Mad families and non-meager filters Haim Horowitz and Saharon Shelah Abstract We prove the consistency of ZF +DC +”there are no mad families”+”there exists a non-meager filter on ω” relative to ZFC, answering a question of Neeman and Norwood. We also introduce a weaker version of madness, and we strengthen the result from [HwSh:1090] by showing that no such families exist in our model.1 7 1 0 Introduction 2 n This paper isacontinuation of[HwSh:1090], which ispart oftheongoing efforttoin- a vestigate thepossible non-existence anddefinability ofmadfamilies. In[HwSh:1090] J 0 we proved that ZF +DC+”there are no mad families” is equiconsistent with ZFC 1 (previous results by Mathias and Toernquist established the consistency of that statement relative to large cardinals, see [Ma1] and [To]). In this paper we ex- ] O tend our results from [HwSh:1090] to address the following question by Neeman and L Norwood: . h Question ([NN]): If there are no mad families, does it follow that every filter is t a meager? m [ By a result of Mathias ([Ma2]), if every set of reals has the Ramsey property, then 1 every filter is meager. v 6 Weshallconstruct amodelofZF+DC where therearenomadfamilies, but there is 0 a non-meager filter on ω. Our proof relies heavily on [HwSh:1090], the main change 8 is that now we’re dealing with a class K consisting of pairs (P,A) such that P is ccc 2 2 0 and forces MA , and in addition, P forces that A is independent (we shall require . ℵ1 1 more, see definition 2). In order to imitate the proof from [HwSh:1090], we need 0 to prove analogous amalgamation results for an appropriate subclass of K . As in 7 2 1 [HwSh:1090], our final model is obtained by forcing with P where (P,A) is a “very : v large” object in a subclass of K2, and the non-meager filter will be constructed from i A, which should contain many Cohen reals. X r Finally, we consider the notion of nearly mad families (see definition 14), which was a also introduced in [NN]. We introduce the notion of a somewhat mad family, which includes both mad and nearly mad families, and we prove that no somewhat mad families exist in our model. 1Date: January 11, 2017 2000 Mathematics Subject Classification: 03E35,03E15,03E25 Keywords: mad families, non-meager filters, amalgamation Publication 1103 of the second author Partially supported by European Research Council grant 338821. 