Advances in Mathematics 321 (2017) 475–485 Contents lists available at ScienceDirect Advances in Mathematics www.elsevier.com/locate/aim A transversal of full outer measure Ashutosh Kumar a,∗,1 , Saharon Shelah a,b,2 a Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J Safra Campus, Givat Ram, Jerusalem 91904, Israel b Department of Mathematics, Rutgers, The State University of New Jersey, Hill Center-Busch Campus, 110 Frelinghuysen Road, Piscataway, NJ 08854-8019, USA a r t i c l e i n f o Article history: Received 21 October 2015 Received in revised form 7 September 2017 Accepted 2 October 2017 Available online xxxx Communicated by Slawomir J. Solecki a b s t r a c t We show that for every partition of a set of reals into countable sets there is a transversal of the same outer measure. © 2017 Elsevier Inc. All rights reserved. Keywords: Outer measure Forcing 1. Introduction Our aim is to prove the following. Theorem 1.1. Suppose Xα : α ∈ S is a partition of X ⊆ [0, 1] into countable sets. Then there exists Y ⊆ X such that |Y ∩ Xα | = 1 for each α ∈ S and μ (Y ) = μ (X). * Corresponding author. E-mail addresses: akumar@math.huji.ac.il (A. Kumar), shelah@math.huji.ac.il (S. Shelah). Supported by a Postdoctoral Fellowship at the Einstein Institute of Mathematics funded by European Research Council grant 338821 and by NSF grant No. DMS 1101597. 2 Partially supported by European Research Council grant 338821 and by NSF grant No. DMS 1101597; Publication No. 1068. 1 https://doi.org/10.1016/j.aim.2017.10.008 0001-8708/© 2017 Elsevier Inc. All rights reserved. 476 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 Here μ denotes Lebesgue outer measure on R. For partitions into ﬁnite sets, this follows from an old result of Lusin [9] which says that any set of reals can be partitioned into two sets of full outer measure (see Lemma 2.2). Another special case of the above theorem was established in [8]: Every set of reals has a subset of full outer measure that avoids rational distances. The proof given there relied on a theorem of Gitik and Shelah [4–6] which says that forcing with a sigma ideal cannot be isomorphic to a product of random and Cohen forcing (we give another proof of this in Theorem A.1). As a byproduct of our proof, we get a generalization of this theorem to a larger class of forcings (see Lemma 6.1) – For example, an ω-length ﬁnite support iteration of random forcing. For background on forcing and generic ultrapowers, we refer the reader to [2,7]. On notation: For a set of reals X, by env(X) (envelope of X), we mean a Gδ set G containing X such that G \ X has zero inner measure. All relations involving envelopes are supposed to hold modulo null sets. A subset Y of X has full outer measure in X if env(X) = env(Y ). If Y ⊆ X and env(X) = env(X \ Y ) we say that Y has positive inner measure in X; otherwise, we say that Y has zero inner measure in X. For T ⊆ <ω 2, deﬁne [T ] = {x ∈ 2ω : (∀n < ω)(x n ∈ T )}. For σ ∈ <ω 2, deﬁne [σ] = {x ∈ 2ω : σ x}. In forcing, we use the convention that a larger condition is the stronger one – So p ≤ q means q extends p. If Q, P are forcing notions, we write Q P if Q ⊆ P and every maximal antichain in Q is also a maximal antichain in P. For an ideal I over a set X, deﬁne the following. • I + = P(X) \ I; / I; • add(I) is the least cardinal κ satisfying: there exists F ⊆ I, |F| ≤ κ and F ∈ • For Y ∈ I + , I Y = {W ⊆ Y : W ∈ I} is the restriction of I to Y . 