On Possible Restrictions of the Null Ideal

On possible restrictions of the null ideal

Ashutosh Kumar,∗ Saharon Shelah†

Abstract We prove that the null ideal restricted to a non-null set of reals could be isomorphic to a variety of sigma ideals. Using this, we show that the following are consistent: (1) There is a non-null subset of plane each of whose non-null subsets contains three collinear points. (2) There is a partition of a non-null set of reals into null sets, each of size ℵ1, such that every transversal of this partition is null.

1 Introduction

In [9], starting with a measurable cardinal, Shelah constructed a model in which the null ideal restricted to a non-null set of reals is ℵ1-saturated. We would like to prove the following generalization of this. Theorem 1.1. Suppose κ is a cardinal of uncountable coﬁnality and I is a sigma ideal on κ that contains all bounded subsets of κ. Then there is a ccc forcing P such that in V P, the null ideal restricted to some non-null set is isomorphic to the ideal generated by I. An immediate corollary of Theorem 1.1 (see Lemma 5.1 below) is that the null ideal restricted to some non-null set of reals can be isomorphic to the non-stationary ideal on ω1. This answers a question of Fremlin (Problem 9 in [2] and Problem DX in [3]). We also give the following applications.

Theorem 1.2. The following is consistent: There is a non-null X ⊆ R2 such that for every Y ⊆ X if Y is non-null, then Y contains three collinear points.

Theorem 1.3. The following is consistent: There exists a non-null X ⊆ R and a partition hXi : i < ω1i of X into null sets of size ℵ1 such that for every Y ⊆ X, if |Y ∩ Xi| = 1 for every i < ω1, then Y is null. ∗Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J Safra Campus, Givat Ram, Jerusalem 91904, Israel; email: [email protected]; Supported by a Postdoctoral Fellowship at the Einstein Insititute of Mathematics funded by European Research Council grant 338821 and by NSF grant no. 136974 †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 08854-8019, USA; email: [email protected]; Partially supported by European Research Council grant 338821 and by NSF grant no. 136974; Publication no. 1102

1 Recently, we were able to combine the methods of this paper with that of Komjath’s in [4] to construct a model where the null and the meager ideals are both somewhere ℵ1-saturated. This will appear in [7].

On notation: µ denotes the standard Lebesgue measure on 2ω. A subset W ⊆ 2ω is fat if for every clopen set C, either W ∩ C = φ or µ(W ∩ C) > 0. A subtree T ⊆ <ω2 is fat if ω ω [T ] = {x ∈ 2 :(∀n < ω)(x n ∈ T )} is fat. For a clopen subset C ⊆ 2 , deﬁne supp(C) to be ω the smallest (ﬁnite) set F such that (∀x, y ∈ 2 )((x F = y F ) =⇒ (x ∈ C ⇐⇒ y ∈ C)). Random denotes the random real forcing. Note that {[T ]: T ⊆ <ω2 is a fat tree} is dense in Random. Cohenκ denotes the forcing for adding κ Cohen reals. In forcing we use the convention that a larger condition is the stronger one - p ≥ q means p extends q. If P, Q are forcing notions and Q ⊆ P, we write Q l P if every maximal antichain in Q is also a maximal antichain in P.

2 On category

The category analogue of Theorem 1.1 follows from the argument in [4] where Komj´ath proved the following: Let κ be measurable. Then there is a ccc forcing P such that in V P, the meager ideal restricted to some non-meager set of reals is ℵ1-saturated.

Let us sketch this argument. Suppose κ ≥ ω1 and I is a sigma ideal on κ that contains ω P all singletons. Let P = Cohenκ adding κ Cohen reals hci : i < κi where each ci ∈ 2 . In V , Q let Q be the finite support product {QA : A ∈ I} where QA is defined as follows: p ∈ QA iff p = (Fp,Np, n¯p, σ¯p) = (F,N, n,¯ σ¯) where

• F is a ﬁnite subset of A • N < ω

• n¯ = hnk : k ≤ Ni is a strictly increasing sequence of integers with n0 = 0

[nk,nk+1) • σ¯ = hσk : k < Ni where each σk ∈ 2

For p, q ∈ QA, define p ≤ q iff Fp ⊆ Fq, Np ≤ Nq,n ¯p n¯q,σ ¯p σ¯q and for every Np ≤ k < Nq, for every i ∈ Fp, σq,k 6= ci [nq,k, nq,k+1). Note that QA is a sigma centered forcing making {ci : i ∈ A} meager. The set of conditions (p, q) ∈ P ? Q where p ∈ P and for each A ∈ dom(q), p forces an actual value to q(A) is dense in P ? Q. Put S = P ? Q. S We claim that, in V the meager ideal restricted to W = {ci : i < κ} is isomorphic to J = {X ⊆ κ :(∃A ∈ I)(X ⊆ A)} - the ideal generated by I. It is clear that the for each S X ∈ J , {ci : i ∈ X} is meager. Also, in V , for every X ⊆ κ, if X/∈ J , then {ci : i ∈ X} ω S is non-meager. To see this, let B ⊆ 2 be a meager Fσ-set coded in V . Since S is ccc, we S Q can find a countable F ⊆ I such that B is coded in V [hci : i ∈ Fi][ {G : A ∈ F}]. S QA Choose i? ∈ X \ ( F). Note that for each A ∈ F, QA ∈ V [hci : i ∈ Ai]. It follows that ci? S Q is Cohen over V [hci : i ∈ Fi][ {GQA : A ∈ F}]. Hence ci? ∈/ B and therefore {ci : i ∈ X} is non-meager.

