Borel sets without perfectly many overlapping translations, III
aa r X i v : . [ m a t h . L O ] S e p BOREL SETS WITHOUT PERFECTLY MANY OVERLAPPINGTRANSLATIONS, III
ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH
Abstract.
We expand the results of Ros lanowski and Shelah [11, 10] to allperfect Abelian Polish groups ( H , +). We show that if α < ω and 4 ≤ k < ω ,then there is a ccc forcing notion adding a Σ set B ⊆ H which has ℵ α manypairwise k –overlapping translations but not a perfect set of such translations. Introduction
For a Polish space X and a set B ⊆ X × X we say that B contains a µ –square(perfect square, respectively), if there is a set Z of cardinality µ (a perfect set Z ,respectively) such that Z × Z ⊆ B . The problem of Borel sets with large squaresbut no perfect squares was studied and resolved in Shelah [14].Several questions can be phrased in a manner involving µ –squares and/or perfectsquares with some additional structure on them. For instance, looking at a Polishgroup ( H , +) we may ask for its Borel subsets with many, but not too many disjointtranslations (or just translations with small overlaps). This leads to considering the spectrum of translation κ –disjointness of a set A ⊆ H , std κ ( A ) = { ( x, y ) ∈ H × H : | ( A + x ) ∩ ( A + y ) | ≤ κ } , and asking if this set may contain a µ –square but not a perfect square. For κ = 0this is asking for µ many pairwise disjoint translations of A without a perfect setof such translations. This direction is related to works of Balcerzak, Ros lanowskiand Shelah [1], Udayan and Keleti [3], Elekes and Stepr¯ans [5], Zakrzewski [15] andElekes and Keleti [4].It is still unresolved if we may repeat the results of [14] for the disjointnesscontext, but there is some promising work in progress [12]. However a lot of progresshas been made in the dual direction.For a set A ⊆ H we consider its spectrum of translation κ –non-disjointness, stnd κ ( A ) = { ( x, y ) ∈ H × H : | ( A + x ) ∩ ( A + y ) | ≥ κ } . Then a µ –square included in stnd κ ( A ) determines a family of µ many pairwise κ –overlapping translations. These were studied extensively for the context of theCantor space in Ros lanowski and Rykov [9], and Ros lanowski and Shelah [11, 10].Those works fully utilized the algebraic properties of ( ω , +), leaving the generalcase of Polish groups unresolved. Date : August, 2020.1991
Mathematics Subject Classification.
Primary 03E35; Secondary: 03E15, 03E50.Publication 1187 of the second author.The first author thanks the National Science Fundation for supporting his visit to Rutgers Uni-versity where this research was carried out , and the Rutgers University for their hospitality.
In the current paper we aim at generalizing their results to perfect Abelian Polishgroups. The main difficulty in this more general case lies in quite algebraic problem( ♠ ) given below. Suppose S ⊆ H and X ⊆ H is a set of k –intersecting translations,i.e.,( ♦ ) SX | ( S + x ) ∩ ( S + y ) | ≥ k for all x, y ∈ X .Then for all c ∈ H the property ( ♦ ) SX + c also holds true. Thus the properties ofobjects added by our forcing should reflect some “translation invariance”. How canwe know that a set Y is included in a translation of X ? Clearly, if Y ⊆ X + c or Y ⊆ c − X , then Y − Y ⊆ X − X . It would be helpful in our forcing if we knew( ♠ ) when does Y − Y ⊆ X − X imply that Y is included in a (small) neighbor-hood of a translation X + c of X or of a translation c − X of − X ?In the third section we introduce the main algebraic ingredient of our forcing no-tion: qifs and quasi independent sets. In forcing, we will use them in conjunctionwith differences of elements of the group, but a relative result for sums also seemsinteresting, so we present it in Section 4. The third and fourth section might be ofinterest independently from the rest of the paper, as they address the question ( ♠ )giving interesting (though technical) properties of perfect Abelian Polish groupswith few elements of rank 2.Like in [14], the “no perfect set” property of the forcing extension results fromthe use of a “splitting rank” rk sp . We remind its definition and basic properties inthe second section. For the relevant proofs we refer the reader to [14, 11].In the fifth section we prove our main consistency result for groups with fewelements of rank 2. The remaining case when H has many elements of rank 2 istreated in Section 6. We close the paper with summary of our results and a list ofopen problems.The general case of Polish groups will be investigated in a subsequent work [13]. Notation : Our notation is rather standard and compatible with that of classi-cal textbooks (like Jech [7] or Bartoszy´nski and Judah [2]). However, in forcing wekeep the older convention that a stronger condition is the larger one .(1) For a set u we let u h i = { ( x, y ) ∈ u × u : x = y } . (2) Ordinal numbers will be denoted be the lower case initial letters of theGreek alphabet α, β, γ, δ, ε, ζ . Finite ordinals (non-negative integers) willbe denoted by letters i, j, k, ℓ, m, n, J, K, L, M, N and ι . The Greek letters λ and µ will stand for an uncountable cardinals.(3) Finite sequences will be denoted σ, ς (4) For a forcing notion P , all P –names for objects in the extension via P will bedenoted with a tilde below (e.g., τ ˜ , X ˜ ), and G ˜ P will stand for the canonical P –name for the generic filter in P .In the paper we will keep the following notation/assumptions pertaining toAbelian Polish groups. Assumption 1.1. (1) ( H , + ,
0) is an Abelian perfect Polish group with thetopology generated by a complete metric ρ ∗ . The elements of H will becalled a, b, c, d (with possible indices). For an integer ι and a ∈ H , we use OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 3 the notation ιa to denote the element of H obtained by repeated additionof a (or − a ) | ι | many times in the usual way.(2) D ⊆ H is a countable dense subset and ρ : H × H −→ [0 , ∞ ) is a translationinvariant metric compatible with the topology of H . (The metric ρ doesnot have to be complete; it exists by the Birkhoff–Kakutani theorem.)(3) The open ball in the metric ρ with radius 2 − n and center at 0 is denoted B n and we let U = (cid:8) d + B n : d ∈ D ∧ n < ω (cid:9) . By the invariance of themetric ρ , the family U is a countable base of the topology of H .(4) For sets A, B ⊆ H we will write − A = {− a : a ∈ H } , A + B = { a + b : a ∈ A ∧ b ∈ B } and A − B = { a − b : a ∈ A ∧ b ∈ B } . Splitting rank rk sp Let us recall a rank used in previous papers which will be central for the resultshere too. We quote some definitions and theorems from [11, Section 2], howeverthey were first given in [14, Section 1].Let λ be a cardinal and M be a model with the universe λ and a countablevocabulary τ . Definition 2.1. (1) By induction on ordinals δ , for finite non-empty sets w ⊆ λ we define when rk( w, M ) ≥ δ . Let w = { α , . . . , α n } ⊆ λ , | w | = n + 1.(a) rk( w ) ≥ ϕ = ϕ ( x , . . . , x n ) ∈L ( τ ) and each k ≤ n , if M | = ϕ [ α , . . . , α k , . . . , α n ] then the set (cid:8) α ∈ λ : M | = ϕ [ α , . . . , α k − , α, α k +1 , . . . , α n ] (cid:9) is uncountable;(b) if δ is limit, then rk( w, M ) ≥ δ if and only if rk( w, M ) ≥ γ for all γ < δ ;(c) rk( w, M ) ≥ δ + 1 if and only if for every quantifier free formula ϕ = ϕ ( x , . . . , x n ) ∈ L ( τ ) and each k ≤ n , if M | = ϕ [ α , . . . , α k , . . . , α n ]then there is α ∗ ∈ λ \ w such thatrk( w ∪ { α ∗ } , M ) ≥ δ and M | = ϕ [ α , . . . , α k − , α ∗ , α k +1 , . . . , α n ] . By a straightforward induction on δ one easily shows that if ∅ 6 = v ⊆ w thenrk( w, M ) ≥ δ ≥ γ = ⇒ rk( v, M ) ≥ γ. Hence we may define the rank functions on finite non-empty subsets of λ . Definition 2.2.
The rank rk( w, M ) of a finite non-empty set w ⊆ λ is defined as: • rk( w, M ) = − ¬ (rk( w, M ) ≥ • rk( w, M ) = ∞ if rk( w, M ) ≥ δ for all ordinals δ , • for an ordinal δ : rk( w, M ) = δ if rk( w, M ) ≥ δ but ¬ (rk( w, M ) ≥ δ + 1). Definition 2.3.
For an ordinal ε and a cardinal λ let NPr ε ( λ ) be the followingstatement:“there is a model M ∗ with the universe λ and a countable vocab-ulary τ ∗ such that 1 + rk( w, M ∗ ) ≤ ε for all w ∈ [ λ ] <ω \ {∅} .”Let Pr ε ( λ ) be the negation of NPr ε ( λ ).Note that NPr ε of [11, Definition 2.4] differs from our NPr ε : “sup { rk( w, M ∗ ) : ∅ 6 = w ∈ [ λ ] <ω } < ε ” there is replaced by “1 + rk( w, M ∗ ) ≤ ε ” here. However, theproofs for [11, Propositions 2.6, 2.7] show the following results. ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH
Proposition 2.4. (1) NPr ( ω ) . (2) If NPr ε ( λ ) , then NPr ε +1 ( λ + ) . (3) If NPr ε ( µ ) for µ < λ and cf( λ ) = ω , then NPr ε ( λ ) . (4) If α < ω , then NPr α ( ℵ α ) but Pr α ( i ω ) holds. Definition 2.5.
Let τ ⊗ = { R n,j : n, j < ω } be a fixed relational vocabulary where R n,j is an n –ary relational symbol (for n, j < ω ). Definition 2.6.
Assume that ε < ω and λ is an uncountable cardinal such thatNPr ε ( λ ). By this assumption, we may fix a model M ( ε, λ ) = M = ( λ, { R M n,j } n,j<ω )in the vocabulary τ ⊗ with the universe λ such that:( ⊛ ) a for every n and a quantifier free formula ϕ ( x , . . . , x n − ) ∈ L ( τ ⊗ ) there is j < ω such that for all α , . . . , α n − ∈ λ , M | = ϕ [ α , . . . , α n − ] ⇔ R n,j [ α , . . . , α n − ] , ( ⊛ ) b the rank of every singleton is at least 0,( ⊛ ) c v, M ) ≤ ε for every v ∈ [ λ ] <ω \ {∅} ,( ⊛ ) d M | = R , [ α , α ] if and only if α < α < λ .For a nonempty finite set v ⊆ λ let rk sp ( v ) = rk( v, M ), and we fix witnesses j ( v ) < ω and k ( v ) < | v | for the rank of v , so that the following demands ( ⊛ ) e –( ⊛ ) g are satisfied. If { α , . . . , α k , . . . α n − } is the increasing enumeration of v and k = k ( v ) and j = j ( v ), then( ⊛ ) e if rk sp ( v ) ≥
0, then M | = R n,j [ α , . . . , α k , . . . , α n − ] but there is no α ∈ λ \ v such thatrk sp ( v ∪ { α } ) ≥ rk sp ( v ) and M | = R n,j [ α , . . . , α k − , α, α k +1 , . . . , α n − ] , ( ⊛ ) f if rk sp ( v ) = −
1, then M | = R n,j [ α , . . . , α k , . . . , α n − ] but the set (cid:8) α ∈ λ : M | = R n,j [ α , . . . , α k − , α, α k +1 , . . . , α n − ] (cid:9) is countable,( ⊛ ) g for every β , . . . , β n − < λ , if M | = R n,j [ β , . . . , β n − ] then β < . . . <β n − .The choices above define functions j : [ λ ] <ω \ {∅} −→ ω , k : [ λ ] <ω \ {∅} −→ ω , andrk sp : [ λ ] <ω \ {∅} −→ {− } ∪ ( ε + 1).3. QIFs and differences
As declared in Assumption 1.1, ( H , + ,
0) is a perfect Abelian Polish group.
Definition 3.1. (1) Let B ⊆ H . A (2,n)–combination from B is any sum ofthe form ι b + ι b + ι b + . . . + ι n − b n − where b , b , . . . , b n − ∈ B are pairwise distinct and ι , ι , ι , . . . , ι n − ∈{− , − , , , } . The (2 , n )–combination is said to be nontrivial when notall ι , . . . , ι n − are equal 0.(2) We say that the set B is quasi independent in H if | B | ≥ , B equals to 0. OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 5 (3) We say that a family V of non-empty subsets of H is an n –good qif if |V| ≥ n , the sets in V are pairwise disjoint and for distinct V , . . . , V n − ∈ V ,for each choice of b i , b ′ i ∈ V i (for i < n ) and every ι , ι ′ , . . . , ι n − , ι ′ n − ∈{− , , } such that P n − i =0 ( ι i + ι ′ i ) = 0 we have ι b + ι ′ b ′ + ι b + ι ′ b ′ + . . . + ι n − b n − + ι ′ n − b ′ n − = 0 . An expression as on the left hand side above will be called a nontrivial (2 , V , n ) –combination (or a nontrivial (2 , n ) –combination from V ).(4) Let V , W ⊆ P ( H ) \ {∅} . We will say that W is immersed in V if there is abijection π : W − −→ V such that • W ⊆ π ( W ) for all W ∈ W , and • if W , W ∈ W , and a, a ′ ∈ W , b ∈ W , then ( a − a ′ ) + b ∈ π ( W ). Observation 3.2. (1) If B is quasi independent then all elements of B haveorder at least 3 and (cid:8) { b } : b ∈ B (cid:9) is an –good qif. (2) If V is an –good qif and b V ∈ V (for V ∈ V ) then { b V : V ∈ V} is quasiindependent. (3) If V i ⊆ H are disjoint open sets and b i ∈ V i for i < N then there are opensets W i such that b i ∈ W i ⊆ V i for i < N , and { W i : i < N } is immersedin { V i : i < N } . Proposition 3.3.
Assume that (i) ( H , + , is an Abelian perfect Polish group, (ii) the set of elements of H of order larger than 2 is dense in H , (iii) U , . . . , U n − are nonempty open subsets of H .Then there are disjoint open sets V i ⊆ U i (for i < n ) such that { V i : i < n } is an n –good qif.Proof. Let H consists of all elements of H of order ≤
