conclusion follows from the law of large numbers. Corollary Let V 3 be as in Claim In V 3, let P 4 be the measure algebra on 2 µ. Then in V 4 = V P 4 3, we have (1) κ < µ < λ = 2 ℵ 0 (2) κ is RVM (3) For every regular < µ +, inf S µ+ holds. Proof: Follows from Claims 3.22 and RVC and Lusin sets Recall that an uncountable set of reals X is Lusin if every meager subset of X is countable. Fact 4.1 (Gitik-Shelah). Suppose κ is RVC. Then for each uncountable < κ, there exists a Lusin set of size. Proof: Let I be a witnessing ideal over κ. Let G be P(κ)/I-generic over V. In V [G], let M be the transitive collapse of the well founded G-ultrapower of V and j : V M V [G], the generic elementary embedding. By [4], G is generic for adding > κ Cohen reals over V so V [G] and therefore M contains a κ-sequence of Cohen reals over V. Hence for each < κ, M and therefore V (by elementarity) contains a Lusin set of size. It is natural to ask: Suppose κ is RVC. Must there exist a Lusin set of size κ? The usual way of getting a model of κ is RVC is to start with a measurable cardinal κ and add κ Cohen reals. The Cohen reals added constitute a Lusin set of size κ. So this is consistent. On the other hand, we ll show the following. Theorem 4.2. It is consistent that 2 ℵ 0 is RV C and there is no Luzin set of size continuum. The proof uses iterations with restricted memory as defined in [5]. The main difference here is that instead of adding a continuous function, we add a meager set by finite approximation. 4.1 Cohen forcing We describe another way of looking at adding ω 1 Cohen reals. Compare with [7]. Definition 4.3. Let σ k : k < ω be an enumeration of 2 <ω. For X = x α : α < ω 1 2 ω, let Q X be the forcing whose conditions are p = (F, g, N) where (1) N < ω, F = F i : i < n, n < ω and each F i is a finite subset of ω 1. 14

15 (2) g = g i : i < n and each g i : {σ k : k < N} 2 <ω satisfies: For each k < N, g i (σ k ) σ k and [g i (σ k )] {x α : α F i } = φ. Here, for a finite partial function σ from ω to 2, [σ] = {x 2 ω : σ x}. (3) Ordering is defined naturally: F i s and g i s increase, F, g lengthen, N increases. Fact 4.4. Let X, Q X be as above. (a) Q X satisfies ω 1 -Knaster condition: For every uncountable A Q X, there exists B [A] ℵ 1 such that the elements of B are pairwise compatible. (b) In V Q X, X is meager. Let C X = C(X) = {C : i X} (finite support product) denote Cohen forcing for adding X many Cohen reals. So C is 2 <ω ordered by end extension. We also write C(X, Y ) = C(X Y ) = C(X) C(Y ) when X, Y are disjoint. Lemma 4.5. Let P = C ω1 add ω 1 Cohen reals c α : α < ω 1. Let τ be a C-name for a real in 2 ω. Let X = X τ = τ[c α ] : α < ω 1. Let Q X be as above. Then, P Q X is forcing equivalent to P. Moreover, if ω < ω 1, then (P Q X )/C is also forcing equivalent to P. Proof: For ω λ ω 1, consider the forcing A λ of all pairs (p, q) where p C λ, p = ρ α : α u p, ρ α 2 <ω, u p [λ] <ℵ 0 {ρ α : α u p } = {ρ n : n u p ω} q = (F, g, N), F = F i : i < n, g = g i : i < n, N < ω For every i < n, F i u p and g i : {σ k : k < N} 2 <ω For every k < N, σ k g i (σ k ) and p Cλ [g i (σ k )] {τ[c α ] : α F i } = φ A λ is ordered in the natural way. Observe that for any (p, q) C λ Q X λ, there exists (p, q ) A λ such that p p, p P q QX λ q. Hence A λ is dense in C λ Q X λ. For ω < λ ω 1, (p, q) A λ, define (p, q) = (p, q ) A in the natural way. Note that if (p, q) A λ, (p, q ) A and (p, q ) (p, q), then (p, q) and (p, q ) are compatible. It follows that A A λ. Hence A λ : λ [ω, ω 1 ] is -increasing and at limit stages λ, A λ is the finite support limit of A β : β < λ. Moreover, since each quotient A λ+1 /A λ is Cohen (being countable atomless), it follows that A ω1 and A ω1 /A are forcing equivalent to C ω1 for every ω < ω 1. Hence (P Q X )/C is equivalent to A ω1 /A A /C so it is also forcing equivalent to C ω1. 15