2 This proof does not have an obvious analogue in the case of the null ideal since, e.g., if we start by adding a set X = {ri : i < κ} of κ random reals, and do a finite support iteration (for ccc) to make certain subsets of X null, then we’d inevitably add Cohen reals at stages of countable cofinality making X null. To get around this difficulty, Shelah came up with the following idea in [8, 9].

κ Let hXα : α < λi be a list of members of I, each occurring λ = 2 times. Perform a ﬁnite support iteration hPα, Qα : α < λ + κi with limit P where for α < λ, Qα is V [hτα:α∈Aλ+ξi] Cohen forcing with generic real τα and for ξ < κ, Qλ+ξ = (Random) , where Aλ+ξ = {α < λ : ξ∈ / Xα} ∪ [λ, λ + ξ), with generic partial random real τλ+ξ. Using the fact that P is ccc, it is easy to show that if A ⊆ κ is not in the ideal generated by I, then P {τλ+ξ : ξ ∈ A} is not null in V . The other direction is supposed to follow from the fact that for α < λ, τα codes a null Gδ-set that should cover {τλ+ξ : ξ ∈ Xα} because for ξ ∈ Xα, the memory Aλ+ξ of the partial random τλ+ξ does not contain α. But Aλ+ξ contains [λ, λ + ξ) and the partial random added at these stages can use τα so this is not a simple product forcing argument as in the category case.

Some remarks on the structure of the proof of Theorem 1.1 follow. The next section introduces the model witnessing the theorem and proves some preliminary facts about the iteration used in defining it. The main difficulty in the verification appears in Claim 3.8 whose proof is reduced to constructing a condition p? satisfying the hypothesis of Lemma 3.9. The concluding remarks in Section 3 explain the difficulty in an inductive construction of p? due to the use of partial memories for the random reals. Section 4 introduces the main tools to get around this difficulty via Definition 4.6 and Claim 4.7 where the existence of p? is reduced to the existence of certain finitely additive measures on P(ω) ∩ V P. Lemma 4.8 completes the proof by showing that such measures exist. The use of blueprints is to allow a sufficiently general statement in this lemma so that an inductive proof using automorphisms of higher iterations can be given.

3 Forcing

κ +ω Let κ, I be as in the hypothesis of Theorem 1.1. Put λ0 = 2 . For λ0 ≤ λ < λ0 , deﬁne the following.

+ω (1) hXα : α < λ0 i is a sequence of members of I.

+n +n (2) For every n < ω and X ∈ I, |{α < λ0 : Xα = X}| = λ0 .

λ (3) For ξ < κ, Aλ+ξ = Aλ+ξ = {α < λ : ξ∈ / Xα} ∪ [λ, λ + ξ). ¯ (4) Pλ = hPα, Qα : α < λ + κi is a ﬁnite support iteration with limit Pλ+κ such that ω (a) For α < λ, Qα is Cohen forcing with generic real τα ∈ ω .

V [hτi:i∈Aλ+ξi] ω (b) For ξ < κ, Qλ+ξ = (Random) with generic partial random τλ+ξ ∈ 2 .

3 ¯ The reason for considering iterations Pλ for λ > λ0 will become clear during the proof of Lemma 4.8 where we use automorphisms of Pλ+κ for λ > λ0 to construct certain ﬁnitely additive measures on P(ω) ∩ V Pλ0+κ .

The following is easily proved by induction on ξ ≤ κ using the standard properties of Cohen and random forcings.

Claim 3.1. For every ξ ≤ κ, ˚x ∈ 2ω ∩ V Pλ+ξ , there are a Borel function B : ωω → 2ω and h(nk, γk): k < ωi such that every γk < λ + ξ and nk < ω, and P ˚x = B(hτγk (nk): k < ωi). 0 Deﬁnition 3.2. (1) Let Pλ+κ be the set of conditions p ∈ Pλ+κ satisfying the following requirements. (a) For each α ∈ λ ∩ dom(p), p(α) ∈ <ωω. In this case, deﬁne supp(p(α)) = φ.

(b) Let dom(p) ∩ [λ, λ + κ) = {α0 < α1 < . . . αn−1}. For each j < n, there are <ω a Borel function B, ρ ∈ 2 and h(nk, γk): k < ωi such that nk < ω, γk ∈ Aαj , <ω the range of B consists of fat trees in 2 and Pα p(αj) = [B(hτγk (nk): k < ωi)] is a subset of [ρ] of relative measure more than 1 − 2−(n−j+10). In this case, deﬁne supp(p(αj)) = {γk : k < ω}.

0 0 (2) For ξ ≤ κ, deﬁne Pλ+ξ = {p ∈ Pλ+κ : dom(p) ⊆ λ + ξ}.

0 0 0 (3) For A ⊆ λ+κ, deﬁne Pλ,A = PA = {p ∈ Pλ+κ : dom(p) ⊆ A, (∀α ∈ dom(p))(supp(p(α)) ⊆ A)}. Note that if A = λ + ξ for some ξ ≤ κ, then this agrees with the notation in (2) above.

0 Using Claim 3.1 and the Lebesgue density theorem, it is easy to see that Pλ+ξ is dense in Pλ+ξ.

+ω Suppose λ0 ≤ λ < λ0 , ξ? ≤ κ. Let h : λ + ξ? → λ + ξ? be a bijection satisfying the following:

(1) h [λ, λ + ξ?) is the identity

(2) For each ξ < ξ? and α < λ, α ∈ Aλ+ξ ⇐⇒ h(α) ∈ Aλ+ξ

ˆ 0 0 0 ˆ 0 Deﬁne h : Pλ+ξ? → Pλ+ξ? as follows. For p ∈ Pλ+ξ? , put h(p) = p where (a) dom(p0) = {h(α): α ∈ dom(p)}

(b) For α ∈ dom(p) ∩ λ, p0(h(α)) = p(α)

0 (c) For α ∈ dom(p) ∩ [λ, λ + ξ?), Pα p (α) = B(hτh(γk)(nk): k < ωi) where B, h(nk, γk): k < ωi are as in Deﬁnition 3.2 for coordinate α.