2. A suﬃcient condition Without loss of generality, (∀α ∈ S)(|Xα | = ℵ0 ). For each α ∈ S, let Xα = {xα,n : n < ω}. Put Yn = {xα,n : α ∈ S}. For W ⊆ S, write Yn W = {xα,n : α ∈ W }. Note that Yn depends on the speciﬁc enumeration of Xα we ﬁxed. Claim 2.1. It is enough to show the following. (): For every X ⊆ [0, 1], for every partition Xα : α ∈ S of X into ℵ0 -sized subsets, for every enumeration Xα = {xα,n : n < ω} (so we can speak of Yn ’s w.r.t. this enumeration), there is a subset W of S such that either (a) Y0 is null or (b) Y0 W has positive outer measure and for all n ≥ 1, Yn W has zero inner measure in Yn . Proof of Claim 2.1. Assume (). It is enough to show that we can strengthen “Y0 W has positive outer measure” to “μ (Y0 W ) ≥ 0.5(μ (Y0 ))” in () above. For then we can inductively construct a sequence (Wi , ni ) : i < ω such that A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 477 • • • • the Wi ’s are pairwise disjoint subsets of S, each n < ω equals ni for inﬁnitely many i’s, if m = ni = nj , i < j, then env(Ym Wi ) ∩ env(Ym Wj ) = ∅, for each i, letting m = ni and Si = S\ {Wj : j < i}, we have μ (Ym Wi ) ≥ 0.5μ [(Ym Si )\Bi ], where Bi = env( {Ym Wj : j < i, nj = m}), • if n = ni , then μ (Yn (Si \Wi )) = μ (Yn Si ). Claim 2.1 will immediately follow by taking Y = {Yni Wi : i < ω}. Call a sequence Wi : i < δ (where δ ≤ ω) of pairwise disjoint subsets of S greedy, if for each i < δ the following “greediness condition” holds: μ (Y0 Wi ) ≥ 0.5(sup{μ (Y0 W ) : W ⊆ Si , env(Y0 W ) ∩ env(Y0 S\Si ) = 0, (∀n ≥ 1)(Yn W has zero inner measure in Yn Si )}) where Si = S\ {Wj : j < i} and Yn Wi has zero inner measure in Yn Si for every n ≥ 1. Clearly such sequences exist by () so ﬁx a maximal one and let W = {Wi : i < δ}. If possible, suppose Y0 W does not have full outer measure in Y0 . So δ = ω. Using (), pick W ⊆ S\W such that Y0 W is non null, its envelope is disjoint with the envelope of Y0 W and for every n ≥ 1, μ (Yn S\(W ∪ W )) = μ (Yn S\W ). Since μ (Y0 Wi ) goes to zero as i → ∞, we conclude that at some stage i < ω, our sequence ceased to be greedy: A contradiction. Hence Y0 W has full outer measure in Y0 and by taking the union of suﬃciently many Wi ’s we get the required strengthening of () above. 2 Lemma 2.2. Given n ≥ 1, A ⊆ [0, 1] and a partition {xα,k : k < n} : α ∈ T of A into ordered sets of size n, there exists a partition Tk : k < n of T such that for every k < n, {xα,k : α ∈ Tk } has full outer measure in {xα,k : α ∈ T }. Proof of Lemma 2.2. Use induction on n. If n = 1, there is nothing to show. So assume that the result holds for n ≥ 1. Arguing as in the beginning of the proof of Claim 2.1, it is enough to show the following: For every A ⊆ [0, 1], for every partition {xα,k : k < n + 1} : α ∈ T of A into ordered sets of size n + 1, writing Zk = {xα,k : α ∈ T }, there exists W ⊆ T such that • μ (Z0 W ) ≥ 0.5μ (Z0 ) (here Zk W = {xα,k : α ∈ W }), • Zk W has zero inner measure in Zk for every 1 ≤ k ≤ n. By inductive hypothesis, we can choose a partition Tk : 1 ≤ k ≤ n of T such that Zk Tk has full outer measure in Zk . Using Lusin’s result [9], for each 1 ≤ k ≤ n choose a partition Tk = Tk,0 Tk,1 such that each one of Zk Tk,0 and Zk Tk,1 has full outer measure in Zk Tk . Deﬁne W = {Tk,i(k) : 1 ≤ k ≤ n} where i(k) ∈ {0, 1} is such that μ (Z0 Tk,i(k) ) ≥ 0.5μ (Z0 Tk ). It is easily checked that W is as required. 