2. Then H is a closedsubgroup of H and, by the assumption (ii), it has empty interior. Consequently, foreach a ∈ H and i ≤ n the set ( a + H ) ∩ U i is meager. Therefore, for each i < n ,( ⊗ ) i the set (cid:8) a + H : a ∈ H and ( a + H ) ∩ U i = ∅ (cid:9) is infinite.Let m = 10 and m i +1 = 10 i +1 · Q j ≤ i m j + 10 (for i < n ). For each i < n choose aset A i ⊆ U i \ H such that( ⊕ ) | A i | = m i and( ⊕ ) if a, b ∈ A i and a = b , then 2 a = 2 b .(The choice is possible by ( ⊗ ) i for each i < n .) For 0 < i < n let X i = (cid:8) ι a + . . . + ι i − a i − : a ∈ A , . . . , a i − ∈ A i − ∧ ι , . . . , ι i − ∈ {− , − , , , } (cid:9) . By the choice of m j ’s we know that 2 · | X i | < m i = | A i | , so we may choose b ∗ i ∈ A i such that 2 b ∗ i , b ∗ i / ∈ X i . Let b ∗ ∈ A be arbitrary. One easily verifies that (cid:8) { b ∗ i } : i For each such combination we may choose open sets V iι ,ι ′ ,...,ι n − ,ι ′ n − such that b ∗ i ∈ V iι ,ι ′ ,...,ι n − ,ι ′ n − ⊆ U i and for every b i , b ′ i ∈ V iι ,ι ′ ,...,ι n − ,ι ′ n − , i < n , we have ι b + ι ′ b ′ + ι b + ι ′ b ′ + . . . + ι n − b n − + ι ′ n − b ′ n − = 0 . Now, for i < n we set V i = T (cid:8) V iι ,ι ′ ,...,ι n − ,ι ′ n − : ι , ι ′ , . . . , ι n − , ι ′ n − ∈ {− , , } ∧ ( ι − ι ′ ) + . . . + ( ι n − − ι ′ n − ) > (cid:9) . It is clear that the sets V i (for i < n ) are as required. (cid:3) Lemma 3.4. Suppose that ( H , + , is an Abelian Polish group and W ⊆ P ( H ) isa finite 8–good qif. Assume that (a) W is immersed in V , V ⊆ P ( H ) , (b) A ′ ⊆ A ⊆ H , | A ′ | = 8 , (c) A − A ⊆ S (cid:8) W − W ′ : W, W ′ ∈ W (cid:9) , (d) if a, b ∈ A , a = b , then ρ ( a, b ) > diam ρ ( W ) ( = diam ρ ( − W ) ) for all W ∈ W .(1) If there is c ∈ H such that A ′ + c ⊆ S W , then A + c ⊆ S V .(2) If there is c ∈ H such that c − A ′ ⊆ S W , then also c − A ⊆ S V .Proof. (1) Suppose that W , V , A ′ ⊆ A ⊆ H satisfy the assumptions of the Lemmaand c ∈ H is such that A ′ + c ⊆ S W .Assume a ∈ A \ A ′ and let us argue that a + c ∈ S V .Let h a i : i < i list the elements of A ′ . For i < b i = a i + c ∈ W i ∈ W and notethat all W i ’s are pairwise distinct (by assumption (d); remember ρ is translationinvariant). It follows from assumption (c) that we may choose b ′ i ∈ W ′ i ∈ W and b ′′ i ∈ W ′′ i ∈ W such that a − a i = b ′ i − b ′′ i . Then, for each i < 8, we have a + c = a + ( b i − a i ) = ( b ′ i − b ′′ i + a i ) + ( b i − a i ) = b ′ i − b ′′ i + b i . Claim 3.4.1. There are distinct i ∗ , j ∗ < such that ( ♥ ) i ∗ ,j ∗ W i ∗ / ∈ { W ′ j ∗ , W ′′ j ∗ } and W j ∗ / ∈ { W ′ i ∗ , W ′′ i ∗ } .Proof of the Claim. If for some i < |{ j < W i = W ′′ j ∧ j = i }| ≥ j < j < j < i and such that W ′′ j = W ′′ j = W ′′ j = W i . Since all W i ’s are distinct, we may pick i ∗ < i ∗ / ∈ { i , j , j , j } and W i ∗ / ∈ { W ′ j , W ′ j , W ′ j } . Next let j ∗ ∈ { j , j , j } be such that W j ∗ / ∈ { W ′ i ∗ , W ′′ i ∗ } .Then also W i ∗ = W i = W ′′ j ∗ and clearly ( ♥ ) i ∗ ,j ∗ holds true.If for some i < |{ j < W i = W ′ j ∧ j = i }| ≥ 3, then by the sameargument (just interchanging W ′′ j ’s and W ′ j ’s) we find i ∗ , j ∗ so that ( ♥ ) i ∗ ,j ∗ holdstrue.So now suppose that both |{ j < W = W ′ j }| ≤ |{ j < W = W ′′ j }| ≤ 2. Then there are j < j < j < W / ∈ { W ′ j , W ′′ j , W ′ j , W ′′ j , W ′ j , W ′′ j } .Take j ∗ ∈ { j , j , j } such that W j ∗ / ∈ { W ′ , W ′′ } and note that then ( ♥ ) ,j ∗ holdstrue. (cid:3) Let distinct i ∗ , j ∗ < ♥ ) i ∗ ,j ∗ holds.It follows from assumption (d) that W ′ i ∗ = W ′′ i ∗ and W ′ j ∗ = W ′′ j ∗ (remember a i ∗ = a = a j ∗ ). Now, if W i ∗ = W ′′ i ∗ , then a + c = b ′ i ∗ + ( b i ∗ − b ′′ i ∗ ) ∈ (cid:0) W ′ i ∗ + ( W ′′ i ∗ − W ′′ i ∗ ) (cid:1) ⊆ V ′ i ∗ where W ′ i ∗ ⊆ V ′ i ∗ ∈ V (so we are done). Similarly, if W j ∗ = W ′′ j ∗ . OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 7 So suppose towards contradiction that both W i ∗ = W ′′ i ∗ and W j ∗ = W ′′ j ∗ . Now, b ′ i ∗ − b ′′ i ∗ + b i ∗ = a + c = b ′ j ∗ − b ′′ j ∗ + b j ∗ , so ( ⊗ ) ( b i ∗ + b ′ i ∗ + b ′′ j ∗ ) − ( b j ∗ + b ′ j ∗ + b ′′ i ∗ ) = 0.Considering known inequalities among W i ∗ , W ′ i ∗ , W ′′ i ∗ , W j ∗ , W ′ j ∗ , W ′′ j ∗ , we notice thatno equality between them involves more than two sets. Also W i ∗ / ∈ { W j ∗ , W ′ j ∗ , W ′′ i ∗ } ,so the expression on the left hand side of ( ⊗ ) can be written as a nontrivial (2 , W , W is an 8–good qif.(2) The proof is parallel to that of the first part. Assume A ′ = { a i : i < } , A, W , V are as in the assumptions and c − A ′ ⊆ S W . Suppose a ∈ A \ A ′ .For i < b i ∈ W i ∈ W , b ′ i ∈ W ′ i ∈ W and b ′′ i ∈ W ′′ i ∈ W such that a − a i = b ′ i − b ′′ i and c − a i = b i . Then c − a = b i − b ′ i + b ′′ i for all i < Claim 3.4.2. There are distinct i ∗ , j ∗ < such that ( ♠ ) i ∗ ,j ∗ W i ∗ / ∈ { W ′ j ∗ , W ′′ j ∗ } and W j ∗ / ∈ { W ′ i ∗ , W ′′ i ∗ } .Proof of the Claim. Same as for Claim 3.4.1. (cid:3) Let i ∗ < j ∗ < ♠ ) i ∗ ,j ∗ holds true.If W i ∗ = W ′ i ∗ and W ′′ i ∗ ⊆ V ′′ i ∗ ∈ V , then c − a = b i ∗ − b ′ i ∗ + b ′′ i ∗ ∈ ( W i ∗ − W i ∗ )+ W ′′ i ∗ ⊆ V ′′ i ∗ ⊆ S V and we are done. Similarly if W j ∗ = W ′ j ∗ .Suppose towards contradiction that W i ∗ = W ′ i ∗ and W j ∗ = W ′ j ∗ . Since b i ∗ − b ′ i ∗ + b ′′ i ∗ = c − a = b j ∗ − b ′ j ∗ + b ′′ j ∗ we have (cid:0) b i ∗ + b ′ j ∗ + b ′′ i ∗ (cid:1) − (cid:0) b j ∗ + b ′ i ∗ + b ′′ j ∗ (cid:1) = 0 . Like in the first part we may argue now that the expression on the left hand sideabove can be written as a nontrivial (2 , W , contradicting theassumption that W is an 8–good qif. (cid:3) Theorem 3.5. Suppose that ( H , + , is a perfect Abelian Polish group. Assumealso that (a) W , V , Q ⊆ P ( H ) are finite 8–good qifs, and W is immersed in V and V isimmersed in Q , (b) m −→ (cid:0) (cid:1) (Erd˝os–Rado notation see [6] ), (c) A ⊆ H , | A | ≥ m and (d) A − A ⊆ S (cid:8) W − W ′ : W, W ′ ∈ W (cid:9) , and (e) if a, b ∈ A , a = b , then ρ ( a, b ) > diam ρ ( Q ) ( = diam ρ ( − Q ) ) for all Q ∈ Q .Then exactly one of (A), (B) below holds true: (A) There is a c ∈ H such that A + c ⊆ S Q . (B) There is a c ∈ H such that c − A ⊆ S Q .Proof. Let h a i : i < m i be a sequence of pairwise distinct elements of A . Since A − A ⊆ S (cid:8) W − W ′ : W, W ′ ∈ W (cid:9) , we may choose functions b , b : m × m −→ S W and ¯ W , ¯ W : m × m −→ W such that for all i, j < ma i − a j = b ( i, j ) − b ( i, j ) , b ( i, j ) ∈ ¯ W ( i, j ) , b ( i, j ) ∈ ¯ W ( i, j ) , ANDRZEJ ROS LANOWSKI AND SAHARON SHELAH and b ( i, j ) = b ( j, i ), and b ( i, j ) = b ( j, i ). Let h ϕ ℓ ( i , i , i , i ) : ℓ < i listall formulas of the form ¯ W j ( i x , i y ) = ¯ W j ′ ( i x ′ , i y ′ )for j, j ′ < x, y, x ′ , y ′ < x < y , x ′ < y ′ .Let µ : (cid:2) m (cid:3) −→ m such that if i < i
Let i, j, k < be pairwise distinct. Then (1) ¯ W ( i, j ) = ¯ W ( i, j ) and (2) b ( i, k ) − b ( i, k ) = b ( i, j ) − b ( i, j ) + b ( j, k ) − b ( j, k ) and hence (cid:16) ¯ W ( i, k ) − ¯ W ( i, k ) (cid:17) ∩ (cid:16)(cid:0) ¯ W ( i, j ) − ¯ W ( i, j ) (cid:1) + (cid:0) ¯ W ( j, k ) − ¯ W ( j, k ) (cid:1)(cid:17) = ∅ . Proof of the Claim. (1) Follows from assumption (e) of the Proposition (remem-ber every set from W is a subset of a member of Q ).(2) This follows by the equality ( a i − a j ) + ( a j − a k ) = a i − a k and the choice of b ( i, j ) , ¯ W ( i, j ) , b ( i, j ) , ¯ W ( i, j ). (cid:3) Claim 3.5.2. If { ¯ W ( i, j ) : i < j < } ∩ { ¯ W ( i, j ) : i < j < } 6 = ∅ , then either ¯ W (0 , 1) = ¯ W (1 , , or ¯ W (0 , 1) = ¯ W (1 , .Proof of the Claim. Suppose i < j < 10 and i < j < 10 are such that ¯ W ( i , j ) =¯ W ( i , j ). We shall consider all possible orders of i , j , i , j and use the homo-geneity to conclude one of the clauses in the assertion.(a) If i < j < i < j , then (by the homogeneity) ¯ W (0 , 1) = ¯ W (2 , 3) =¯ W (4 , 5) = ¯ W (2 , W (2 , 3) = ¯ W (2 , i < j = i < j then also ¯ W (0 , 1) = ¯ W (1 , i < i < j < j , then ¯ W (1 , 4) = ¯ W (2 , 5) = ¯ W (0 , 3) = ¯ W (1 , i < i < j = j , then ¯ W (0 , 3) = ¯ W (1 , 3) = ¯ W (2 , 3) = ¯ W (1 , i < i < j < j , then ¯ W (1 , 4) = ¯ W (2 , 3) = ¯ W (0 , 5) = ¯ W (1 , i = i < j < j , then ¯ W (0 , 1) = ¯ W (0 , 2) = ¯ W (0 , 3) = ¯ W (0 , i = i < j = j contradicts Claim 3.5.1(1).(h) If i = i < j < j , then ¯ W (0 , 2) = ¯ W (0 , 1) = ¯ W (0 , 3) = ¯ W (0 , i < i < j < j is not possible similarly to (e).(j) The configuration i < i < j = j is not possible similarly to (d).(k) The configuration i < i < j < j is not possible similarly to (c).(l) If i < i = j < j , then ¯ W (0 , 1) = ¯ W (1 , OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 9 (m) The configuration i < j < i < j , is not possible similarly to (a). (cid:3) Now, we will consider three cases, showing that the first one is not possible. Inthe second case we will find c ∈ H such that { c − a i : i < } ⊆ S V . Then by Lemma3.4 we will also have c − A ⊆ S Q . Finally in the last case we will find c ∈ H suchthat { a i + c : i < } ⊆ S V , so by Lemma 3.4 we will also have A + c ⊆ S Q .For ℓ < i < j < 10 let ¯ V ℓ ( i, j ) ∈ V be the unique set such that ¯ W ℓ ( i, j ) ⊆ ¯ V ℓ ( i, j ). Also, let V ℓ = { ¯ V ℓ ( i, j ) : i < j < } . Case 1 : { ¯ W ( i, j ) : i < j < } ∩ { ¯ W ( i, j ) : i < j < } = ∅ .By Claim 3.5.1(2) we have b (0 , − b (0 , 2) = b (0 , − b (0 , 1) + b (1 , − b (1 , (cid:0) b (0 , − b (0 , 2) + b (1 , (cid:1) + (cid:0) b (0 , − b (0 , − b (1 , (cid:1) = 0 . If (cid:12)(cid:12) { ¯ W (0 , , ¯ W (0 , , ¯ W (1 , } (cid:12)(cid:12) ≤ 2, theneither ¯ W (0 , 1) = ¯ W (1 , 2) and by the homogeneity ¯ W (0 , 1) = ¯ W ( i, j ) for all i 9, so b (0 , 1) + b (1 , − b (0 , ∈ ¯ W (0 , 1) + ( ¯ W (0 , − ¯ W (0 , ⊆ ¯ V (0 , W (0 , 2) = ¯ W (0 , 1) and then b (0 , − b (0 , 2) + b (1 , ∈ ( ¯ W (0 , − ¯ W (0 , W (1 , ⊆ ¯ V (1 , W (0 , 2) = ¯ W (1 , 2) and then b (1 , − b (0 , 2) + b (0 , ∈ ( ¯ W (0 , − ¯ W (0 , W (0 , ⊆ ¯ V (0 , (cid:12)(cid:12) { ¯ W (0 , , ¯ W (0 , , ¯ W (1 , } (cid:12)(cid:12) ≤ b (0 , − b (0 , b (1 , ∈ S V . If elements of { ¯ W (0 , , ¯ W (0 , , ¯ W (1 , } are all distinct, then they arerespectively included in disjoint sets ¯ V (0 , , ¯ V (0 , , ¯ V (1 , b (0 , − b (0 , 2) + b (1 , 2) equals to a nontrivial (2 , V , (cid:12)(cid:12) { ¯ W (0 , , ¯ W (0 , , ¯ W (1 , } (cid:12)(cid:12) ≤ 2, theneither ¯ W (0 , 1) = ¯ W (1 , 2) and then − (cid:0) ( b (0 , − b (0 , b (1 , (cid:1) ∈ − ¯ V (1 , W (0 , 1) = ¯ W (0 , 2) and then − (cid:0) ( b (0 , − b (0 , b (1 , (cid:1) ∈ − ¯ V (1 , W (0 , 2) = ¯ W (1 , 2) and then − (cid:0) ( b (1 , − b (0 , b (0 , (cid:1) ∈ − ¯ V (0 , b (0 , − b (0 , − b (1 , 2) equals to a nontrivial(2 , V , V ∩ V = ∅ , so we may conclude that 0 = (cid:0) b (0 , − b (0 , b (1 , (cid:1) + (cid:0) b (0 , − b (0 , − b (1 , (cid:1) is equal to a nontrivial(2 , V , V is an 8–good qif.Thus Case 1 cannot happen and by Claim 3.5.2 either ¯ W (0 , 1) = ¯ W (1 , W (0 , 1) = ¯ W (1 , Case 2 : ¯ W (0 , 1) = ¯ W (1 , W ( j, 8) = ¯ W (8 , 9) for each j < 8. By Claim 3.5.1(2), forevery j < a j − a = b ( j, − b ( j, 8) + b (8 , − b (8 , a + b (8 , − a j = (cid:0) b (8 , − b ( j, (cid:1) + b ( j, ∈ (cid:0) ¯ W ( j, − ¯ W ( j, (cid:1) + ¯ W ( j, W is immersed in V , the set on the far right above is included in ¯ V ( j, c = a + b (8 , 9) and A ′ = { a j : j < } we have c − A ′ ⊆ S V . UsingLemma 3.4(2) we may conclude that c − A ⊆ S Q . Case 3 : ¯ W (0 , 1) = ¯ W (1 , W ( j, 8) = ¯ W (8 , 9) for each j < 8. As before we use Claim3.5.1(2) to get( b (8 , − a )+ a j = (cid:0) b (8 , − b ( j, (cid:1) + b ( j, ∈ (cid:0) ¯ W (8 , − ¯ W (8 , (cid:1) + ¯ W ( j, W is immersed in V , the set on the far right above is included in ¯ V ( j, c = b (8 , − a and A ′ = { a j : j < } we have A ′ + c ⊆ S V . By Lemma3.4(1) we get A + c ⊆ S Q .Finally, to show that only one of (A) and (B) may take place, suppose A + c ⊆ S Q and d − A ⊆ S Q for some c, d ∈ H . For a ∈ A let Q a , Y a ∈ Q be such that a + c ∈ Q a and d − a ∈ Y a .Fix any a ∈ A and choose b ∈ A \ (cid:0) { a } ∪ ( Y a − c ) ∪ ( d − Q a ) (cid:1) (it is possible asby the assumption 3.5(e), | A ∩ ( Y a − c ) | < | A ∩ ( d − Q a ) | < a + c ) + ( d − a ) = c + d = ( b + c ) + ( d − b ) , so 0 ∈ Q a + Y a − Q b − Y b . By the choice of b we have Q b = Y a , Q a = Y b and also(by 3.5(e)) Q a = Q b and Y a = Y b . Therefore some nontrivial (2 , Q , Q is a good qif. (cid:3) Quasi independence and sums In a special case when Q , V , W are all families consisting of singletons, Proposi-tion 3.5 gives the following result of its own interest. Corollary 4.1. Suppose that ( H , + , is an Abelian group and B ⊆ H is quasiindependent. Assume also that (a) m −→ (cid:0) (cid:1) , (b) A ⊆ H , | A | ≥ m and A − A ⊆ B − B .Then exactly one of (A), (B) below holds true: (A) There is a unique c ∈ H such that A + c ⊆ B . (B) There is a unique c ∈ H such that c − A ⊆ B . The above Corollary inspired our interest in its dual version when A − A and B − B are replaced by A + A and B + B . This result is not used in the proof ofour independence theorem, but we find it interesting. Lemma 4.2. Suppose that ( H , + , is an Abelian group and B ⊆ H is quasiindependent. Assume that A ′ ⊆ A ⊆ H and c ∈ H are such that (a) A + A ⊆ B + B , (b) A ′ + c ⊆ B and | A ′ | = 4 .Then A − c ⊆ B .Proof. Suppose that A ′ ⊆ A ⊆ H satisfy the assumptions (a) and (b). Assume a ∈ A and let us argue that a − c ∈ B .Let h a i : i < i list the elements of A ′ . For i < b i = a i + c ∈ B and note thatall b i ’s are pairwise distinct. Since a i + a ∈ B + B we may also choose b ′ i , b ′′ i ∈ B such that a i + a = b ′ i + b ′′ i . Then, for each i < 4, we have a − c = a − ( b i − a i ) = b ′ i + b ′′ i − a i − ( b i − a i ) = b ′ i + b ′′ i − b i . Thus for i < j < OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 11 ( ∗ ) b ′ i + b ′′ i + b j ) − ( b ′ j + b ′′ j + b i ).If for some i < j < { b ′ i , b ′′ i , b j } and { b ′ j , b ′′ j , b i } had at least 2 elements,then the right hand side of ( ∗ ) would give a (2 , B with thevalue 0. So the combination would have to be a trivial one, and hence( ∗ ) for each i < j < b ′ i = b ′′ i = b j ,or (ii) b ′ j = b ′′ j = b i ,or (iii) { b ′ i , b ′′ i , b j } = { b ′ j , b ′′ j , b i } .Suppose that i < j < ∗ ) (iii) holds true. Since b i = b j , we get b i ∈ { b ′ i , b ′′ i } and hence a − c = b ′ i + b ′′ i − b i ∈ { b ′ i , b ′′ i } ⊆ B , and we are done.Assume towards contradiction that( ∗ ) for each i < j < 4, either ( ∗ ) (i) or ( ∗ ) (ii) holds true.Then for some i < b ′ j = b ′′ j whenever j = i . Necessarily, (cid:16) j = j ∧ i / ∈ { j , j } (cid:17) ⇒ b ′ j = b ′ j (as a + a j = a + a j ). Since there are no repetitions among b j ’s, we may nowchoose j = i such that b j = b ′ i , b ′ j = b i getting immediate contradiction with ourassumption ( ∗ ) . (cid:3) Lemma 4.3. Suppose that ( H , + , is an Abelian group and B ⊆ H is quasiindependent. Assume that A ′ ⊆ A ⊆ H are such that (a) A + A ⊆ B + B , (b) | A ′ | ≥ , and A ′ + c ⊆ B for some c ∈ H .Then A + c ⊆ B and the order of c is ≤ .Proof. Let A ′ + c ⊆ B . It follows from Lemma 4.2 that A − c ⊆ B . Applying thatlemma again for A ′ , A, B and − c we get A + c ⊆ B .Concerning the second part of the assertion, suppose towards contradiction that c + c = 0. Let a , a , a , a be distinct elements of A . Then for distinct i, j < a i + c = a i − c, a i + c = a j + c, and a i − c = a j − c, and consequently we may find i < { a + c, a − c } ∩ { a i + c, a i − c } = ∅ .Then, by the first paragraph of this proof, a + c, a − c, a i + c, a i − c ∈ B are alldistinct and ( a + c ) − ( a − c ) − ( a i + c ) + ( a i − c ) = 0, contradicting the quasiindependence of B . (cid:3) Theorem 4.4. Suppose that ( H , + , is an Abelian group and B ⊆ H is quasiindependent. Assume also that (a) m −→ (cid:0) (cid:1) , (b) A ⊆ H , | A | ≥ m and A + A ⊆ B + B .Then there is a unique c ∈ H of order ≤ such that A + c ⊆ B .Proof. Let h a i : i < m i be a sequence of pairwise distinct elements of A . Since A + A ⊆ B + B , we may choose symmetric functions b , b : m × m −→ B suchthat a i + a j = b ( i, j ) + b ( i, j ) for all i, j < m. Let h ϕ ℓ ( i , i , i , i ) : ℓ < i list all formulas of the form b j ( i x , i y y ) = b j ′ ( i x ′ , i y ′ ) for j, j ′ < x < y < x ′ < y ′ < µ : (cid:2) m (cid:3) −→ m such that if i < i