16 4.2 Iterations with restricted memory For proofs and details on this section, see Section 3 in [5]. In the definition of the class K δ (S) below, we assume that each forcing in K δ (S) is a finite support iteration (P α, Q α ) : α < δ satisfying: There are posets R α H ω2 of size ω 1 such that each Q α is a P α name for an upward closed subordering (not necessarily complete) of R α. Members of P α are sequences of length α with finite support ordered coordinatewise - So if p, q P α, then p q iff for all β < α, p β Pβ p(β) Rβ q(β). For u α, P u is the suborder of all conditions in P α whose supports are in u. We now introduce the class K δ (S). (1) A memory template of length δ is a sequence u = u α : α < δ such that (i) u α α (ii) β u α = u β u α (2) Let u be a template of length δ. A subset u of δ is u-closed if for every α u, u α u. (3) Let u be a template of length δ. A finite support iteration (P α, Q α ) : α < δ is a u-iteration if for all α < δ, Q α is a P uα -name for a subset of R α. (4) Let S S δ ω 1. A memory template u of length δ is an S-template if (i) For all α δ\s, u α = φ (ii) For all α, u α ω 1 (iii) For all α S, if β S α is a limit point of u α, then β u α (5) Let u be an S-template of length δ where S S δ ω 1. (i) For u v δ, both u-closed, we say that v is a u-straight extension of u if for every β v S, if u β is unbounded in β, then β u. (ii) A u-closed set u is u-straight if every u-closed set v u is a u-straight extension of u. (6) Let K δ (S) consist of finite support u-iterations P = (P α, Q α ) : α < δ where u is an S-template and the following hold (i) If α δ\s, Q α = C adds a Cohen real (ii) For all α < δ, R α ω 1 and R α H ω2 (iii) For all β < α < δ, β / S, α S, P uα {α}/p uα β is isomorphic to Cohen forcing Fact 4.6. Suppose u is a template of length δ and u v δ are u-closed. Let P = (P α, Q α ) : α < δ be a u-iteration. Then, P u P v witnessed by the function p p u. Fact 4.7. Let P K δ (S) with witnessing template u. Let ρ be a P δ -name for a real. Then there is a u-straight set u of size ω 1 such that δ is a P u name. 16

17 Fact 4.8. Let P K δ (S) with witnessing template u. Let v be a u-straight extension of u and suppose S does not reflect at any limit point of v\u. Then P v /P u is isomorphic to Cohen forcing. In particular, if v is a u-straight extension of u and v\u ω 1, then P v /P u is isomorphic to Cohen forcing. 4.3 No Lusin set of size κ The aim of this section is to prove the following. Theorem 4.9. Let κ be inaccessible and S Sω κ 1 be stationary. Assume S. Then there exists some P K κ (S) such that, in V Pκ, every set of reals of size κ has a meager subset of size ω 1. P along with its witnessing template u will be constructed inductively using a S sequence. To see how this construction should go let us suppose P has already been constructed and let x α : α κ\s be a κ-sized set of reals in V P δ. We d like to have added a meager set containing uncountably many x α s by some stage of the iteration. Let v α be a u-straight set of size ω 1 such that x α is a P vα -name. We have P vα = P vα α (P vα /P vα α) Using Fact 4.8, we can identify P vα α = C(X α ), for some X α with X α ω 1. Also, in V Pvα α, we can identify P vα /P vα α = C(Y α ) for some Y α with Y α ω 1. So P vα = C(Xα, Y α ) Since κ is inaccessible, there are a stationary S 1 κ\s and f α : α S 1, φ, v, X, Y, τ, such that (a) α S 1 = v α α = v (c) X, Y are disjoint sets and for each α S 1, f α : P vα C(X, Y ), φ : P v C(X) are dense embeddings satisfying f α P v = φ (b) τ is a C(X, Y ) name for a real such that for each α S 1, f α ( x α ) = τ For α S 1, let γ α > α be a successor ordinal such that v α γ α. Let E = {α < κ : α = sup(s 1 α) and α S 1 α = γ α < α}. Then E is a club in κ. Let δ E S. Construct α i : i < ω 1 increasing such that for every i < ω 1, α i S 1, i < j = γ i < α j and sup{α i : i < ω 1 } = δ. Put A = {α i : i < ω 1 }. The next lemma says that we can make { x α : α A} meager in V P δ+1. Lemma Let S S κ ω 1, δ S and P K δ (S δ) with u as its witnessing template. Suppose that α i, v i, x i : i < ω 1, X, Y, τ satisfy the following. (1) α i : i < ω 1 is an increasing unbounded subset of δ disjoint with S (2) For each i < ω 1, v i is a u-straight subset of δ of size ω 1, v i α i+1, v i α i = v does not depend on i 17