ˆ 0 Claim 3.3. h is an automorphism of Pλ+ξ? .

Proof: By induction on ξ?.

4 Lemma 3.4. Let ξ? ≤ κ, A ⊆ λ+ξ? and [λ, λ+ξ?) ⊆ A. Suppose for every countable B ⊆ λ, there is a bijection h : λ + ξ? → λ + ξ? such that h ((B ∩ A) ∪ [λ, λ + ξ?)) is the identity,

(∀ξ < ξ?)(∀α < λ)(α ∈ Aλ+ξ ⇐⇒ h(α) ∈ Aλ+ξ), h[B] ⊆ A and h[B ∩ Aλ+ξ? ] ⊆ A ∩ Aλ+ξ? . 0 0 Then PA l Pλ+ξ? .

Proof of Lemma 3.4: By induction on ξ?. If ξ? = 0 or limit this is clear. So assume 0 0 ξ? = ξ + 1 and put α = λ + ξ. By inductive hypothesis, PA∩α l Pα so it suﬃces to check 0 0 the following: If {p : n < ω} ⊆ , p ∈ and p 0 {p (α): n < ω, p α ∈ G 0 } n PA PA∩α PA∩α n n PA∩α V [hτ :β∈A∩A i] β α 0 0 is predense in (Random) , then p Pα {pn(α): n < ω, pn α ∈ GPα } is predense V [hτβ :β∈Aαi] 0 in (Random) . Suppose this fails and choose q ∈ Pα, D, h(nk, γk): k < ωi such 0 that q ≥ p, D is a Borel function whose range consists of fat trees, γk ∈ Aα and q Pα r =

[D(hτγk (nk): k < ωi)] is incompatible with every member of {pn(α): n < ω, pn α ∈ GPα }. S S S Let W = dom(q)∪ {supp(q(β)) : β ∈ dom(q)}∪ {supp(p(β)) : β ∈ dom(p)}∪ {dom(pn): S n < ω} ∪ {supp(pn(β)) : n < ω, β ∈ dom(pn)} ∪ {γk : k < ω}. Put B = W ∩ λ. Choose ˆ 0 h : α → α satisfying the hypothesis for this B. So h is an automorphism of Pα. As h[B] ⊆ A, ˆ 0 ˆ h(q) ∈ PA. Since h (B ∩ A) is identity, it follows that h(p) = p and for every n < ω, 0 ˆ 0 ˆ h(pn) = pn. Since {γk : k < ω} ⊆ W and h[B ∩ Aα] ⊆ A ∩ Aα, we have that Pα r = h(r) = V [hτ :β∈A∩A i] 0 β α ˆ 0 [D(hτh(γk)(nk): k < ωi)] ∈ (Random) . It follows that h(q) Pα r is incompatible 0 0 ˆ ˆ 0 with every condition in {pn(α): n < ω, pn α ∈ GPα }. Hence also p = h(p) ≤ h(q) P r A∩α is incompatible with every condition in p (α): n < ω, p α ∈ G 0 – Contradiction. n n PA∩α

0 0 Corollary 3.5. For every ξ? < κ, . PAλ+ξ? l Pλ+ξ? Proof of Corollary 3.5: Let B ⊆ λ be countable. By Lemma 3.4, it suﬃces to con-

struct a bijection h : λ + ξ? → λ + ξ? such that h ((B ∩ Aλ+ξ? ) ∪ [λ, λ + ξ?)) is iden-

tity, (∀ξ < ξ?)(∀α < λ)(α ∈ Aλ+ξ ⇐⇒ h(α) ∈ Aλ+ξ) and h[B] ⊆ Aλ+ξ? . For each x ⊆ ξ?, let Wx = {α < λ : Xα ∩ ξ? = x}. Let Wx,0 = {α ∈ Wx : ξ? ∈ Xα} and Wx,1 = {α ∈ Wx : ξ? ∈/ Xα} so Wx = Wx,0 t Wx,1. Since every bounded subset of κ is in I, it follows that for every x ⊆ ξ?, |Wx,0| = |Wx,1| = λ. So for each x ⊆ ξ?, we can choose a bijection hx : Wx → Wx such that hx[Wx,0 ∩ B] ⊆ Wx,1 and hx (Wx,1 ∩ B) is identity. Put S h λ = {hx : x ⊆ ξ?}.

P Put P = Pλ+κ. In V , let J = {X ⊆ κ :(∃Y ∈ I)(X ⊆ Y )} be the ideal generated by I. Since P is ccc, J is a sigma ideal.

P + Claim 3.6. In V , for every A ∈ J , {τλ+ξ : ξ ∈ A} is not null.

Proof of Claim 3.6: Let N be a null Borel set coded in V P. Since P is ccc we can ﬁnd a countable family {pk : k < ω} such that {k < ω : pk ∈ G } determines N. Let S P W = {dom(pk): k < ω} ∪ {supp(pk(α)) : k < ω, α ∈ dom(pk)}. Note that W is countable. Since cf(κ) ≥ ℵ1 and I contains all bounded subsets of κ, we can ﬁnd ξ ∈ A such that S λ + ξ > sup(W ) and ξ∈ / {Xi : i ∈ W ∩ λ}. It follows that N is coded in V [hτα : α ∈ Aλ+ξi] and hence τλ+ξ ∈/ N.

