Covering of the Null Ideal May Have Countable Cofinality

FUNDAMENTA MATHEMATICAE 166 (2000)

Covering of the null ideal may have countable coﬁnality

Saharon S h e l a h (Jerusalem and New Brunswick, NJ)

Abstract. We prove that it is consistent that the covering number of the ideal of measure zero sets has countable coﬁnality.

0. Introduction. In the present paper we show that it is consistent that the covering of the null ideal has countable cofinality. Recall that the covering number of the null ideal (i.e. the ideal of measure zero sets) is defined as n [ ω o cov(null) = min |P| : P ⊆ null and A = R (= 2) . A∈P The question whether the cofinality of cov(null) is uncountable has been raised by D. Fremlin and has been around since the late seventies. It appears in Fremlin’s list of problems, [Fe94], as problem CO. Recall that for the ideal of meagre sets the answer is positive: A. Miller [Mi82] proved that the cofinality of the covering of category is uncountable. T. Bartoszyński [Ba88] saw that b < ℵω is necessary (see [BaJu95, Ch. 5] for more results related to this problem). It should be noted that most people thought cf(cov(null)) = ℵ0 was impossible. The main result of this paper is the following:

Theorem 0.1. Con(cov(null) = ℵω + MAℵn ) for each n < ω. The presentation of the proof of 0.1 sacriﬁces generality for hopeful trans- parency. We ﬁnish by some further remarks, e.g. the exact cardinal assumption for 0.1. We try to make the paper self-contained for readers with basic knowledge of forcing.

2000 Mathematics Subject Classiﬁcation: 03E17, 03E35. Key words and phrases: null sets, cardinal invariants of the continuum, iterated forcing, ccc forcing. Research supported by “The Israel Science Foundation” founded by The Israel Academy of Sciences and Humanities. Publication no. 592.

[109] 110 S. Shelah

In a subsequent paper, [Sh 619], we deal with the question: “can every non-null set be partitioned into uncountably many non-null sets”, equiva- lently: “can the ideal of null sets which are subsets of a non-null subset of R be ℵ1-saturated”. P. Komj´ath[Ko89] proved that it is consistent that there is a non-meagre set A such that the ideal of meagre subsets of A is ℵ1-saturated. The question whether a similar fact may hold for measure dates back to Ulam; see also Prikry’s thesis. It appears as question EL(a) on Fremlin’s list. In [Sh 619] we prove the following: Theorem 1. It is consistent that there is a non-null set A ⊆ R such that the ideal of null subsets of A is ℵ1-saturated (of course, provided that “ZFC + ∃ measurable” is consistent).

In [Sh 619] we also prove the following. Theorem 2. It is consistent that: (⊕) there is a non-null A ⊆ R such that for every f : A → R, the function f as a subset of the plane R × R is null provided that “ZFC + there is a measurable cardinal” is consistent.

Notation 0.2. We denote: • natural numbers by k, l, m, n and also i, j, • ordinals by α, β, γ, δ, ζ, ξ (δ always limit), • cardinals by λ, κ, χ, µ, • reals by a, b and positive reals (normally small) by ε, • subsets of ω or ω≥2 or Ord by A, B, C, X, Y , Z but B is a Borel function, • finitely additive measures by Ξ, • sequences of natural numbers or ordinals by η, ν, %, s is used for various things, T is as in Definition 2.9, t is a member of T . We denote • forcing notions by P , Q, • forcing conditions by p, q and use r to denote members of Random (see below) except in Definition 2.2. Leb is the Lebesgue measure (on ω2), and Random will be the family ω> ω> ω> {r ⊆ 2 : r is a subtree of ( 2, C) (i.e. non-empty subset of 2 closed under initial segments) with no C-maximal element ω (so lim(r) := {η ∈ 2 : (∀n ∈ ω)(ηn ∈ t)} is a closed subset of ω2) and Leb(lim(r)) > 0} ordered by inverse inclusion. We may sometimes use instead {B : B is a Borel non-null subset of ω2}. Covering of the null ideal 111

For η ∈ ω>2 and A ⊆ ω≥2 let [η] A = {ν ∈ A : ν E η ∨ η E ν}. Let H(χ) denote the family of sets with transitive closure of cardinality ∗ < χ, and let <χ denote a well ordering of H(χ). We thank Tomek Bartoszy´nski and Mariusz Rabus for reading and com- menting and correcting.

1. Preliminaries. We review various facts on finitely additive measures. Definition 1.1. (1) M is the set of functions Ξ from some Boolean subalgebra P of P(ω) including the finite sets to [0, 1]R such that: • Ξ(∅) = 0, Ξ(ω) = 1, • if Y,Z ∈ P are disjoint, then Ξ(Y ∪ Z) = Ξ(Y ) + Ξ(Z), • Ξ({n}) = 0 for n ∈ ω. (2) Mfull is the set of Ξ ∈ M with domain P(ω) and members are called finitely additive measures (on ω). (3) We say A has Ξ-measure a (or > a, or whatever) if A ∈ dom(Ξ) and Ξ(A) is a (or > a, or whatever). ∗ Proposition 1.2. Let aα, bα (α < α ) be reals, 0 ≤ aα ≤ bα ≤ 1, and ∗ let Aα ⊆ ω (α < α ) be given. The following conditions are equivalent: ∗ (A) There exists Ξ ∈ M which satisfies Ξ(Aα) ∈ [aα, bα] for α < α . ∗ (B) For every ε > 0, m < ω and n < ω, and α0 < α1 < . . . < αn−1 < α we can find a finite, non-empty u ⊆ [m, ω) such that for l < n,

aαl − ε ≤ |Aαl ∩ u|/|u| ≤ bαl + ε. ∗ (C) For every real ε > 0, n < ω and α0 < α1 < . . . < αn−1 < α there n are cl ∈ [aαl − ε, bαl + ε] such that in the vector space R , hc0, . . . , cn−1i is in the convex hull of {% ∈ n{0, 1} : for inﬁnitely many m ∈ ω we have

(∀l < n)[%(l) = 1 ⇔ m ∈ Aαl ]}. (D) Like part (A) with Ξ ∈ Mfull. (E) Like part (B) demanding u ⊆ ω, |u| ≥ m. P r o o f. Straightforward (in fact, in clause (E) we can omit “u⊆[m, ω)”). On (C) see 2.17. ∗ Proposition 1.3. (1) Assume that Ξ0 ∈ M and for α < α , Aα ⊆ ω and 0 ≤ aα ≤ bα ≤ 1, aα, bα reals. The following are equivalent: full ∗ (A) There is Ξ ∈ M extending Ξ0 such that α < α ⇒ Ξ(Aα) ∈ [aα, bα]. (B) For every partition hB0,...,Bm−1i of ω with Bi ∈ dom(Ξ0) and ∗ ε > 0, n < ω and α0 < . . . < αn−1 < α we can ﬁnd a ﬁnite set u ⊆ ω such that Ξ(Bi)−ε ≤ |u∩Bi|/|u| ≤ Ξ(Bi)+ε and aαl −ε ≤ |u∩Aαl |/|u| ≤ bαl +ε. 112 S. Shelah

(C) For every partition hB0,...,Bm−1i of ω with Bi ∈ dom(Ξ0) and ∗ ε > 0, n < ω and α0 < . . . < αn−1 < α we can ﬁnd cl,k ∈ [0, 1]R for l < n, k < m such that: P (a) k

l < n ⇒ cl,k − ε < (|u ∩ Al ∩ Bk|/|u|)Ξ(Bk) < cl,k + ε.

(D) for every partition hB0,...,Bm−1i of ω with Bi ∈ dom(Ξ0), ε > 0, ∗ n < ω, and α0, . . . , αn−1 < α we can ﬁnd cl,k ∈ [0, 1]R for l < n, k < m such that: P (a) k

(∀l < n)[%(l) = 1 ⇔ i ∈ Aαl ]}. (2) The following are sufficient conditions for (A)–(D) above: ∗ ∗ (E) For every ε > 0, A ∈ dom(Ξ0) such that Ξ0(A ) > 0, n < ω, ∗ ∗ α0 < . . . < αn−1 < α , we can find a finite, non-empty u ⊆ A such that aαl − ε ≤ |Aαl ∩ u|/|u| ≤ bαl + ε for l < n. ∗ ∗ (F) For every ε > 0, n < ω, α0 < α1 < . . . < αn−1 < α and A ∈ ∗ Q n dom(Ξ0) such that Ξ0(A ) > 0, the set l

(∀l < n)[%(l) = 1 ⇔ m ∈ Aαl ]}. ∗ (3) If in addition bα = 1 for α < α then a suﬃcient condition for (A)–(E) above is

∗ ∗ (G) if A ∈ dom(Ξ0) and Ξ(A ) > 0 and n < ω and α0 < . . . < αn−1 < ∗ ∗ T α then A ∩ l

full Definition 1.4. (1) For Ξ ∈ M and a sequencea ¯ = hal : l < ωi of reals in [0, 1]R (or just supl<ω |al| < ∞), deﬁne AvΞ (¯a) to be n X ∗ o sup Ξ(Ak) inf{al : l ∈ Ak} : hAk : k < k i is a partition of ω k

(2) For Ξ ∈ M and A ⊆ ω such that Ξ(A) > 0 we deﬁne

ΞA(B) = Ξ(A ∩ B)/Ξ(A).