2 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 478 3. Forcing Assume () fails. Fix a witnessing X ⊆ [0, 1], a partition Xα : α ∈ S, enumerations Xα = {xα,n : n ∈ ω} and the corresponding Yn ’s. For each n ≥ 1, let An ⊆ S be such that Y0 An is null and for every A ⊆ S\An , if Y0 A is null, then Yn A ⊆ env(Yn An ). Let W0 = {An : n ≥ 1}. Let Cn = env(Yn W0 ), Bn = env(Yn ) \ Cn . We can assume that μ(Bn ) > 0 for inﬁnitely many n – Otherwise we can use Lemma 2.2 to get a contradiction to the failure of (). By ignoring the Yn ’s for which Bn is null, we can also assume that μ(Bn ) > 0 for every n ≥ 1. Replace S by S\W0 . By modifying the Yn ’s, we can assume that env(Yn ) = Bn . This ensures the following: If W ⊆ S is such that Y0 W is null, then for every n ≥ 1, Yn W has zero inner measure in Yn . For n ≥ 1, let Wn ⊆ S be such that Yn Wn is null and if W ⊆ S\Wn is such that Yn W is null, then for all m ≥ n + 1, Ym W has zero inner measure in Ym . Let I = {W ⊆ S : μ(Y0 W ) = 0}. Then I is a sigma ideal on S. Let P = P(S)/I be forcing with I. Observe that Z ∈ P iﬀ (∃n ≥ 1)(Yn Z has positive inner measure in Yn ). In fact, by Lemma 2.2, we can assume that Z ∈ P iﬀ (∃∞ n)(Yn Z has positive inner measure in Yn ). For n ≥ 1, let Qn = {Z ⊆ S : (∃B ⊆ Bn ) (B is Borel positive and Z = {α ∈ S \ Wn : xα,n ∈ B ∩ Yn })}. Note that if Z ∈ Qn , then Yn Z has positive inner measure in Yn and hence Y0 Z is non null. So Qn ⊆ P for every n ≥ 1. Let Pn = {Z ∈ P : Z ∩ Wn = 0}. Lemma 3.1. The following hold: (1) (2) (3) (4) For each n ≥ 1, Qn Pn ; {Qn : n ≥ 1} is dense in P; Each Qn is isomorphic to random forcing; P adds a Cohen real. Proof of Lemma 3.1. (1) Suppose {Zk : k < θ ≤ ω} is a maximal antichain in Qn but not in Pn . Let Z ∈ Pn be incompatible with Zk for every k < θ. So Z ⊆ S \ Wn , Yn Z is null and Y0 Z is non null. Choose m > n such that Ym Z has positive inner measure in Ym . But now Z ⊆ S \ Wn , Yn Z is null and Ym Z has positive inner measure in Ym which is impossible by our choice of Wn . (2) Let Z ∈ P. Choose n ≥ 1 such that Yn Z has positive inner measure in Yn . Let B ⊆ Bn be Borel positive such that B ∩ (Yn (S \ Z)) = ∅. Let Z = {α ∈ (Z \ Wn ) : xα,n ∈ B}. Then Z ∈ Qn and Z ⊆ Z. (3) Should be clear. (4) Construct a tree Aσ : σ ∈ <ω 2 of subsets of S such that the following hold. (a) A∅ = S, and for every σ ∈ <ω 2, (b) Aσ is a disjoint union of Aσ0 and Aσ1 , (c) for every 0 ≤ k ≤ |σ| + 1, μ (Yk Aσ0 ) = μ (Yk Aσ1 ) = μ (Yk Aσ ). A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 479 For Clause (c), we make use of Lemma 2.2 and Lusin’s result [9]. We claim that if G is P-generic over V , then in V [G], the real x = {σ ∈ <ω 2 : Aσ ∈ G} is Cohen over V . To see this, suppose that D ⊆ <ω 2 is dense and x ∈ / {[σ] : σ ∈ D}. Then S\ {Aσ : σ ∈ D} = Z ∈ G so that Y0 Z is non null. But then, for some n ≥ 1, Yn Z has positive inner measure in Yn . So Z meets Aσ for every σ extending some σ0 , |σ0 | = n, and hence also meets some condition in {Aσ : σ ∈ D}: A contradiction. 