1) = b (1 , , or b (0 , 1) = b (1 , , or b (0 , 1) = b (0 , .Proof of the Claim. Suppose i < j < i < j < b ( i , j ) = b ( i , j ). We shall consider all possible orders of i , j , i , j and use the homo-geneity to conclude one of the clauses in the assertion.(a) If i < j < i < j , then (by the homogeneity) b (0 , 1) = b (2 , 3) = b (4 , 5) = b (2 , b (0 , 1) = b (0 , i < j = i < j then also b (0 , 1) = b (1 , i < i < j < j , then b (0 , 3) = b (2 , 4) = b (1 , 4) = b (0 , 2) and also b (0 , 1) = b (1 , i < i < j = j , then b (0 , 3) = b (1 , 3) = b (2 , 3) = b (1 , 3) and also b (0 , 1) = b (0 , i < i < j < j , then b (0 , 5) = b (3 , 4) = b (1 , 2) = b (0 , 3) and also b (0 , 1) = b (1 , i = i < j < j , then b (0 , 1) = b (0 , 2) = b (0 , 3) = b (0 , b (0 , 1) = b (0 , i = i < j = j then b (0 , 1) = b (0 , i = i < j < j , then b (0 , 2) = b (0 , 1) = b (0 , 3) = b (0 , b (0 , 1) = b (0 , i < i < j < j , then b (0 , 1) = b (1 , 2) similarly to (e).(j) If i < i < j = j , then b (0 , 1) = b (0 , 1) similarly to (d).(k) If i < i < j < j , then b (0 , 1) = b (1 , 2) similarly to (c).(l) If i < i = j < j , then b (0 , 1) = b (1 , i < j < i < j , then b (0 , 1) = b (0 , 1) similarly to (a). (cid:3) Claim 4.4.2. If b (0 , 3) = b (1 , , then b (0 , 1) = b (1 , 2) = b (2 , 3) = b (0 , .Similarly if b is replaced by b .Proof of the Claim. Straightforward by homogeneity. (cid:3) Claim 4.4.3.b (0 , 1) + b (0 , − b (1 , − b (1 , 2) + b (2 , 3) + b (2 , 3) = b (0 , 3) + b (0 , . Proof of the Claim. Follows by the choice of b ( i, j ) , b ( i, j ) and( a + a ) − ( a + a ) + ( a + a ) = a + a . (cid:3) OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 13 Now, we will consider six cases, showing that the first four of them are notpossible. In the remaining two cases we will find c ∈ H such that { a i + c : i < } ⊆ B .Then by Lemma 4.3 we will also have A + c ⊆ B . Case 1 : { b ( i, j ) : i < j < } ∩ { b ( i, j ) : i < j < } = ∅ and b (0 , / ∈{ b (0 , , b (1 , , b (2 , } .Then b (0 , / ∈ { b (0 , , b (0 , , b (1 , , b (1 , , b (2 , , b (2 , , b (0 , } andby Claim 4.4.3 b (0 , 3) = b (0 , 1) + b (0 , − b (1 , − b (1 , 2) + b (2 , 3) + b (2 , − b (0 , , contradicting quasi independence of B . Case 2 : { b ( i, j ) : i < j < } ∩ { b ( i, j ) : i < j < } = ∅ and b (0 , / ∈{ b (0 , , b (1 , , b (2 , } .By an argument similar to Case 1, one shows that this case is not possible as well. Case 3 : { b ( i, j ) : i < j < } ∩ { b ( i, j ) : i < j < } = ∅ and b (0 , ∈{ b (0 , , b (1 , , b (2 , } and b (0 , ∈ { b (0 , , b (1 , , b (2 , } . Subase 3A : b (0 , 3) = b (1 , b (0 , 3) = b (0 , 1) = b (1 , 2) = b (2 , b (0 , 3) = b (0 , a + a = a + a and a = a , a contradiction.If b (0 , 3) = b (2 , a + a = a + a and a = a , a contradiction.If b (0 , 3) = b (1 , b (0 , 3) = b (0 , 1) and we alreadyknow that this leads to a contradiction.Consequently Subcase 3A is not possible. Subase 3B : b (0 , 3) = b (1 , Subase 3C : b (0 , 3) = b (0 , 1) and b (0 , 3) = b (0 , a + a = a + a and a = a giving a contradiction. Subase 3D : b (0 , 3) = b (2 , 3) and b (0 , 3) = b (2 , Subase 3E : b (0 , 3) = b (0 , 1) and b (0 , 3) = b (2 , b (0 , 1) = b (1 , b (1 , 2) = b (2 , b (1 , 2) = b (2 , 3) then also b (2 , 3) = b (0 , 1) and we get a contradictionlike in Subcase 3D.Consequently, there must be no repetitions in { b (1 , , b (2 , , b (0 , , b (1 , } .By Claim 4.4.3 and the assumption of the current subcase we have b (0 , 1) + b (2 , − b (1 , − b (1 , 2) = 0 , a contradiction with the quasi independence of B . Subase 3F : b (0 , 3) = b (2 , 3) and b (0 , 3) = b (0 , { b ( i, j ) : i < j < } ∩ { b ( i, j ) : i < j < } 6 = ∅ . By Claim 4.4.1, this implies that either b (0 , 1) = b (1 , b (0 , 1) = b (1 , b (0 , 1) = b (0 , Case 4 : b (0 , 1) = b (0 , i < j < b ( i, j ) = b ( i, j ). If for some i < j ≤ i < j we had b ( i , j ) = b ( i , j ), then by the homo-geneity we would have had b (0 , 1) = b ( i, j ) = b ( i, j ) for all i < j < b (0 , 1) = ( a + a ) + ( a + a ) = 2 a + 2 b (0 , . Hence 2 b (0 , a + a ) = 4 b (0 , 1) = 2 a +2 b (0 , 1) and a = a , a contradiction.Therefore, b ( i , j ) = b ( i , j ) whenever i < j ≤ i < j ≤ 3. Now, byClaim 4.4.3,2 b (0 , 3) = b (0 , 3) + b (0 , 3) = b (0 , 1) + b (0 , − b (1 , − b (1 , 2) + b (2 , 3) + b (2 , 3) =2 b (0 , − b (1 , 2) + 2 b (2 , . If we had b (0 , 3) = b (1 , b (1 , 2) = b (0 , 5) = b (2 , b (0 , = b (1 , 2) and b (0 , , b (1 , , b (2 , 3) are pairwise distinct. Hence2 b (0 , − b (1 , 2) + 2 b (2 , − b (0 , , B .Consequently, Case 4 is also impossible. Case 5 : b (0 , 1) = b (1 , j < b ( j, 4) = b (4 , j < a j + a = b ( j, 4) + b ( j, 4) = b (4 , 5) + b ( j, , and consequently a j + ( a − b (4 , b ( j, ∈ B . Thus letting c = a − b (4 , 5) we will have { a i + c : i < } ⊆ B . By Lemma 4.3 wealso have A + c ⊆ B . Case 6 : b (0 , 1) = b (1 , j < b ( j, 4) = b (4 , 5) and a j + a = b ( j, 4) + b ( j, 4) = b ( j, 4) + b (4 , . Hence a j + ( a − b (4 , b ( j, ∈ B and the rest is clear.Concerning the uniqueness of c , suppose towards contradiction that c = d aresuch that A + c ⊆ B and A + d ⊆ B . Let a , a , a , a be distinct elements of A .Then for distinct i, j < a i + c = a i + d, a i + c = a j + c, and a i + d = a j + d, and we may find i < { a + c, a + d } ∩ { a i + c, a i + d } = ∅ . Thenthe elements a + c, a + d, a i + c, a i + d belong to B , they are all distinct and( a + c ) − ( a + d ) − ( a i + c ) + ( a i + d ) = 0, contradicting the quasi independenceof B .Finally, Lemma 4.3 gives that c must be of order at most 2. (cid:3) OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 15 Forcing for Abelian groups with few elements of order two In this section, for an Abelian Polish group H with few elements of rank 2 weintroduce a forcing notion adding a Borel set B ⊆ H which has many pairwise k –overlapping translations but no perfect set of such translations. Note that if P ⊆ B then x + y ∈ ( B + x ) ∩ ( B + y ) for each x, y ∈ P . Consequently if P ⊆ B is perfect,then it witnesses that B has a perfect set of pairwise non-disjoint translations.Hence necessarily k ≥ Assumption 5.1. (1) The set of elements of H of order larger than 2 is densein H .(2) 1 < k < ω .(3) ε is a countable ordinal and λ is an uncountable cardinal such that NPr ε ( λ )holds true. The model M ( ε, λ ) and functions rk sp , j and k on [ λ ] <ω \ {∅} are as fixed in Definition 2.6.We will define a forcing notion P adding λ many elements h η α : α < λ i of thegroup H as well as a sequence h F m : m < ω i of closed subsets of H . The Σ subset S = S m<ω F m of H will have the property that (in the forcing extension)( ♥ ) there is no perfect set P ⊆ H satisfying( (cid:0) ∀ x, y ∈ P (cid:1)(cid:0)(cid:12)(cid:12) ( x + S ) ∩ ( y + S ) (cid:12)(cid:12) ≥ k (cid:1) . At the same we will make sure that( ♥ ) (cid:12)(cid:12) ( − η α + S ) ∩ ( − η β + S ) (cid:12)(cid:12) ≥ k for all α, β < λ .To ensure ( ♥ ) holds, the forcing will also add witnesses for it: group elements ν i,α,β = ν i,β,α ∈ H and integers h α,β < ω such that η α + ν i,α,β ∈ F h α,β (for i < k , α, β < λ ).A condition p ∈ P will give a “finite information” on objects mentioned above.Thus for some finite w p ⊆ λ , for all distinct α, β ∈ w p , the condition p providesa basic open neighborhood U pα ( n p ) of η α , basic open neighborhood W pi,α,β of ν i,α,β and the values of h α,β = h p ( α, β ). An approximation to the closed set F m ⊆ H willbe given as its open neighborhood F ( p, m ) = [ (cid:8) U pα ( n p ) + W pi,α,β : ( α, β ) ∈ ( w p ) h i ∧ i < k ∧ h p ( α, β ) = m (cid:9) . Clause ( ♥ ) as well as the ccc of the forcing P will result from the involvement ofthe rank rk sp and additional technical pieces of information carried by conditions p ∈ P : basic open sets Q pi,α,β , V pi,α,β and integers r pm . Definition 5.2. (A) Let P be the collection of all tuples p = (cid:0) w p , M p , ¯ r p , n p , ¯Υ p , ¯ V p , h p (cid:1) = (cid:0) w, M, ¯ r, n, ¯Υ , ¯ V , h (cid:1) such that the following demands ( ⊠ ) –( ⊠ ) are satisfied.( ⊠ ) w ∈ [ λ ] <ω , | w | ≥ 4, 0 < M < ω , 3 ≤ n < ω and ¯ r = h r m : m < M i where r m ≤ n − m < M .( ⊠ ) ¯Υ = h ¯ U α : α ∈ w i where each ¯ U α = h U α ( ℓ ) : ℓ ≤ n i is a ⊆ –decreasingsequence of elements of U .( ⊠ ) ¯ V = h Q i,α,β , V i,α,β , W i,α,β : i < k, ( α, β ) ∈ w h i i ⊆ U and Q i,α,β = Q i,β,α ⊇ V i,α,β = V i,β,α ⊇ W i,α,β = W i,β,α for all i < k and ( α, β ) ∈ w h i . ( ⊠ ) (a) The indexed family h U α ( n − 2) : α ∈ w i ⌢ h Q i,α,β : i < k, α, β ∈ w, α <β i is an 8–good qif (so in particular the sets in this system are pairwisedisjoint), and(b) h U α ( n ) : α ∈ w i ⌢ h W i,α,β : i < k, α, β ∈ w, α < β i is immersedin h U α ( n − 1) : α ∈ w i ⌢ h V i,α,β : i < k, α, β ∈ w, α < β i and h U α ( n − 1) : α ∈ w i ⌢ h V i,α,β : i < k, α, β ∈ w, α < β i is immersed in h U α ( n − 2) : α ∈ w i ⌢ h Q i,α,β : i < k, α, β ∈ w, α < β i ; see Definition3.1(4) (so all these families are 8–good qifs).( ⊠ ) (a) If α, β ∈ w , ℓ ≤ n and U α ( ℓ ) ∩ U β ( ℓ ) = ∅ , then U α ( ℓ ) = U β ( ℓ ), and(b) if α, β, γ ∈ w , ℓ ≤ n , U α ( ℓ ) = U β ( ℓ ) and a ∈ U α ( ℓ ), b ∈ U β ( ℓ ), then ρ ( a, b ) > diam ρ (cid:0) U γ ( ℓ ) (cid:1) (= diam ρ (cid:0) − U γ ( ℓ ) (cid:1) ).( ⊠ ) h : w h i onto −→ M is such that h ( α, β ) = h ( β, α ) for ( α, β ) ∈ w h i .( ⊠ ) Assume that u, u ′ ⊆ w , π and ℓ ≤ n are such that • ≤ | u | = | u ′ | and π : u −→ u ′ is a bijection, • r h ( α,β ) ≤ ℓ for all ( α, β ) ∈ u h i , • U α ( ℓ ) ∩ U β ( ℓ ) = ∅ and h ( α, β ) = h ( π ( α ) , π ( β )) for all distinct α, β ∈ u , • for some c ∈ H ,either for all α ∈ u , we have (cid:0) U α ( ℓ ) + c (cid:1) ∩ U π ( α ) ( ℓ ) = ∅ or for all α ∈ u , we have (cid:0) c − U α ( ℓ ) (cid:1) ∩ U π ( α ) ( ℓ ) = ∅ .Then rk sp ( u ) = rk sp ( u ′ ), j ( u ) = j ( u ′ ), k ( u ) = k ( u ′ ) and for α ∈ u | α ∩ u | = k ( u ) ⇔ | π ( α ) ∩ u ′ | = k ( u ) . ( ⊠ ) Assume that • ∅ 6 = u ⊆ w , rk sp ( u ) = − ℓ ≤ n and • α ∈ u is such that | α ∩ u | = k ( u ), and • r h ( β,β ′ ) ≤ ℓ and U β ( ℓ ) ∩ U β ′ ( ℓ ) = ∅ for all ( β, β ′ ) ∈ u h i .Then there is no α ′ ∈ w \ u such that U α ( ℓ ) = U α ′ ( ℓ ) and h ( α, β ) = h ( α ′ , β )for all β ∈ u \ { α } . (B) For p ∈ P and m < M p we define F ( p, m ) = [ (cid:8) U pα ( n p ) + W pi,α,β : ( α, β ) ∈ ( w p ) h i ∧ i < k ∧ h p ( α, β ) = m (cid:9) . (C) For p, q ∈ P we declare that p ≤ q if and only if • w p ⊆ w q , M p ≤ M q , ¯ r q ↾ M p = ¯ r p , n p ≤ n q , h q ↾ ( w p ) h i = h p , and • if α ∈ w p and ℓ ≤ n p then U qα ( ℓ ) = U pα ( ℓ ), and • if ( α, β ) ∈ ( w p ) h i , i < k , then Q qi,α,β ⊆ Q pi,α,β , V qi,α,β ⊆ V pi,α,β , and W qi,α,β ⊆ W pi,α,β , and • if m < M p , then F ( q, m ) ⊆ F ( p, m ). Lemma 5.3. (1) ( P , ≤ ) is a partial order of size λ . (2) The following sets are dense in P : (i) D γ,M,n = (cid:8) p ∈ P : γ ∈ u p ∧ M p > M ∧ n p > n (cid:9) for γ < λ and M, n < ω . (ii) D N = (cid:8) p ∈ P : diam ρ ∗ ( U pα ( n p − < − N ∧ diam ρ ∗ ( Q pi,α,β ) < − N ∧ diam ρ ∗ ( U pα ( n p − Q pi,α,β ) < − N for all i < k, ( α, β ) ∈ ( w p ) h i (cid:9) for N < ω . (iii) D N = (cid:8) p ∈ P : for all i, j < k and ( α, β ) , ( γ, δ ) ∈ ( w p ) h i it holds that diam ρ ( U pα ( n p − < − N and diam ρ ( Q pi,α,β ) < − N and diam ρ ( U pα ( n p − 2) + Q pi,α,β ) < − N and OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 17 if ( i, α ∗ , α, β ) = ( j, γ ∗ , γ, δ ) then (cid:0) U pα ∗ ( n p ) + W pi,α,β (cid:1) ∩ (cid:0) U pγ ∗ ( n p ) + W pi,γ,δ (cid:1) = ∅ (cid:9) .for N < ω . (3) Assume p ∈ P . Then there is q ≥ p such that n q ≥ n p + 3 , w q = w p and • for all α ∈ w p , cl (cid:0) U qα ( n q − (cid:1) ⊆ U pα ( n p ) , and • for all i < k and ( α, β ) ∈ ( w p ) h i , cl (cid:0) U qα ( n q − 2) + Q qi,α,β (cid:1) ⊆ U pα ( n p ) + W pi,α,β and cl (cid:0) Q qi,α,β (cid:1) ⊆ W pi,α,β . Proof. (2)(i) Suppose p ∈ P and γ ∈ λ \ w p . Let α ∗ = min( w p ) and let w = w p ∪ { γ } and n = n p + 3. Using Proposition 3.3 we may choose U α ( n − ∈ U (for α ∈ w ) and Q i,α,β ∈ U (for i < k , α < β , α, β ∈ w ) such that • U α ( n − ⊆ U pα ( n p ) and Q i,α,β ⊆ W pi,α,β when α, β ∈ w p , • U γ ( n − ⊆ U pα ∗ ( n p ), • h U α ( n − 2) : α ∈ w i ⌢ h Q i,α,β : i < k, α < β, α, β ∈ w i is an 8–good qif, • diam ρ (cid:0) U δ ( n − (cid:1) = diam ρ (cid:0) − U δ ( n − (cid:1) < ρ ( a, b ) for all δ ∈ w , ( α, β ) ∈ w h i , a ∈ U α ( n − 2) and b ∈ U β ( n − U α ( n − , U α ( n ) , V i,α,β , W i,α,β ∈ U (for α < β from w and i < k ) such that U α ( n ) ⊆ U α ( n − ⊆ U α ( n − W i,α,β ⊆ V i,α,β ⊆ Q i,α,β and • h U α ( n − 1) : α ∈ w i ⌢ h V i,α,β : i < k, α, β ∈ w, α < β i is immersed in h U α ( n − 2) : α ∈ w i ⌢ h Q i,α,β : i < k, α, β ∈ w, α < β i , and • h U α ( n ) : α ∈ w i ⌢ h W i,α,β : i < k, α, β ∈ w, α < β i is immersed in h U α ( n − 1) : α ∈ w i ⌢ h V i,α,β : i < k, α, β ∈ w, α < β i .Put ¯Υ = h ¯ U α : α ∈ w i , where ¯ U α = ¯ U pα⌢ h U α ( n − , U α ( n − , U α ( n ) i if α ∈ w p and ¯ U γ = ¯ U pα ∗ ⌢ h U γ ( n − , U γ ( n − , U γ ( n ) i . Let Q i,β,α = Q i,α,β , V i,β,α = V i,α,β and W i,β,α = W i,α,β (for i < k , α < β from w ), and let ¯ V = h Q i,α,β , V i,α,β , W i,α,β : i < k, ( α, β ) ∈ w h i i . Let M = M p + | w p | and let h : w h i −→ M be such that • h ( α, β ) = h p ( α, β ) when ( α, β ) ∈ ( w p ) h i , • h ( α, γ ) = h ( γ, α ) = M p + j when α ∈ w p and j = | w p ∩ α | .We also define ¯ r : M −→ ( n − 1) so that ¯ r ↾ M p = ¯ r p and r m = n − m ∈ [ M p , M )Put q = ( w, M, r, n, ¯Υ , ¯ V , h ). Let us argue that q ∈ P . To this end we have toverify conditions ( ⊠ ) –( ⊠ ) of Definition 5.2. Of these the first six demands followimmediately by our choices. To show ( ⊠ ) , suppose u, u ′ ⊆ w and π : u −→ u ′ and ℓ ≤ n and c ∈ H satisfy the assumptions there. If α ∈ w p then h ( γ, α ) ≥ M p andtherefore γ ∈ u if and only if γ ∈ u ′ . If γ / ∈ u , then u ∪ u ′ ⊆ w p and clause ( ⊠ ) for p (applied to min( n p , ℓ ) instead of ℓ ) gives the needed conclusion. If γ ∈ u , then γ ∈ u ′ too and we look at h ( β, γ ) for β ∈ u ∩ w p . Each of these values is taken by h exactly one time, so h ( π ( β ) , π ( γ )) = h ( β, γ ) for all β ∈ w p implies that π ( γ ) = γ and π ( β ) = β for β ∈ u ∩ w p . Hence u = u ′ and π is the identity, so the desiredconclusion follows.Now suppose ℓ ≤ n , α ∈ u ⊆ w are as in the assumptions of ( ⊠ ) (so by 2.6( ⊛ ) b also u ≥ γ / ∈ u , then applying ( ⊠ ) for p to α, u and ℓ ′ = min( ℓ, n p ) we seethat there is no α ′ ∈ w p \ u with U α ( ℓ ) = U α ′ ( ℓ ), and h ( α, β ) = h ( α ′ , β ) for all β ∈ u \ { α } . The values of h ( β, γ ) (for β ∈ u ) are above M p , so they cannot beequal to h ( β, α ) either. Consequently, the conclusion of ( ⊠ ) holds in this case. Soassume now that γ ∈ u \ { α } . The value of h ( γ, α ) is taken exactly once, so no α ′ ∈ w \ { γ, α } satisfies h ( γ, α ) = h ( γ, α ′ ) and the desired conclusion should be clear now. Finally, assume γ = α . As we said, | u | ≥ β ∈ u \ { γ } and look at h ( γ, β ). There is no α ′ ∈ w \ { γ } with h ( α ′ , β ) = h ( γ, β ), so desiredconclusion follows, finishing the proof of ( ⊠ ) .Now one easily deduces (2)(i).