Clearly ΞA ∈ M with the same domain, ΞA(A) = 1. If B ⊆ ω and Ξ(B) > 0 then we let Av (ha : k ∈ Bi) = Av (ha0 : k < ωi)/Ξ(B) ΞB k Ξ k where a if k ∈ B, a0 = k k 0 if k 6∈ B. full i ∗ Proposition 1.5. Assume Ξ ∈ M and al ∈ [0, 1]R for i 0 and AvΞB (hal : l < ωi) = bi for i 0. Then for some ﬁnite u ⊆ B \ m∗ we have: if i < i∗, then P i bi − ε < {al : l ∈ u}/|u| < bi + ε. S ∗ P r o o f. Let B = j

≤ bi + ε/2 + ε/2 = bi + ε,

X i X X i X i ∗ al/|u| = {al : l ∈ uj}/|u| ≥ inf{al : l ∈ Bj}kj/k l∈u j bi − ε. j

Ql “Ξl is a ﬁnitely additive measure extending Ξ0”. Then e

Q1×Q2 “there is a ﬁnitely additive measure extending Ξ1 and Ξ2 e e (hence Ξ0)”. 114 S. Shelah

P r o o f. Straightforward by 1.2 as

(∗) if Ql “Al ⊆ ω” and Q1×Q2 “A1 ∩ A2 is finite” then e e e Q1×Q2 “for some m and A ⊆ ω, A ∈ V we have A1 \ m ⊆ A, (A2 \ m) ∩ A = ∅”. Fact 1.7. Assume Ξ is a partiale finitely additivee measure, and a¯α = α ∗ α hak : k < ωi a sequence of reals for α < α such that lim supk |ak | < ∞ for each α. Then (B)⇒(A) where ∗ ∗ full α (A) there is Ξ with Ξ ⊆ Ξ ∈ M such that AvΞ∗ (¯a ) ≥ bα for α < α∗, (B) for every partition hB0,...,Bm∗−1i of ω with Bm ∈ dom(Ξ) and ∗ ∗ ∗ ε > 0, k > 0 and α0 < . . . < αn∗−1 < α , there is a finite u ⊆ ω \ k such that:

(i) Ξ(Bm) − ε < |u ∩ Bm|/|u| < Ξ(Bm) + ε, −1 P αl ∗ (ii) |u| k∈u ak > bαl − ε for l < n . α Remark 1.8. If in (A) we demand AvΞ (¯a ) = bα, then in (B)(ii) add −1 P αl |u| k∈u ak ≤ bαl + ε.

2. The iteration. Ignoring MA<κ (which anyhow was a side issue) a quite natural approach in order to get 0.1 (i.e. cov(null) = λ, say λ = ℵω) ∗ is to use finite support iteration, Q¯ = hPα,Qα : α < α i, add in the first λ S ω steps null sets Nα (the intention is that α<λ Nα = 2 in the final model), 0 VPα 0 0 and then iterate with Qα being Random where Pα <◦ Pα and |Pα| < λ. Say, for some Aα ⊆ α, 0 Pα = {p ∈ Pα : dom(p) ⊆ Aα and this holds for the conditions

involved in the Pγ -name p(γ) for γ ∈ dom(p) etc.}

(so each Qα is a partial random; see Definition 2.2). If every set of < λ null 0 P ∗ P P ∗ sets from V α is included in some V α , clearly V α |= cov(null) ≥ λ; but we need the other inequality too. So we are using “non-transitive” memory, i.e. α ∈ Aβ 6⇒ Aα ⊆ Aβ; this makes our life hard. The problem is: why does hNα : α < λi continue to cover? For Pλ+n such that α ∈ [λ, λ + n) ⇒ Aα = α this is very clear (we get iteration of Random forcing) and if α ∈ [λ, λ + n) ⇒ Aα ⊆ λ this is clear (we get product). But ∗ necessarily we get a quite chaotic sequence hAαm ∩ {αl : l < m} : m < m i for some α0 < . . . < αm∗−1. More concretely this is the problem of why there are no perfect sets of random reals (see 2.7) or even just no dominating reals. We need to “let the partial randoms whisper secrets to one another”, in other words to pass information in some way. This is done by the finitely t additive measures Ξα. We had tried with thinking of using ℵε-support (see [Sh 538]), the ideae is still clear in the proof of 3.3. In this proof we start Covering of the null ideal 115 with “no dominating reals” for which we can just use ultrafilters (rather than finitely additive measures). Let us start with a ground model V satisfying the following hypothesis: P Hypothesis 2.1. (a) λ = ζ<δ(∗) λζ , ℵ0 < κ = cf(κ), κ < λζ < λγ for κ ℵ0 ζ < γ < δ(∗), 2 = χ and ζ < δ(∗) ⇒ (λζ ) < λ. (b) We have one of the following (1): (α) cf(χ) > λ, the length of the final iteration is χ, (β) the length of the final iteration is χ×χ×λ+ and cf([χ]<λ, ⊆) = χ. We speak mainly on (α). In case (β) we should be careful to have no ∗ ∗ repetitions inη ¯ = hηα : α < δ i (see below) or hηα/≈κ : α < δ i with no κ repetitions, where η ≈κ ν iff η, ν ∈ 2 and |{i < κ : η(i) 6= ν(i)}| < κ. The reader may choose to restrict himself and start with V satisfying: GCH, λ = ℵω, δ(∗) = ω, λn = ℵn(∗)+n, κ = ℵn(∗) > ℵ1 and χ = ℵω+1. Now add ℵω+1 generic subsets of κ, i.e., force with a product of χ copies of κ> ( 2, C) with support < κ. This model satisfies the hypothesis. We intend to define a forcing P such that

P ℵ0 V |= 2 = χ + cov(null) = λ + MA<κ. Definition 2.2. (1) K is the family of sequences ∗ Q¯ = (Pα, Qα,Aα, µα, τα : α < α ) ˜ satisfying: e ∗ (A) (Pα, Qα : α < α ) is a ﬁnite support iteration of c.c.c. forcing no- ∗ ¯ ¯ tions, we set eα = lg(Q) (the length of Q), Pα∗ is the limit. Pα (B) τα ⊆ µα < κ is the generic of Qα (i.e. over V from GQ we can ˜ α compute τα and vice versa). e ˜ e (C) Aα ⊆ α (for proving Theorem 0.1 we use |Aα| < λ). (D) Qα is a Pα-name of a c.c.c. forcing notion but computable from hτγ [GPα ]:e γ ∈ Aαi; in particular it belongs to Vα = V[hτγ [GPα ]: γ ∈ Aαi]. ˜ ∗ ω> ˜ (E)e α ≥ λ and for α < λ we have Qα = ( 2, C) (the Cohen forcing) ω> and µα = ℵ0 (well, identiﬁes 2 with ω). (F) For each α < α∗ one of the following holds (and the case is deter- mined in V, not just a Pα-name):

(α) |Qα| < κ, |Aα| < κ and (just for notational simplicity) the set ofe elements of Qα is µα < κ (but the order not necessarily the

order of the ordinals)e and Qα is separative (i.e. ζ ξ ∈ GQα ⇔ Qα |= ξ ≤ ζ). e e e In this case let τα = GQ . ˜ α e e (1) Actually, any ordinal α∗ of cardinality χ, divisible by χ and of coﬁnality > λ is O.K. 116 S. Shelah

Vα ω> (β) Essentially Qα = Random (= {r ∈ Vα : r ⊆ 2, perfect tree, Leb(lim(r))e> 0}) and |Aα| ≥ κ; but for simplicity Qα = ¯ RandomAα,Qα where for A ⊆ lg(Q¯), e A,Q¯ Random = {p : there is (in V) a Borel function B = B(x0, x1,...), with variables ranging on {true, false} and range perfect subtrees r of ω>2 with Leb(lim(r)) > 0 such that (∀η ∈ r)[Leb(lim(r[η])) > 0] (recall r[η] = {ν ∈ r : ν E η ∨η E ν}), and there are pairs (γl, ζl) for l < ω,

with γl ∈ A and ζl < µγl , such that p = B(..., truth value(ζl ∈ τγ ),...)l<ω} ˜ l (in other notation, p = B(truth value(ζl ∈ τγ ): l < ω)) and then we let ˜ l supp(p) = {γl : l < ω}. In this case µα = ω and τα is the random real, i.e. ˜ n ω> [η_hli] τα(n) = l ⇔ (∃η ∈ 2)[( 2) ∈ GQα ]. ˜ e (2) Let 0 Pα = {p ∈ Pα : for every γ ∈ dom(p), if |Aγ | < κ then p(γ) is an ordinal < µγ (not just a Pγ -name) and if |Aγ | ≥ κ then p(γ) has the form mentioned in clause (F)(β) above (and not just a Pγ -name of such an object)}