1 A non-meager filter without mad families Hypothesis 1: We fix µ and λ such that ℵ ≤ µ, λ = λ<µ, µ = cf(µ) and 2 α < µ → |α|ℵ1 < µ. Definition 2: A. Let K be the class of k such that: 2 a. Each k has the form (P,A) = (Pk,Ak). b. P is a ccc forcing such that (cid:13) MA . P ℵ1 c. A is a set of canonical P-names of subsets of ω. d. (cid:13) ”A is independent, i.e. every finite non-trivial Boolean combination of ele- P ments of A is infinite”. B. For k ∈ K2 and a Pk-name ∼b, let (cid:13)Pk ”∼b ∈ pos(k)” mean (cid:13)Pk ”∼b ∈ [ω]ωand there is no non-trivial Boolean combination of sets from Ak that is almost disjoint to b”. ∼ C. Let ≤ be the following partial order on K : k ≤ k if and only if: 1 2 1 1 2 a. Pk ⋖Pk . 1 2 b. Ak ⊆ Ak . 1 2 D. Let ≤ be the following partial order on K : 2 2 k ≤ k if and only if: 1 2 2 a. As in B(a). b. As in B(b). c. If ∼b is a Pk1-name then (cid:13)Pk1 ”∼b ∈ pos(Ak1)” implies (cid:13)Pk2 ”∼b ∈ pos(Ak2)”. E. Let K2+ be the class of k ∈ K2 such that (cid:13)Pk ”Ak is a maximal independent set everywhere”, where A is a maximal independent set everywhere if for every a ,...,a ∈ A without repetition, b := ∩ aifliseven ∈ [ω]ω and {a ∩ b : a ∈ 0 n−1 l l ℵ , then the union is an 1 upper bound in K+. 2 Proof: Given an increasing continuous sequence (kα : α < β), we choose Pkβ as in [HwSh:1090] and we let Akβ = α∪<βAkα. This is enough for ≤1, so by claim 5 we’re done. (cid:3) Claim 8: A. If k ∈ K then there are k and a such that: 1 2 2 ∼ a. k ≤ k . 1 1 2 b. Ak ∪{a} ⊆ Ak . 1 ∼ 2 c. a is Cohen over VPk1. ∼ d. |Pk | ≤ (2+|Pk |)ℵ1. 2 1 B. Moreover, we may require that k ∈ K+. 2 2 Proof: A. Let P = Pk ⋆C where C is Cohen forcing, now let Pk be a ccc forcing 1 2 such that P⋖Pk2, (cid:13)Pk2 ”MAℵ1” and |Pk2| ≤ (2+|P|)ℵ1. Finally, let Ak2 = Ak1∪{∼a} where a is a name for a Cohen real added by Pk , it’s easy to see that (Pk ,Ak ) are ∼ 2 2 2 as required. B. By claim 5. (cid:3) Definition 9: WedefinetheamalgamationpropertyinthecontextofK+ asfollows: 2 K+ has the amalgamation property if A implies B where: 2 A. a. k ∈ K+ (l = 0,1,2). l 2 b. k ≤ k (l = 1,2). 0 2 l c. Pk ∩Pk = Pk . 1 2 0 B. There exists k3 = (Pk3,Ak3) ∈ K2+ such that kl ≤2 k3 (l = 1,2). Claim 10: a. (K+,≤ ) has the amalgamation property. 2 2 4 b. Suppose that k0,k1 and k2 ∈ K2+, g : Pk0 → Pk1 is an embedding such that (g(Pk0),g(Ak0)) ≤2 k1 and k0 ≤2 k2, then there exist k,k2′ ∈ K2+ and f such that: 1. k1 ≤2 k and |Pk| ≤ (2+|Pk1|+|Pk2|)ℵ1. 2. (g(Pk0),g(Ak0)) ≤2 k2′ ≤2 k. 3. f : Pk → Pk′ is an isomorphism mapping Ak to Ak′ 2 2 2 2 4. g ⊆ f. Proof: a. We shall first prove that Ak ∩Ak = Ak . Note that Ak ⊆ Ak ∩Ak 1 2 0 0 1 2 is true by the definition of ≤2, so suppose that ∼a ∈ Ak1\Ak0, we need to show that ∼a ∈/ Ak2. As k0 ≤2 k1, ∼a is not a Pk0-name. Therefore, it’s not a Pk2-name, hence a ∈/ Ak . ∼ 2 Now construct P as in [HwSh:1090], i.e. we take the amalgamation P′ = Pk × Pk 1 2 Pk0 and then we take P ∈ K such that P′ ⋖ P and |P| ≤ (2 + |P′|)ℵ1. Now let A := Ak ∪Ak . We need to show that A is as required, i.e. we need to prove that (P,A) 1 2 satisfyrequirements (A)(d)and(D)(c)inDefinition2(intheend, wewilluseclaim5 for the requirement in Definition (2)(D)(c)). By symmetry, it’s enough to show that if(cid:13)Pk1 ”∼b ∈ pos(Ak1)”anda∼0,...,an∼−1 ∈ A,then(cid:13)P ”∼b∩(l<∩nailf(l∼iseven)) ∈ [ω]ω”. Let n2 := n, wlogtherearen0 < n1 < n2 suchthat∼al ∈ Ak0 ⇐⇒ l < n0,∼al ∈ Ak1 ⇐⇒ l ∈ [0,n1) and ∼al ∈ Ak2 ⇐⇒ l ∈ [0,n0)∪[n1,n2). It’s enough to show that the last statement is forced by P′, so let k < ω and p = (p ,p ) ∈ P′, we shall find q ∈ P′ and 1 2 m < ω such that p ≤ q, k ≤ m and q (cid:13)P′ ”m ∈ ∼b ∩(l<∩nailf(l∼iseven))”. Let p0 ∈ Pk0 witness ”(p1,p2) ∈ P′”, i.e. p0 (cid:13)Pk0 ”l=∧1,2pl ∈ Pkl/Pk0”. Let b∼∗ = {m : p1 1Pk1/Pk0 ”m ∈/ ∼b ∩(∩{∼ailfliseven : l ∈ [n0,n1)})”} (so b∼∗ is a Pk0-name) and let p0 ∈ G0 ⊆ Pk0 be generic over V, then b∼∗ = b∼∗[G0] ∈ V[G0] and as p1 (cid:13)Pk1 ”∼b ∈ pos(Ak1)”, it follows that p0 (cid:13)Pk0 ”b∼∗ ∈ [ω]ω”, moreover, p0 (cid:13)Pk0 ”∼b∗ ∈ pos(Ak0)”. Let b∗∗ be the Pk0-name defined as b∗ if p0 is in the generic set, and as ω otherwise. As k0 ≤2 k2, it follows that p (cid:13) ”b∗∗ ∩( ∩ aif(liseven)) ∈ [ω]ω”. 2 Pk2/G0 ∼ l∈[0,n0)∪[n1,n2) l ∼ Therefore, in V[G ] there are (p′,m) such that: 0 2 a. p2 ≤ p′2 ∈ Pk2/G0. b. m > k. c. p′ (cid:13) ”m ∈ b∗∗ ∩( ∩ aif(liseven))”. 2 Pk2/G0 ∼ l∈[0,n0)∪[n1.n2) l ∼ Note that as b∗∗ ∈ V[G ], V[G ] |= ”m ∈ b∗∗[G ] = b∗[G ]”. Therefore, by the 0 0 0 0 ∼ ∼ ∼ definitions of ∼b and b∼∗, there is p′1 ∈ Pk1/G0 above p1 such that p′1 (cid:13)Pk1/G0 ”m ∈ ∼b ∩(∩{∼ailfliseven : l ∈ [n0,n1)})”. Therefore, there is p0 ≤ p′0 ∈ G0 forcing (in Pk0) all of the aforementioned statements about (p′,p′) in V[G ]. Now it’s easy to check 1 2 0 5 that q = (p′,p′) is as required. Finally, extend (P,A) (with respect to ≤ ) to a 1 2 1 member of K+. By observation 3, we’re done. 2 b. Follows from (a) by changing names. (cid:3) Claim 11: There exists k = (Pk,Ak) = (P,A) ∈ K2+ such that |Pk| = λ and: 1. For every X ⊆ P of cardinality < µ, there exists k′ = (Q,A′) ∈ K+ such that 2 X ⊆ Q, k′ ≤ k and |Q| < µ. 2 2. If k1,k2 ∈ K2+, |Pk1|,|Pk2| < µ, k1 ≤2 k2 and f1 : Pk1 → P is a complete embedding such that (f1(Pk1),f1(Ak1)) ≤2 (P,A), then there is a complete embedding f2 such that f1 ⊆ f2 and (f2(Pk2),f2(Ak2)) ≤2 (P,A). Proof: The first property is satisfied by every k ∈ K+ by observation 4. The proof 2 of (2) is as in [HwSh:1090]. (cid:3) Claim 12: A implies B where: A. a. k0,k1,k2 ∈ K2+, k0 ≤2 kl (l = 1,2) and Pk0 = Pk1 ∩Pk2. b. D is a Pk -name of a nonprincipal ultrafilter on ω. ∼ 0 c. For l = 1,2, ∼al and b∼l are canonical Pkl-names of a member of [ω]ω. d. For l = 1,2, (cid:13)Pkl ”∼al∩∼bl is infinite and ∼al contains no members of D∼ from VPk0. e. Pk ∩Pk = Pk . 