(ii) Assume p ∈ P and N < ω . For ( α, β ) ∈ ( w p ) h i and i < k first choose U α ( n p +1) , Q i,α,β ∈ U such that U α ( n p +1) ⊆ U pα ( n p ), Q i,α,β = Q i,β,α ⊆ W pi,α,β and ρ ∗ –diameters of U α ( n p + 1) , Q i,α,β and U α ( n p + 1)+ Q i,α,β are all smaller than 2 − N .Note that h U α ( n p + 1) : α ∈ w p i ⌢ h Q i,α,β : i < k, α < β, α, β ∈ w p i is an 8–goodqif. Next, use Proposition 3.3 to choose U α ( n p + 2) , U α ( n p + 3) , V i,α,β , W i,α,β ∈ U such that U α ( n p + 3) ⊆ U α ( n p + 2) ⊆ U α ( n p + 1), and W i,α,β = W i,β,α ⊆ V i,α,β = V i,β,α ⊆ Q i,α,β (for ( α, β ) ∈ ( w p ) h i and i < k ), and • h U α ( n p + 3) : α ∈ w p i ⌢ h W i,α,β : i < k, α < β, α, β ∈ w p i is immersed in h U α ( n p + 2) : α ∈ w p i ⌢ h V i,α,β : i < k, α < β, α, β ∈ w p i , and • h U α ( n p + 2) : α ∈ w p i ⌢ h V i,α,β : i < k, α < β, α, β ∈ w p i is immersed in h U α ( n p + 1) : α ∈ w p i ⌢ h Q i,α,β : i < k, α < β, α, β ∈ w p i .Now, for α ∈ w p let ¯ U α = ¯ U pα⌢ h U α ( n p + 1) , U α ( n p + 2) , U α ( n p + 3) i and then let¯Υ = h ¯ U α : α ∈ w p i and ¯ V = h Q i,α,β , V i,α,β , W i,α,β : i < k, ( α, β ) ∈ ( w p ) h i i .These choices clearly determine a condition q = ( w p , M p , ¯ r p , n p + 3 , ¯Υ , ¯ V , h p ) ∈ D N stronger than p .(iii) Similarly to (ii), we just make U α ( n p + 1), U α ( n p + 2), U α ( n p + 3), V i,α,β , Q i,α,β and W i,α,β suitably small.(3) Analogous. (cid:3) Lemma 5.4. Suppose that p ∈ P and α, β, γ, δ ∈ w p are such that α = β . If (cid:16) U pα ( n p − − U pβ ( n p − (cid:17) ∩ (cid:16) U pγ ( n p − − U pδ ( n p − (cid:17) = ∅ , then α = γ and β = δ .Proof. Let n = n p . Suppose that a ∈ U pα ( n − b ∈ U pβ ( n − c ∈ U pγ ( n − 2) and d ∈ U pδ ( n − 2) are such that a − b = c − d . Then a + ( c − a ) = c and b + ( c − a ) = d ,so as ρ is invariant we have ρ ( a, b ) = ρ ( c, d ). Demand 5.2(A)( ⊠ ) (b) implies that ρ ( a, b ) > diam ρ ( U pγ ( n − γ = δ . Now look at a + d − b − c : since α = β and γ = δ it is a (2,4)–combination from an 8–good qif h U pζ ( n − 2) : ζ ∈ w i . Sincethe value of the combination is 0, it has to be trivial. Hence immediately α = γ and β = δ . (cid:3) Lemma 5.5. The forcing notion P has the Knaster property.Proof. Suppose h p ε : ε < ω i is a sequence of pairwise distinct conditions from P .Applying standard ∆–lemma based cleaning procedure we may find w ⊆ λ and A ∈ [ ω ] ω such that for distinct ξ, ζ ∈ A the following demands ( ∗ ) + ( ∗ ) aresatisfied.( ∗ ) | w p ξ | = | w p ζ | , w = w p ξ ∩ w p ζ , M p ξ = M p ζ , n p ξ = n p ζ , ¯ r p ξ = ¯ r p ζ .( ∗ ) If π ∗ : w p ζ −→ w p ξ is the order isomorphism, then • π ∗ ↾ w is the identity, • ¯ U p ζ α ( ℓ ) = ¯ U p ξ π ∗ ( α ) ( ℓ ) whenever α ∈ w p ζ , ℓ ≤ n p ζ , OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 19 • if ( α, β ) ∈ (cid:0) w p ζ (cid:1) h i , i < k , then h p ζ ( α, β ) = h p ξ ( π ∗ ( α ) , π ∗ ( β )), and Q p ζ i,α,β = Q p ξ i,π ∗ ( α ) ,π ∗ ( β ) , V p ζ i,α,β = V p ξ i,π ∗ ( α ) ,π ∗ ( β ) and W p ζ i,α,β = W p ξ i,π ∗ ( α ) ,π ∗ ( β ) , • if ∅ 6 = u ⊆ w p ζ , then rk sp ( u ) = rk sp ( π ∗ [ u ]), j ( u ) = j ( π ∗ [ u ]) and k ( u ) = k ( π ∗ [ u ]).Note that then for all ξ ∈ A we have( ∗ ) if u ⊆ w , α ∈ w p ξ \ w and rk sp (cid:0) u ∪ { α } (cid:1) = − 1, then k (cid:0) u ∪ { α } (cid:1) = | u ∩ α | .[Why? Suppose towards contradiction that k (cid:0) u ∪ { α } (cid:1) = | u ∩ α | . For ζ ∈ A let α ζ ∈ w p ζ be such that | α ζ ∩ w p ζ | = | α ∩ w p ξ | . By ( ∗ ) we have j def = j (cid:0) u ∪{ α } (cid:1) = j (cid:0) u ∪{ α ζ } (cid:1) and k (cid:0) u ∪{ α ζ } (cid:1) = k (cid:0) u ∪{ α } (cid:1) = | u ∩ α | = | u ∩ α ζ | def = k. Therefore, letting u ∪ { α } = { α , . . . , α ℓ − } be the increasing enumeration, we have α k = α and M | = R ℓ,j [ α , . . . , α k − , α ζ , α k +1 , . . . , α ℓ − ] for all ζ ∈ A. However, this contradicts the choice of j , k in Definition 2.6 and the assumptionrk sp (cid:0) u ∪ { α } (cid:1) = − ξ, ζ ∈ A the conditions p ξ , p ζ are compatible. Solet ξ < ζ be from A and let π ∗ : w p ζ −→ w p ξ be the order isomorphism. Set w = w p ξ ∪ w p ζ , M = M p ξ + (cid:12)(cid:12) w p ξ \ w p ζ (cid:12)(cid:12) , n = n p ξ + 3 and let ¯ r = h r m : m < M i be such that r m = r p ξ m if m < M p ξ , and r m = n − M p ξ ≤ m < M .Use Proposition 3.3 and Observation 3.2(iii) to choose U α ( n − U α ( n − U α ( n ), Q i,α,β , V i,α,β and W i,α,β from U for i < k and ( α, β ) ∈ w h i so that( ∗ ) (a) demands 5.2( ⊠ ) –( ⊠ ) are satisfied and(b) if ( α, β ) ∈ (cid:0) w p ξ (cid:1) h i , i < k , then U α ( n − ⊆ U p ξ α ( n p ξ ) and Q i,α,β ⊆ W p ξ i,α,β , and(c) if ( α, β ) ∈ (cid:0) w p ζ (cid:1) h i , i < k , then U α ( n − ⊆ U p ζ α ( n p ζ ) and Q i,α,β ⊆ W p ζ i,α,β .Let ¯ U α = ¯ U p ξ α ⌢ h U α ( n − , U α ( n − , U α ( n ) i if α ∈ w p ξ and ¯ U α = ¯ U p ζ α ⌢ h U α ( n − , U α ( n − , U α ( n ) i if α ∈ w p ζ , and let ¯Υ , ¯ V be defined naturally. Choose h : w h i −→ M extending both h p ξ and h p ζ in such a manner that h ( α, β ) = h ( β, α )for ( α, β ) ∈ w h i and the mapping (cid:0) w p ξ \ w (cid:1) × (cid:0) w p ζ \ w (cid:1) ∋ ( α, β ) h ( α, β )is a bijection onto [ M p ξ , M ). Finally we set q = ( w, M, ¯ r, n, ¯Υ , ¯ V , h ).Let us argue that q ∈ P (once we are done with that, it should be clear that q isstronger than both p ξ and p ζ ). The only potentially unclear demands to verify are( ⊠ ) and ( ⊠ ) of 5.2.First, to demonstrate ( ⊠ ) , suppose that u, u ′ ⊆ w and π : u −→ u ′ and ℓ ≤ n and c ∈ H are as in the assumptions there. Let us consider the following threecases. Case 1: u ⊆ w p ξ .Then for each ( α, β ) ∈ u h i we have h ( α, β ) < M p ξ , so this also holds for all( γ, δ ) ∈ ( u ′ ) h i . Consequently, either u ′ ⊆ w p ξ or u ′ ⊆ w p ζ .If u ′ ⊆ w p ξ , then let ℓ ′ = min( ℓ, n p ξ ) and consider u, u ′ , π, ℓ ′ . Using clause ( ⊠ ) for p ξ we immediately obtain the desired conclusion. If u ′ ⊆ w p ζ , then we let ℓ ′ = min( ℓ, n p ξ ) and we consider u, π ∗ [ u ′ ] , ℓ ′ and π ∗ ◦ π (where, remember, π ∗ : w p ζ −→ w p ξ is the order isomorphism). By ( ∗ ) + ( ∗ ) ,clause ( ⊠ ) for p ξ applies to them and we get • rk sp ( u ) = rk sp (cid:0) π ∗ [ u ′ ] (cid:1) , j ( u ) = j (cid:0) π ∗ [ u ′ ] (cid:1) , k ( u ) = k (cid:0) π ∗ [ u ′ ] (cid:1) and • for α ∈ u , | α ∩ u | = k ( u ) ⇔ (cid:12)(cid:12) ( π ∗ ◦ π )( α ) ∩ π ∗ [ u ′ ] (cid:12)(cid:12) = k ( u ).Now ( ∗ ) , ( ∗ ) immediately imply the desired conclusion. Case 2: u ⊆ w p ζ .Same as the previous case, just interchanging ξ and ζ . Case 3: u \ w p ξ = ∅ 6 = u \ w p ζ .Choose α ∈ u \ w p ξ and β ∈ u \ w p ζ . Then h ( α, β ) ≥ M p ξ and therefore n − r h ( α,β ) ≤ ℓ .If π is the identity on u (and so u = u ′ ), then the needed assertion is immediate.So suppose towards contradiction that we got a γ ∈ u such that π ( γ ) = γ . Since | u | ≥ γ ′ ∈ u such that { γ, π ( γ ) } ∩ { γ ′ , π ( γ ′ ) } = ∅ . Now weconsider two subcases determined by the property of c ∈ H .Suppose (cid:0) U δ ( ℓ ) + c (cid:1) ∩ U π ( δ ) ( ℓ ) = ∅ for all δ ∈ u . Then for some b ∈ U γ ( ℓ ), b ′ ∈ U π ( γ ) ( ℓ ), b ′′ ∈ U γ ′ ( ℓ ) and b ′′′ ∈ U π ( γ ′ ) ( ℓ ) we have b ′ − b = c = b ′′′ − b ′′ .However, this (and the choice of γ and γ ′ ) gives immediate contradiction with h U δ ( ℓ ) : δ ∈ w i being a good qif (remember ℓ ≥ n − (cid:0) c − U δ ( ℓ ) (cid:1) ∩ U π ( δ ) ( ℓ ) = ∅ for all δ ∈ u . Then for some b ∈ U γ ( ℓ ), b ′ ∈ U π ( γ ) ( ℓ ), b ′′ ∈ U γ ′ ( ℓ ) and b ′′′ ∈ U π ( γ ′ ) ( ℓ ) we have b ′ + b = c = b ′′′ + b ′′ , gettingimmediate contradiction with h U δ ( ℓ ) : δ ∈ w i being a good qif.Now, concerning ( ⊠ ) , suppose that u ⊆ w , ℓ ≤ n and α ∈ u are such that • | α ∩ u | = k ( u ) and rk sp ( u ) = − • r h ( β,β ′ ) ≤ ℓ and U β ( ℓ ) ∩ U β ′ ( ℓ ) = ∅ for all ( β, β ′ ) ∈ u h i .We want to argue that there is no α ′ ∈ w such that( z ) α ′ α ′ / ∈ u , h ( α, β ) = h ( α ′ , β ) for all β ∈ u \ { α } , and U α ( ℓ ) = U α ′ ( ℓ ).This is immediate if ℓ ≥ n − 2, so let us assume ℓ ≤ n p ζ . Then we must also have r h ( β,β ′ ) ≤ n p ζ for all ( β, β ′ ) ∈ u h i , so either u ⊆ u p ξ or u ⊆ u p ζ . By the symmetry,we may assume that u ⊆ u p ξ .If u ⊆ w p ξ ∩ w p ζ then we may first use ( ⊠ ) for p ξ to assert that there is no α ′ ∈ w p ξ satisfying ( z ) α ′ and then in the same manner argue that no α ′ ∈ w p ζ satisfies ( z ) α ′ .If u ⊆ w p ξ but u \ w p ζ = ∅ and α ∈ w p ξ ∩ w p ζ , then ( ⊠ ) for p ξ implies there isno α ′ ∈ w p ξ satisfying ( z ) α ′ . Also if α ′ ∈ w p ζ \ w p ξ then for β ∈ u \ w p ζ we have h ( α, β ) < M p ξ ≤ h ( α ′ , β ), so ( z ) α ′ fails then too.Thus we are left only with the possibility that α ∈ u p ξ \ u p ζ . Like before, ( ⊠ ) for p ξ implies there is no α ′ ∈ w p ξ satisfying ( z ) α ′ . So suppose now α ′ ∈ w p ζ \ w p ξ .By ( ∗ ) we know that ( u \ { α } ) \ u p ζ = ∅ , so let β ∈ u \ u p ζ , β = α . Then we have h ( α, β ) < M p ξ ≤ h ( α ′ , β ), so ( z ) α ′ fails. The proof of ( ⊠ ) is complete now. (cid:3) Lemma 5.6. For each ( α, β ) ∈ λ h i and i < k , (cid:13) P “ the sets \ (cid:8) U pα ( n p ) : p ∈ G ˜ P ∧ α ∈ w p (cid:9) and \ (cid:8) W pi,α,β : p ∈ G ˜ P ∧ α, β ∈ w p (cid:9) have exactly one element each. ” OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 21 Proof. Follows from Lemma 5.3. (cid:3) Definition 5.7. (1) For ( α, β ) ∈ λ h i and i < k let η ˜ α , ν ˜ i,α,β and h ˜ α,β be P –names such that (cid:13) P “ { η ˜ α } = \ (cid:8) U pα ( n p ) : p ∈ G ˜ P ∧ α ∈ w p (cid:9) , { ν ˜ i,α,β } = \ (cid:8) W pi,α,β : p ∈ G ˜ P ∧ α, β ∈ w p (cid:9) h ˜ α,β = h p ( α, β ) for some (all) p ∈ G ˜ P such that α, β ∈ w p . ”(2) For m < ω let F ˜ m be a P –name such that (cid:13) P “ F ˜ m = \ (cid:8) F ( p, m ) : m < M p ∧ p ∈ G ˜ P (cid:9) . ”(Remember F ( p, m ) was defined in Definition 5.2 (B) .) Lemma 5.8. (1) For each m < ω , (cid:13) P “ F ˜ m is a closed subset of H . ” (2) For i < k and ( α, β ) ∈ λ h i we have (cid:13) P “ η ˜ α , ν ˜ i,α,β ∈ H , h ˜ α,β < ω, ν ˜ i,α,β = ν ˜ i,β,α and η ˜ α + ν ˜ i,α,β ∈ F ˜ h ˜ α,β . ” (3) (cid:13) P “ h η ˜ α , ν ˜ i,α,β : i < k, α < β < λ i is quasi independent (so they are alsodistinct) .” (4) (cid:13) P “ (cid:12)(cid:12)(cid:12)(cid:0) − η ˜ α + S m<ω F ˜ m (cid:1) ∩ (cid:0) − η ˜ β + S m<ω F ˜ m (cid:1)(cid:12)(cid:12)(cid:12) ≥ k . ”Proof. Should be clear (remember Lemma 5.3). (cid:3) Lemma 5.9. Let p = ( w, M, ¯ r, n, ¯Υ , ¯ V , h ) ∈ D ⊆ P (cf. 5.3(iii)) and a ℓ , b ℓ ∈ H and U ℓ , W ℓ ∈ U (for ℓ < ) be such that the following conditions are satisfied. ( ⊛ ) U ℓ ∈ { U α ( n ) : α ∈ w } , W ℓ ∈ { W i,α,β : i < k, ( α, β ) ∈ w h i } (for ℓ < ). ( ⊛ ) • ( U + W ) ∩ ( U + W ) = ∅ , • ( U + W ) ∩ ( U + W ) = ∅ , • ( U + W ) ∩ ( U + W ) = ∅ , • ( U + W ) ∩ ( U + W ) = ∅ . ( ⊛ ) a ℓ ∈ U ℓ and b ℓ ∈ W ℓ and a ℓ + b ℓ ∈ S m 1. If forsome U we have |{ ℓ < U ℓ = U }| = 3, then we may use (d) to claim thatLHS a is a (nontrivial) (2 , U ∗− and then LHS is a nontrivial(2 , U ∗− ∪ V ∗ , contradicting ( ⊛ ) (remember 5.2(A)( ⊠ ) ).So suppose that for each U we have |{ ℓ < U ℓ = U }| ≤ 2. Then LHS a is a(possibly trivial) (2 , U ∗ and consequently LHS is a nontrivial(2 , U ∗ ∪ V ∗ , so also from U ∗− ∪ V ∗ , again contradicting ( ⊛ ) .(f) For each U , |{ ℓ < U ℓ = U }| < p ∈ D , it follows from our assumption ( ⊛ ) that for each ℓ < 4, for some α = α ( ℓ ) , β = β ( ℓ ), and i = i ( ℓ ) we have U ℓ = U α ( n ) and W ℓ = W i,α,β . It followsfrom (e)+(f) that LHS is a (2 , U ∗ ∪ W ∗ . Necessarily it is atrivial combination (as LHS = 0 by ( ⊛ ) ). Consequently ,( ⊙ ) either U = U = U = U , or U = U = U = U , and( ⊙ ) either W = W = W = W , or W = W = W = W .Suppose U = U = U = U . Then by ( ⊛ ) we must have W = W , W = W and by ( ⊙ ) we get W = W and W = W . Thus for some ( α, β ) ∈ w h i and i, j < k , i = j , we have U = U = U α ( n ) , U = U = U β ( n ) , W = W = W i,α,β , W = W = W j,α,β . OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 23 Suppose now that U = U and U = U . By ( ⊛ ) we must have then W = W and W = W . Therefore, by ( ⊙ ) , we may conclude that W = W and W = W .Consequently, for some ( α, β ) ∈ w h i and distinct i, j < k we have U = U = U α ( n ) , U = U = U β ( n ) , W = W = W i,α,β , W = W = W j,α,β . (cid:3) Lemma 5.10. Let p = ( w, M, ¯ r, n, ¯Υ , ¯ V , h ) ∈ D and X ⊆ H , | X | ≥ . Supposethat a i ( x, y ) , b i ( x, y ) , U i ( x, y ) and W i ( x, y ) for x, y ∈ X , x = y and i < k satisfythe following demands (i)–(iv) (for all x = y , i = i ′ ). (i) U i ( x, y ) ∈ { U α ( n ) : α ∈ w } , W i ( x, y ) ∈ { W j,α,β : j < k, ( α, β ) ∈ w h i } . (ii) • (cid:0) U i ( x, y ) + W i ( x, y ) (cid:1) ∩ (cid:0) U i ( y, x ) + W i ( y, x ) (cid:1) = ∅ , • (cid:0) U i ( x, y ) + W i ( x, y ) (cid:1) ∩ (cid:0) U i ′ ( x, y ) + W i ′ ( x, y ) (cid:1) = ∅ . (iii) a i ( x, y ) ∈ U i ( x, y ) and b i ( x, y ) ∈ W i ( x, y ) , and a i ( x, y ) + b i ( x, y ) ∈ S m 2) : α, β ∈ w (cid:9) . (2) If ( x, y ) ∈ X h i and x − y ∈ U α ( n − − U β ( n − , α, β ∈ w , then α = β andfor each i < k we have a i ( x, y ) + b i ( x, y ) , a i ( y, x ) + b i ( y, x ) ∈ F ( p, h ( α, β )) .Proof. (1) Fix x, y ∈ X , x = y , for a moment.Let i = i ′ , i, i ′ < k . We may apply Lemma 5.9 for U i ( x, y ), W i ( x, y ), U i ( y, x ), W i ( y, x ), a i ( x, y ), b i ( x, y ), a i ( y, x ), b i ( y, x ) here as U , W , U , W , a , b , a , b thereand for similar objects with i ′ in place of i as U , W , U , W , a , b , a , b there.This will produce distinct α = α ( x, y, i, i ′ ) , β = β ( x, y, i, i ′ ) ∈ w and distinct j = j ( x, y, i, i ′ ) , j ′ = j ′ ( x, y, i, i ′ ) < k