(this is a dense subset of Pα). (3) For A ⊆ α let 0 0 PA = {p ∈ Pα : dom(p) ⊆ A and γ ∈ dom(p) ⇒ supp(p(γ)) ⊆ A}. Fact 2.3. Suppose Q¯ ∈ K with lg(Q¯) = α∗. ∗ 0 (1) For α ≤ α , Pα is a dense subset of Pα and Pα satisﬁes the c.c.c. (2) Suppose (a) cf(α∗) > λ, (b) for every A ⊆ α∗, if |A| < λ, then there is β < α∗ such that A ⊆ Aβ (and |Aβ| ≥ κ). Then, in the extension, ω2 is not the union of < λ null sets. Pα (3) In V , from τα[GQα ] we can reconstruct GQα and vice versa. From ˜ Pα hτγ : γ < αi[GPα ] we can reconstruct GPα and vice versa. So V = V[hτβ : β˜ < αi]. ˜ (4) If µ < λ, and X is a Pα∗ -name of a subset of µ, then there is a set ∗ A ⊆ α such that |A|e ≤ µ and P ∗ “X ∈ V[hτγ : γ ∈ Ai]”. Moreover for α ˜ each ζ < µ there is in V a Borel functione B(x0, . . . , xn,...)n<ω with domain and range the set {true, false} and γl ∈ A, ζl < µγl for l < ω such that

Pα∗ “ζ ∈ X iﬀ true = B(..., “truth value of ζl ∈ τγl [GQγ ]”,...)l<ω”. ˜ l e Covering of the null ideal 117

∗ (5) For Q¯ ∈ K and A ⊆ α , every real in V[hτγ : γ ∈ Ai] has the form ∗ ˜ B(..., truth value(ζl ∈ τγl ),...)l<ω where γl < α , ζl < µγl and B is a Borel function (from V) from ω{true, false} to ω2 (the set of reals). P ∗ (6) If condition (c) below holds then V α |= MA<κ, where

Pα “Q = Qα”. e P e e V Aβ (7) If |A | ≥ κ and P 0 <◦ P then Q is actually Random . β Aβ β β 0 (8) If |Aα| < λ then Pα “Qα < λ”, in fact |{p(α): p ∈ Pα∗ }| < λ. P r o o f. (1) Easy, by inductione on α. ∗ (2) Easy using parts (3)–(7). Note that for any β < α satisfying |Aβ| ≥ [hτγ :γ∈Aβ i] ω P ∗ κ the null sets from V ˜ do not cover 2 in V α as we have random [hτγ :γ∈Aβ i] reals over V ˜ . So, by clause (b) of the assumption, it is enough to ω note that if y is a Pα∗ -name of a member of 2, then there is a countable ∗ A ⊆ α suche that y[G] ∈ V[hτβ : β ∈ Ai]. This follows by part (4). ˜ (3) By inductione e on α. (4) Let χ∗ be such that {Q,¯ λ} ∈ H(χ∗), and let ζ < µ; let M be a count- ∗ ∗ ¯ able elementary submodel of (H(χ ), ∈, <χ∗ ) to which {Q, λ, κ, µ, X, ζ} be- ∗ Pα longs, so Pα∗ “M[GPα∗ ] ∩ H(χ ) = M”. Hence by 2.3(3) (i.e. ase V = ∗ V[hτβ : β < αi]) we havee M[GPα∗ ] = M[hτi : i ∈ α ∩Mi] and the conclusion should˜ be clear. e ˜ (5) By 2.3(4). (6) Straight. (7) Easy (see more [Sh 619, §3]). ℵ0 (8) By the deﬁnitions (and since (λζ ) < λ and λ = sup{λζ : ζ < δ(∗)} by 2.1).

Definition 2.4. (1) Suppose thata ¯ = hal : l < ωi and hnl : l < ωi are such that: n (a) al ⊆ l 2, (b) nl < nl+1 < ω for l < ω, n l (c) |al|/2 l > 1 − 1/10 . ω ∞ Let N[¯a] =: {η ∈ 2 : (∃ l)(∀ν ∈ al)(ν C6 η)}. (2) Fora ¯ as above and n ∈ ω, let ω> treen(¯a) = {ν ∈ 2 : nl > n ⇒ νnl ∈ al}. It is well known that fora ¯ as above the set N[¯a] is null (and N[¯a] = ω S 2 \ n<ω lim(treen(¯a))).

Definition 2.5. For α < λ we identify Qα (the Cohen forcing) with

nl nl l {h(nl, al): l < ki : k < ω, nl < nl+1 < ω, al ⊆ 2, |al|/2 > 1 − 1/10 }, 118 S. Shelah

α α ordered by end extension. If GQα is Qα-generic, let h(¯al , nl ): l <ωi[GQα ] be the ω-sequence such that every p ∈ GQα is an initiale segmente of it. So α α α we have deﬁned the Qα-namea ¯ = hal : l < ωi and similarly hnl : l < ωi. α Let Nα = N[¯a ]. e e e e e Our aim is to prove (∗)Q¯ , where Definition 2.6. For Q¯ ∈ K with α∗ = lg(Q¯) let

ω Pα∗ ¯ (∗)Q¯ hNα : α < λi cover 2 in V , where Pα∗ = Lim(Q). We shall eventually prove it not for every Q¯, but for enough Q¯’s (basically asking the Aα of cardinality ≥ κ to be closed enough). ¯ ¯ Lemma 2.7. For Q ∈ K with γ = lg(Q), a suﬃcient condition for (∗)Q¯ is:

+ Pγ ω> κ (∗∗)Q¯ In V , there is no perfect tree T ⊆ 2 and E ∈ [λ] such that, α for some n < ω, T ⊆ treen[¯a ] for all α ∈ E. P r o o f. By induction on γ ≥ λ. For γ = λ, trivial by properties of the Cohen forcing. Suppose γ > λ is limit. Assume toward contradiction that [ α p Pγ “η 6∈ N[¯a ]”. e α<λ e Without loss of generality,

Pβ p Pγ “η 6∈ V ” for every β < γ, hence by properties ofe FS iteration of c.c.c. forcing notions, cf(γ) = ℵ0. So for each α < λ there are pα, mα such that α p ≤ pα ∈ Pγ , pα “η ∈ lim(treemα (¯a ))”. e e Note that (by properties of c.c.c. forcing notions) h{α < λ : pα ∈ Pβ} : β < γi is an increasing sequence of subsets of λ of length γ, so for some κ+ γ1 < γ there is E ∈ [λ] such that pα ∈ Pγ1 for every α ∈ E and we can + assume mα = m for α ∈ E. Note that for all but < κ of the ordinals α ∈ E we have p “|{β ∈ E : p ∈ G }| = κ+”. α β Pγ1 Fix such an α, and let G be a P -generic (over V) subset of P to which Pγ1 γ1 γ1 p belongs. Now in V[G ] let E0 = {β ∈ E : p ∈ G } so |E0| = κ+. α Pγ1 β Pγ1 ∗ T β Pγ1 ∗ ω> Let T = β∈E0 treem(¯a ). Note that, in V , T is a subtree of 2 and ∗ ∗ by (∗∗)Q¯ , T contains no perfect subtree. Hence lim(T ) is countable, so ∗ Pγ1 absolute. But pα Pγ “η ∈ lim(T )”, so pα “η ∈ V ”, a contradiction. P Assume now that γe= β + 1e> λ and worke in V β . Choose p, pα ∈ Qβ as before. By 2.3(7), Qβ has a dense subset of cardinality < λ, so there are Covering of the null ideal 119 q ∈ Qβ and m such that E = {α < λ : mα = n, pα ≤ q} has cardinality ≥ κ+. Continue as above. As we have covered the cases γ = λ, γ > λ limit and γ > λ successor, we have ﬁnished the proof.