1 2 0 f. For l = 1,2, (cid:13) ”b is a pseudo intersection of D”. Pkl ∼l ∼ B. There is k ∈ K+ such that: 2 a. |Pk| ≤ (2+|Pk |+|Pk |)ℵ1. 1 2 b. k ≤ k (l = 1,2). l 2 c. (cid:13) ”a \a and a \a are infinite”. Pk 2 1 1 2 ∼ ∼ ∼ ∼ Proof: Let k ∈ K+ be the object constructed by the proof of claim 10. We need to 2 prove that k satisfies clause (B)(c). For l = 0,1,2, let Pl = Pkl and let P′ be as in the proof of claim 10, so it suffices to prove that (cid:13)P′ ”a2\a1 and a1\a2 are infinite”. ∼ ∼ ∼ ∼ Let p = (p ,p ) ∈ P′, k < ω and let p ∈ P be a witness of ”p = (p ,p ) ∈ P′”. 1 2 0 0 1 2 Now let G ⊆ P′ be generic over V such that p ∈ G and let D = D[G ]. By 0 0 0 0 ∼ the assumptions, for l = 1,2, p (cid:13) ”a ∩b is an infinite pseudo intersection of l P1/G0 l l ∼ ∼ D”. In V[G ], let b∗ = {m : p 1 ”m ∈/ a ∩ b ”}, then p (cid:13) ”a ∩b ⊆ b∗, 0 l l Pl/G0 ∼l ∼l l Pl/G0 ∼l ∼l ∼l hence b∗ infinite”. As p (cid:13) ”a ∩ b is a pseudo intersection of D”, necessarily l l Pl/G0 ∼l ∼l V[G ] |= b∗ ∈ D. 0 l Let a∗ = {m : p (cid:13) m ∈ a }, so a∗ ∈ V[G ] and p (cid:13) ”a∗ ⊆ a ”. Now recall l l Pl/G0 ∼l l 0 l Pl/G0 l ∼l that (cid:13)Pkl ”∼al contains no member of D∼ from VPk0”, therefore pl (cid:13)Pl/G0 ”a∗l ∈/ D∼”. 6 Hence in V[G ] (recalling D is an ultrafilter), b := (b∗ ∩ b∗) \ (a∗ ∪ a∗) ∈ D. Let 0 1 2 1 2 m ∈ b be such that k < m. By the definition of b∗, there is p′ ∈ P /G above p l l l 0 l such that p′ (cid:13) ”m ∈ a ∩b ”. By the definition of a∗, there is p′′ ∈ P /G above l Pl/G0 ∼l ∼l l l l 0 p such that p′′ (cid:13) ”m ∈/ a ”. Let p′ ∈ G be a condition above p forcing the l l Pl/G0 ∼l 0 0 0 above statements, so p′ is witnessing the fact that (p′,p′′),(p′′,p′) ∈ P′ are above 0 1 2 1 2 p = (p ,p ). Now m > k, (p′,p′′) (cid:13) ”m ∈ a \ a ” and (p′′,p′) (cid:13) ”m ∈ a \ a ”, 1 2 1 2 1 2 1 2 2 1 ∼ ∼ ∼ ∼ which completes the proof. (cid:3) Definition 13: Let P = Pk be the forcing from claim 11, let G ⊆ P be generic over V and in V[G], let V1 = HOD(R<µ∪{Ak}). Definition 14 ([NN]): A family F ⊆ [ω]ω is nearly mad if |A ∩ B| < ℵ or 0 |A∆B| < ℵ for every A 6= B ∈ F, and F is maximal with respect to this property. 