5 n Deﬁnition 3.7. For each n < ω, let hCk : k < ωi be a one-one listing of all clopen subsets of 2ω of measure 2−n. For α < λ, deﬁne N˚ = T S Cn . So N˚ is a null G -set coded α k n>k τα(n) α δ by τα.

To complete the proof of Theorem 1.1, we need to show that in V P, for every A ∈ I, {τλ+ξ : ξ ∈ A} is null. For this it suﬃces to show the following. Its proof will be completed at the end of Section 4 with the proof of Lemma 4.8. ˚ Claim 3.8. Suppose ξ < κ and α ∈ λ \ Aλ+ξ. Then P τλ+ξ ∈ Nα.

0 Proof of Claim 3.8: Suppose not. Choose p ∈ Pλ+ξ+1, k? < ω such that p (∀k ≥ k )(τ ∈/ Ck ). We can assume that α ∈ dom(p) and p(α) = σ ∈ l? ω for some l > k . ? λ+ξ τα(k) ? ? ? Choose a Borel function B and h(nj, γj): j < ωi such that γj ∈ Aλ+ξ (see Deﬁnition ˚ ˚ 3.2(1)(b)), range of B consists of fat trees and Pλ+ξ B(hτγj (nj): j < ωi) = T and [T ] = p(λ + ξ). It follows that p (λ + ξ) (∀k ≥ k )([T˚] ∩ Ck = φ). Choose hα : i < λi Pλ+ξ ? τα(k) i such that the following hold.

• For all i < j < λ, αi < αj < λ

• Xαi = Xα S • αi ∈/ dom(p) {supp(p(β)) : β ∈ dom(p)}

0 For i < λ, deﬁne qi ∈ Pλ+ξ as follows.

• dom(qi) = (dom(p (λ + ξ)) \{α}) ∪ {αi}

• qi(αi) = p(α) = σ?

• If β ∈ dom(p) ∩ (λ \{α}), then qi(β) = p(β)

• If β ∈ dom(p) ∩ [λ, λ + ξ), and B, h(nj, γj): j < ωi are such that Pβ [B(hτγj (nj): 0 j < ωi)] = p(β), then q (β) = [B(hτ 0 (n ): j < ωi)] where γ = γ if γ 6= α and α i γj j j j j i otherwise.

0 Since interchanging α, αi coordinates gives rise to an automorphism of Pλ+ξ that ﬁxes ˚ ˚ k the name T (as α, αi ∈/ Aλ+ξ), it follows that, for each i < λ, qi (∀k ≥ k?)([T ]∩C = φ). ταi (k)

Let dom(p) ∩ λ = {α} t {β0 < β1 < ··· < βr?−1} and dom(p) ∩ [λ, λ + ξ) = {λ + ξ0 <

λ + ξ1 < ··· < λ + ξn?−1}. For k < n?, let ρk be as in Clause (1)(b) of Deﬁnition 3.2 for the (λ + ξk)th coordinate of p. It follows that {(qj, αj): j < ω} satisﬁes the following.

(a) dom(qj) = {αj} t {βk : k < r?} t {λ + ξk : k < n?}

(b) For every j < ω, k < r?, qj(βk) = p(βk) = σk

6 (d) For every j < ω, k < n?, qj(λ + ξk) is a fat subset of [ρk] of fractional measure Pλ+ξk more than 1 − 2−(n?−k+10)

˚ k (e) For every j < ω and qj λ+ξ (∀k ≥ k?)([T ] ∩ C = φ) P ταj (k)

Recall that |σ?| = l? > k?. We’ll extend each qj on the αjth coordinate to get pj as l? follows. For each n < ω, let Kn = {k < ω : supp(Ck ) ⊆ n}. Note that for all n ≥ l?, 2n 2n |Kn| = 2n−l? . Deﬁne hkn : n < ωi by: k0 = 0, kn+1 − kn = 2n−l? . Let f : ω → ω be such that f[[kn, kn+1)] = Kn. For each j < ω, γ ∈ dom(qj) deﬁne ( qj(γ) if γ 6= αj pj(γ) = σ? ∪ {(l?, f(j))} if γ = αj

Our plan to construct a condition p? which will force a “large number” of pj’s in the generic ﬁlter which will imply that [T˚] is ﬁnite giving us the desired contradiction. The following lemma makes this precise.

Lemma 3.9. Suppose for some p? ∈ P and ε > 0,

∞ |{j ∈ [kn, kn+1): pj ∈ GP}| p? (∃ n) ≥ ε kn+1 − kn ˚ Then, p? [T ] is ﬁnite. ˚ ˚ n Proof of Lemma 3.9: For n < ω, let Wn = {j ∈ [kn, kn+1): pj ∈ GP} and ˚an = |T ∩ 2|. ˚ ˚ n l? Note that p? (∀j ∈ Wn)(∀σ ∈ T ∩ 2)(Cf(j) ∩ [σ] = φ) because l? > k?, pj(αj)(l?) = f(j) n k ˚ ˚ 2 −˚an and pj (∀k ≥ k?)(C ∩ [T ] = φ). It follows that |Wn| ≤ n−l? . Hence ταj (k) 2

n 2 −˚a ˚an ˚an ˚ n n−l? n−l? |Wn| n−l? Y 2 Y 2 ≤ 2 = 1 − ≤ 1 − k − k 2n 2n −˚a + i 2n n+1 n 2n−l? i=1 n i=1 Hence

|W˚ | ˚a n ≤ 1 − 2−l? n kn+1 − kn ˚ As ˚an is increasing with n, it follows that p? forces that limn ˚an < ∞ and hence [T ] is ﬁnite.

n ˚ ˚ Note that ˚an ≈ 2 µ([T ]) which is much better asymptotic behaviour than just T being a perfect tree (which was what was used in [8]) but this doesn’t seem to help simplify the construction of p?. Towards constructing p?, we can further reﬁne hpj : j < ωi as follows.