Discussion 2.8. Note that by Lemma 2.7 and Fact 2.3 it is enough to show that there is Q¯ ∈ K such that α∗ = lg(Q¯) (where α∗ is chosen as the length of the final iteration from 2.1(b)), satisfying clauses (a)+(b)+(c) of Fact 2.3(2)+(6) and (∗∗)Q¯ . To prove the latter we need to impose more restrictions on the iteration. Now when Haim Judah asked me on the problem, whereas 2.1–2.7 were quite immediate, arriving at 2.9–2.11 has taken me many years and much effort. Definition 2.9. T , the set of blueprints, is the set of tuples t t t t t t t t t = (w , n , m , η¯ , h0, h1, h2, n¯ ) where: (a) wt ∈ [κ]ℵ0 , (b) 0 < nt < ω, mt ≤ nt, t t t t wt t (c)η ¯ = hηn,k : n < n , k < ωi, ηn,k ∈ 2 for n < n , k < ω, t t ω (d) h0 is a partial function from [0, n ) to κ, its domain includes the set t 2 {0,..., m − 1} (here we consider members of Qα for α < λ as integers ( )), t t (e) h1 is a partial function from [0, n ) to (0, 1)Q (rationals), but for n ∈ t t t P p t [0, n ) \ dom(h1) we stipulate h1(n) = 0 and we assume n (f) h2 is a partial function from [0, n ) to 2, t t t (g) dom(h0), dom(h1) are disjoint with union [0, n ), t t (h) dom(h2) = dom(h1), (i) ηt = ηt ⇒ n = n , n1,k1 n2,k2 1 2 t t (j) for each n < n we have: hηn,k : k < ωi is constant or with no t t repetitions; if it is constant and n ∈ dom(h0) then h0(n) is constant, t t t t t (k)n ¯ = hnk : k < ωi where n0 = 0, nk < nk+1 < ω and the sequence t t hnk+1 − nk : k < ωi goes to infinity. For l < ω and suchn ¯ let kn¯ (l) = k(l, n¯) be the unique k such that nk ≤ l < nk+1. Discussion 2.10. The definitions of a blueprint t ∈ T (in Definition 2.9) and of iterations Q¯ ∈ K3 (defined in Definition 2.11(c) below; the reader may first read it) contain the main idea of the proof, so though they have many clauses, the reader is advised to try to understand them.

2 t ( ) Actually the case where each h0(n) is a constant function from ω to κ suﬃces, and so κ < λ suﬃces instead of κℵ0 < λ. 120 S. Shelah

+ Pα∗ κ In order to prove (∗∗)Q¯ we will show in V that if E ∈ [λ] and T α n < ω, then α∈E treen(¯a ) is a tree with finitely many branches. So let + p be given, let p ≤ pζ e“βζ ∈ E” for ζ < κ , βζ 6∈ {βξ : ξ < ζ}, we can + assume pζ is in some pregiven densee set, and hpζ : ζ < κ i form a ∆-system ∗ (with some more “thinning” demands), dom(pζ ) = {αn,ζ : n < n }, αn,ζ is ∗ 0 increasing with n, and αn,ζ < λ iff n < m . Let pζ be pζ when pζ (αn,ζ ) is increased a little, as described below. ∗ ∗ 0 It suffices to find p ≥ p such that p “A := {ζ < ω : pζ ∈ G} is large T βζ e enough such that ζ∈A treem(¯a ) has only finitelye many branches”. Because of “communicatione problems” the “large enough” is interpreted t as of Ξα-measure (again defined in 2.11 below). ∗ Thee natural numbers n < n such that Qαn,ζ is a forcing notion of t cardinality < κ do not cause problems, as h0(n) tells us exactly what the condition pζ (αn,ζ ) is. Still there are many cases of such hpζ : ζ < ωi which fall into the same t; we possibly will get contradictory demands if αn1,ζ1 = t t αn2,ζ2 , n1 6= n2. But the w ,η ¯ are exactly built to make this case not to happen. That is, we have to assume 2κ = χ (= |α∗|) in order to be able for ∗ ∗ κ our iteration hPα, Qα : α < α i to choose hηα : α < α i, ηα ∈ 2, with no t repetitions, so thate if v ⊆ χ, |v| ≤ ℵ0 (e.g. v = {αn,ζ : n < n , ζ < ω}) then t ℵ0 for some w = w ∈ [κ] we have hηαw : α ∈ vi with no repetitions. So the blueprint t describes such a situation, giving as much information as possible, as long as the number of blueprints is not too large, κℵ0 = κ in our case.

If Qαn,ζ is a partial random, we may get many candidates for pζ (αn,ζ ) ∈ Random and they are not all the same ones. We want them in many cases t t to be in the generic set. Well, we can (using h1(n), h2(n)) know that in ω [ht (n)] some interval ( 2) 2 the set lim(pζ (αn,ζ )) is large, say of relative measure t t t ≥ 1 − h1(n), and we could have chosen the pζ ’s such that hh1(n): n < n i t is small enough, still the number of candidates is not bounded by 1/h1(n). Here taking limit along ultrafilters is not good enough, but using finitely additive measures is. t t t t t t Well, we have explained w ,η ¯ , h0, h1, h2, but what about then ¯ = t 0 hnk : k < ωi? In the end (in §3) the specific demand on {ζ : pζ ∈ G} being large, is that for infinitely many k < ω,

t t t t |{l : nk ≤ l < nk+1 and pl ∈ G}|/(nk+1 − nk) t is large, the nk will be chosen such that it is increasing fast enough and 0 t t hpl(βl): l ∈ [nk, nk+1)i will be chosen such that for each ε > 0 for some s < ω, for k large enough if the above fraction is ≥ ε then essentially k βl t t 0 2 ∩ {treem(¯a ): nk ≤ l < nk+1 and pl ∈ G} has ≤ s members; this suﬃces. Covering of the null ideal 121

What is our plan? We define K3, the class of suitable expanded iterations Q¯ by choice of ηα (for α < lg(Q¯)) and names for finitely additive measures t Ξα satisfying the demands natural in this context. You may wonder why we usee Ξ-averages; this is like integral or expected value, and so they “behave nicely” making the “probability computations” simpler. Then we show that we can find Q¯ ∈ K3 in which all obligations toward “cov(null) ≥ λ” and MA<κ hold. The main point of §3 will be that we can carry out the argument of ∗ ∗ 0 “for some p we have p {l < ω : pl ∈ G} is large” and why it gives κ+ T ζ n < ω & E ∈ [χ] ⇒ ζ∈A treen(¯a ) has finitely many branches, thus proving Theorem 0.1. t The reader may wonder how much the Ξα are actually needed. As explained above they are just a transparente way to express the property; this will be utilized in [Sh 619]. Definition 2.11. K3 is the class of sequences t ∗ ∗ Q¯ = hPα, Qβ,Aβ, µβ, τβ, ηβ, (Ξ )t∈T : α ≤ α , β < α i ˜ α (we write α∗ = lg(Q¯e)) such that: e ∗ ∗ (a) hPα, Qβ,Aβ, µβ, τβ : α ≤ α , β < α i is in K, κ ˜ ∗ (b) ηβ ∈e2 and (∀β < α < α )[ηα 6= ηβ], t (c) T is from Definition 2.9, and Ξα is a Pα-name of a finitely additive Pα measure on ω (in VPα , i.e. ∈ (Mfull)eV ), increasing with α, (d) we say thatα ¯ = hαl : l < ωi satisfies (t, n) (for Q¯) if:

• hαl : l < ωi ∈ V (of course), • t ∈ T , n < nt, ∗ • αl ≤ αl+1 < α , t • n < m ⇔ (∀l)(αl < λ) ⇔ (∃k)(αk < λ), t • ηn,l ⊆ ηαl (as functions), t • if n ∈ dom(h0), then µαl < κ and “|Q | < κ and (ht (n))(l) ∈ Q , i.e. (ht (n))(l) < µ ”, Pαl αl 0 αl 0 αl • if n ∈ dom(ht ), then µ ≥ κ so e “Q has cardinality ≥ κ” 1 αl Pαl αl (so it is a partial random), t t • if hηn,k : k ∈ ωi is constant, then (∀l)[αl = α0]; if hηn,k : k < ωi is not constant, then (∀l)[αl < αl+1], ¯ V (e) ifα ¯ = hαl : l < ωi satisﬁes (t, n) for Q, l<ω(αl < αl+1), n ∈ t dom(h0) then t Pα∗ “the following set has Ξα∗ -measure 1: {k < ω : if le∈ [nt , nt ) then (ht (n))(l) ∈ G }”, k k+1 0 Qαl e 122 S. Shelah