0 Theorem 15: V |= ZF+DC +”there are no mad families”+”there are no nearly mad families”+ 1 <µ ”there exists a non-meager filter on ω”. Proof: 1. In order to see that there exists a non-meagre filter in V , let D be the 1 ∼ filter generated by Ak and the cofinite sets. By claim 8 and the choice of k, D ∼ contains many Cohen reals and therefore is non-meager. 2. Theproofofthenon-existence of madfamilies isexactly asin[HwSh:1090], where (K+,≤ ) here replaces (K,⋖) there, and claim 10 is used for the amalgamation 2 2 arguments. Alternatively, see the proof of (3) below. 3. The non-existence of a nearly mad family in V will follow from the proofs below. 1 (cid:3) Somewhat mad families Definition 16: A family F ⊆ [ω]ω is somewhat mad if: a. For every a ,a ∈ F, |a ∩a | < ℵ or a ⊆∗ a or a ⊆∗ a . 1 2 1 2 0 1 2 2 1 b. If b ∈ [ω]ω then for some a ∈ F, |a∩b| = ℵ . 0 Observation 17: Nearly mad families are somewhat mad. (cid:3) Definition 18: Let Pr(k ,k ,D,b ) mean: 1 2 2 ∼ ∼ a. k1 ≤2 k2, b∼2 is a Pk2-name and D∼ is a Pk1-name such that (cid:13)Pk2 ”b∼2 ∈ [ω]ω” and (cid:13) ”D is a nonprincipal ultrafilter on ω”. Pk1 ∼ b. If G1 ⊆ Pk1 is generic over V, p1 ∈ Pk2/G1, b∗0 = {n : p1 (cid:13)Pk1G1 ”n ∈ b∼2”} and b∗ = {n : p 1 ”n ∈/ b ”}, then V[G ] |= b∗ \b∗ ∈ D. 1 1 Pk1/G1 ∼2 1 1 0 Claim 19: (A) implies (B) where: A. a. k ∈ K+. 1 2 7 b. (cid:13) ”D is a nonprincipal ultrafilter on ω”. Pk1 ∼ c. (cid:13) ”S is somewhat mad”. Pk1 ∼1 d. (cid:13) ”S ∩D = ∅”. Pk1 ∼1 ∼ B. There is k such that: 2 a. k ∈ K+. 2 2 b. k ≤ k . 1 2 2 c. |Pk | ≤ (2+|Pk |)ℵ1. 2 1 d. (α) implies (β) where: α. (k ,S ) satisfy the following properties: 3 2 ∼ 1. k ≤ k ∈ K+. 2 2 3 2 2. (cid:13) ”S is somewhat mad and S ⊆ S ”. Pk3 ∼2 ∼1 ∼2 3. (cid:13) ”no member of S \S contains a member of D”. Pk3 ∼2 ∼1 ∼ β. For some Pk3-name ∼a, (cid:13)Pk3 ”∼a ∈ S∼2” and Pr(k1,k3,D∼,∼a). Proof: Using Mathias forcing restricted to D, it’s easy to see that there is k and a 2 ∼ Pk2 name ∼b such that k1 ≤2 k2 ∈ K2+, |Pk2| ≤ (2+|Pk1|)ℵ1 and (cid:13)Pk2 ”∼b is a pseudo intersection of D”. Therefore, (cid:13) ”b ∈ [ω]ω is almost disjoint to every a ∈ S ” (by ∼ Pk2 ∼ ∼ ∼1 the fact that (cid:13) ”S ∩D = ∅”). Pk1 ∼1 ∼ We shall now prove that k satisfies (B)(d). Suppose that (k ,S ) are as there. By 2 3 2 ∼ the somewhat madness of S , there is a such that (cid:13) ”a ∈ S and |a∩ b| = ℵ ”. ∼2 ∼ Pk3 ∼ ∼2 ∼ ∼ 0 Therefore, (cid:13)Pk3 ”∼a ∩∼b ∈ [ω]ω is a pseudo intersection of D∼”. Now let G1 ⊆ Pk1 be genericoverV. Ifp1 ∈ Pk3/G1 thenb∗ = {n : p1 1Pk3/G1 ”n ∈/ ∼a∩∼b”} ∈ D∼[G1]bythe fact that a∩b is a pseudo intersection of D. Now let a∗ = {n : p (cid:13) ”n ∈ a”}, ∼ ∼ ∼ 1 Pk3/G1 ∼ then p (cid:13) ”a∗ ⊆ a is infinite”. If a∗ ∈ D[G ], then p forces that a (which 1 PK3/G1 ∼ ∼ 1 1 ∼ belongs to S \ S ) contains a member of D[G ], contradicting (α)(3). Therefore, 2 1 1 ∼ ∼ ∼ a∗ ∈/ D[G ], and a is as required in the definition of Pr(k ,k ,D,a). (cid:3) 1 1 3 ∼ ∼ ∼ ∼ Claim 20: There is no somewhat mad family in V . 