<ω Lemma 3.10. Suppose K < ω, F ⊆ [λ, λ + κ) is ﬁnite, hρα : α ∈ F i is a sequence in 2, haα : α ∈ F i is a sequence in (0.5, 1) and hqj : j < Ki is a sequence of conditions in P such that for every j < K, dom(qj) = F and for each α ∈ F , Pα qj(α) is a subset of [ρα] ? 0 ? of relative measure ≥ aα. Then there exists q ∈ P with dom(q ) = F such that for every ? α ∈ F , Pα q (α) is a subset of [ρα] of relative measure ≥ 2aα − 1 and

7 ? −|F | Y q P |{j < K : qj ∈ GP}| ≥ K2 aα α∈F

Proof of Lemma 3.10: By induction on |F |. Suppose F = {α}. Work in V Pα . Deﬁne P φ = j µ([ρα])(2aα − 1). We have

Z Z Z Kaα Kaαµ([ρα]) ≤ φdµ = φdµ + φdµ ≤ Kµ(A) + (µ([ρα]) − µ(A)) A 2ω\A 2

µ(A) aα Solving gives ≥ > 2aα − 1. µ([ρα]) 2 − aα 0 0 0 Now suppose |F | ≥ 2 and β is the largest member of F . Let F = F \{β}, qj = qj F . 0 0 0 0 0 Choose q ∈ P with domain F such that for every α ∈ F , Pα q (α) is a subset of [ρα] 0 0 −|F 0| Q of relative measure 2aα − 1 and q P |{j < K : qj ∈ GP}| ≥ K2 α∈F 0 aα. Let ˚ 0 −|F 0| Q W = {j < K : qj ∈ GP}. Let {Wi : i < N} list all subsets of K of size ≥ K2 α∈F 0 aα. 0 ˚ Choose a maximal antichain {ri : i < N} in Pβ above q such that each ri P W = Wi. Pβ Work in V . For each i < N, arguing as above, we can get a condition si ∈ Qβ such that |W |a r µ(s ) ≥ 2a − 1 and s |{j ∈ W : q (β) ∈ G }| ≥ i β . Choose q? such that i Pβ i β i Qβ i j Qβ 2 ? 0 0 ? q (α) = q (α) if α ∈ F and for each i < N, ri Pβ q (β) = si.

? Using Lemma 3.10, construct hqn : n < ωi such that

? 0 ? (a) For every n < ω, qn ∈ Pλ+ξ and dom(qn) = {λ + ξk : k < n?}

? (b) For every n < ω and k < n?, q (λ + ξk) is a subset of [ρk] of relative measure Pλ+ξk n ≥ 2(1 − 2−(n?−k+10)) − 1 = 1 − 2−(n?−k+9)

? (c) For every n < ω, qn Pλ+ξ |{j ∈ [kn, kn+1): pj [λ, λ + ξ) ∈ GPλ+ξ }| ≥ (kn+1 − k )2−n? Q (1 − 2−(n?−k+10)) > (k − k )4−n? n k

? ? Deﬁnition 3.11. Deﬁne p¯ = hpj : j < ωi as follows

? 0 ? (1) For every j < ω, pj ∈ Pλ+ξ and dom(pj ) = dom(pj)

? (2) For every k < r?, pj (βk) = pj(βk) = σk

? (2) For every j < ω, pj (αj) = pj(αj) = σ? ∪ {(l?, f(j))}

? ? (3) For every n < ω, k < n? and j ∈ [kn, kn+1), pj (λ + ξk) = pj(λ + ξk) ∩ qn(λ + ξk). Hence ? −(n?−k+8) µ(p (λ + ξk)) ≥ 1 − 2 Pλ+ξk j

8 ˚ ? Let A = {i < ω :(∃n < ω)(i ∈ [kn, kn+1) ∧ (∀k ∈ [kn, kn+1))(pk λ ∈ GP))}. Note that ˚ q = {(βk, σk): k < r?} P A is infinite. By Lemma 3.9, it suffices to construct a condition ˚ ? p? ≥ q that forces that A ∩ {j < ω : pj ∈ GP} is infinite. What prevents us from arguing as we did in Lemma 3.10 to construct p?? If n? = 1, there is no problem. But the inductive step ? fails. Having constructed p (λ+ξk), we must somehow pass the information about the the ˚ ? set A ∩ {j < ω : p (λ + ξk) ∈ G } to V [hτα : α ∈ Aλ+ξ i]. This will be materialized by j Pλ+ξk k introducing a coherent family of finitely additive measures on P(ω) that interact with the partial random reals appearing at stages {λ + ξk : k < n?} of the iteration in a sense made precise in the next section (Definition 4.6).

4 Measures and blueprints

An algebra A is a family of subsets of ω that contains all finite subsets of ω and is closed under complementation and finite union. A finitely additive measure on an algebra A is a function m : A → [0, 1] that satisfies the following. • For every finite F ⊆ ω, m(F ) = 0. • m(ω) = 1.

• If A1,A2 ∈ A and A1 ∩ A2 = φ, then m(A1 ∪ A2) = m(A1) + m(A2). Suppose m : P(ω) → [0, 1] is a ﬁnitely additive measure and f : ω → [0, 1]. Deﬁne

Z 2n X kak fdm = lim n→∞ 2n k=0 n n where ak = m({n < ω : k/2 ≤ f(n) < (k + 1)/2 }).