¯ t (f) ifα ¯ = hαl : l < ωi satisﬁes (t, n) for Q, n ∈ dom(h1), (∀l < ω)[αl <

αl+1], andr ¯ = hrl : l < ωi where for l < ω, rl is a Pαl -name of a member of

Qαl such that (ite is forced that) e t e t ω t lg(h2(n)) (∗) 1 − h1(n) ≤ Leb({η ∈ 2 : h2(n) C η ∈ lim(rl)})/2 , then for each ε > 0 we have e t Pα∗ “the following set has Ξα∗ -measure 1: {k < ω : in the set {l ∈e[nt , nt ): r ∈ G } there are k k+1 l Qαl t t e t at least (nk+1 − nk)(1 − h1(n))(1 − ε) elements}”, ¯ t (g) ifα ¯ = hαl : l < ωi satisﬁes (t, n) for Q, n ∈ dom(h1), (∀l)[αl = α], 0 Q¯ and r, rl are P -names of members of Qα satisfying (∗∗) (see below Aα r,hrl:l<ωi for thee e deﬁnition of (∗∗)) then ˜ ˜

Pα∗ “if r ∈ GQα then t e e t t t t 1 − h (n) ≤ Av t (h|{l ∈ [n , n ): rl ∈ GQ }|/(n − n ): k < ωi)”, 1 Ξα∗ k k+1 α k+1 k where e e e Q¯ 0 (∗∗) r, rl are P -names of members of Qα, h rl : l < ωi ∈ V r,hrl:l<ωi Aα ˜ ˜ Pα 0 0 eand,e in V , for every r ∈ Qα satisfying re≤ r we have 0 t AvΞt (hak(r ): k < ωi) ≥ 1 − h1(n) where α 0 0 0 t ) ak(r ) = ak(r , r¯) = ak(r , r,¯ n¯ )( 0 X Leb(lim(r ) ∩ lim(rl)) 1 = 0 t t t t Leb(lim(r )) nk+1 − nk l∈[nk,nk+1) 0 (so ak(r , r,¯ n¯) ∈ [0, 1]R is well deﬁned for k < ω,r ¯ = hrl : l < ωi, {r, rl} ⊆ Random,n ¯ = hnl : l < ωi, nl < nl+1 < ω), (h) |A | ≥ κ ⇒ P 0 <◦ P , and (3) β ∈ A & |A | < κ ⇒ A ⊆ A , α Aα α α β β α P t V Aα 4 (i) if Pα “|Qα| ≥ κ”, then ΞαP(ω) is a PAα -name ( ) for every t ∈ T . e e Definition 2.12. (1) For Q¯ ∈ K3 and for α∗ ≤ lg(Q¯) let ∗ t ∗ ∗ Q¯ α = hPα, Qβ,Aβ, µβ, τβ, ηβ, (Ξ )t∈T : α ≤ α , β < α i. ˜ α ¯1 ¯2 e3 ¯1 ¯2 ¯1 ¯2 ¯1 (2) For Q , Q ∈ K we say: Q ≤ Q if Q = Q lg(Q ). ¯ 3 ¯ ¯ 3 Fact 2.13. (1) If Q ∈ K and α ≤ lg(Q), then Qα ∈ K . (2) (K3, ≤) is a partial order.

(3) If we add “(∀α < κ)(|α|ℵ0 < κ)” to 2.1, we can simplify and assume α < α∗ ⇒ P 0 <◦ P . Aα α (4) Here the secret was whispered. Covering of the null ideal 123

β 3 (3) If a sequence hQ¯ : β < δi ⊆ K is increasing and cf(δ) > ℵ0, then there is a unique Q¯ ∈ K3 which is the least upper bound, lg(Q¯) = S ¯β ¯β ¯ β<δ lg(Q ) and Q ≤ Q for all β < δ. ω VPδ S ω VPβ P r o o f. Easy (recall that it is well known that ( 2) = β<δ( 2) , t S t so Ξδ = β<δ Ξβ is a legal choice; this is the use of cf(δ) > ℵ0). e n 3 n n+1 n Lemma 2.14. Suppose that Q¯ ∈ K , Q¯ < Q¯ , αn = lg(Q¯ ), δ = ¯ 3 ¯ ¯n ¯ supn<ω αn. Then there is Q ∈ K such that lg(Q) = δ and Q ≤ Q for n ∈ ω. t P r o o f. Note that the only problem is to define Ξδ for t ∈ T , i.e., we S t have to extend α<δ Ξα so that the following two conditionse are satisfied (they correspond to clausese (f) and (e) of Definition 2.11). 5 t t (a) We are given ( ) n < n , n ∈ dom(h1), hαl : l < ωi (from V of course) satisfies (t, n) for Q¯ and is strictly increasing with limit δ and we are given hp : l < ωi such that “p ∈ Q and 1 − ht (n) ≤ Leb({η ∈ l Pαl l αl 1 t ω t lg(h2(n)) e 2 : h2(n) eC η ∈ lim(pl)})/2 ”.e The demand is: for each ε > 0 we t have Pδ “Ξδ(C) = 1”,e where C = {ke < ωe: in the set {l : l ∈ [nt , nt ) and p ∈ G } there are k k+1 l Qαl e t t t e at least (nk+1 − nk)(1 − h1(n))(1 − ε) elements}. 4 t t ¯ (b) If ( ) n < n , n ∈ dom(h0), hαl : l < ωi satisfies (t, n) for Q and is t strictly increasing with limit δ, and pl ∈ Qαl satisfy pl = h0(n)(l) for l < ω t (an ordinal < µαl ), then Pδ “Ξδ(C) = 1” where C = {k < ω : for everyel ∈e[n , n ) we have p ∈ G }. k k+1 l Qαl

Se t Peδ As α<δ Ξα is a (Pδ-name of a) member of M in V , by 1.3(3) it suﬃces to provee

S t S V[Pα] t (∗) Pδ “if B ∈ α<δ dom(Ξα) = α<δ P(ω) and Ξα(B) > 0 ∗ ∗ and j 0, and n(j) < n , heαl : l < ωi, hpl : l < ωi involved in the deﬁnition S t S V[Pα] of Cj (in (a) or (b) above), q force: B ∈ α<δ dom(Ξα) = α<δ P(ω) S t ∗ ande ( Ξα)(B)>0 and Cj (for je α(∗) and moreover such that n − n is largee j α(∗)) (possibly j(1) j(2) αl(1) = αl(2) &(j(1), l(1)) 6= (j(2), l(2))). Now we choose by induction ∗ on m ≤ m a condition qm ∈ Pβm above q, increasing with m, where we stipulate βm∗ = δ. During this deﬁnition we “throw a dice” and prove that the probability ∗ of success (i.e. qm∗ “k ∈ Cj” for j < j ) is positive, so there is qm∗ as desired and hence we get thee desired contradiction.

Case A: m = 0. Let q0 = q. t t Case B: m + 1, and for some n < n , we have n ∈ dom(h0) and ζ and: ∗ j j t if j < j and l < ω then αl = βm ⇒ n(j) = n & pl = ζ (= h0(n(j))(l)) ∈

Qβm ). In this case dom(qm+1) = dom(qm) ∪ {βm}, and qm(β) if β < βm (so β ∈ dom(qm)), qm+1(β) = ζ if β = βm.

t t j Case C: m + 1 and for some n < n , we have n ∈ dom(h1) and: αl = β ⇒ n(j) = n. Work first in V[G ], q ∈ G , G generic over V. m Pβm m Pβm Pβm The sets j ω [ht (n)] j t t ∗ {lim(p [G ]) ∩ ( 2) 2 : α = β (and l ∈ [n , n ), j < j )} l Pβm l m k k+1 e ω [ht (n)] ω t are subsets of ( 2) 2 = {η ∈ 2 : h2(n) C η}, and we can define an ω [ht (n)] equivalence relation Em on ( 2) 2 : ν E ν iff ν ∈ lim(pj[G ]) ≡ ν ∈ lim(pj[G ]) 1 m 2 1 l Pβm 2 l Pβm e j e whenever αl = βm. m ∗ Clearly Em has finitely many equivalence classes, call them hZi : i < imi; all are Borel (sets of reals), hence measurable; without loss of generality, m ⊗ ∗ ⊗ ⊗ Leb(Zi ) = 0 ⇔ i ∈ [im, im), so clearly im > 0. For each i < im there is r = r ∈ Q [G ] such that m,i βm Pβm e lim(pj[G ]) ⊇ Zm ⇒ r ≥ pj[G ], l Pβm i l Pβm lim(pi[G ]) ∩ Zm = ∅ ⇒ lim(r) ∩ lim(pi[G ]) = ∅. l Pβm i l Pβm

We can also ﬁnd a rational am,i ∈ (0, 1)R such that t m − lg(h2(n)) ∗ am,i < Leb(Zi )/2 < am,i + ε/(2im). Covering of the null ideal 125