1 Proof: Suppose towards contradiction that S is a P-name of a somewhat mad ∼ family. As in [HwSh:1090], let D be a P-name of a Ramsey ultrafilter on ω such ∼ that (cid:13)P ”S∼∩D∼ = ∅”. By claim 11(a), there is k1 ≤2 k such that k1 ∈ K2+, |Pk1| < µ and S∼ is definable using a Pk1−name. Let KP+ be the set of k′ ∈ K2+ such that k′ ≤2 k, |Pk′| < µ, S ↾ Pk′ is a canonical Pk′-name of a somewhat mad family in ∼ VPk′ and D ↾ Pk′ is a Pk′-name of a Ramsey ultrafilter on ω. As in [HwSh:1090], ∼ 8 K+ is ≤ -dense in K+, so there exists k ∈ K+ such that k ≤ k . Let k ∈ K+ be P 2 2 2 P 1 2 3 2 as in claim 19 for (k2,S∼ ↾ Pk2), wlog k3 ≤2 k (see claim 11). Choose k4 ∈ KP+ such that k3 ≤2 k4 and let S∼4 := S∼ ↾ Pk4. Let ∼a be a Pk4-name such that Pr(k3,k4,D∼,∼a) holds, as guaranteed by claim 19. As in [HwSh:1090], there are k ,k ,∈ K+ and an isomorphism f from k to k 5 6 P 4 5 over k2 such that Pk5 adds a generic for MD∼↾Pk4 (Mathias forcing restricted to the ultrafilter D∼ ↾ Pk4) and (k2,k4,k5,k6) here are as (k0,k1,k2,k3) in claim 10, and wlog k ≤ k. By the choice of f, (cid:13) ”a,f(a) ∈ S”. 6 2 P ∼ ∼ ∼ By claim 12, with (k ,k ,k ,k ,a,f(a)) standing for (k ,k ,k ,ka ,a ) there, it’s 2 4 5 6 0 1 2 , ∼ ∼ ∼1 ∼2 forced by Pk , and hence by P, that a \ f(a) and f(a) \ a are infinite. As in 5 ∼ ∼ ∼ ∼ [HwSh:1090], (cid:13) ”|a∩f(a)| = ℵ ”. As (cid:13) ”a,f(a) ∈ S”, we get a contradiction. (cid:3) P 0 P ∼ ∼ ∼ ∼ ∼ References [HwSh:1090] Haim Horowitz and Saharon Shelah, Can you take Toernquist’s inac- cessible away? arXiv:1605.02419 [Ma1] A. R. D. Mathias, Happy families, Ann. Math. Logic 12 (1977), no. 1, 59-111. [Ma2] A. R. D.. Mathias, A remark on rare filters. [NN] Itay Neeman and Zach Norwood, Happy and mad families in L(R), preprint. [To] Asger Toernquist, Definability and almost disjoint families, arXiv:1503.07577. (Haim Horowitz) Einstein Institute of Mathematics Edmond J. Safra campus, The Hebrew University of Jerusalem. Givat Ram, Jerusalem, 91904, Israel. E-mail address: haim.horowitz@mail.huji.ac.il (Saharon Shelah) Einstein Institute of Mathematics Edmond J. Safra campus, The Hebrew University of Jerusalem. Givat Ram, Jerusalem, 91904, Israel. Department of Mathematics Hill Center - Busch Campus, 9 Rutgers, The State University of New Jersey. 110 Frelinghuysen road, Piscataway, NJ 08854-8019 USA E-mail address: shelah@math.huji.ac.il 10

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