The following is a standard application of the Hahn-Banach theorem. Lemma 4.1. Suppose m : A → [0, 1] is a ﬁnitely additive measure on an algebra A and X ⊆ ω. Let a ∈ [0, 1] be such that for every A, B ∈ A, if A ⊆ X ⊆ B, then m(A) ≤ a ≤ m(B). Then, there exists a ﬁnitely additive measure m0 : P(ω) → [0, 1] that extends m and m0(X) = a. The proofs of the next two lemmas can be found in [1].

Lemma 4.2. Suppose m : P(ω) → [0, 1] is a finitely additive measure. For i = 1, 2, let Ri Ri be a forcing notion and m˚i ∈ V be such that Ri m˚i : P(ω) → [0, 1] is a finitely additive R1×R2 measure extending m. Then, there exists m˚3 ∈ V such that R1×R2 m˚3 : P(ω) → [0, 1] is a finitely additive measure extending both m˚1 and m˚2. Lemma 4.3. Suppose that m : P(ω) → [0, 1] is a finitely additive measure. Let B = Random, B ˚ B r ∈ B. Define m˚r ∈ V as follows. For X ∈ P(ω) ∩ V , define ( ( ) ) Z µ(q ∩ [[n ∈ X˚]] m˚ (X˚) = sup inf B dm : q ≥ p : p ≥ r, p ∈ G r µ(q) B Then, the following hold.

9 (1) r m˚r : P(ω) → [0, 1] is a ﬁnitely additive measure extending m. µ(r ∩ [[n ∈ X˚]] ) (2) If X˚ ∈ P(ω) ∩ V B and a > 0 satisfy for every n < ω, B ≥ a, then there µ(r) ˚ exists s ≥ r such that s m˚r(X) ≥ a. +ω ¯ ¯ Deﬁnition 4.4. For λ0 ≤ λ < λ0 , let Tλ be the set of tuples t = (¯α, η,¯ β, r, ξ, n, σ,¯ ρ,¯ ε¯) = (¯αt, η¯t, β¯t, rt, ξ¯t, nt, σ¯t, ρ¯t, ε¯t) where

(i) n, r < ω

(ii) α¯ = hαj : j < ωi where each αj < λ and they are pairwise distinct

<ω (iii) η¯ = hηj : j < ωi where each ηj ∈ ω ¯ (iv) β = hβk : k < ri is a sequence of pairwise distinct ordinals in λ \{αj : j < ω}

<ω (v) σ¯ = hσk : k < ri where each σk ∈ ω ¯ (vi) ξ = hξk : k < ni is an increasing sequence in κ

<ω (vii) ρ¯ = hρk : k < ni where each ρk ∈ 2

−8 (viii) ε¯ = hεk : k < ni, where εn−1 ∈ (0, 2 ) and 2εk ≤ εk+1 for every k < n − 1

We call members of Tλ blueprints. They are intended to code information about certain 0 ? sequences of conditions in Pλ that look likep ¯ from Deﬁnition 3.11 in the following sense. 0 ¯ ¯ Deﬁnition 4.5. Suppose p¯ = hpj : j < ωi is a sequence in Pλ and t = (¯α, η,¯ β, r, ξ, n, σ,¯ ρ,¯ ε¯) ∈ Tλ. We say that p¯ is of type t if the following hold.

(a) For every j < ω, dom(pj) = {αj} t {βk : k < r} t {λ + ξk : k < n}

(b) For every j < ω and k < r, pj(βk) = σk and pj(αj) = ηj

For t ∈ Tλ, ξ < κ, we write t ξ for the blueprint which is obtained by restricting the sequence ξ¯t to ordinals below ξ and modifyingρ ¯t, ε¯t and nt accordingly.

¯ ¯ Pλ+ξ Deﬁnition 4.6. Suppose t = (¯α, η,¯ β, r, ξ, n, σ,¯ ρ,¯ ε¯) ∈ Tλ, ξn−1 < ξ ≤ κ and m˚ ∈ V . We say that m˚ satisﬁes t if the following hold.

(1) Pλ+ξ m˚ : P(ω) → [0, 1] is a ﬁnitely additive measure

(2) For every k < n, letting Vk = V [hτα : α ∈ Aλ+ξk i], we have Pλ+ξ m˚ (P(ω) ∩ Vk) ∈ Vk

10 0 (3) For every p¯ = hpj : j < ωi of type t, there exists pp¯ ∈ Pλ+ξ such that

(a) dom(pp¯) = {βk : k < r} ∪ {λ + ξk : k < n}

(b) For every k < r, pp¯(βk) = σk

˚ Ap¯ = {i < ω :(∃n < ω)(i ∈ [kn, kn+1) ∧ (∀j ∈ [kn, kn+1))(pj λ ∈ GP))} ˚ (e) For every k < n, pp¯ Pλ+ξ m˚(Xp,k¯ ) ≥ 1 − 2εk > 0 where ˚ Xp,k¯ = {j < ω : pj [λ, λ + ξk + 1) ∈ GP}

? ? Letp ¯ = hpj : j < ωi be as in Deﬁnition 3.11. The next claim says that it is enough to construct a measurem ˚ ∈ V Pλ+ξ that satisﬁes the blueprint associated withp ¯?.

Pλ+ξ Claim 4.7. Suppose for every t ∈ Tλ, for some ξn−1 < ξ < κ, there exists m˚ ∈ V such that m˚ satisﬁes t. Then there exists p? ∈ P such that p? satisﬁes the hypothesis of Lemma 3.9.