We can find q0 ∈ G , q ≤ q0 , such that q0 forces all this information m Pβm m m m m ⊗ ∗ (so for Zi , rm,i we shall have Pβm -names, but am,i, im, im are actual objects).e Wee can then find rationals bm,i ∈ (am,i, am,i + ε/2) such that P ⊗ b = 1. i 1 − 1/j , successes meaning qm∗ “k ∈ Cj” (remember that if j comes from clause (b) we always succeed). e t wt Remark 2.15. In the definition of t ∈ T (i.e. 2.9) we can add ηn,ω ∈ 2 t t (i.e. replace hηn,l : l < ωi by hηn,l : l ≤ ωi) and demand t t t (l) if ζ ∈ w then for every n < ω large enough, ηn,n(ζ) = ηn,ω(ζ), and in Definition 2.11(d) useα ¯ = hαl : l < ωi but this does not help here. Lemma 2.16. (1) Assume: ¯ 3 ¯ t ∗ ∗ (a) Q ∈ K , Q = hPα, Qβ,Aβ, µβ, rβ, ηβ, (Ξα)t∈T : α ≤ α , β < α i, (b) A ⊆ α∗, κ ≤ |A|

t t ηn∗,l = ηw for l < ω. Let Γ (∈ V) be the set of all pairs (r, hrl : l < ωi) Q¯ which satisfy the assumption (∗∗) of 2.11(g). In VPeα∗e+1 we have r,hrl:l<ωi t ˜ ˜ Pα∗ to choose Ξα∗+1 taking care of all these obligations. We work in V . P 0 By assumptione (d) and Claim 1.6 it suﬃces to prove it for V A so Qα∗ is P 0 RandomV A (see 2.3(7)). By 1.7 it is enough to prove condition (B) of 1.7. Suppose it fails. Then there are hBm : m < m(∗)i, a partition of ω from 0 PA t i i V , for simplicity Ξα∗ (Bm) > 0 for m < m(∗), and (r , hrl : l < ωi) ∈ Γ ∗ t ∗ ∗ and n(i) = n < n for i 0 and r e∈ Qeα∗ which forces the failure (of (B) of 1.7) for these parameters (the ε∗ comes from 1.7). We i ∗ can assume that r forces r ∈ GQα∗ for i < i (otherwise we can ignore such ri as nothing is demandede one them in 2.11(g)). So r ≥ ri for i < i∗ (see e2.2(F)(β)). By assumption, for each i < i∗ we have: for each r0 ≥ r (hence r0 ≥ ri and r0 ∈ Random) and i < i∗ we have i 0 t ∗ Av t (ha (r ): k < ωi) ≥ 1 − h (n ) Ξα∗ k 1 where (see 2.11(g)()) we let 0 i i 0 t 1 X Leb(lim(r ) ∩ lim(rl )) ak(r ) = ak(r, hrl : l < ωi, n¯ ) = t t 0 . nk+1 − nk t t Leb(lim(r )) e e l∈[nk,nk+1) By 1.7 it suﬃces to prove

Lemma 2.17. Assume Ξ is a ﬁnitely additive measure, hB0,...,Bm∗−1i ∗ i ∗ a partition of ω, Ξ(Bm) = am, i < ω and r, rl ∈ Random for i < i , l < ω ∗ ∗ ∗ ∗ and n¯ = hni : i < ωi, ni < ni+1 < ω, are such that 0 0 ∗ i 0 (∗) for every r ∈ Random, r ≥ r and i 0 and k∗ < ω there are a ﬁnite u ⊆ ω \ k∗ and r0 ≥ r such that: ∗ (1) am − ε < |u ∩ Bm|/|u| < am + ε for m < m , (2) for each i < i∗ we have

1 X |{l : n∗ ≤ l < n∗ and r0 ≥ ri}| k k+1 l ≥ b − ε. |u| n∗ − n∗ i k∈u k+1 k Covering of the null ideal 127

P r o o f. For i < i∗, m < m∗ and r0 ≥ r (from Random) let c (r0) = Av (hai (r0): k ∈ B i) ∈ [0, 1] . i,m ΞBm k m R So clearly 0 (∗)1 for r ≥ r (in Random) i 0 bi ≤ AvΞ (hak(r ): k < ωi) X X = Av (hai (r0): k ∈ B i)Ξ(B ) = c (r0)a . ΞBm k m m i,m m m

(∗)2 (a) ci,m ∈ [0, 1] , P R (b) m 0). Let ∗ ∗ ∗ Γ = {c¯ :c ¯ = hci,m : i ks,

(e) ks ∈ Bms . 128 S. Shelah

∗ In stage s, given rs, we define rs+1, is, ms, ks as follows: choose ms < m randomly with the probability of ms = m being am. Next we can find a ∗ finite set us ⊆ Bms with min(us) > max{k + 1, ks1 + 1 : s1 < s} such that

∗ 1 X i (∗) if i < i then ci,ms − ε/2 < ak(rs) < ci,ms + ε/2. |us| k∈us

We deﬁne an equivalence relation es on lim(rs): ∗ ∗ ∗ i i η1esη2 iﬀ (∀i < i )(∀k ∈ us)(∀l ∈ [nk, nk+1))[η1 ∈ lim(rl ) ≡ η2 ∈ lim(rl )].

The number of es-equivalence classes is finite, and if Y ∈ lim(rs)/es satisfies Leb(Y ) > 0 choose rs,Y ∈ Random such that lim(rs,Y ) ⊆ Y and rs,Y satisfies clause (c) of (∗)3 (possible by clause (c) of (∗)2). Now choose rs+1 among {rs,Y : Y ∈ lim(rs)/es and Leb(Y ) > 0}, with the probability of rs,Y being Leb(Y )/ Leb(lim(rs)). Lastly choose ks ∈ us, with all k ∈ us having the same probability. Now the expected value, assuming ms = m, of

1 ∗ ∗ i |{l : n ≤ l < n and rs+1 ≥ r }| n∗ − n∗ ks ks+1 l ks+1 ks belongs to the interval (ci,m −ε/2, ci,m +ε/2), because the expected value of

1 X 1 ∗ ∗ i ∗ ∗ |{l : nk ≤ l < nk+1 and rs+1 ≥ rl }| |us| nk+1 − nk k∈us is in this interval (as i X i Leb(lim(rs) ∩ lim(rl )) {Leb(Y ): Y ∈ lim(rs)/es, rs,Y ≥ rl } = Leb(lim(rs)) i and see the choice of ak(−)). 0 ∗ Let r = rs∗ , u = {ks : s ≤ s }. Hence the expected value of 1 X 1 |{l : n∗ ≤ l < n∗ and r0 ≥ ri}| |u| n∗ − n∗ k k+1 l k∈u k+1 k P is ≥ am(ci,m − ε/2) ≥ bi − ε/2. As s∗ is large enough with high probability (though just positive prob- ∗ 0 ability suﬃces), (rs∗ , {ks : s < s }) are as required for (r , u); note: we do not know the variance but we have a bound for it not depending on s. (2) Straightforward.

This completes the proof of Lemma 2.16.

The following is needed later to show that there are enough cases of the deﬁnition of t with clause (g) of Deﬁnition 2.11 being non-trivial (i.e. (∗∗) there holds). Covering of the null ideal 129

Lemma 2.18. Assume:

(a) Ξ is a ﬁnitely additive measure on ω and b ∈ (0, 1]R, ∗ ∗ ∗ ∗ ∗ (b) nk < ω (for k < ω), nk < nk+1, and lim(nk+1 − nk) = ∞, ∗ (c) r , rl ∈ Random are such that ∗ ∗ (∗)(∀l < ω)[Leb(lim(r ) ∩ lim(rl))/ Leb(lim(r )) ≥ b]. Then for some r⊗ ≥ r∗ we have: 0 ⊗ 0 0 ⊗r⊗ for every r ≥ r we have AvΞ (ha(r , k): k < ωi) ≥ b where ak(r ) = 0 0 ω a(r , k) = ak(lim r ) and for any Borel set X ⊆ 2 we let

1 X Leb(X ∩ lim(rl)) ak(X) = ∗ ∗ . nk+1 − nk ∗ ∗ Leb(X) l∈[nk,nk+1) P r o o f. Let ∗ 0 I = {r ∈ Random : r ≥ r and AvΞ (hak(r ): k < ωi) < b}. If I is not dense above r∗ there is r⊗ ≥ r∗ (in Random) such that for every r ≥ r⊗, r 6∈ I so r⊗ is as required; so assume toward contradiction ∗ ∗ that I is dense above r . There is a maximal antichain I1 = {si : i < i } ⊆ I (maximal among those ⊆ I); it is a maximal antichain above r∗ as r ∈ I ⇒ r ≥ r∗ and by the previous sentence. Hence Leb(lim(r∗)) = P ∗ i

X Leb(si) = Av (ha (s ): k < ωi) Leb(S s ) Ξ k i i

Leb(lim(s0)) X Leb(lim(si)) ≤ (b − ε) + b Leb(lim(S s )) Leb(lim(S s )) i

= b − Leb(lim(s0))ε where ε = b − AvΞ (hak(s0): k < ωi) so ε > 0. Let j be large enough such that Leb(lim(r∗) \ lim(sj)) < Leb(lim(s ))ε. Leb(lim(r∗)) 0 130 S. Shelah