2 Note that Lemma 3.1 implies that P satisﬁes ccc. 4. A dichotomy Claim 4.1. Let Q P be atomless. Then forcing with Q adds a new real. Proof of Claim 4.1. Since Q satisﬁes ccc (as Q P and P satisﬁes ccc), it is enough to show that every generic extension contains a new ω-sequence of members of V . Towards a contradiction, suppose p ∈ Q forces that no such sequence appears in the extension. Let α be the least ordinal such that for some ν ∈ V Q and q ∈ Q, q ≥ p and q Q ν : α → V ∧ν ∈ / V . It is clear that α is regular uncountable. Choose q ≥ p, ν ∈ V Q such that q Q ν : α → V ∧ ν ∈ / V . For each β < α, choose (qβ,n , νβ,n ) : n < ω such that Aβ = {qβ,n : n < ω} is a maximal antichain in Q above q, νβ,n ∈ V and qβ,n Q ν β = νβ,n . Choose N < ω such that W = {β < α : Aβ ∩ QN = ∅} has size α. For each β ∈ W , let nβ < ω be such that qβ,nβ ∈ QN . Since QN is isomorphic to random forcing we can ﬁnd W ∈ [W ]α such that {qβ,nβ : β ∈ W } consists of pairwise compatible conditions. Put ν = {νβ,nβ : β ∈ W }. Since Q satisﬁes ccc, there exists q ≥ q such that q Q |GQ ∩ {qβ,nβ : β ∈ W }| = α. But now q ν = ν ∈ V : A contradiction. 2 Lemma 4.2. Suppose ν is a P-name such that P ν ∈ 2ω \ V . Then for every p ∈ P there exist q ≥ p and a Borel function B : 2ω → ω ω such that either q P B(ν) is an inﬁnitely often equal real over V or q P V [ν] is a random real extension of V . Proof of Lemma 4.2. For simplicity, assume that p is the empty condition. For each n ≥ 1, let T̊n ∈ V Qn be such that for every Qn -generic Gn , T̊n [Gn ] = {σ ∈ <ω 2 : (∀p ∈ Gn )(∃q ∈ P)(q ≥ p and q P σ ≺ ν)}. Clearly, Qn T̊n is a leaﬂess subtree of <ω 2. We break our proof into two cases. Claim 4.3. Suppose for some n ≥ 1 and q ∈ Qn , q Qn T̊n is not a perfect tree. Then there exists p ∈ P, p ≥ q such that p P V [ν] is a random real extension of V . Proof of Claim 4.3. Choose q1 ∈ Qn , q1 ≥ q and σ ∈ <ω 2 such that q1 Qn σ ∈ T̊n and T̊n has a unique branch above σ. Choose p ∈ P, p ≥ q1 such that p P σ ≺ ν. Let GP be P-generic over V with p ∈ GP . Put GQn = GP ∩ Qn . Then, ν[GP ] ∈ V [GQn ] since it is the unique branch through T̊n [GQn ] above σ. Since intermediate models in a random 480 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 real extension are also random real extensions (as a complete subalgebra of a measure algebra is also a measure algebra), the claim follows. 2 Claim 4.4. Suppose for every n ≥ 1, Qn T̊n is a perfect tree. Then for some Borel function B, P (∀x ∈ ω ω ∩ V )(∃∞ l)(B(ν)(l) = x(l)). Proof of Claim 4.4. We will think of conditions in Qn as the set of branches through a perfect subtree S ⊆ <ω 2 such that for each σ ∈ S, μ([S] ∩ [σ]) > 0 – This is a dense set of conditions in Qn . Let us call such trees fat. Let (pn,k , Sn,k , gn,k , Gn,k , fn,k ) : k < ω satisfy the following. • • • • • • • • • {pn,k : k < ω} is a maximal antichain in Qn ; Sn,k is a fat perfect subtree of <ω 2 and pn,k = [Sn,k ]; gn,k is a function with domain Sn,k ; For each σ ∈ Sn,k , gn,k (σ) is a ﬁnite subtree of ≤h 2 for some h < ω such that each terminal node of gn,k (σ) has length h = hn,k (|σ|) – So hn,k depends only on |σ|; Whenever σ τ are from Sn,k , gn,k (σ) is a subtree of gn,k (τ ); Gn,k is a function from pn,k to perfect subtrees of <ω 2; For every x ∈ pn,k , the height of gn,k (x m) goes to inﬁnity as m goes to inﬁnity and the limiting tree is Gn,k (x); pn,k Qn “Gn,k (r̊n ) = T̊n is a perfect tree” where r̊n is the Qn -name for the random real added by Qn ; fn,k : ω → ω is increasing such that if σ τ are from Sn,k and |τ | ≥ fn,k (|σ|), then for every maximal node ν1 ∈ gn,k (σ) there are ≥ 2|σ|+3 + hn,k (|σ|) maximal nodes ν2 ν1 in gn,k (τ ) – fn,k is well deﬁned because for every x ∈ [Sn,k ], Gn,k (x) is a perfect tree. The existence of such objects follows from the following well known fact about random forcing. Fact 4.5. Let B denote the random forcing whose conditions are viewed as compact positive measure subsets of 2ω . Let τ be a B-name for a real in 2ω . Then for every p ∈ B, there exist a fat tree S ⊆ 2<ω and a (uniformly) continuous function G : [S] → 2ω such that letting q = [S], q ≥ p and q B G(r˚B ) = τ where r̊B is the canonical name for the random real added by B. Claim 4.6. Given k, m < ω and n ≥ 1 there exist m , l = lH and H where m > m, l < ω and H : l 2 → {0, 1} such that () holds: For every σ ∈ Sn,k ∩ m 2, the set {τ ∈ Sn,k ∩ m 2 : τ σ and (∀η ∈ gn,k (σ))(range(H {η η : η ∈ l 2 ∩ gn,k (τ )} = {0, 1})} m has size at least |{τ ∈ Sn,k ∩ m 2 : τ σ}|(1 − 2−2 ). Proof of Claim 4.6. Fix n ≥ 1 and k, m < ω. Let N0 = max{|gn,k (σ) ∩ hn,k (m) 2| : σ ∈ Sn,k ∩ m 2}. Then N0 ≤ 2hn,k (m) . Put m = fn,k (m). Let l = hn,k (m ) be the common A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 481 height of gn,k (τ ) for τ ∈ Sn,k ∩ m 2. Let p = min{|{η η : η ∈ l 2 ∩ gn,k (τ )}| : τ ∈ Sn,k ∩ m 2 ∧ η ∈ gn,k (τ m) ∩ hn,k (m) 2}. Note that p ≥ 2m+3 + hn,k (m). For each ρ ∈ l 2, we choose H(ρ) ∈ {0, 1} with equal probability – So for l A ⊆ {H : H : l 2 → {0, 1}}, the probability P r(H ∈ A) = |A|2−2 . For each σ ∈ Sn,k ∩ m 2, τ ∈ Sn,k ∩ m 2 with τ σ, say that H satisﬁes (σ, τ ) if for every η ∈ gn,k (σ), the range of H {η η : η ∈ l 2 ∩ gn,k (τ )} is −p+1 {0, 1}. Note that P r(H satisﬁes (σ, τ )) . It follows that for each ≥ 1 − N0 2 m m σ ∈ Sn,k ∩ 2, the probability that τ σ : τ ∈ Sn,k ∩ 2 ∧ H satisﬁes (σ, τ ) ≥ m m 1 − 2−2 τ σ : τ ∈ Sn,k ∩ m 2 is at least 1 − N0 2−p+1 22 . So for an appropriate H to exist, it suﬃces to have 2m N0 2−p+1 22 p ≥ hn,k (m) + 2m+3 > log(N0 ) + 2m + m + 1. 2 m < 1. But this is clear since For each n ≥ 1, k < ω, choose (mn,k,j , Hn,k,j , sn,k,j ) = (mj , Hj , sj ) : j < ω satisfying the following. • The mj ’s are increasing with j; • For each j, () of Claim 4.6 holds with m = mj , m = mj+1 and H = Hj+1 ; • sj : j < ω lists ω inﬁnitely often. Deﬁne a Borel function B : 2ω → ω ω as follows: For each η ∈ 2ω , l < ω, letting (n, k, j) = l (so l codes a triplet via some bijection between 3 ω and ω) if there exists some j1 > j such that Hn,k,j1 (η lHn,k,j1 ) = 1, then pick the least such j1 and deﬁne B(η)(l) = sn,k,j1 . If there is no such