18 (3) X, Y are disjoint and there are dense embeddings f i : P vi C(X, Y ), φ : P v C(X) such that f i P v = φ (4) τ is a C(X, Y )-name for a real, x i is a P vi -name for a real and f i ( x i ) = τ Let v = {v i : i < ω 1 }. Then v {δ} is u-straight and there is a P v -name Q for a forcing such that P Q K δ+1 (S δ + 1) witnessed by u v and in V P δ+1, the set { xi : i < ω 1 } is meager. Proof: v {δ} is u-straight because every limit point of v S, other than δ, is a limit point of some v i, i < ω 1. The forcing P v is isomorphic to C(X) {C(Y i ) : i < ω 1 } where Y i s are disjoint copies of Y as witnessed by the function that takes p P v to (q, r i : i < ω 1 ) where φ(p v ) = q C(X) and f i (p v i ) = (q, r i ) C(X, Y i ). Let τ = τ/g C(X) so that τ[g C(X,Y ) ] = τ [G C(X) ][G C(Y ) ]. Let Y Y be infinite countable such that τ is a C(Y )-name (so essentially C-name). Let P = C(X) {C(Y i \Y i ) : i < ω 1 }. Then we can write P v as P {C(Y i ) : i < ω 1 } Let c i to be the Cohen real added by C(Y i ). In V Pv, let X = {τ [c i ] : i < ω 1 }. Put Q δ = Q X. Clearly, { x i : i < ω 1 } = X is meager in V P δ+1. To show that Pδ+1 K δ+1 (S) with witnessing template u v, it is enough to show that for each β δ\s, (P v Q X ) /P v β is isomorphic to Cohen forcing. Pick ω k < ω 1 large enough so that α k > β and write P v αk as C(X) {C(Y i ) : i < k} Hence (P v Q X ) /P v αk is forcing isomorphic to [( ) {C(Yi \Y i ) : k i < ω 1 } {C(Y i ) : i < ω 1 } Q X / ] {C(Y i ) : i < k} which is isomorphic to Cohen forcing by Lemma 4.5. By Fact 4.8, P v αk /P v β is isomorphic to Cohen forcing. It follows that (P v Q X ) /P v β is also isomorphic to Cohen forcing. Proof of Theorem 4.9: We can now define P, u. Using S, get f δ : δ S such that for each δ S, f δ is a function, dom(f δ ) is an unbounded subset of δ, range(f δ ) V δ and for every unbounded T κ, F : T V κ, there are stationary many δ S such that F dom(f δ ) = f δ. Suppose P K δ (S δ), u δ have already been defined. If δ / S, put u δ = φ, Q δ = C. If δ S, dom(f δ ) = {α i : i < ω 1 } unbounded in δ, for every i, f δ (α i ) = (v i, X, Y, τ) where α i, v i : i < ω 1, X, Y, τ satisfy the hypothesis of the Lemma 4.10, then define u δ, Q δ as given in the conclusion of this lemma. We need the following facts from [5]. Fact Suppose κ is measurable and 2 κ = κ +. Then there is a forcing extension W of V such that, in W, κ is measurable and there is a stationary S Sω κ 1 such that S holds and j(s) reflects nowhere in the interval (κ, j(κ)]. 18

19 Fact Let P K δ (S). If α < δ, α / S, S reflects nowhere in the interval (α, δ], then P δ /P α is isomorphic to Cohen forcing. Proof of Theorem 4.2: Let W, κ, S Sω κ Then, W Pκ is as required. be as above. Let P K κ (S) be as in Theorem References [1] B. Balcar, P. Simon, Disjoint refinement, Handbook of Boolean algebras Vol. 2 (J. D. Monk and R. Bonnett, Eds.), North-Holland Amsterdam (1989), pp [2] S. Fuchino, S. Shelah and L. Soukup, Sticks and clubs, Annals Pure and Applied Logic 90 (1997), [3] M. Gitik and S. Shelah, On certain indestructibility of strong cardinals and a question of Hajnal, Archive for Math. Logic 28 (1989), [4] M. Gitik and S. Shelah, Forcings with ideals and simple forcing notions, Israel J. Math. 68 (1989), [5] N. Greenberg, S. Shelah, Models of Cohen measurability, Annals Pure and Applied Logic 165 (2014), [6] K. Kunen, Set Theory - An introduction to independence proofs, North-Holland 1980 [7] S. Shelah, Possibly every real function is continuous on a non-meagre set, Publications de L Institute Mathematique - Beograd, Nouvelle Series 57(71) (1995), [8] S. Shelah, Not collapsing cardinals κ in (< κ)-support iterations, Israel J. Math 136 (2003), [9] S. Shelah, Many forcing axioms for all regular uncountable cardinals, Publication 832 on Shelah s archive [10] S. Shelah, Diamonds, Proc. Amer. Math. Soc. 138 (2010), [11] S. Shelah, Many partition relations below density, Israel J. Math 192 (2012), [12] S. Shelah, Proper and improper forcing, Perspectives in mathematical logic, Springer- Verlag Berlin, Second edition, 1998 [13] R. Solovay, Real valued measurable cardinals, Axiomatic set theory Proc. Sympos. Pure Math. Vol. 13 Part I Amer. Math. Soc. 1971,