So ∗ AvΞ (hak(r ): k < ωi) Leb(lim(r∗) \ lim(sj)) = Av (ha (lim(r∗) \ lim(sj)) : k < ωi) Leb(lim(r∗)) Ξ k Leb(lim(sj)) + Av (ha (sj): k < ωi) Leb(lim(r∗)) Ξ k Leb(lim(r∗) \ lim(sj)) Leb(lim(sj)) ≤ 1 + (b − Leb(lim(s ))ε) Leb(r∗) Leb(lim(r∗)) 0

< Leb(lim(s0))ε + (b − Leb(lim(s0))ε) = b, contradicting assumption (c). Claim 2.19. Assume: ¯ 3 ¯ t ∗ ∗ (a) Q ∈ K , Q = hPα, Qβ,Aβ, µβ, rβ, ηβ, (Ξα)t∈T : α ≤ α , β < α i, ∗ 6 (b) A ⊆ α and ( ) |Ae| < κ and µe < κ aree such that β ∈ A & |Aβ| < κ ⇒ Aβ ⊆ A, κ ∗ (c) η ∈ 2 \{ηβ : β < α }, (d) Q is a Pα∗ -name of a forcing notion with set of elements µ, and is really deﬁnablee in V[hτα : α ∈ Ai] from hτα : α ∈ Ai and parameters from V ˜ ˜ (we can even assume the truth value of ζ1 <θ ζ2 for ζ1 < ζ2 is as in 2.3(5)). Then there is + t ∗ ∗ Q¯ = hP α, Qα,Aβ, µβ, τβ, ηβ, (Ξ )t∈T : α ≤ α + 1, β < α + 1i ˜ α 3 e e from K extending Q¯ such that Qα∗ = Q, Aα∗ = A, ηα∗ = η and µα∗ = µ. P r o o f. Straight. e e Remark 2.20. If Q is the Cohen forcing we can make one step toward t {A ⊆ ω : Ξα∗+1(A) = 1} being a selective ﬁlter, even simultaneously for all ω-sequences of members, but not needed at present.

3. Continuation of the proof of Theorem 0.1. We need the following lemma.

Lemma 3.1. Suppose that ε¯ = hεl : l < ωi is a sequence of positive reals 3 and Q¯ ∈ K has length α. The following set Iε¯ ⊆ Pα is dense: 0 Iε¯ = {p ∈ Pα : there are m and αl, νl (for l < m) such that (a) dom(p) = {α0, . . . , αm−1}, α0 > α1 > . . . > αm−1,

(b) if |Aαl | < κ, then p(αl) is an ordinal,

(6) If “(∀α < κ)(|α|ℵ0 < κ)” is added to 2.1, it is natural to restrict ourselves to the 0 case PA <◦ Pα∗ as the parallel of 2.16(2) holds. Covering of the null ideal 131

(c) if Q is a partial random, i.e. |A | ≥ κ, then αl αl Pαl ω [ν ] lg(ν ) “p(αl) ⊆ ( 2) l and Leb(lim(p(αl))) ≥ (1−εl)/2 l ”}. P r o o f. By induction on α for all possibleε ¯.

Discussion 3.2. (1) By the previous sections it follows that it is enough 3 Pα to prove that if Q¯ ∈ K , Pα = Lim(Q¯), then in V the following suﬃcient condition holds: + Pα ω> κ (∗∗)Q¯ In V , there is no perfect tree T ⊆ 2, m ∈ ω and E ∈ [λ] α such that T ⊆ treem[¯a ] for all α ∈ E. t (2) Note that if we just want to prove Pα “b ≤ κ” life is easier: Ξα can be chosen a zero-one measure (so essentially an ultraﬁlter) and for αe < λ we ω> interpret the forcing notion Qα as ( ω, C) with generic real ηα and replace below (∗∗)Q¯ by e

+ Pα ∗ ω (∗∗)Q¯ in V , there is no η ∈ ω such that {α < λ :(∀l < ω)(ηα(l) ≤ η∗(l))} has cardinality ≥ κ+. e 0 ∗_ In the proof below, T is replaced by η, and pζ (αζ ) is s hζi. (3) We can makee the requirementse on the ∆-system stronger: make it indiscernible also over some A ⊆ α of cardinality < κ, where T is a 0 ∗ 0 PA-name, p ∈ PA, and assume the heart is ⊆ A. e t P (4) Here the existence of h2 helps; we can use 3.1 with l<ω εl very small. ¯ ¯ ¯ 3 Lemma 3.3. If Pα = Lim(Q), α = lg(Q) and Q ∈ K , then (∗∗)Q¯ from 2.7 holds. ∗ P r o o f. Suppose p Pα “T , m, E form a counterexample to (∗∗)Q¯”; we ∗ 0 can assume p ∈ Pα. Letε ¯ =e hεl e: l < ωi be such that εl ∈ (0, 1)R and P √ + ∗ 0 l<ω εl < 1/10. For each ζ < κ let pζ ≥ p be such that pζ ∈ Iε¯ (⊆ Pα) ζ witnessed by hνβ : β ∈ dom(pζ ) and |Aβ| ≥ κi (on Iε¯ see 3.1) and

pζ Pα “αζ is the ζth element of E”. e So clearly αζ < λ. By thinning out, we can assume that: ζ ∗ ζ ζ ∗ • dom(pζ ) = {γi : i < i } with γi increasing with i; let v0 = {i < i : ζ + ∗ |A ζ | < κ}; then v = v0 is fixed for all ζ < κ , and let v1 = i \ v0, γi 0 + ∗ • dom(pζ )(ζ < κ ) form a ∆-system with heart ∆, so ∆ ⊇ dom(p ), ζ ∗ • αζ ∈ dom(pζ ), αζ = γz for a fixed z • if i ∈ v1, then ν ζ = νi (recall that ν ζ ∈ 2 is given by the definition γi γi of Iε¯), 132 S. Shelah

∗ + ∗ ∗ • pζ (αζ ) = s for ζ < κ with s = h(nl, al): l < m i; we can assume ∗ ∗ m > m (where m is from the “counterexample to (∗∗)Q¯ ”) and m > 10, ∗ ζ + • for each i < i the sequence hγi : ζ < κ i is constant or strictly increasing, + • the sequence hαζ : ζ < κ i is with no repetitions (since if pζ1 , pζ2 are compatible and ζ1 < ζ2 < λ then αζ1 6= αζ2 ), ζ(1) ζ(2) + ∗ • if γi(1) =γi(2) then i(1)=i(2) (when ζ(1), ζ(2)<κ3 and i(1), i(2) 0 we deﬁne a Pα-name by 0 ∗ ∗ Aε = {k < ω : |{ζ ∈ [sk, sk+1): pζ ∈ GPα }|/(sk+1 − sk) > ε }. e e For the proof of 3.3 we need:

Claim 3.4. There is a condition p⊗ ≥ p∗ which forces that for some ∗ ε > 0 the set Aε∗ is infinite. e P r o o f. Choose ε∗ > 0 small enough. First we define a suitable blueprint t ∈ T , t t t t t t t t t = (w , n , m , η¯ , h0, h1, h2, n¯ ). Let t w = {min{β < κ : η ζ(1) (β) 6= η ζ(2) (β)} : ζ(1), ζ(2) < ω and γi(1) γi(2) ∗ ζ(1) ζ(2) i(1), i(2) < i and γi(1) 6= γi(2) }. t ∗ t t t t Let n = i , dom(h0) = v0, dom(h1) = dom(h2) = v1 and nl = sl. If t t t t n ∈ v , then h (n)(l) = γ and η = η ζ w . If n ∈ v , then h (n) = ε 0 0 n n,ζ γn 1 1 n t and h2(n) = νn. ⊗ ⊗ ∗ ⊗ We now define a condition p ; it will be in Pα, dom(p ) = ∆, p ≤ p ; ∗ ∗ remember dom(p ) ⊆ ∆ as for each ζ we have p ≤ pζ . If γ ∈ ∆ then for Covering of the null ideal 133