j1 , put B(η)(l) = 0. Claim 4.7. For every p ∈ P, x ∈ ω ω , l < ω, there exist q ≥ p and l1 > l such that q P B(ν)(l1 ) = x(l1 ). Proof of Claim 4.7. Choose n ≥ 1, k < ω, q ≥ p such that q ∈ Qn and q, pn,k are compatible. Let S ⊆ <ω 2 be a fat perfect subtree such that [S] ⊆ q ∩ pn,k . Choose j < ω m+i (1 −2−2 ) > 0.5, suﬃciently large and σ ∈ S such that, letting m = mn,k,j , we have i≥0 (n, k, j) = l1 > l, σ ∈ m 2 and [S] has more than 99 percent measure in [σ]. Choose j1 > j such that sn,k,j1 = x(l1 ). Now using (), we can ﬁnd τ ∈ S, τ σ, τ : mn,k,j1 → 2, l η ∈ gn,k (τ ) ∩ 2 Hn,k,j1 such that for each j < j < j1 , Hn,k,j (η lHn,k,j ) = 0 and Hn,k,j1 (η) = 1. Choose q1 ∈ P such that q1 ≥ [S] ∩ [τ ] and q1 P η ν. It follows that for each j < j < j1 , q1 P Hn,k,j (ν lHn,k,j ) = 0 and q1 P Hn,k,j1 (ν lHn,k,j1 ) = 1. Hence q1 P B(ν)(l1 ) = sn,k,j1 = x(l1 ). 2 This ﬁnishes the proof of Claim 4.4 and Lemma 4.2. 482 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 5. A non meager set in V P Recall that non(Meager) denotes the least cardinality of a non meager set of reals. Lemma 5.1. Suppose V |= non(Meager) ≤ κ. Then V P |= non(Meager) ≤ κ. Proof of Lemma 5.1. Let A be the collection of all quadruples s = (m, n, k, h) where m = mi : i < ω, n = ni : i < ω and k = ki : i < ω are sequences in ω, m0 = 0, the mi ’s are strictly increasing and h = hi : i < ω where hi : ki 2 → [mi ,mi+1 ) 2. Let {xα : α < κ} be non meager in A where the topology over A is generated by declaring ﬁnite restrictions of members of A clopen. For α < κ, deﬁne ẙα ∈ 2ω ∩ V P by ẙα [mα,i , mα,i+1 ) = hα,i (r̊nα,i kα,i ) where mα,i , nα,i , kα,i , hα,i correspond to xα and r̊j is the random real added by Qj . It suﬃces to show the following. Claim 5.2. P Y̊ = {ẙα : α < κ} is non meager. Proof of Claim 5.2. Suppose not. Then we can ﬁnd p ∈ P, T̊ ∈ V P , W ⊆ κ, n ≥ 1 such that, p P T̊ is a nowhere dense leaﬂess subtree of <ω 2, {xα : α ∈ W } is non meager and for every α ∈ W , ∃qα ≥ p such that qα ∈ Qn and qα P ẙα ∈ [T̊ ]. Let T̊n ∈ V Qn be such that T̊n [GQn ] = {σ ∈ <ω 2 : (p ∈ P)(p ≥ p , p is compatible with every condition in GQn and p P σ ∈ / T̊ )}. Note that for every q ∈ Qn , if q ≥ p , then q Qn [T̊n ] is nowhere dense. We need the following Claim 5.3. There exist q ∈ Qn , S ⊆ 2<ω , W ⊆ W , H such that the following hold. • S is a fat perfect tree and [S ] = q ≥ p ; • H is a function with domain S ; • For each σ ∈ S , H(σ) = (H0 (σ), H1 (σ)) = (m, t) where m < ω and t ⊆ <m 2 is a subtree; • If σ τ are from S , then H0 (σ) ≤ H0 (τ ) and H1 (σ) = H1 (τ ) ∩ <H0 (σ) 2; • For every k < ω, there exists k < ω such that for every σ ∈ S ∩ k 2, H0 (σ) ≥ k; • For every r ∈ [S ], {H1 (r k) : k < ω} is nowhere dense in <ω 2; • q Qn T̊n = {H1 (r̊n k) : k < ω}; • {xα : α ∈ W } is non meager; • For every α ∈ W , there exists qα ∈ Qn such that qα ≥ q and qα P ẙα ∈ [T̊ ]. Proof of Claim 5.3. First using Fact 4.5, choose a maximal antichain {qi, : i < ω} of conditions qi, ∈ Qn such that for each i < ω, qi, ≥ p and there are Si, and H for this qi, satisfying all but the last two clauses about W . Then note that W = i<ω Wi, where Wi, = {α ∈ W : (∃qα ∈ Qn )(qα ≥ qi, ∧ qα P ẙα ∈ [T̊ ])}. Choose