such thateither (A) U i ( x, y ) = U i ′ ( x, y ) = U α ( n ), U i ( y, x ) = U i ′ ( y, x ) = U β ( n ), W i ( x, y ) = W i ( y, x ) = W j,α,β , W i ′ ( x, y ) = W i ′ ( y, x ) = W j ′ ,α,β ,or (B) U i ( x, y ) = U i ( y, x ) = U α ( n ), U i ′ ( x, y ) = U i ′ ( y, x ) = U β ( n ), W i ( x, y ) = W i ′ ( x, y ) = W j,α,β , W i ( y, x ) = W i ′ ( y, x ) = W j ′ ,α,β .Note that if for some i = i ′ , i, i ′ < k , the possibility (A) above holds, then it holdsfor all i, i ′ < k and x − y = (cid:0) a i ( x, y ) + b i ( x, y ) (cid:1) − (cid:0) a i ( y, x ) + b i ( y, x ) (cid:1) = (cid:16) a i ( x, y ) + (cid:0) b i ( x, y ) − b i ( y, x ) (cid:1)(cid:17) − a i ( y, x )and a i ( x, y ) + (cid:0) b i ( x, y ) − b i ( y, x ) (cid:1) ∈ U α ( n ) + (cid:0) W j,α,β − W j,α,β (cid:1) ⊆ U α ( n − ⊆ U α ( n − x − y ∈ U α ( n − − U β ( n − x, y . By what we have said, the first assertion of the Lemma willfollow once we show that( ♥ ) for all x, y ∈ X , x = y , there are i = i ′ such that possibility (A) aboveholds for them.Here the argument breaks into two cases: k ≥ k = 2, with the former beingsomewhat simpler. Case k ≥ x, y ∈ X , x = y . Suppose towards contradiction that in the previous consid-erations both for x, y, , x, y, , α, β, j, j ′ such that α = β , j = j ′ and( ∗ ) U ( x, y ) = U ( y, x ) = U α ( n ),( ∗ ) U ( x, y ) = U ( y, x ) = U β ( n ),( ∗ ) W ( x, y ) = W ( x, y ) = W j,α,β ,( ∗ ) W ( y, x ) = W ( y, x ) = W j ′ ,α,β ,and we also get γ, δ, ℓ, ℓ ′ such that γ = δ and ℓ = ℓ ′ and( ∗ ) U ( x, y ) = U ( y, x ) = U γ ( n ),( ∗ ) U ( x, y ) = U ( y, x ) = U δ ( n ),( ∗ ) W ( x, y ) = W ( x, y ) = W ℓ,γ,δ ,( ∗ ) W ( y, x ) = W ( y, x ) = W ℓ ′ ,γ,δ .It follows from ( ∗ ) + ( ∗ ) that γ = β and from ( ∗ ) + ( ∗ ) we have ℓ = j and δ = α .Finally, ( ∗ ) + ( ∗ ) imply ℓ ′ = j ′ . Consequently, U ( x, y ) = U ( x, y ) , U ( y, x ) = U ( y, x ) , W ( x, y ) = W ( x, y ) , W ( y, x ) = W ( y, x ) , contradicting assumption (ii). Case k = 2.By the way of contradiction, we will argue that ( ♥ ) holds true in this case as well.First, however, we have to establish some auxiliary facts.For each x, y ∈ X , x = y , we may choose α = α ( x, y ), β = β ( x, y ) and j = j ( x, y )such that α = β andeither ( A ) α,β,jx,y U ( x, y ) = U ( x, y ) = U α ( n ), U ( y, x ) = U ( y, x ) = U β ( n ), W ( x, y ) = W ( y, x ) = W j,α,β , W ( x, y ) = W ( y, x ) = W − j,α,β ,or ( B ) α,β,jx,y U ( x, y ) = U ( y, x ) = U α ( n ), U ( x, y ) = U ( y, x ) = U β ( n ), W ( x, y ) = W ( x, y ) = W j,α,β , W ( y, x ) = W ( y, x ) = W − j,α,β ,Note that for each ( x, y ) ∈ w h i , either there are α, β, j such that ( A ) α,β,jx,y holdstrue or there are α, β, j such that ( B ) α,β,jx,y is true, but not both. Also, remembering5.2(A)( ⊠ ) (b),( △ ) if ( A ) α,β,jx,y holds, then x − y ∈ U α ( n − − U β ( n − 1) and if ( B ) α,β,jx,y issatisfied, then x − y ∈ V j,α,β − V − j,α,β .Define functions χ : X h i −→ X h i −→ [ w ] × x, y ) ∈ X h i , • if for some α, β, j the demand ( A ) α,β,jx,y holds, then χ ( x, y ) = 1 and Θ( x, y ) =( { α, β } , j ), • if for some α, β, j the demand ( B ) α,β,jx,y is satisfied, then χ ( x, y ) = 0 andΘ( x, y ) = ( { α, β } , j ).Note that( △ ) if χ ( x, y ) = 0 and Θ( x, y ) = ( { α, β } , j ), then χ ( y, x ) = 0 and Θ( y, x ) =( { α, β } , − j ), so also Θ( x, y ) = Θ( y, x ).Our goal is to show that the function χ never takes value 0 (as this will imply thatthe assertion ( ♥ ) holds true).( △ ) If x, y, z ∈ X are pairwise distinct and χ ( x, y ) = χ ( y, z ) = 1, then χ ( x, z ) =1. OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 25 Why? Assume χ ( x, z ) = 0. Then, by ( △ ) , for some j, ξ, ζ we have x − z ∈ V j,ξ,ζ − V − j,ξ,ζ . However, x − y ∈ U α ( n − − U β ( n − 1) and y − z ∈ U γ ( n − − U δ ( n − α = β and γ = δ ), so x − z ∈ U α ( n − − U β ( n − 1) + U γ ( n − − U δ ( n − . Thus for some a ∈ U α ( n − b ∈ U β ( n − c ∈ U γ ( n − d ∈ U δ ( n − e ∈ V j,ξ,ζ ,and f ∈ V − j,ξ,ζ we have a − b + c − d + f − e = 0. The left hand side of this equationrepresents a nontrivial (2,8)–combination from h U ζ ( n − 1) : ζ ∈ w i ⌢ h V ,ζ,ζ ′ , V ,ζ,ζ ′ :( ζ, ζ ′ ) ∈ w h i i (remember α = β , γ = δ , ξ = ζ ), a contradiction.( △ ) If x, y, z ∈ X are pairwise distinct and χ ( x, y ) = χ ( y, z ) = 0, then Θ( x, y ) =Θ( y, z ) = Θ( z, x ) and χ ( x, z ) = 0.Why? Let Θ( x, y ) = ( { α, β } , i ), Θ( y, z ) = ( { γ, δ } , j ), and Θ( x, z ) = ( { ξ, ζ } , ℓ ). If { α, β } 6 = { γ, δ } , then x − z = ( x − y ) + ( y − z ) ∈ V i,α,β − V − i,α,β + V j,γ,δ − V − j,γ,δ and { V ,α,β , V ,α,β } ∩ { V ,γ,δ , V ,γ,δ } = ∅ . Since either x − z ∈ V ℓ,ξ,ζ − V − ℓ,ξ,ζ or x − z ∈ (cid:16) U ξ ( n − − U ζ ( n − (cid:17) ∪ (cid:16) U ζ ( n − − U ξ ( n − (cid:17) , we easily get that somenontrivial (2,8)–combination from h U ζ ( n − 1) : ζ ∈ w i ⌢ h V ,ζ,ζ ′ , V ,ζ,ζ ′ : ( ζ, ζ ′ ) ∈ w h i i equals 0, a contradiction. Consequently, { α, β } = { γ, δ } , i.e., Θ( x, y ) =( { α, β } , i ) and Θ( y, z ) = ( { α, β } , j ) for some α, β, i, j .If i = j then x − z ∈ V i,α,β − V − i,α,β + V − i,α,β − V i,α,β . But also either x − z ∈ U ξ ( n − − U ζ ( n − x − z ∈ U ζ ( n − − U ξ ( n − x − z ∈ V ℓ,ξ,ζ − V − ℓ,ξ,ζ .In the first case we get0 ∈ (cid:16)(cid:0) V i,α,β − V i,α,β (cid:1) + U ξ ( n − (cid:17) − (cid:16)(cid:0) V − i,α,β − V − i,α,β (cid:1) + U ζ ( n − (cid:17) ⊆ U ξ ( n − − U ζ ( n − , and symmetrically in the second case. In the last case we have0 ∈ (cid:16)(cid:0) V i,α,β − V i,α,β (cid:1) + V ℓ,ξ,ζ (cid:17) − (cid:16)(cid:0) V − i,α,β − V − i,α,β (cid:1) + V − ℓ,ξ,ζ (cid:17) ⊆ Q ℓ,ξ,ζ − Q − ℓ,ξ,ζ . In any case this gives a contradiction with 5.2(A)( ⊠ ) . Consequently i = j andΘ( x, y ) = Θ( y, z ) = ( { α, β } , i ).By considerations as above we see that necessarily χ ( x, z ) = 0 and Θ( x, z ) =( { α, β } , ℓ ). If ℓ = i , then x − z ∈ V i,α,β − V − i,α,β and x − z ∈ V i,α,β − V − i,α,β + V i,α,β − V − i,α,β . Hence0 ∈ (cid:16)(cid:0) V i,α,β − V i,α,β (cid:1) + V i,α,β (cid:17) − (cid:16)(cid:0) V − i,α,β − V − i,α,β (cid:1) + V − i,α,β (cid:17) ⊆ Q i,α,β − Q − i,α,β , a contradiction.Consequently, ℓ = 1 − i and Θ( z, x ) = ( { α, β } , i ) = Θ( x, y ) (and χ ( x, z ) = 0).Now, suppose towards contradiction that ( ♥ ) is not true and x, y ∈ X are suchthat x = y and χ ( x, y ) = 0. Let z ∈ X \{ x, y } . We cannot have χ ( x, z ) = χ ( y, z ) = 1(as then ( △ ) would give a contradiction with χ ( x, y ) = 0). So one of them is 0,and then ( △ ) implies that the other is 0 as well and χ ( x, y ) = χ ( y, z ) = χ ( x, z ) = 0 and Θ( x, y ) = Θ( y, z ) = Θ( z, x ) . Taking t ∈ X \ { x, y, z } by similar considerations we obtain χ ( x, t ) = χ ( y, t ) = 0 and Θ( x, y ) = Θ( y, t ) = Θ( t, x ) . Now consider x, z, t : since χ ( x, z ) = χ ( x, t ) = 0 we may use ( △ ) to conclude that χ ( z, t ) = 0 and Θ( x, z ) = Θ( z, t ) = Θ( t, x ) . But we have established already that Θ( t, x ) = Θ( x, y ) = Θ( z, x ), a contradiction(remember ( △ ) ). The proof of Lemma 5.10(1) is complete now.(2) Suppose ( x, y ) ∈ X h i . In the previous part we showed that for all i < i ′ < k possibility (A) holds true. More precisely, there are distinct α, β ∈ w such thatfor all i < k for some j < k we have a i ( x, y ) ∈ U α ( n ) and a i ( y, x ) ∈ U β ( n ), and b i ( x, y ) , b i ( y, x ) ∈ W j,α,β . Then also • a i ( x, y ) + b i ( x, y ) ∈ U α ( n ) + W j,α,β ⊆ F ( p, h ( α, β )), • a i ( y, x ) + b i ( y, x ) ∈ U β ( n ) + W j,α,β ⊆ F ( p, h ( α, β )).We also know that for these α, β we have x − y ∈ U α ( n − − U β ( n − α ′ , β ′ (cid:0) U α ( n − − U β ( n − (cid:1) ∩ (cid:0) U α ′ ( n − − U β ′ ( n − (cid:1) = ∅ implies α = α ′ and β = β ′ . (cid:3) Lemma 5.11. (cid:13) P “ there is no perfect set P ⊆ H such that (cid:16) ∀ x, y ∈ P (cid:17)(cid:16)(cid:12)(cid:12)(cid:12)(cid:0) x + S m<ω F ˜ m (cid:1) ∩ (cid:0) y + S m<ω F ˜ m (cid:1)(cid:12)(cid:12)(cid:12) ≥ k (cid:17) . ”Proof. Suppose towards contradiction that G ⊆ P is generic over V and in V [ G ]the following assertion holds true:for some perfect set P ⊆ H we have (cid:12)(cid:12)(cid:12)(cid:0) x + [ m<ω F ˜ Gm (cid:1) ∩ (cid:0) y + [ m<ω F ˜ Gm (cid:1)(cid:12)(cid:12)(cid:12) ≥ k for all x, y ∈ P .Then for any distinct x, y ∈ P there are c , d , . . . , c k − , d k − ∈ S m<ω F ˜ Gm such that c i = c j whenever i = j and x − y = c i − d i (for all i < k ).For ¯ ℓ = h ℓ i : i < k i ⊆ ω , ¯ m = h m i : i < k i ⊆ ω and N < ω let Z N ¯ ℓ, ¯ m = { ( x, y ) ∈ P : there are c i ∈ F ˜ Gℓ i , d i ∈ F ˜ Gm i (for i < k ) such that x − y = c i − d i and 2 − N < min (cid:0) ρ ( c i , c j ) , ρ ( d i , d j ) (cid:1) for all distinct i, j < k } . By our assumption on P we know that( ⊡ ) for all x, y ∈ P , x = y , there are ¯ ℓ, ¯ m and N such that ( x, y ) ∈ Z N ¯ ℓ, ¯ m .The sets Z N ¯ ℓ, ¯ m are Σ , so we may apply Mycielski’s theorem [8, Theorem 1, p. 141](see also [9, Lemma 2.4]) to choose a perfect set P ∗ ⊆ P , an increasing sequence0 < n ∗ < n < n ∗ < n < n ∗ < n < . . . and a system h d σ : σ ∈ ι , ι < ω i ⊆ D such that the following demands ( ⊡ ) a –( ⊡ ) d are satisfied.( ⊡ ) a If ι < ω , σ, σ ′ ∈ ι σ = σ ′ , then ( d σ + B n ι ) ∩ P ∗ = ∅ and ρ ( d σ , d σ ′ ) > − n ι ,and diam ρ ∗ ( d σ + B n ι ) < − ι .( ⊡ ) b If ι < ω , σ ∈ ι 2, then cl (cid:0) d σ ⌢ h i + B n ι +1 (cid:1) ∪ cl (cid:0) d σ ⌢ h i + B n ι +1 (cid:1) ⊆ ( d σ + B n ι ). OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 27 ( ⊡ ) c If ι < ω and σ, σ ′ ∈ ι σ = σ ′ , and x, x ′ ∈ P ∗ ∩ ( d σ + B n ι ), y, y ′ ∈ P ∗ ∩ ( d σ ′ + B n ι ), then for all ¯ ℓ ⊆ n ∗ ι , ¯ m ⊆ n ∗ ι and N < n ∗ ι we have( x, y ) ∈ Z N ¯ ℓ, ¯ m ⇔ ( x ′ , y ′ ) ∈ Z N ¯ ℓ, ¯ m . ( ⊡ ) d If ι < ω and σ, σ ′ ∈ ι σ = σ ′ , and x ∈ P ∗ ∩ ( d σ + B n ι ), y ∈ P ∗ ∩ ( d σ ′ + B n ι ),then there are ¯ ℓ ⊆ n ∗ ι , ¯ m ⊆ n ∗ ι and N < n ∗ ι such that ( x, y ) ∈ Z N ¯ ℓ, ¯ m .It follows from ( ⊡ ) a + ( ⊡ ) b that for each η ∈ ω T ℓ<ω d η ↾ ℓ + B n ℓ is a singletonincluded in P ∗ . For σ ∈ ι ℓ < ω let σ ∗ ℓ σ ⌢ h , . . . , | {z } ℓ i and let x ∗ σ ∈ H besuch that( ⊡ ) { x ∗ σ } = \ ℓ<ω (cid:16) d σ ∗ ℓ + B n ι + ℓ (cid:17) (so x ∗ σ ∈ P ∗ ).Let P ˜ ∗ , n ˜ ∗ ι , n ˜ ι , d ˜ σ , x ˜ ∗ σ be P –names for the objects appearing in ( ⊡ ) –( ⊡ ) . Stillworking in V [ G ], we may choose a sequence h p ι , q ι : ι < ω i ⊆ G such that:( ⊡ ) a p (cid:13) P “ P ˜ ∗ is a perfect subset of H and n ˜ ∗ ι , n ˜ ι , d ˜ σ , x ˜ ∗ σ have the propertiesstated in ( ⊡ ) a –( ⊡ ) d , ( ⊡ ) ”, and( ⊡ ) b p ι decides the values of n ˜ ∗ ι , n ˜ ι and d ˜ σ for σ ∈ ι ι > ⊡ ) c p ι ≤ q ι ≤ p ι +1 and p ι , q ι ∈ D n ι ∩ D ,n ι ,n ι ∩ G (see 5.3(2)) and n p ι + 10 < n q ι and w p ι = w q ι .Note that the properties of conditions from P stated in 5.2(A) are absolute, sothey hold in V [ G ] as well (with B ℓ being B Gℓ etc). Now, still working in V [ G ], for0 < ι < ω let X ι = { x ∗ σ : σ ∈ ι } . Note that x ∗ σ = x ∗ σ ′ when σ, σ ′ ∈ ι X ι ⊆ X ι ′ when ι ≤ ι ′ < ω . It follows from ( ⊡ ) d that for x, y ∈ X ι , x = y , wehave ( x, y ) ∈ Z N ¯ ℓ, ¯ m for some N = N ( x, y ) < n ∗ ι , ¯ ℓ = ¯ ℓ ( x, y ) , ¯ m = ¯ m ( x, y ) ⊆ n ∗ ι . Byclause ( ⊡ ) c , these N ( x, y ) , ¯ ℓ ( x, y ) , ¯ m ( x, y ) may be chosen in such a manner that( ⊡ ) if σ, σ ′ ∈ ι ι ∗ < ι , σ ↾ ι ∗ = σ ′ ↾ ι ∗ but σ ( ι ∗ ) = σ ′ ( ι ∗ ), and ς = σ ↾ ( ι ∗ + 1), ς ′ = σ ′ ↾ ( ι ∗ + 1), then ¯ ℓ ( x σ , x σ ′ ) = ¯ ℓ ( x ς , x ς ′ ), ¯ m ( x σ , x σ ′ ) = ¯ m ( x ς , x ς ′ ), and N ( x σ , x σ ′ ) = N ( x ς , x ς ′ ).Let J < ω be such that the arrow property J −→ (10) holds true and fix a ι ≥ J for a while.Fix x, y ∈ X ι , x = y , and let N = N ( x, y ) < n ∗ ι , ¯ ℓ = ¯ ℓ ( x, y ) , ¯ m = ¯ m ( x, y ) ⊆ n ∗ ι .Then there are c i ∈ F ˜ Gℓ i and d i ∈ F ˜ Gm i (for i < k ) such that for i = i ′ we have x − y = c i − d i and 2 − n ι < − N < ρ ( c i , c i ′ ) , and 2 − n ι < − N < ρ ( d i , d i ′ ) . The reasons for the use of q ι rather than p ι in what follows will become clearat the end. Since n ι < M q ι and F ˜ Gm ⊆ F ( q ι , m ) for all m < M q ι , we get c i ∈ U q ι α ( n q ι ) + W q ι j,α,β for some j < k and ( α, β ) ∈ ( w q ι ) h i = ( w p ι ) h i and similarly for d i . Therefore, for each i < k we may pick • U i ( x, y ) , U i ( y, x ) ∈ { U q ι α ( n q ι ) : α ∈ w q ι } , and • W i ( x, y ) , W i ( y, x ) ∈ { W q ι j,α,β : ( α, β ) ∈ ( w q ι ) h i } , and • a i ( x, y ) ∈ U i ( x, y ), a i ( y, x ) ∈ U i ( y, x ) and b i ( x, y ) ∈ W i ( x, y ), b i ( y, x ) ∈ W i ( y, x )such that x − y = (cid:0) a i ( x, y ) + b i ( x, y ) (cid:1) − (cid:0) a i ( y, x ) + b i ( y, x ) (cid:1) and for i = i ′ − n ι < ρ (cid:16) a i ( x, y ) + b i ( x, y ) , a i ′ ( x, y ) + b i ′ ( x, y ) (cid:17) − n ι < ρ (cid:16) a i ( y, x ) + b i ( y, x ) , a i ′ ( y, x ) + b i ′ ( y, x ) (cid:17) . Since the metric ρ is invariant (and by ( ⊡ ) a ), we also have2 − n ι < ρ ( x, y ) = ρ (cid:16) a i ( x, y ) + b i ( x, y ) , a i ( y, x ) + b i ( y, x ) (cid:17) . Since q ι ∈ D n ι we know that for all relevant j, α, β ,diam ρ (cid:16) U q ι α ( n q ι ) + W q ι j,α,β (cid:17) < − n ι , and consequently each of the sets U q ι α ( n q ι ) + W q ι j,α,β contains at most one ele-ment from each of the sets (cid:8) a i ( x, y ) + b i ( x, y ) , a i ( y, x ) + b i ( y, x ) (cid:9) , (cid:8) a i ( x, y ) + b i ( x, y ) , a i ′ ( x, y ) + b i ′ ( x, y ) (cid:9) and (cid:8) a i ( y, x ) + b i ( y, x ) , a i ′ ( y, x ) + b i ′ ( y, x ) (cid:9) . Since q ι ∈ D n ι , different sets of the form U q ι α ( n q ι ) + W q ι j,α,β are disjoint, and thus we seethat the assumptions (i)–(iv) of Lemma 5.10 are satisfied.Unfixing x, y , we may use Lemma 5.10(1) to conclude that( ⊡ ) X ι − X ι ⊆ S (cid:8) U q ι α ( n q ι − − U q ι β ( n q ι − 2) : α, β ∈ w q ι (cid:9) and hence also X ι − X ι ⊆ [ (cid:8) U p ι α ( n p ι ) − U p ι β ( n p ι ) : α, β ∈ w p ι (cid:9) . Moreover, by 5.10(2), we also conclude that( ⊡ ) if x, y ∈ X ι and 0 = x − y ∈ U q ι α ( n q ι − − U q ι β ( n q ι − α = β and¯ m ( x, y )( i ) = ¯ ℓ ( x, y )( i ) = h q ι ( α, β ) = h p ι ( α, β ) for all i < k .Since (cid:8) U p ι α ( n p ι ) : α ∈ w p ι (cid:9) , (cid:8) U p ι α ( n p ι − 1) : α ∈ w p ι (cid:9) , (cid:8) U p ι α ( n p ι − 2) : α ∈ w p ι (cid:9) and X ι satisfy the assumptions of Theorem 3.5, we get that exactly one of ( A ) ι ,( B ) ι below holds true.( A ) ι There is a c ι ∈ H such that X ι + c ι ⊆ S (cid:8) U p ι α ( n p ι − 2) : α ∈ w p ι (cid:9) .