? ? ? Pλ+ξ Proof of Claim 4.7: Let t ∈ Tλ be such thatp ¯ is of type t . Choosem ˚ ∈ V where ? ˚ ? ξn−1 < ξ < κ such thatm ˚ satisﬁes t . Let p? = pp¯? , Xp¯? = {j < ω : pj [λ, λ + ξ) ∈ GP} and

˚ ? Ap¯? = i < ω :(∃n < ω)(i ∈ [kn, kn+1) ∧ (∀j ∈ [kn, kn+1))(pj λ ∈ GP)) ˚ ˚ ˚ ˚ Then p? forces thatm ˚(Ap¯? ) = 1 andm ˚(Xp¯? ) > 0 and hence that Ap¯? ∩ Xp¯? is inﬁnite. It follows that (see Deﬁnition 3.11(3))

∞ |{j ∈ [kn, kn+1): pj ∈ GP}| −n? p? (∃ n) ≥ 4 kn+1 − kn

The next Lemma completes the proof of Claim 3.8 and therefore of Theorem 1.1. ¯ ¯ Lemma 4.8. Suppose λ0 ≤ λ < λω, t = (¯α, η,¯ β, r, ξ, n, σ,¯ ρ,¯ ε¯) ∈ Tλ and ξn−1 < ξ < κ. Then there exists m˚ ∈ V Pλ+ξ such that m˚ satisﬁes t. Proof of Lemma 4.8: By induction on n = |ξ¯|.

Suppose n = 0. Fix ξ < κ. Put pp¯ = {(βk, σk): k < r}. Note that there is a uniquep ¯ of ˚ type t. Let Ap¯ = {i < ω :(∃n < ω)(i ∈ [kn, kn+1) ∧ (∀j ∈ [kn, kn+1))(pj λ ∈ GP))}. Then ˚ Pλ+ξ pp¯ Pλ Ap¯ is infinite. Hence we can choosem ˚ ∈ V such that Pλ+ξ m˚ : P(ω) → [0, 1] is ˚ a finitely additive measure and q Pλ+ξ m˚(Ap¯) = 1. It follows thatm ˚ satisfies t.

+ω ¯ ¯ Pλ+ξn+1 Next ﬁx λ0 ≤ λ < λ0 , t = (¯α, η,¯ β, r, ξ, n+1, σ,¯ ρ,¯ ε¯) ∈ Tλ. We’ll constructm ˚1 ∈ V such thatm ˚1 satisﬁes t. By the inductive hypothesis, there is a measurem ˚ satisfying t ξn. The next claim says that we can do slightly better.

11 0 Pλ,λ+ξ ¯ ¯ Claim 4.9. There exists m˚ ∈ V n such that m˚ satisﬁes t ξn = (¯α, η,¯ β, r, ξ n, n, σ,¯ ρ¯ 0 0 Pλ,A Pλ,A n, ε¯ n) and 0 m˚ (P(ω) ∩ V λ+ξn ) ∈ V λ+ξn . Pλ,λ+ξn 0 0 0 t0 ¯t0 ¯ Proof of Claim 4.9: Let Tλ+ = {t ∈ Tλ+ : t = (¯α , η,¯ β , r, ξ n, n, σ,¯ ρ¯ n, ε¯ n)}. By 0 0 0 0 t Pλ+,λ++ξn t inductive assumption, for every t ∈ Tλ+ , there existsm ˚ ∈ V such thatm ˚ satisﬁes t0. Fix such a map t0 7→ m˚t0 .

Let χ be suﬃciently large. Choose M0,M1 elementary submodels of (Hχ, ∈, <χ) such ¯ 0 0 t0 that M0 ∈ M1, |M0| = |M1| = λ, and for l ∈ {0, 1}, Pλ+ , Tλ+ and the map t 7→ m˚ are in M , λ + 1 ⊆ M and ≤κM ⊆ M . Note that if B ∈ {λ ∩ Aλ , λ \ Aλ } for l l l l j λ+ξj λ+ξj T + λ+ + λ+ j < n + 1, then | j

+ (i) For every ξ < ξn, h(λ + ξ) = λ + ξ

λ λ+ (ii) For every k < n and α < λ, α ∈ A iﬀ h(α) ∈ A + λ+ξk λ +ξk

λ (iii) For every α < λ, α ∈ Aλ+ξn iﬀ h(α) ∈ M0

0 ¯ Let t = (hh(αj): j < ωi, η,¯ hh(βk): k < ri, r, ξ n, n, σ,¯ ρ¯ n, ε¯ n). As Ml is countably 0 t0 closed, t and thereforem ˚ are in M1.

ˆ 0 0 ˆ 0 0 Deﬁne h : → + + as follows: h(p) = p where dom(p ) = {h(α): α ∈ Pλ,λ+ξn Pλ ,(λ +ξn)∩M1 0 dom(p)}. If α ∈ dom(p) ∩ λ, then p (h(α)) = p(α). If α ∈ dom(p) ∩ [λ, λ + ξn), then 0 p (α) = B(hτh(γk)(nk): k < ωi) where B, h(nk, γk): k < ωi are as in Deﬁnition 3.2(1)(b) for coordinate α.

Claim 4.10. The following hold. ˆ 0 0 (1) h : → + + is an isomorphism Pλ,λ+ξn Pλ ,(λ +ξn)∩M1 0 0 0 (2) + + + + + + Pλ ,(λ +ξn)∩M0 l Pλ ,(λ +ξn)∩M1 l Pλ ,λ +ξn 0 0 + 0 P + P + λ t λ ,Ak λ ,Ak (3) For k < n, put Ak = Aλ++ξ ∩ M1. Then P0 m˚ (P(ω) ∩ V ) ∈ V k λ+,λ++ξn 0 0 0 P + + P + + t λ ,(λ +ξn)∩Ml λ ,(λ +ξn)∩Ml (4) For l ∈ {0, 1}, P0 m˚ (P(ω) ∩ V ) ∈ V λ+,λ++ξn Proof of Claim 4.10: (1) and (4) should be clear. For (2), use Lemma 3.4. For (3), recall thatm ˚t0 satisﬁes t0.