i < ω, such that {xα : α ∈ Wi , } is non meager and take q = qi , , W = Wi , and the corresponding S and H. 2 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 483 Claim 5.4. C = {x ∈ A : (∃i < ω)(∀σ ∈ S ∩ 2kx,i )(∀ρ ∈ mx,i 2)(nx,i = n ∧ H0 (σ) ≥ mx,i+1 ∧ ρ ∪ hx,i (σ) ∈ / H1 (σ))} is open dense in A. Here mx,i , nx,i , kx,i , hx,i correspond to x ∈ A. Proof of Claim 5.4. Let (m, n, l, h) be the restriction of some member of A to some i < ω. Extend (m, n, l, h) to a member x ∈ A as follows. Put nx,i = n . Choose mx,i > mi−1 . For each r ∈ [S ], let τr ∈ 2<ω , kr < ω be such that (∀ρ ∈ mx,i 2), [ρ τr ] ∩ [H1 (r)] = ∅ and H0 (r kr ) > mx,i + |τr | where H1 (r) = {H1 (r k) : k < ω}. Choose F ⊆ [S ] ﬁnite such that {[r kr ] : r ∈ F } covers [S ]. Choose mx,i+1 > max{H0 (r kr ) : r ∈ F }. Choose kx,i such that for every r ∈ F , kx,i > kr and for every σ ∈ S ∩ kx,i 2, H0 (σ) ≥ mx,i+1 . For σ ∈ S ∩ kx,i 2, choose hx,i (σ) ∈ [mx,i ,mx,i+1 ) 2 such that for some r ∈ F , r kr = σ kr and for all t < |τr |, hx,i (σ)(mx,i + t) = τr (t). Extend x arbitrarily to a member of A. It is easily checked that x ∈ C. 2 Now let α ∈ W be such that xα ∈ C as witnessed by i < ω. Choose qα ∈ Qn such that qα ≥ q and qα P ẙα ∈ [T̊ ]. Let σ ∈ kα,i 2 be such that q = qα ∩ [σ] has positive measure. Notice that q P ẙα mα,i+1 ∈ / T̊n hence for some q1 ∈ P, q1 ≥ q ≥ qα and q1 P ẙα ∈ / [T̊ ]: A contradiction. 2 6. Contradiction The following lemma contradicts Lemma 3.1 and thus ﬁnishes the proof of Theorem 1.1. Lemma 6.1. The following is impossible. (1) I is an ideal over S, add(I A) = κ > ℵ0 for every A ∈ I + ; (2) (Qn , pn ) : n < ω satisﬁes the following: For every n < ω, pn ∈ P, Qn P≥pn where P≥pn = {p ∈ P : p ≥ pn }, Qn is forcing isomorphic to random forcing and {Qn : n < ω} is dense in P; (3) P is forcing isomorphic to P(S)/I; (4) Forcing with P adds a Cohen real. Proof of Lemma 6.1. Let G be P(S)/I-generic over V and let j : V → N ⊆ V [G] be the generic ultrapower embedding with critical point κ. Let f : S → κ represent κ. Deﬁne J = {A ⊆ κ : f −1 [A] ∈ I}. Let Q = P(κ)/J . Then Q P – So Q adds a new real by Claim 4.1. Let H be Q-generic over V and let k : V → M ⊆ V [H] be the generic ultrapower embedding. Note that κ N ∩ V [G] ⊆ N and κ M ∩ V [H] ⊆ M (see Proposition 2.14, [2]) and we’ll use this freely below. We can divide our proof into two cases. Case 1: There is a real r ∈ ω ω ∩ M such that r is inﬁnitely often equal to every real in V . Let rα : α < κ represent r. Then for every x ∈ ω ω ∩ V , there exists α < κ such 484 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 that rα and x agree inﬁnitely often. It follows that there is a non meager set of size κ in V . Since P adds a random real, V ∩ 2ω is meager in V [G]. Let B be a meager Fσ -set coded in V [G] that contains V ∩2ω . Then B is also coded in N . So by elementarity of j, it follows that every set of reals in V of size < κ is meager in V . So V |= non(Meager) = κ. Hence N |= non(Meager) = j(κ) > κ. By Lemma 5.1, V [G] and hence N has a non meager set of size κ: A contradiction. Case 2: No real in M is inﬁnitely often equal to every real in V . By Lemma 4.2, for every new real r ∈ M , V [r] is a random real