( B ) ι There is a c ι ∈ H such that c ι − X ι ⊆ S (cid:8) U p ι α ( n p ι − 2) : α ∈ w p ι (cid:9) .Unfixing ι < ω we let A = { ι < ω : J ≤ ι and case ( A ) ι holds true } B = { ι < ω : J ≤ ι and case ( B ) ι holds true } . One of the sets A, B is infinite and this leads us to two very similar cases. Case: The set A is infinite.For ι ∈ A let X ι , c ι be as before. Let w ι = { α ∈ w p ι : U p ι α ( n p ι − ∩ ( X ι + c ι ) = ∅} .Since diam ρ (cid:0) U p ι α ( n p ι − (cid:1) < − n ι < ρ ( x, y ) for α ∈ w ι and distinct x, y ∈ X ι , weget (cid:12)(cid:12) U p ι α ( n p ι − ∩ ( X ι + c ι ) (cid:12)(cid:12) = 1 for α ∈ w ι . Consequently, we have a naturalbijection ϕ ι : X ι −→ w ι such that x + c ι ∈ U p ι ϕ ι ( x ) ( n p ι − ι < ι ′ from A we have X ι ⊆ X ι ′ and the mapping π ι,ι ′ = ϕ ι ′ ◦ ϕ − ι : w ι −→ w ι ′ is an injection. Clearly, if x ∈ X ι , α = ϕ ι ( x ) ∈ w ι then( ⊡ ) x + c ι ′ ∈ (cid:16) U p ι ′ α ( n p ι − 2) + ( c ι ′ − c ι ) (cid:17) ∩ U p ι ′ π ι,ι ′ ( α ) ( n p ι − = ∅ . OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 29 Suppose now that x, y ∈ X ι , x = y . By ( ⊡ ) , there are α, β ∈ w q ι such that x − y ∈ U q ι α ( n q ι − − U q ι β ( n q ι − 2) (and, by ( ⊡ ) , α = β ). Then also x − y ∈ (cid:16) U p ι α ( n p ι − − U p ι β ( n p ι − (cid:17) ∩ (cid:16) U p ι ϕ ι ( x ) ( n p ι − − U p ι ϕ ι ( y ) ( n p ι − (cid:17) . By Lemma 5.4 we conclude that α = ϕ ι ( x ) and β = ϕ ι ( y ). Together with ( ⊡ ) thisgives us that( ⊡ ) ι if ( x, y ) ∈ ( X ι ) h i , then ¯ m ( x, y )( i ) = ¯ ℓ ( x, y )( i ) = h p ι ( ϕ ι ( x ) , ϕ ι ( y )) for all i < k .Putting together ( ⊡ ) ι and ( ⊡ ) ι ′ we see that( ⊡ ) if ι < ι ′ are from A and ( x, y ) ∈ ( X ι ) h i , then h p ι ( ϕ ι ( x ) , ϕ ι ( y )) = h p ι ′ ( ϕ ι ′ ( x ) , ϕ ι ′ ( y )) . In other words, if ( α, β ) ∈ w ι then h p ι ( α, β ) = h p ι ′ ( α, β ) = h p ι ′ ( π ι,ι ′ ( α ) , π ι,ι ′ ( β )) . It follows from ( ⊡ ) + ( ⊡ ) and 5.2(A)( ⊠ ) that for ι < ι ′ from A we have( ⊡ ) rk sp ( w ι ) = rk sp (cid:0) π ι,ι ′ [ w ι ] (cid:1) , j ( w ι ) = j (cid:0) π ι,ι ′ [ w ι ] (cid:1) , k ( w ι ) = k (cid:0) π ι,ι ′ [ w ι ] (cid:1) and | α ∩ w ι | = k ( w ι ) ⇔ | π ι,ι ′ ( α ) ∩ π ι,ι ′ [ w ι ] | = k ( w ι ) for all α ∈ w ι . (Note that r p ι ′ m ≤ n p ι − m = h p ι ′ ( α, β ), α, β ∈ w ι ⊆ w p ι , α = β .)Choose a strictly increasing sequence h ι ( ℓ ) : ℓ < ω i ⊆ A such that( ⊡ ) for each ℓ < ω ,2 − n ι ( ℓ +1) − < diam ρ (cid:16) U p ι ( ℓ ) ( n p ι ( ℓ ) − (cid:17) = diam ρ (cid:16) U p ι ( ℓ +1) ( n p ι ( ℓ ) − (cid:17) (remember n ι ’s were chosen in ( ⊡ ) and p ι ( ℓ ) ∈ D ,n ι ( ℓ ) ,n ι ( ℓ ) so also 0 ∈ w p ι ( ℓ ) ).Fix ℓ < ω for a moment and suppose ς ∈ ι ( ℓ ) (cid:12)(cid:12) ϕ ι ( ℓ ) ( x ∗ ς ) ∩ w ι ( ℓ ) (cid:12)(cid:12) = k ( w ι ( ℓ )) ) . Let ς ∗ ∈ ι ( ℓ +1) ς ⊳ ς ∗ , ς ∗ ( n ) = 0 for n ∈ [ ι ( ℓ ) , ι ( ℓ + 1)), and let σ ∈ ι ( ℓ +1) ς ∗ ↾ ( ι ( ℓ + 1) − ⊳ σ and σ ( ι ( ℓ + 1) − 1) = 1. Then x ∗ ς ∗ = x ∗ ς and ρ ( x ∗ ς ∗ , x ∗ σ ) < − n ι ( ℓ +1) − . By ( ⊡ ) we have then ρ ( x ∗ ς + c ι ( ℓ +1) , x ∗ σ + c ι ( ℓ +1) ) = ρ ( x ∗ ς ∗ , x ∗ σ ) < diam ρ (cid:16) U p ι ( ℓ +1) ( n p ι ( ℓ ) − (cid:17) . Consequently,( ⊡ ) U p ι ( ℓ +1) ϕ ι ( ℓ +1) ( x ∗ ς ) ( n p ι ( ℓ ) − 2) = U p ι ( ℓ +1) ϕ ι ( ℓ +1) ( x ∗ σ ) ( n p ι ( ℓ ) − ⊠ ) (b)). It follows from ( ⊡ ) c,d + ( ⊡ ) that for each x ∈ X ι ( ℓ ) \{ x ∗ ς } we have ¯ ℓ ( x, x ∗ ς ) = ¯ ℓ ( x, x ∗ σ ) and ¯ m ( x, x ∗ ς ) = ¯ m ( x, x ∗ σ ), so by ( ⊡ ) ι ( ℓ +1)9 we alsohave( ⊡ ) h p ι ( ℓ +1) (cid:0) ϕ ι ( ℓ +1) ( x ) , ϕ ι ( ℓ +1) ( x ∗ ς ) (cid:1) = h p ι ( ℓ +1) (cid:0) ϕ ι ( ℓ +1) ( x ) , ϕ ι ( ℓ +1) ( x ∗ σ ) (cid:1) .Condition 5.2(A)( ⊠ ) for p ι ( ℓ +1) together with ( ⊡ ) imply now that, letting w ι ( ℓ ) ,σ = (cid:0) π ι ( ℓ ) ,ι ( ℓ +1) (cid:2) w ι ( ℓ ) (cid:3) \ { ϕ ι ( ℓ +1) ( x ∗ ς ) } ) ∪ { ϕ ι ( ℓ +1) ( x ∗ σ ) } , we have( ⊡ ) rk sp (cid:0) w ι ( ℓ ) ,σ (cid:1) = rk sp (cid:0) π ι ( ℓ ) ,ι ( ℓ +1) [ w ι ( ℓ ) ] (cid:1) = rk sp ( w ι ( ℓ ) ), j (cid:0) w ι ( ℓ ) ,σ (cid:1) = j (cid:0) π ι ( ℓ ) ,ι ( ℓ +1) [ w ι ( ℓ ) ] (cid:1) = j ( w ι ( ℓ ) ), and k (cid:0) w ι ( ℓ ) ,σ (cid:1) = k (cid:0) π ι ( ℓ ) ,ι ( ℓ +1) [ w ι ( ℓ ) ] (cid:1) = k ( w ι ( ℓ ) ) = (cid:12)(cid:12) ϕ ι ( ℓ +1) ( x ∗ σ ) ∩ w ι ( ℓ ) ,σ (cid:12)(cid:12) . (Remember, r p ι ( ℓ +1) m ≤ n p ι ( ℓ ) − m = h p ι ( ℓ +1) ( α, β ), α, β ∈ π ι ( ℓ ) ,ι ( ℓ +1) (cid:2) w ι ( ℓ ) (cid:3) are distinct.) Consequently, if rk sp ( w ι ( ℓ ) ) ≥ sp ( w ι ( ℓ +1) ) ≤ rk sp (cid:16) π ι ( ℓ ) ,ι ( ℓ +1) (cid:2) w ι ( ℓ ) (cid:3) ∪ { ϕ ι ( ℓ +1) ( x ∗ σ ) } (cid:17) < rk sp ( w ι ( ℓ ) ) . Unfixing ℓ < ω , we see that for some ℓ ∗ we have rk sp ( w ι ( ℓ ∗ ) ) = − 1. However, ap-plying to ℓ ∗ the procedure described above we get σ ∈ ι ( ℓ ∗ +1) ϕ ι ( ℓ +1) ( x ∗ σ )contradicts clause 5.2(A)( ⊠ ) for p ι ( ℓ +1) (remember ( ⊡ ) + ( ⊡ ) ). Case: The set B is infinite.Almost identical to the previous case. Defining ϕ ι we use the condition c ι − x ∈ U p ι ϕ ι ( x ) ( n p ι − ⊡ ) we have c ι ′ − x = ( c ι − x ) + ( c ι ′ − c ι ) ∈ (cid:16) U p ι ′ α ( n p ι − 2) + ( c ι ′ − c ι ) (cid:17) ∩ U p ι ′ π ι,ι ′ ( α ) ( n p ι − = ∅ (where α = ϕ ι ( x ) ∈ w ι ). (cid:3) The following theorem is a consequence of results presented in this section. Theorem 5.12. Assume that (1) ( H , + , is an Abelian perfect Polish group, (2) the set of elements of H of order larger than 2 is dense in H , (3) 2 ≤ k < ω and (4) ε < ω and λ is an uncountable cardinal such that NPr ε ( λ ) holds true.Then there is a ccc forcing notion P of cardinality λ such that (cid:13) P “ for some Σ subset B of H we have:there is a set X ⊆ H of cardinality λ such that (cid:0) ∀ x, y ∈ X (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ) ∩ ( y + B ) (cid:12)(cid:12) ≥ k (cid:1) but there is no perfect set P ⊆ H such that (cid:0) ∀ x, y ∈ P (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ) ∩ ( y + B ) (cid:12)(cid:12) ≥ k (cid:1) ”. Forcing for groups with all elements of order ≤ H of order larger than 2 is NOT dense in H .Let H = { a ∈ H : a + a = 0 } , so H is a closed subgroup of H and its complement H \ H is not dense in H . Consequently, the interior of H is not empty. Each cosetof H is closed and has a nonempty interior and consequently H /H is countable.Suppose that T ⊆ H is a Borel set with λ many κ –overlapping translationsbut without a perfect set of such translations. Then T is also a Borel subset of H and it still has λ many κ –overlapping translations. If P ⊆ H is a perfect set,then (as | H /H | ≤ ω ) for some a ∈ H the intersection P ∩ ( H + a ) is uncountable.Consider Q = (cid:0) P ∩ ( H + a ) (cid:1) − a ⊆ H — it is a closed uncountable subset of H (socontains a perfect set) and by the assumptions on T there are c, d ∈ Q such that (cid:12)(cid:12) ( T + c ) ∩ ( T + d ) (cid:12)(cid:12) < κ . Then c + a, d + a ∈ P and (cid:12)(cid:12) ( T + ( c + a )) ∩ ( T + ( d + a )) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:0) ( T + c ) ∩ ( T + d ) (cid:1) + a (cid:12)(cid:12) < κ .Consequently, to completely answer the problem of Borel sets with non–disjointtranslations it is enough to deal with the case of all elements of H being of order ≤ 2. The arguments in this case are similar to those from Section 5, but they aresimpler. However, there is one substantial difference. If H is a Polish group with OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 31 all elements of order ≤ B ⊆ H is an uncountable Borel set, then B has aperfect set of pairwise 2–overlapping translations. Namely, choosing a perfect set P ⊆ B we will have x + y, ∈ ( B + x ) ∩ ( B + y ) for each x, y ∈ P . Moreover, if x + b = y + b , then also x + b = y + b . Therefore, if x = y and ( B + x ) ∩ ( B + y )is finite, then | ( B + x ) ∩ ( B + y ) | must be even. For that reason the meaning of k in our forcing here will be slightly different: the translations of the new Borel setwill have at least 2 k elements. Assumption 6.1. In the rest of the section we assume the following:(1) ( H , + , D , ρ, ρ ∗ and U are as in Assumption 1.1.(2) All elements of H have orders at most 2.(3) 1 < k < ω .(4) ε is a countable ordinal and λ is an uncountable cardinal such that NPr ε ( λ )holds true. The model M ( ε, λ ) and functions rk sp , j and k on [ λ ] <ω \ {∅} are as fixed in Definition 2.6.In groups with all elements of order two we should use a weaker notion of inde-pendence. Definition 6.2. Let ( H , + , 0) be an Abelian group(1) A set B ⊆ H is quasi − independent in H if | B | ≥ b , b , b , . . . , b ∈ B and any e , e , e , . . . , e ∈ { , } not all equal 0, wehave e b + e b + e b + e b + e b + e b + e b + e b = 0 . (2) A family { V i : i ≤ n } of disjoint subsets of H is a qif − if for each choice of b i ∈ V i , i ≤ n , the set { b i : i ≤ n } is quasi − independent. Proposition 6.3. Assume that (i) ( H , + , is an Abelian perfect Polish group, (ii) U , . . . , U n are nonempty open subsets of H , n ≥ .Then there are non-empty open sets V i ⊆ U i (for i ≤ n ) such that { V i : i ≤ n } is aqif − .Proof. Similar to Proposition 3.3. (cid:3) The forcing notion used in the case of groups with all elements of order ≤ − . (There are no 8–goodqifs in the current case.) Since in the current case, a − b = a + b for a, b ∈ H , westill can repeat all needed ingredients of Section 5. To stress the importance of thisproperty we will consistently use the addition + rather than subtraction − . Definition 6.4. (A) Let Q be the collection of all tuples p = (cid:0) w p , M p , ¯ r p , n p , ¯Υ p , ¯ V p , h p (cid:1) = (cid:0) w, M, ¯ r, n, ¯Υ , ¯ V , h (cid:1) such that the following demands ( ⊗ ) –( ⊗ ) are satisfied.( ⊗ ) w ∈ [ λ ] <ω , | w | ≥ 4, 0 < M < ω , 3 ≤ n < ω and ¯ r = h r m : m < M i where r m ≤ n − m < M .( ⊗ ) ¯Υ = h ¯ U α : α ∈ w i where each ¯ U α = h U α ( ℓ ) : ℓ ≤ n i is a ⊆ –decreasingsequence of elements of U . ( ⊗ ) ¯ V = h Q i,α,β , V i,α,β , W i,α,β : i < k, ( α, β ) ∈ w h i i ⊆ U and Q i,α,β = Q i,β,α ⊇ V i,α,β = V i,β,α ⊇ W i,α,β = W i,β,α for all i < k and ( α, β ) ∈ w h i .( ⊗ ) (a) The indexed family h U α ( n − 2) : α ∈ w i ⌢ h Q i,α,β : i < k, α, β ∈ w, α < β i is a qif − (so in particular the sets in this system are pairwisedisjoint), and(b) h U α ( n ) : α ∈ w i ⌢ h W i,α,β : i < k, α, β ∈ w, α < β i is immersedin h U α ( n − 1) : α ∈ w i ⌢ h V i,α,β : i < k, α, β ∈ w, α < β i and h U α ( n − 1) : α ∈ w i ⌢ h V i,α,β : i < k, α, β ∈ w, α < β i is immersed in h U α ( n − 2) : α ∈ w i ⌢ h Q i,α,β : i < k, α, β ∈ w, α < β i .( ⊗ ) (a) If α, β ∈ w , ℓ ≤ n and U α ( ℓ ) ∩ U β ( ℓ ) = ∅ , then U α ( ℓ ) = U β ( ℓ ), and(b) if α, β, γ ∈ w , ℓ ≤ n , U α ( ℓ ) = U β ( ℓ ) and a ∈ U α ( ℓ ), b ∈ U β ( ℓ ), then ρ ( a, b ) > diam ρ (cid:0) U γ ( ℓ ) (cid:1) .( ⊗ ) h : w h i onto −→ M is such that h ( α, β ) = h ( β, α ) for ( α, β ) ∈ w h i .( ⊗ ) Assume that u, u ′ ⊆ w , π and ℓ ≤ n are such that • ≤ | u | = | u ′ | and π : u −→ u ′ is a bijection, • r h ( α,β ) ≤ ℓ for all ( α, β ) ∈ u h i , • U α ( ℓ ) ∩ U β ( ℓ ) = ∅ and h ( α, β ) = h ( π ( α ) , π ( β )) for all distinct α, β ∈ u , • for some c ∈ H , for all α ∈ u , we have (cid:0) U α ( ℓ ) + c (cid:1) ∩ U π ( α ) ( ℓ ) = ∅ .Then rk sp ( u ) = rk sp ( u ′ ), j ( u ) = j ( u ′ ), k ( u ) = k ( u ′ ) and for α ∈ u | α ∩ u | = k ( u ) ⇔ | π ( α ) ∩ u ′ | = k ( u ) . ( ⊗ ) Assume that • ∅ 6 = u ⊆ w , rk sp ( u ) = − ℓ ≤ n and • α ∈ u is such that | α ∩ u | = k ( u ), and • r h ( β,β ′ ) ≤ ℓ and U β ( ℓ ) ∩ U β ′ ( ℓ ) = ∅ for all ( β, β ′ ) ∈ u h i .Then there is no α ′ ∈ w \ u such that U α ( ℓ ) = U α ′ ( ℓ ) and h ( α, β ) = h ( α ′ , β )for all β ∈ u \ { α } . (B) For p ∈ Q and m < M p we define F ( p, m ) = [ (cid:8) U pα ( n p ) + W pi,α,β : ( α, β ) ∈ ( w p ) h i ∧ i < k ∧ h p ( α, β ) = m (cid:9) . (C) For p, q ∈ Q we declare that p ≤ q if and only if • w p ⊆ w q , M p ≤ M q , ¯ r q ↾ M p = ¯ r p , n p ≤ n q , h q ↾ ( w p ) h i = h p , and • if α ∈ w p and ℓ ≤ n p then U qα ( ℓ ) = U pα ( ℓ ), and • if ( α, β ) ∈ ( w p ) h i , i < k , then Q qi,α,β ⊆ Q pi,α,β , V qi,α,β ⊆ V pi,α,β , and W qi,α,β ⊆ W pi,α,β , and • if m < M p , then F ( q, m ) ⊆ F ( p, m ). Lemma 6.5. (1) ( Q , ≤ ) is a partial order of size λ . (2) The following sets are dense in Q : (i) D γ,M,n = (cid:8) p ∈ Q : γ ∈ u p ∧ M p > M ∧ n p > n (cid:9) for γ < λ and M, n < ω . (ii) D N = (cid:8) p ∈ Q : diam ρ ∗ ( U pα ( n p − < − N ∧ diam ρ ∗ ( Q pi,α,β ) < − N ∧ diam ρ ∗ ( U pα ( n p − Q pi,α,β ) < − N for all i < k, ( α, β ) ∈ ( w p ) h i (cid:9) for N < ω . (iii) D N = (cid:8) p ∈ Q : for all i, j < k and ( α, β ) , ( γ, δ ) ∈ ( w p ) h i it holds that diam ρ ( U pα ( n p − < − N and diam ρ ( Q pi,α,β ) < − N and diam ρ ( U pα ( n p − 2) + Q pi,α,β ) < − N andif ( i, α ∗ , α, β ) = ( j, γ ∗ , γ, δ ) then OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 33 (cid:0) U pα ∗ ( n p ) + W pi,α,β (cid:1) ∩ (cid:0) U pγ ∗ ( n p ) + W pi,γ,δ (cid:1) = ∅ (cid:9) .for N < ω . (iv) D N = (cid:8) p ∈ D N : h U pα ( n − 3) : α ∈ w p i is a qif − and h U pα ( n − 2) : α ∈ w p i is immersed in it (cid:9) . (3) Assume p ∈ Q . Then there is q ≥ p such that n q ≥ n p + 3 , w q = w p and • for all α ∈ w p , cl (cid:0) U qα ( n q − (cid:1) ⊆ U pα ( n p ) , and • for all i < k and ( α, β ) ∈ ( w p ) h i , cl (cid:0) U qα ( n q − 2) + Q qi,α,β (cid:1) ⊆ U pα ( n p ) + W pi,α,β and cl (cid:0) Q qi,α,β (cid:1) ⊆ W pi,α,β . Proof. Same as for 5.3 (just using Proposition 6.3). (cid:3) Lemma 6.6. Suppose that p ∈ D and α, β, γ, δ ∈ w p are such that α = β . If (cid:16) U pα ( n p − 2) + U pβ ( n p − (cid:17) ∩ (cid:16) U pγ ( n p − 2) + U pδ ( n p − (cid:17) = ∅ , then { α, β } = { γ, δ } .Proof. Similar to 5.4, remembering h U pα ( n − 2) : α ∈ w p i is immersed in a qif − h U pα ( n − 3) : α ∈ w p i ; see 6.5(2)(iv). (cid:3) Lemma 6.7. The forcing notion Q has the Knaster property.Proof. Same as 5.5. (cid:3) Lemma 6.8. For each ( α, β ) ∈ λ h i and i < k , (cid:13) Q “ the sets \ (cid:8) U pα ( n p ) : p ∈ G ˜ Q ∧ α ∈ w p (cid:9) and \ (cid:8) W pi,α,β : p ∈ G ˜ Q ∧ α, β ∈ w p (cid:9) have exactly one element each. ”Proof. Follows from Lemma 6.5. (cid:3) Definition 6.9. (1) For ( α, β ) ∈ λ h i and i < k let η ˜ α , ν ˜ i,α,β and h ˜ α,β be Q –names such that (cid:13) Q “ { η ˜ α } = \ (cid:8) U pα ( n p ) : p ∈ G ˜ Q ∧ α ∈ w p (cid:9) , { ν ˜ i,α,β } = \ (cid:8) W pi,α,β : p ∈ G ˜ Q ∧ α, β ∈ w p (cid:9) h ˜ α,β = h p ( α, β ) for some (all) p ∈ G ˜ Q such that α, β ∈ w p . ”(2) For m < ω let F ˜ m be a Q –name such that (cid:13) Q “ F ˜ m = \ (cid:8) F ( p, m ) : m < M p ∧ p ∈ G ˜ Q (cid:9) . ” Lemma 6.10. (1) For each m < ω , (cid:13) Q “ F ˜ m is a closed subset of H . ” (2) For i < k and ( α, β ) ∈ λ h i we have (cid:13) Q “ η ˜ α , ν ˜ i,α,β ∈ H , h ˜ α,β < ω, ν ˜ i,α,β = ν ˜ i,β,α and η ˜ α + ν ˜ i,α,β ∈ F ˜ h ˜ α,β . ” (3) (cid:13) Q “ h η ˜ α , ν ˜ i,α,β : i < k, α < β < λ i is quasi − independent.” (4) (cid:13) Q “ ν ˜ ,α,β , . . . , ν ˜ k − ,α,β , ( η α + η β + ν ˜ ,α,β ) , . . . , ( η α + η β + ν ˜ k − ,α,β ) aredistinct elements of (cid:0) η ˜ α + S m<ω F ˜ m (cid:1) ∩ (cid:0) η ˜ β + S m<ω F ˜ m (cid:1) . ”Proof. Should be clear. (cid:3) Lemma 6.11. Let p = ( w, M, ¯ r, n, ¯Υ , ¯ V , h ) ∈ D ⊆ Q (cf. 6.5(iii)) and a ℓ , b ℓ ∈ H and U ℓ , W ℓ ∈ U (for ℓ < ) be such that the following conditions are satisfied. ( ⊛ ) U ℓ ∈ { U α ( n ) : α ∈ w } , W ℓ ∈ { W i,α,β : i < k, ( α, β ) ∈ w h i } (for ℓ < ). ( ⊛ ) ( U ℓ + W ℓ ) ∩ ( U ℓ ′ + W ℓ ′ ) = ∅ for ℓ < ℓ ′ < . ( ⊛ ) a ℓ ∈ U ℓ and b ℓ ∈ W ℓ and a ℓ + b ℓ ∈ S m 4, for some α = α ( ℓ ) , β = β ( ℓ ), and i = i ( ℓ ) we have U ℓ = U α ( n ) and W ℓ = W i,α,β .By assumption ( ⊛ ) we know that0 ∈ U + U + U + U + W + W + W + W . If all of U i ’s are distinct, then 0 ∈ U − + U − + U − + U − + X , where X = { } or X = W i + W j for some i < j < W i = W j or X = W + W + W + W withall W i ’s distinct (remember (*)). This contradicts 6.4(A)( ⊗ ) . Similarly if all W i ’sare distinct.So suppose |{ U , U , U , U }| = 3. Then for some ℓ < ℓ ′ < ∈ U − ℓ + U − ℓ ′ + W + W + W + W ⊆ U −− ℓ + U −− ℓ ′ + X, where X = { } or X = W i + W j for some i < j < W i = W j or X = W + W + W + W with all W i ’s distinct (remember (*)). This again contradicts6.4(A)( ⊗ ) . Similarly if |{ W , W , W , W }| = 3.Consequently, |{ U , U , U , U }| = 2 = |{ W , W , W , W }| . Moreover for somedistinct α, β ∈ w we have |{ ℓ < U ℓ = U α ( n ) }| = |{ ℓ < U ℓ = U β ( n ) }| = 2and for some ( i, γ, δ ) = ( j, ε, ζ ) we have |{ ℓ < W ℓ = W i,γ,δ }| = |{ ℓ < W ℓ = W j,ε,ζ }| = 2 . Now we consider all possible configurations . OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 35 Case 1 U = U , U = U , say they are respectively U α ( n ) and U β ( n ).Necessarily W = W and W = W .If W = W , W = W then recalling (**) above, we also get { γ, δ } = { ε, ζ } = { α, β } and (possibly after interchanging i and j ) W = W i,α,β , W = W j,α,β . Thisgives conclusion (B).If W = W , W = W then again by (**) above, we get { γ, δ } = { ε, ζ } = { α, β } and (possibly after interchanging i and j ) W = W i,α,β , W = W j,α,β . This alsogives conclusion (B). Case 2 U = U , U = U , say they are respectively U α ( n ) and U β ( n ).Necessarily W = W and W = W .If W = W , W = W then recalling (**) above, we also get { γ, δ } = { ε, ζ } = { α, β } . After possibly interchanging i and j , W = W i,α,β , W = W j,α,β and weget conclusion (A).If W = W , W = W then again by (**) , we have { γ, δ } = { ε, ζ } = { α, β } .After possibly interchanging i and j , W = W i,α,β , W = W j,α,β . This leads toconclusion (C). Case 3 U = U , U = U , say they are respectively U α ( n ) and U β ( n ).Necessarily W = W and W = W .If W = W , W = W then like above we get { γ, δ } = { ε, ζ } = { α, β } . Afterpossibly interchanging i and j , W = W i,α,β , W = W j,α,β and we get conclusion(A).If W = W , W = W then we also have { γ, δ } = { ε, ζ } = { α, β } and afterpossibly interchanging i and j , W = W i,α,β , W = W j,α,β . This leads to conclusion(C). (cid:3) Lemma 6.12. Let p = ( w, M, ¯ r, n, ¯Υ , ¯ V , h ) ∈ D and X ⊆ H , | X | ≥ . Supposethat a i ( x, y ) , b i ( x, y ) , U i ( x, y ) and W i ( x, y ) for x, y ∈ X , x = y and i < k satisfythe following demands (i)–(iv) (for all x = y , i = i ′ ). (i) U i ( x, y ) ∈ { U α ( n ) : α ∈ w } , W i ( x, y ) ∈ { W j,α,β : j < k, ( α, β ) ∈ w h i } . (ii) • (cid:0) U i ( x, y ) + W i ( x, y ) (cid:1) ∩ (cid:0) U i ( y, x ) + W i ( y, x ) (cid:1) = ∅ , • (cid:0) U i ( x, y ) + W i ( x, y ) (cid:1) ∩ (cid:0) U i ′ ( x, y ) + W i ′ ( x, y ) (cid:1) = ∅ , • (cid:0) U i ( x, y ) + W i ( x, y ) (cid:1) ∩ (cid:0) U i ′ ( y, x ) + W i ′ ( y, x ) (cid:1) = ∅ . (iii) a i ( x, y ) ∈ U i ( x, y ) and b i ( x, y ) ∈ W i ( x, y ) , and a i ( x, y ) + b i ( x, y ) ∈ S m 2) + U β ( n − 2) : α, β ∈ w (cid:9) . (2) If ( x, y ) ∈ X h i and x + y ∈ U α ( n − U β ( n − , α, β ∈ w , then α = β andfor each i < k we have a i ( x, y ) + b i ( x, y ) , a i ( y, x ) + b i ( y, x ) ∈ F ( p, h ( α, β )) .Proof. (1) Fix x, y ∈ X , x = y , for a moment.Let i = i ′ , i, i ′ < k . We may apply Lemma 6.11 for U i ( x, y ), W i ( x, y ), U i ( y, x ), W i ( y, x ), a i ( x, y ), b i ( x, y ), a i ( y, x ), b i ( y, x ) here as U , W , U , W , a , b , a , b thereand for similar objects with i ′ in place of i as U , W , U , W , a , b , a , b there.This will produce distinct α = α ( x, y, i, i ′ ) , β = β ( x, y, i, i ′ ) ∈ w and distinct j = j ( x, y, i, i ′ ) , j ′ = j ′ ( x, y, i, i ′ ) < k such thateither ( A ) α,β,j,j ′ x,y,i,i ′ : (cid:8) { U i ( x, y ) + W i ( x, y ) , U i ( y, x ) + W i ( y, x ) } , { U i ′ ( x, y ) + W i ′ ( x, y ) , U i ′ ( y, x ) + W i ′ ( y, x ) } (cid:9) = (cid:8) { U α ( n ) + W j,α,β , U β ( n ) + W j,α,β } , { U α ( n ) + W j ′ ,α,β , U β ( n ) + W j ′ ,α,β } (cid:9) , or ( B ) α,β,j,j ′ x,y,i,i ′ : (cid:8) { U i ( x, y ) + W i ( x, y ) , U i ( y, x ) + W i ( y, x ) } , { U i ′ ( x, y ) + W i ′ ( x, y ) , U i ′ ( y, x ) + W i ′ ( y, x ) } (cid:9) = (cid:8) { U α ( n ) + W j,α,β , U α ( n ) + W j ′ ,α,β } , { U β ( n ) + W j,α,β , U β ( n ) + W j ′ ,α,β } (cid:9) , or ( C ) α,β,j,j ′ x,y,i,i ′ : (cid:8) { U i ( x, y ) + W i ( x, y ) , U i ( y, x ) + W i ( y, x ) } , { U i ′ ( x, y ) + W i ′ ( x, y ) , U i ′ ( y, x ) + W i ′ ( y, x ) } (cid:9) = (cid:8) { U α ( n ) + W j,α,β , U β ( n ) + W j ′ ,α,β } , { U α ( n ) + W j ′ ,α,β , U β ( n ) + W j,α,β } (cid:9) . Plainly,( ⊙ ) x,y if for some i = i ′ and α, β, j, j ′ the clause ( A ) α,β,j,j ′ x,y,i,i ′ holds true,then x + y ∈ U α ( n − 2) + U β ( n − x = y and distinct i, i ′ , i ′′ ,( ⊙ ) if ( B ) α,β,j,j ′ x,y,i,i ′ holds true, then also ( B ) α,β,j,j ′ x,y,i,i ′′ holds true,and( ⊙ ) if ( C ) α,β,j,j ′ x,y,i,i ′ holds true, then also ( C ) α,β,j,j ′ x,y,i,i ′′ holds true,Consequently, if k ≥ k ≥ 3) for any x = y from X neither of possibilities ( B ) α,β,j,j ′ x,y,i,i ′ nor ( C ) α,β,j,j ′ x,y,i,i ′ can hold. Therefore wemay easily finish the proof of Lemma 6.12 (when k ≥ k = 2. For each x = y from X we fix α = α ( x, y ) and β = β ( x, y ) suchthat either ( A ) α,β, , x,y, , or ( B ) α,β, , x,y, , or ( C ) α,β, , x,y, , . Let χ ( x, y ) = χ ( y, x ) ∈ { A, B, C } and θ ( x, y ) = θ ( y, x ) ∈ [ w ]2 be such that (cid:0) χ ( x, y ) (cid:1) θ ( x,y ) , , x,y, , holds true. Claim 6.12.1. If x, y, z ∈ X are distinct and χ ( x, y ) = χ ( y, z ) = A , then χ ( x, z ) = A .Proof of the Claim. Let χ ( x, y ) = A = χ ( y, z ) and θ ( x, y ) = { α, β } , θ ( y, z ) = { γ, δ } . Assume towards contradiction that χ ( x, z ) ∈ { B, C } and let θ ( x, z ) = { ξ, ζ } .Then for some ξ ′ , ζ ′ ∈ { ξ, ζ } we have x + z ∈ U ξ ′ ( n ) + U ζ ′ ( n ) + W ,ξ,ζ + W ,ξ,ζ . • If |{ α, β } ∩ { γ, δ }| = 1, say α = γ , β = δ , and { ξ, ζ } = { ξ ′ , ζ ′ } = { β, δ } , then x + z ∈ U α ( n ) + U α ( n ) + U β ( n ) + U δ ( n ) + W i,α,β + W i,α,β + W j,α,δ + W j,α,δ but also x + z ∈ U β ( n ) + U δ ( n ) + W ,β,δ + W ,β,δ . Consequently,0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U β ( n ) (cid:1) + (cid:0) U β ( n ) + ( U δ ( n ) + U δ ( n )) (cid:1) + (cid:0) ( W i,α,β + W i,α,β ) + W ,β,δ (cid:1) + (cid:0) ( W j,α,δ + W j,α,δ ) + W ,β,δ (cid:1) ⊆ U β ( n − 1) + U β ( n − 1) + V ,β,δ + V ,β,δ ⊆ Q ,β,δ + Q ,β,δ . This immediately contradicts 6.4(A)( ⊗ ) . • If |{ α, β } ∩ { γ, δ }| = 1, say α = γ , β = δ , and ξ ′ = ζ ′ , |{ ξ ′ , ζ ′ } ∩ { β, δ }| = 1, say ξ ′ = β , then by similar considerations we arrive to0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U δ ( n ) (cid:1) + (cid:0) ( U β ( n ) + U β ( n )) + U ζ ′ ( n ) (cid:1) + (cid:0) ( W i,α,β + W i,α,β ) + W ,β,ζ ′ (cid:1) + (cid:0) ( W j,α,δ + W j,α,δ ) + W ,β,ζ ′ (cid:1) ⊆ U δ ( n − 1) + U ζ ′ ( n − 1) + V ,β,ζ ′ + V ,β,ζ ′ . OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 37 In our case necessarily δ = ζ ′ so we easily get contradiction with 6.4(A)( ⊗ ) . • If |{ α, β } ∩ { γ, δ }| = 1, say α = γ , β = δ , and ξ ′ = ζ ′ and { ξ ′ , ζ ′ } ∩ { β, δ } = ∅ ,then 0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U β ( n ) (cid:1) + U δ ( n ) + U ξ ( n ) + U ζ ( n )+ (cid:0) ( W i,α,β + W i,α,β ) + W ,ξ,ζ (cid:1) + (cid:0) ( W j,α,δ + W j,α,δ ) + W ,ξ,ζ (cid:1) ⊆ U β ( n − 1) + U δ ( n ) + U ξ ( n ) + U ζ ( n ) + V ,ξ,ζ + V ,ξ,ζ , and β, δ, ξ, ζ are all pairwise distinct. This again contradicts 6.4(A)( ⊗ ) . • If |{ α, β } ∩ { γ, δ }| = 1, say α = γ , β = δ , and ξ ′ = ζ ′ , then0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U β ( n ) (cid:1) + (cid:0) ( U ξ ′ ( n ) + U ξ ′ ( n )) + U δ ( n ) (cid:1) + (cid:0) ( W i,α,β + W i,α,β ) + W ,ξ,ζ (cid:1) + (cid:0) ( W j,α,δ + W j,α,δ ) + W ,ξ,ζ (cid:1) ⊆ U β ( n − 1) + U δ ( n − 1) + V ,ξ,ζ + V ,ξ,ζ , and β = δ . Again contradiction with 6.4(A)( ⊗ ) . • If { α, β } = { γ, δ } , then we arrive to0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U ξ ′ ( n ) (cid:1) + (cid:0) ( U β ( n ) + U β ( n )) + U ζ ′ ( n ) (cid:1) + (cid:0) ( W i,α,β + W i,α,β ) + W ,ξ,ζ (cid:1) + (cid:0) ( W j,α,β + W j,α,β ) + W ,ξ,ζ (cid:1) ⊆ U ξ ′ ( n − 1) + U ζ ′ ( n − 1) + V ,ξ,ζ + V ,ξ,ζ . Considering cases ξ ′ = ζ ′ and ξ ′ = ζ ′ separately we easily get a contradiction with6.4(A)( ⊗ ) . • If { α, β } ∩ { γ, δ } = ∅ , then0 ∈ (cid:0) ( W i,α,β + W i,α,β ) + U α ( n ) (cid:1) + (cid:0) ( W j,α,δ + W j,α,δ ) + U β ( n ) (cid:1) + U γ ( n ) + U δ ( n ) + U ξ ′ ( n ) + U ζ ′ ( n ) + W ,ξ,ζ + W ,ξ,ζ ⊆ U α ( n − 1) + U β ( n − 1) + U γ ( n ) + U δ ( n ) + U ξ ′ ( n ) + U ζ ′ ( n ) + W ,ξ,ζ + W ,ξ,ζ . If ξ ′ = ζ ′ then this gives0 ∈ U α ( n − 1) + U β ( n − 1) + U γ ( n ) + U δ ( n − 1) + W ,ξ,ζ + W ,ξ,ζ , a contradiction. So ξ ′ = ζ ′ and we ask what is the intersection { ξ ′ , ζ ′ } ∩ { α, β, γ, δ } .In each possible case we also get a contradiction. (cid:3) Claim 6.12.2. If χ ( x, y ) = A and z ∈ X \ { x, y } , then either χ ( x, z ) = A or θ ( x, z ) = θ ( x, y ) .Proof of the Claim. Suppose χ ( x, y ) = χ ( x, z ) = A and θ ( x, y ) = θ ( x, z ) = { α, β } .By 6.12.1 we know that χ ( y, z ) = A . Hence for some ξ = ζ and i < y + z ∈ U ξ ( n ) + U ζ ( n ) + W i,ξ,ζ + W i,ξ,ζ . Also, y + z = y + x + x + z ∈ U α ( n )+ U β ( n )+ W ,α,β + W ,α,β + U α ( n )+ U β ( n )+ W ,α,β + W ,α,β . Hence 0 ∈ U ξ ( n − 1) + U ζ ( n − 1) + V i,ξ,ζ + V i,ξ,ζ , and we get contradiction asusual. (cid:3) Claim 6.12.3. χ ( x, y ) = B for any distinct x, y ∈ X .Proof of the Claim. Suppose χ ( x, y ) = B , θ ( x, y ) = { α, β } . By 6.12.2 we maychoose z ∈ X \ { x, y } such that (cid:0) χ ( x, z ) , θ ( x, z ) (cid:1) = (cid:0) A, { α, β } (cid:1) = (cid:0) χ ( y, z ) , θ ( y, z ) (cid:1) (remember | X | ≥ χ ( x, z ) = A or χ ( y, z ) = A ; bythe symmetry we may assume χ ( x, z ) = A . Now we consider the other possibilities for the value of χ ( x, z ).(i) If χ ( x, z ) = B and θ ( x, z ) = { α, β } , then x + y, x + z ∈ U α ( n ) + U α ( n ) + W ,α,β + W ,α,β . Hence y + z ∈ U α ( n − 1) + U α ( n − 1) + U α ( n ) + U α ( n ). Also, for some ξ ′ , ζ ′ ∈ θ ( y, z ) = { ξ, ζ } and i, j < y + z ∈ U ξ ′ ( n ) + U ζ ′ ( n ) + W i,ξ,ζ + W j,ξ,ζ , where either ξ ′ = ζ ′ or i = j . Thus0 ∈ U α ( n − 1) + U α ( n − 1) + U α ( n ) + U α ( n ) + U ξ ′ ( n ) + U ζ ′ ( n ) + W i,ξ,ζ + W j,ξ,ζ def = Y. If ξ ′ = ζ ′ then i = j and Y ⊆ U α ( n − 1) + U α ( n − 1) + U α ( n − 1) + U α ( n ) + W i,ξ,ζ + W j,ξ,ζ ⊆ Q ,ξ,ζ + Q ,ξ,ζ , and we get a contradiction with 6.4(A)( ⊗ ) . If ξ ′ = ζ ′ then Y ⊆ U ξ ′ ( n − 2) + U ζ ′ ( n − 1) + W i,ξ,ζ + W j,ξ,ζ and regardless of i being equal to j or not, we may get a contradiction too.(ii) If χ ( x, z ) = B and θ ( x, z ) = { γ, δ } 6 = { α, β } , then x + z ∈ U γ ( n ) + U γ ( n ) + W ,γ,δ + W ,γ,δ and x + y ∈ U α ( n ) + U α ( n ) + W ,α,β + W ,α,β . Hence y + z ∈ V ,γ,δ + W ,γ,δ + V ,α,β + W ,α,β . Like before, for some ξ ′ , ζ ′ ∈ θ ( y, z ) = { ξ, ζ } and i, j < y + z ∈ U ξ ′ ( n ) + U ζ ′ ( n ) + W i,ξ,ζ + W j,ξ,ζ , where either ξ ′ = ζ ′ or i = j . Since { V ,γ,δ , V ,γ,δ } ∩ { V ,α,β , V ,α,β } = ∅ , like beforewe get a contradiction with 6.4(A)( ⊗ ) .(iii) If χ ( x, z ) = C and θ ( x, z ) = { α, β } , then x + y ∈ U α ( n ) + U α ( n ) + W ,α,β + W ,α,β x + z ∈ U α ( n ) + U β ( n ) + W ,α,β + W ,α,β . Also, y + z ∈ U ξ ′ ( n )+ U ζ ′ ( n )+ W i,ξ,ζ + W j,ξ,ζ , where ξ ′ , ζ ′ ∈ θ ( y, z ) = { ξ, ζ } , i, j < ξ ′ = ζ ′ or i = j . We consider 2 subcases now.If i = j then ( ξ ′ = ζ ′ and) χ ( y, z ) = A so by the choice of z at the beginning weknow that θ ( y, z ) = { α, β } . So we arrive to0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U α ( n ) (cid:1) + (cid:0) ( W i,ξ,ζ + W i,ξ,ζ ) + U β ( n ) (cid:1) + (cid:0) ( W ,α,β + W ,α,β ) + U ξ ( n ) (cid:1) + (cid:0) ( W ,α,β + W ,α,β ) + U ζ ( n ) (cid:1) ⊆ U α ( n − 1) + U β ( n − 1) + U ξ ( n − 1) + U ζ ( n − { ξ, ζ } 6 = { α, β } a contradiction follows.If i = j then we get0 ∈ (cid:0) ( U α ( n ) + U α ( n )) + U α ( n ) (cid:1) + (cid:0) ( W ,ξ,ζ + W ,ξ,ζ ) + U β ( n ) (cid:1) + (cid:0) ( W ,α,β + W ,α,β ) + U ξ ′ ( n ) (cid:1) + U ζ ′ ( n ) + W ,ξ,ζ + W ,ξ,ζ ⊆ U α ( n − 1) + U β ( n − 1) + U ξ ′ ( n − 1) + U ζ ′ ( n ) + W ,ξ,ζ + W ,ξ,ζ , OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 39 and again a contradiction.