t V ζ ⊗ t some n < n , we have ζ<ω γn = γ. If n ∈ v0 we let p (γ) = h0(n), so trivially in VPγ , ⊗ t t t p (γ) Qγ “Ξγ+1({ζ < ω : h0(n) ∈ GQγ }) = 1 if n ∈ dom(h0) (= v0)”. e If n ∈ v1, then define a Pγ -name for a member of Qγ as follows. Consider t n 0 ω [h2(n)] rζ = pζ (γ) for ζ < ω. Let r be the member ( 2) of Qγ . If we work in 0 e PA e ∗ 0 ∗ V αe, by Lemma 2.18 there is rγ ≥ r from Qγ such that for every r ≥ rγ in Qγ we have e e e n 0 t (∗∗) 0 Av t (ha (r ): k < ωi) ≥ 1 − h (n) = 1 − ε where r ,ε Ξα k 1 n 0 n n 0 1 X Leb(lim(r ) ∩ lim(rl )) ak (r ) := t t 0 . nk+1 − nk t t Leb(lim(r )) l∈[nk,nk+1) Hence the assumption of condition (g) in Definition 2.11 holds, and so in VPγ we have ∗ t t t t r “Av t (h|{l ∈ [n − n ): p (γ) ∈ G }|/(n − n ): k ∈ ωi) γ Qγ Ξγ+1 k+1 k l Qγ k+1 k e ≥ 1 − εn”. ∗ ⊗ ∗ So there is a Pγ -name rγ of such a condition. In this case let p (γ) = rγ , so we have finished defininge p⊗; clearly it has the right domain. e t ζ Now suppose that n < n with n ∈ v1 is such that γn 6∈ ∆. Define ¯ ζ ¯ ¯ β = hβζ : ζ < ωi, βζ = γn. Then β satisfies (t, n) for Q. By our assumption the assumption of clause (f) in Definition 2.11 is satisfied, hence in VPα , for any ε > 0, t t l |{l ∈ [nk, nk+1): pl(γn) ∈ GQ l }| “Ξt k : γn ≥ (1 − ε )(1 − ε) = 1”. Pα α nt − nt e n e k+1 k Now for each n ∈ v1, as t t n |{l : nk ≤ l < nk+1 and rl ∈ GPα }| (1 − ε )(1 − ε) ≤ Av t : k < ω n Ξα t t nk+1 − nk and ε > 0 was arbitrary, clearly

Pα (∗)n in V , √ √ t t n t |{l : nk ≤ l < nk+1 and rl ∈ GPα }| 2εn ≥ Ξα k < ω : 1 − 2εn ≥ t t . nk+1 − nk Let 0 ζ Aε∗ = {k < ω : if ζ ∈ [sk, sk+1) and i ∈ v0 then pζ {γi } ∈ GPα }. √ e t 0 ∗ P ∗ e Clearly Ξα(Aε∗ ) = 1. Let ε < 1 − n 2εn and ε > 0, so e 134 S. Shelah

0 Aε∗ ∪ (ω \ Aε∗ ) e | T {l ∈ [nt , nt ) and p0 (γζ ) ∈ G }| ∗ n∈v1 k k+1 ζ n Qγn = k < ω : ε < t t nk+1 − nk t t n |{l : nk ≤l < nk+1 and rl ∈ GPα }| √ ⊇ k < ω : if n ∈ v1 then t t ≥1− 2εn nk+1 − nk t t n [ |{l : n ≤ l < n and r ∈ GPα }| √ = ω \ k < ω : k k+1 l < 1 − 2ε , nt − nt n n∈v1 k+1 k √ t 0 P ∗ hence Ξα(Aε∗ ∪ (ω \ Aε∗ )) ≥ 1 − n∈v 2εn ≥ ε > 0; but e t e 0 t 0 Ξα(ω \ Aε∗ ) = 1 − Ξα(Aε∗ ) = 1 − 1 = 0 e e e e and hence necessarily Aε∗ is inﬁnite. This suﬃces for 3.4.

⊗ Continuation of the proof of 3.3. Let p be as in Claim 3.4, let GPα be ⊗ a generic subset of Pα to which p belongs and we shall work in V[GPα ]. So A = Aε∗ [G] is inﬁnite. For k ∈ A, let e 0 bk = {ζ ∈ [sk, sk+1): pζ ∈ GPα }.

∗ j We know that |bk| > (sk+1 − sk)ε . Note that if k ∈ A, then T ∩ k 2 ⊆ T a as lg(s∗) = m∗ > m. To reach a contradiction it is enough to show ζ∈bk ζ j that for inﬁnitely many k ∈ A there is a bound on the size of ck = T ∩ k 2 which does not depend on k.

Now |bk|/(sk+1 −sk) is at most the probability that if we choose a subset ∗ of (jk)2 with 2jk (1 − 8−m ) elements, it will include T ∩ (jk)2; now if k ∈ A this probability has a lower bound ε∗ not depending on k, and this implies a bound on |T ∩ (jk)2| not depending on k. More formally, for a ﬁxed k < ω we have

|bk| = |{aζ : ζ ∈ [sk, sk+1), ζ ∈ bk}|

(jk) ≤ |{aζ : ζ ∈ [sk, sk+1),T ∩ 2 ⊆ aζ }| ∗ ≤ |{a ⊆ (jk)2 : T ∩ (jk)2 ⊆ a and |a| = 2jk (1 − 8−m )}| ∗ = |{a ⊆ (jk)2 \ (T ∩ (jk)2) : |a| = 2jk 8−m }| 2jk − |T ∩ (jk)2| = ∗ . 2jk 8−m Similarly 2jk sk+1 − sk = ∗ . 2jk 8−m Covering of the null ideal 135

(j ) j −3m∗ Let ik(∗) = min{|T ∩ k 2|, 2 k }. Hence 2jk − |T ∩ (jk)2| 2jk |bk|/(sk+1 − sk) ≤ ∗ ∗ 2jk 8−m 2jk 8−m j j 2 k − ik(∗) 2 k ≤ ∗ ∗ 2jk 8−m 2jk 8−m Y ∗ Y = (2jk − 2jk 8−m − i)/ (2jk − i)

So we can find a bound on ik(∗) not depending on k: ∗ −m∗ ik(∗) ≤ log(1/ε )/log(1/(1 − 8 )), ∗ −m∗ remember m > 10 so 1 − 8 ∈ (0, 1)R. So for k large enough, ∗ ∗ 2jk−3m < log(1/ε∗)/log(1/(1 − 8−m )), (j ) j −3m∗ hence if |T ∩ k 2| ≤ 2 k we are done. Otherwise, by the choice of ik(∗), ∗ (jk) ∗ −m |T ∩ 2| = ik(∗) ≤ log(1/ε )/log(1/(1 − 8 )). This finishes the proof. Theorem 3.5. Under Hypothesis 2.1 there is Q¯ ∈ K3 with lg(Q¯) = χ = δ∗ (if clause (α) of 2.1(b) holds) or lg(Q¯) = χ × χ × λ+ (if clause (β) of P ¯ 2.1(b) holds) such that in V lim Q we have MA<κ + cov(null) = λ. P r o o f. First assume clause (α) of 2.1(b). By 2.3(2) and 2.3(6) it suffices to find an iteration t 3 hPα, Qβ,Aβ, µβ, τβ, ηβ, (Ξ )t∈T : α ≤ χ, β < χi ∈ K ˜ α (see Definition 2.11)e satisfying clausese (a)+(b)+(c) of 2.3(2)+(6) (as the only property missing, cov(null) ≤ λ, holds by 2.7 + 3.3). − ¯ ¯ Let K3 = {Q ∈ K3 : lg(Q) < χ}. ¯ξ − Now choose Q ∈ K3 for ξ < χ increasing with ξ (see Definition 2.12) by induction on χ. If cf(ξ) > ℵ0 use 2.13(3), and if cf(ξ) = ℵ0 use 2.14. So we deal with ξ + 1. Bookkeeping gives us sometimes a case Q of 2.3(6)(c) as assignment; we can find suitable A ⊆ lg(Q¯ξ) by 2.3(4) ande then apply 136 S. Shelah

2.19 to get Q¯ξ+1. For other ξ, bookkeeping gives us a case of 2.3(2)(b) as assignment A ⊆ lg(Q¯ξ) such that |A| < λ. Now we apply 2.16(2) (with Q¯, A there standing for Q¯ξ, A here) and get A0 as there. Now apply 2.16(1) with Q¯0, A0 here standing for Q¯, A0 here standing for Q¯, A there (and η any κ 0 0 0 3 member of 2 \{ηβ : β < lg(Q¯ )}, possible as lg(Q¯ ) < χ as Q¯ ∈ K ) and get Q¯ξ+1 (corresponding to Q¯+ there). Second, assume clause (β) of (b). We just should be more careful in our + bookkeeping, particularly at the beginning let hηα : α < χ × χ × λ i be an enumeration of κ2 with no repetition and note cf([χ]<λ, ⊆) = χ suﬃces instead of χ = χλ.

References

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Institute of Mathematics Department of Mathematics The Hebrew University of Jerusalem Rutgers University 91904 Jerusalem, Israel New Brunswick, NJ 08854, U.S.A. E-mail: [email protected] URL: http://www.math.rutgers.edu/∼shelah

Received 25 May 1996; in revised form 17 June 1999