extension of V . Using Claim 4.1, choose r ∈ V [H] random over V . Then r ∈ M . Let rα : α < κ represent r. Then {rα : α < κ} is a non null set in V . Since P adds a Cohen real, V ∩ 2ω is null in V [G]. Let B be a null Gδ -set coded in V [G] that contains V ∩ 2ω . Then B is also coded in N . So by elementarity of j, it follows that every set of reals in V of size < κ is null in V . In particular, for every γ < κ, {rα : α < γ} is null in V . By considering k(rα : α < κ), it follows that {rα : α < κ} is null in M . Let r be a null Borel set coded in M witnessing this. It follows that V [r ] is not a random real extension of V : A contradiction. 2 Appendix A Let Randomθ (resp. Cohenθ ) denote the forcing for adding θ random (resp. Cohen) reals. Randomθ is the measure algebra on the product measure space [0, 1]θ (with Lebesgue measure on [0, 1]) and Cohenθ is the forcing for adding a function in ω×θ 2 using ﬁnite approximations. Gitik and Shelah [4–6] proved that forcing with a sigma ideal cannot be isomorphic to Random1 × Cohen1 . In [3] (Section 546), Fremlin asked if the result generalizes to a product that adds more of these reals. We show that this is the case. Theorem A.1. Let θ1 , θ2 be positive cardinals. Let I be a sigma ideal over X. Then forcing with P(X)/I is not isomorphic to P = Randomθ1 × Cohenθ2 . Proof of Theorem A.1. Suppose not. By restricting I, we can assume add(I A) = κ for every A ∈ I + . Note that since P is ccc, κ > ℵ1 . Let G be P-generic over V and let j : V → M ⊆ V [G] be the generic ultrapower embedding with critical point κ. The following is well known [1]. Fact A.2. V Cohen1 |= cov(Null) = ℵ1 . Here, cov(Null) is the least cardinality of a family of null sets whose union covers R. So V [G] and therefore M (as κ M ∩ V [G] ⊆ M ) thinks that cov(Null) = ℵ1 . Hence also V |= cov(Null) = ℵ1 . Let Ni : i < ω1 be a sequence of null Gδ -sets witnessing this in V . By elementarity of j, M |= {Ni : i < ω1 } = R. But this is impossible as V [G] and therefore M contains a random real over V which cannot belong to any of the Ni ’s. 2 A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485 485 References [1] T. Bartoszynski, H. Judah, Set Theory: On the Structure of the Real Line, A K Peters, Wellesley MA, 1995. [2] M. Foreman, Ideals and generic elementary embeddings, in: M. Foreman, A. Kanamori (Eds.), Handbook of Set Theory, vol. 2, Springer, 2010, pp. 885–1147. [3] D.H. Fremlin, Measure Theory, vol. 5: Set-Theoretic Measure Theory, Part II, 2009. [4] M. Gitik, S. Shelah, Forcing with ideals and simple forcing notions, Israel J. Math. 68 (1989) 129–160. [5] M. Gitik, S. Shelah, More on simple forcing notions and forcings with ideals, Ann. Pure Appl. Logic 59 (1993) 219–238. [6] M. Gitik, S. Shelah, More on real-valued measurable cardinals and forcing with ideals, Israel J. Math. 124 (2001) 221–242. [7] A. Kanamori, The Higher Inﬁnite, 2nd edition, Springer Monogr. Math., Springer, Berlin, 2003. [8] A. Kumar, Avoiding rational distances, Real Anal. Exchange 38 (2) (2012/2013) 493–498. [9] N. Lusin, Sur la decomposition des ensembles, C. R. Acad. Sci. Paris 198 (1934) 1671–1674.

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