0 0 P + + 0 P + + 0 λ ,(λ +ξn)∩M1 0 t λ ,(λ +ξn)∩M1 Choosem ˚ ∈ V such that P0 m˚ =m ˚ (P(ω) ∩ V ) λ+,λ++ξn 0 Pλ,λ+ξ ˆ 0 ¯ ¯ and deﬁnem ˚ ∈ V n by h(˚m) =m ˚ . By Claim 4.10,m ˚ satisﬁes t ξn = (¯α, η,¯ β, r, ξ 0 0 Pλ,A Pλ,A n, n, σ,¯ ρ¯ n, ε¯ n) and 0 m˚ (P(ω) ∩ V λ+ξn ) ∈ V λ+ξn . This completes the Pλ,λ+ξn proof of Claim 4.9.

12 0 Pλ,A Vn Put Vn = V λ+ξn and B = (Random) . Working in Vn, apply Lemma 4.3 tom ˚ B (P(ω) ∩ Vn), with r = [ρn] to obtain the extensionm ˚r ∈ (Vn) as defined there. Since 0 0 0 Pλ+ξ Q , we can write V n = (Vn) for some ∈ Vn. By Lemma 4.2, it follows Pλ,Aλ+ξn l Pλ+ξn Q 0 B Q Q×B Pλ+ξ +1 thatm ˚r ∈ (Vn) andm ˚ ∈ (Vn) have a common extensionm ˚1 ∈ (Vn) = V n . It suffices to check thatm ˚1 satisfies t. So fixp ¯ = hpj : j < ωi of type t and construct pp¯ as follows. Putq ¯ = hp (λ + ξ ): j < ωi. Sincem ˚ satisfies t ξ , we can find p ∈ 0 j n n q¯ Pλ+ξn

satisfying clauses (3)(a)-(d) in Defintion 4.6. Working in Vn = V [hτα : α ∈ Aλ+ξn i], let Vn ˚ B ˚ B = (Random) , r = [ρn] and X ∈ P(ω)∩(Vn) be such that [[j ∈ X]]B = pj(n). By Lemma ˚ 4.3, we can choose s ≥ r, such that s B m˚r(X) ≥ 1 − εn. Define pp¯ = pq¯ ∪ {(λ + ξn, s)} and note that it satisfies clauses (3)(a)-(e) in Definition 4.6.

5 Avoiding collinear points

Note that under the continuum hypothesis (or just add(Null) ≤ c), every non-null subset X of the plane has a subset Y of the same two-dimensional Lebesgue outer measure which omits collinear points. In [5], Komj´athshowed the following. It is consistent that there is a non-meager A ⊆ R such that every injective f : A → A is meager in the plane. It follows that A2 is a non-meager subset of the plane each of whose non-meager subsets contains three collinear points. To see this, note that if Y ⊆ A2 is such that every vertical and horizontal section of Y has ≤ 2 points, then Y can be covered by four injective functions from A to A. Let us prove the measure analogue of this.

Lemma 5.1. It is consistent that there is a non-null A ⊆ R such that the null ideal restricted to A is isomorphic to the non-stationary ideal on ω1.

Proof of Lemma 5.1: Let κ = ω1 and I be the non-stationary ideal on κ. Apply Theorem 1.1 and use the fact that the forcing is ccc.

Proof of Theorem 1.2: Let A ⊆ R be as in Lemma 5.1. List A = hxα : α < ω1i such that 2 for every W ⊆ ω1, W is non-stationary iﬀ {xα : α ∈ W } is null. Put X = A . Then X is non-null. As noted above, it suﬃces to show that if f : A → A is injective, then f is null in the plane. Let A1 = {xα :(∃β < α)(f(xα) = xβ)}, A2 = {xα :(∃β > α)(f(xα) = xβ)} and A3 = {xα : f(xα) = xα}. Note that f A3 is null. Towards a contradiction, suppose f A1 is non-null in the plane. Then A1 is non-null in R. Hence {α : xα ∈ A1} is stationary. But f is injective so this contradicts Fodor’s lemma. Similarly the inverse of f A2 is null in the plane. Hence f is null in the plane.

6 On transversals

In [6], the following was shown: For X ⊆ [0, 1] and for every partition {Xi : i ∈ S} of X into countable sets, there exists Y ⊆ X such that for every i ∈ S, |Y ∩ Xi| = 1 and µ?(Y ) = µ?(X). Bill Weiss asked what happens if we consider partitions into null sets of size ℵ1. Note that under the continuum hypothesis the result continues to hold. Theorem

13 1.3 says that it can consistently fail too.

Proof of Theorem 1.3: Deﬁne I = {A ⊆ ω1 × ω1 :(∃i0 < ω1)(∀i > i0)(|Ai| ≤ ℵ0)} where Ai = {j :(i, j) ∈ A} and apply Theorem 1.1.

7 A question

The diligent reader might have noticed that there are sigma ideals for which our method doesn’t work. This is because of the requirement in Theorem 1.1 that the sigma ideal contain all bounded subsets of κ. So we can ask the following.

Question 7.1. Can the null ideal restricted to some non null set of reals be isomorphic to ≤ℵ1 (ω3, [ω3] )?

References

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[3] D. H. Fremlin, Problems to add to the gaiety of nations, Version of 10.10.2018, https: //www1.essex.ac.uk/maths/people/fremlin/problems.pdf

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[5] P. Komj´ath, A second category set with only ﬁrst category functions, Proc. Amer. Math. Soc. 112(1991), 11291136

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[7] A. Kumar, S. Shelah, Saturated null and meager ideal, Trans. Amer. Math. Soc., To appear

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