(iv) If χ ( x, z ) = C and θ ( x, z ) = { γ, δ } 6 = { α, β } , then x + z ∈ U γ ( n ) + U δ ( n ) + W ,γ,δ + W ,γ,δ and x + y ∈ U α ( n ) + U α ( n ) + W ,α,β + W ,α,β , and y + z ∈ U ξ ′ ( n ) + U ζ ′ ( n ) + W i,ξ,ζ + W j,ξ,ζ , where ξ ′ , ζ ′ ∈ θ ( y, z ) = { ξ, ζ } , i, j < ξ ′ = ζ ′ or i = j . Thus0 ∈ (cid:0) U γ ( n ) + ( U α ( n ) + U α ( n )) (cid:1) + U δ ( n ) + W ,γ,δ + W ,γ,δ + W ,α,β + W ,α,β + U ξ ′ ( n ) + U ζ ′ ( n ) + W i,ξ,ζ + W j,ξ,ζ ⊆ U α ( n − 1) + U δ ( n ) + U ξ ′ ( n ) + U ζ ′ ( n )+ W ,γ,δ + W ,γ,δ + W ,α,β + W ,α,β + W i,ξ,ζ + W j,ξ,ζ . Since W ,γ,δ , W ,γ,δ , W ,α,β and W ,α,β are all distinct we get a contradiction in theusual manner. (cid:3) Claim 6.12.4. χ ( x, y ) = C for any distinct x, y ∈ X .Proof of the Claim. Suppose towards contradiction χ ( x, y ) = C and let θ ( x, y ) = { α, β } . Let z ∈ X \{ x, y } . By 6.12.1 we know that either χ ( x, z ) = A or χ ( y, z ) = A ;by the symmetry we may assume χ ( x, z ) = A . By 6.12.3 we know that χ ( x, z ) = B ,so χ ( x, z ) = C If θ ( x, y ) = θ ( x, z ) = { α, β } , then y + z ∈ U α ( n − U α ( n )+ U β ( n − U β ( n ). Weknow that χ ( y, z ) ∈ { A, C } and in both cases y + z ∈ U γ ( n )+ U δ ( n )+ W i,γ,δ + W j,γ,δ ,where θ ( y, z ) = { γ, δ } and i, j < 2. Now we may conclude0 ∈ U α ( n − 1) + U α ( n ) + U β ( n − 1) + U β ( n ) + U γ ( n ) + U δ ( n ) + W i,γδ + W j,γδ def = S. If i = j then S ⊆ U γ ( n − 2) + U δ ( n − 2) + W ,γδ + W ,γδ and an immediatecontradiction with 6.4(A)( ⊗ ) follows. If i = j then S ⊆ U γ ( n − 2) + U δ ( n − 2) + W i,γδ + W i,γδ ⊆ U γ ( n − 3) + U δ ( n − 3) and we get a contradiction with p ∈ D .If θ ( x, z ) = { ξ, ζ } 6 = θ ( x, y ) = { α, β } , then { W ,ξ,ζ , W ,ξ,ζ } ∩ { W ,α,β , W ,α,β } = ∅ and y + z ∈ U α ( n ) + U β ( n ) + W ,α,β + W ,α,β + U ξ ( n ) + U ζ ( n ) + W ,ξ,ζ + W ,ξ,ζ . Now, by considerations as before, we get a contradiction with 6.4(A)( ⊗ ) . (cid:3) Therefore,( ⊙ ) χ ( x, y ) = A for all distinct x, y ∈ X .Hence, if x = y are from X and θ ( x, y ) = { α, β } , then x + y ∈ U α ( n − U β ( n − (cid:3) Lemma 6.13. Let p = ( w, M, ¯ r, n, ¯Υ , ¯ V , h ) ∈ D and X ⊆ H , | X | ≥ . Supposethat (a) X + X ⊆ S { U α ( n ) + U β ( n ) : α, β ∈ w } , and (b) diam ρ (cid:0) U α ( n ) (cid:1) < ρ ( x, y ) for all α ∈ w , ( x, y ) ∈ X h i .Then there is a c ∈ H such that X + c ⊆ [ { U α ( n − 1) : α ∈ w } . Proof. By assumption (b), if x, y ∈ X are distinct and x + y ∈ U α ( n )+ U β ( n ), α, β ∈ w , then α = β . Also, if ( x, y ) ∈ X h i and x + y ∈ (cid:0) U α ( n )+ U β ( n ) (cid:1) ∩ (cid:0) U γ ( n )+ U δ ( n ) (cid:1) ,then { α, β } = { γ, δ } (by 6.4(A)( ⊗ ) ). Consequently, for each ( x, y ) ∈ X h i we maylet θ ( x, y ) be the unique { α, β } ∈ [ w ]2 such that x + y ∈ U α ( n ) + U β ( n ). Claim 6.13.1. | θ ( x, y ) ∩ θ ( x, z ) | = 1 whenever x, y, z ∈ X are distinct.Proof of the Claim. Let θ ( x, y ) = { α, β } , θ ( x, z ) = { γ, δ } and θ ( y, z ) = { ξ, ζ } .Then y + z ∈ (cid:16) U α ( n ) + U β ( n ) + U γ ( n ) + U δ ( n ) (cid:17) ∩ (cid:0) U ξ ( n ) + U ζ ( n ) (cid:1) . Hence 0 ∈ U α ( n ) + U β ( n ) + U γ ( n ) + U δ ( n ) + U ξ ( n ) + U ζ ( n ). Since α = β , γ = δ and ξ = ζ we conclude that { α, β } ∩ { γ, δ } 6 = ∅ (remember 6.4(A)( ⊗ ) ). If we had { α, β } = { γ, δ } , then 0 ∈ U ξ ( n − U ζ ( n − |{ α, β } ∩ { γ, δ }| = 1. (cid:3) Fix distinct x , y , z ∈ X . Let θ ( x , y ) = { α , β } , θ ( x , z ) = { γ , α } andlet a ′ , a ′′ ∈ U α ( n ), b ∈ U β ( n ), c ∈ U γ ( n ) be such that x + y = a ′ + b and x + z = a ′′ + c .Let c = a ′ + x . We will show that x + c ∈ S { U α ( n − 1) : α ∈ w } for all x ∈ X .To this end, first note that • x + c = x + a ′ + x = a ′ ∈ U α ( n ), • y + c = y + a ′ + x = a ′ + b + a ′ = b ∈ U β ( n ), • z + c = z + a ′ + x = a ′′ + c + a ′ ∈ U γ ( n ) + ( U α + U α ( n )) ⊆ U γ ( n − x ∈ X \ { x , y , z } . Let θ ( x, x ) = { δ, ζ } , x + x = d + e , d ∈ U δ ( n ), e ∈ U ζ ( n ).(*) α ∈ { δ, ζ } .Why? By Claim 6.13.1 we have | θ ( x , x ) ∩ θ ( x , y ) | = | θ ( x , x ) ∩ θ ( x , z ) | = 1.Hence if α / ∈ { δ, ζ } , then θ ( x, x ) = { β , γ } . Take x ′ ∈ X \ { x , y , z , x } and notethat (again by Claim 6.13.1) (cid:12)(cid:12) θ ( x , x ′ ) ∩ { α , β } (cid:12)(cid:12) = (cid:12)(cid:12) θ ( x , x ′ ) ∩ { α , γ } (cid:12)(cid:12) = (cid:12)(cid:12) θ ( x , x ′ ) ∩ { γ , β } (cid:12)(cid:12) = 1 , and this is clearly impossible.By symmetry we may assume α = δ . But now x + c = x + x + a ′ = ( d + a ′ ) + e ∈ U ζ ( n − , so we are done. (cid:3) Lemma 6.14. (cid:13) Q “ there is no perfect set P ⊆ H such that (cid:16) ∀ x, y ∈ P (cid:17)(cid:16)(cid:12)(cid:12)(cid:12)(cid:0) x + S m<ω F ˜ m (cid:1) ∩ (cid:0) y + S m<ω F ˜ m (cid:1)(cid:12)(cid:12)(cid:12) ≥ k (cid:17) . ”Proof. Suppose towards contradiction that G ⊆ Q is generic over V and in V [ G ]the following assertion holds true: OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 41 for some perfect set P ⊆ H we have (cid:12)(cid:12)(cid:12)(cid:0) x + [ m<ω F ˜ Gm (cid:1) ∩ (cid:0) y + [ m<ω F ˜ Gm (cid:1)(cid:12)(cid:12)(cid:12) ≥ k for all x, y ∈ P .Then for any distinct x, y ∈ P there are c , d , . . . , c k − , d k − ∈ S m<ω F ˜ Gm such that x + y = c i + d i (for all i < k ) and { c i , d i } ∩ { c i ′ , d i ′ } = ∅ (for i < i ′ < k );remember x + y = c i + d i implies that x + c i , x + d i are distinct elements of (cid:0) x + S m<ω F ˜ Gm (cid:1) ∩ (cid:0) y + S m<ω F ˜ Gm (cid:1) . For ¯ ℓ = h ℓ i : i < k i ⊆ ω , ¯ m = h m i : i < k i ⊆ ω and N < ω let Z N ¯ ℓ, ¯ m = (cid:8) ( x, y ) ∈ P : there are c i ∈ F ˜ Gℓ i , d i ∈ F ˜ Gm i (for i < k ) such that x + y = c i + d i and 2 − N < min (cid:0) ρ ( c i , c j ) , ρ ( d i , d j ) , ρ ( c i , d j ) (cid:1) for all distinct i, j < k (cid:9) . Now we continue as in 5.11, but instead of 3.5 we use 6.13. In ( ⊡ ) c ) as there wedemand p ι , q ι ∈ D n ι . Also under current assumptions on H , X ι + c ι = c ι − X ι , sowe have only one case. Otherwise the same proof works. (cid:3) The following theorem is a consequence of results presented in this section. Theorem 6.15. Assume that (1) ( H , + , is an Abelian perfect Polish group, (2) all elements of H have order at most 2, (3) 2 ≤ k < ω and (4) ε < ω and λ is an uncountable cardinal such that NPr ε ( λ ) holds true.Then there is a ccc forcing notion Q of cardinality λ such that (cid:13) Q “ for some Σ subset B of H we have:there is a set X ⊆ H of cardinality λ such that (cid:0) ∀ x, y ∈ X (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ) ∩ ( y + B ) (cid:12)(cid:12) ≥ k (cid:1) but there is no perfect set P ⊆ H such that (cid:0) ∀ x, y ∈ P (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ) ∩ ( y + B ) (cid:12)(cid:12) ≥ k (cid:1) ”. Conclusions and Questions Let us recall from the Introduction, that the spectrum of translation κ –non-disjointness of a set A ⊆ H isstnd κ ( A ) = stnd κ ( A, H ) = { ( x, y ) ∈ H × H : | ( A + x ) ∩ ( A + y ) | ≥ κ } . By the definition, X × X ⊆ stnd κ ( A ) if and only if (cid:0) ∀ x, y ∈ X (cid:1)(cid:0)(cid:12)(cid:12) ( x + A ) ∩ ( y + A ) (cid:12)(cid:12) ≥ κ (cid:1) . In particular, there is a perfect square P × P included in stnd κ ( A ) if and only if A has a perfect set P of κ –overlapping translations. Conclusion . Assume that(a) H = ( H , , +) is an Abelian perfect Polish group,(b) 1 < ι < ω and • k = ι if { c ∈ H : c + c = 0 } is dense in H , and • k = 2 ι otherwise, (c) λ is an uncountable cardinal such that NPr ε ( λ ) holds true for some count-able ordinal ε , and(d) λ = λ ℵ ≤ µ = µ ℵ .Then there is a ccc forcing notion P ∗ and a P ∗ –name B ˜ for a Σ subset of H suchthat(1) (cid:13) P ∗ “ 2 ℵ = µ ”,(2) (cid:13) P ∗ “ there is a set X ⊆ H of cardinality λ such that X × X ⊆ stnd k ( B ˜ ) ”,but(3) (cid:13) P ∗ “ there is no set X ⊆ H of cardinality λ + such that X × X ⊆ stnd k ( B ˜ )”, and(4) (cid:13) P ∗ “ there is no perfect set P ⊆ H such that P × P ⊆ stnd k ( B ˜ ) ”. Proof. Let us consider the case when (in assumption (b) of the Corollary) the set { c ∈ H : c + c = 0 } is dense in H . The other case is fully parallel. So we assume • ( H , + , D , ρ, ρ ∗ and U are as in Assumption 1.1 and Assumption 5.1, • k, ε, λ, rk sp , j , k and µ satisfy Assumption 5.1 and assumption (d) of theCorollary.Let P be the forcing notion discussed in Section 5 (cf Theorem 5.12) and let C µ bethe forcing notion adding µ Cohen reals, where conditions are finite functions withdomains included in µ and values 0 , P ∗ = P × C µ .By standard arguments, P ∗ is a ccc forcing notion and (cid:13) P ∗ ℵ = µ . Let B ˜ be a P –name for the Σ subset of H added by P ⋖ P ∗ . Claim 7.1.1. (2) (cid:13) P ∗ “ there is a set X ⊆ H of cardinality λ such that (cid:0) ∀ x, y ∈ X (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ˜ ) ∩ ( y + B ˜ ) (cid:12)(cid:12) ≥ k (cid:1) ”,but (4) (cid:13) P ∗ “ there is no perfect set P ⊆ H such that (cid:0) ∀ x, y ∈ P (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ˜ ) ∩ ( y + B ˜ ) (cid:12)(cid:12) ≥ k (cid:1) ”.Proof of the Claim. If H ⊆ C µ is generic over V , then in V [ H ] we may look atthe definition of the forcing notion P as all the ingredients still have the requiredproperties. Identifying B V n with B V [ H ] n we easily see that P V = P V [ H ] . Hence P ∗ is equivalent to the iteration C µ ∗ P and consequently the results of Section 5 givethe desired conclusion. (cid:3) Claim 7.1.2. (3) (cid:13) P ∗ “ there is no set X ⊆ H of cardinality λ + such that (cid:0) ∀ x, y ∈ X (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ˜ ) ∩ ( y + B ˜ ) (cid:12)(cid:12) ≥ k (cid:1) ”.Proof of the Claim. Assume λ < µ (otherwise clear). Suppose towards contradic-tion that G = G × G ⊆ P × C µ is generic over V and in V [ G ][ G ] there aredistinct x α ∈ H (for α < λ + ) such that (cid:12)(cid:12) ( x α + B ˜ G ) ∩ ( x β + B ˜ G ) (cid:12)(cid:12) ≥ k for α, β < λ + . Then in V [ G ] we may find a condition q ∈ G and C µ –names x ˜ α , α < λ + , forelements of the group H such that q (cid:13) C µ “ x ˜ α = x ˜ β and (cid:12)(cid:12) ( x ˜ α + B ˜ ) ∩ ( x ˜ β + B ˜ ) (cid:12)(cid:12) ≥ k ” OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 43 for all α < β < λ + . Each of the names x ˜ α is actually a C A α –name for somecountable set A α ⊆ µ . Since V [ G ] | = 2 ℵ = λ , we may choose a set I ∈ [ λ + ] λ + and a set u ⊆ µ such that the following two demands are satisfied (in V [ G ]).( ♣ ) otp( A α ) = otp( A β ) for α, β ∈ I .( ♣ ) For each α < β from I , letting π α,β : A α −→ A β be the order isomorphism,we have u = A α ∩ A β , π α,β ↾ u = id u and A α \ u is infinite.Let u ∗ = u ∪ dom( q ) ⊆ µ . Dismissing finitely many elements of I we may assumethat A α \ u = A α \ u ∗ for all α ∈ I .Let G ∗ = G ∩ C u ∗ and let us work in V [ G ][ G ∗ ] for a moment. Each name x ˜ α (for α ∈ I ) can be thought of as a C A α \ u ∗ –name now. Let ξ = otp( A α \ u ∗ ) forsome (equivalently, all) α ∈ I . Since V [ G ][ G ∗ ] | = 2 ℵ = λ , we may find I ∗ ∈ [ I ] λ + and a Borel function τ : ξ −→ H such that( ♣ ) (cid:13) x ˜ α = τ (cid:0) c ˜ α ◦ π α (cid:1) , where π α : ξ −→ A α \ u ∗ is the order isomorphism and c ˜ α is (a name for) the Cohen real added by C A α \ u ∗ .Consequently, if α = β are from I ∗ , then( ♣ ) (cid:13) C Aα \ u ∗ × C Aβ \ u ∗ “ (cid:12)(cid:12) ( τ ( c ˜ α ◦ π α ) + B ˜ G ) ∩ ( τ ( c ˜ β ◦ π β ) + B ˜ G ) (cid:12)(cid:12) ≥ k and τ (cid:0) c ˜ α ◦ π α (cid:1) = τ (cid:0) c ˜ β ◦ π β (cid:1) ”Therefore,( ♣ ) if d , d ∈ ξ V [ G ][ G ∗ ], then V [ G ][ G ∗ ][ d , d ] | = (cid:12)(cid:12) ( τ ( d ) + B ˜ G ) ∩ ( τ ( d ) + B ˜ G ) (cid:12)(cid:12) ≥ k and τ ( d ) = τ ( d ) . Take α ∈ I and note that in V ∗ = V [ G ][ G ∗ ][ G ∩ C A α \ u ∗ ] there is a perfect set P ⊆ ξ V [ G ][ G ∗ ]. By ( ♣ ) we know V ∗ | = τ ↾ P is one-to-one and (cid:12)(cid:12) ( τ ( x ) + B ˜ G ) ∩ ( τ ( y ) + B ˜ G ) (cid:12)(cid:12) ≥ k for x, y ∈ P. By upward absoluteness of Σ sentences we may assert now that V [ G × G ] | = there is a perfect set P ∗ ⊆ H such that (cid:0) ∀ x, y ∈ P ∗ (cid:1)(cid:0)(cid:12)(cid:12) ( x + B ˜ G ) ∩ ( y + B ˜ G ) (cid:12)(cid:12) ≥ k (cid:1) . This, however, contradicts Claim 7.1.1. (cid:3)(cid:3) Conclusion . Assume that(1) H is a perfect Polish group and B ⊆ H is a Borel set,(2) a cardinal λ is such that Pr ε ( λ ) holds true for every ε < ω , and(3) 1 < k < ω , and(4) there is a set X ⊆ H of cardinality λ such that X × X ⊆ stnd k ( B ).Then there is a perfect set P ⊆ H such that P × P ⊆ stnd k ( B ). Proof. Under our assumptions on λ , if an analytic set B ⊆ ω × ω λ –square, it includes a perfect square (see [14, Claim 1.12(1)]).The space H is Borel isomorphic with ω 2; let f : H −→ ω f : H × H −→ ω × ω x, y ) ( f ( x ) , f ( y )). Then the set f [stnd k ( B )]is analytic and f [ X ] × f [ X ] ⊆ f [stnd k ( B )]. Consequently there is a perfect set P ∗ ⊆ ω P ∗ × P ∗ ⊆ f [stnd k ( B )]. We may choose a perfect set P ⊆ f − [ P ∗ ] ⊆ H – it will also satisfy P × P ⊆ stnd k ( B ). (cid:3) Now, in Claim 7.1.1 we used the upward absoluteness to show (cid:13) P ∗ “ (2) ”. Ifthe group H is compact and B ⊆ H is Σ , then the set stnd k ( B ) is Σ and hencethe assertion in (4) of 7.1 is Π , so also absolute. However, in the case of general H the corresponding assertion appears to be Π so not so obviously absolute. Itsabsoluteness could be establish if we can introduce corresponding rank. (This wouldbe helpful for natural consequences under MA.) Problem 7.3. Develop the rank and the results parallel to ndrk ι and cute YZR –systems presented in [10] for the case of general Abelian perfect Polish groups.The forcing notions presented in this article for various Abelian Polish groupslook similar, but the particular group structures may have different impacts. Problem 7.4. Is it consistent that for some Abelian Polish perfect groups H , H and 2 < k < ω and an uncountable cardinal λ we have:(1) for some Borel set B ⊆ H ,(a) there is a set X ⊆ H of cardinality λ such that X × X ⊆ stnd k ( B , H )(i.e., stnd k ( B , H ) includes a λ –square) , but(b) there is no perfect set P ⊆ H such that P × P ⊆ stnd k ( B , H ) (i.e.,stnd k ( B , H ) does not include any perfect square)and(2) for every Borel set B ⊆ H , if stnd k ( B, H ) includes a λ –square, then itincludes a perfect square ?Considering differences caused by various choices of parameters, it is natural toask about the impact of k . Problem 7.5. Is it consistent that for some Abelian Polish perfect group H and2 < k < ℓ < ω and an uncountable cardinal λ we have:(1) For some Borel set B ⊆ H ,(a) there is a set X ⊆ H of cardinality λ such that X × X ⊆ stnd ℓ ( B , H ),but(b) there is no perfect set P ⊆ H such that P × P ⊆ stnd ℓ ( B , H ).(2) For every Borel set B ⊆ H , if stnd k ( B, H ) includes a λ –square, then itincludes a perfect square.Of course, the next steps could be to investigate stnd ω and stnd ω : Problem 7.6. Let H be a perfect Abelian Polish group. Is it consistent that forsome Borel set B ⊆ H : • there is an uncountable set X ⊆ H such that ( B + x ) ∩ ( B + y ) is uncountablefor every x, y ∈ X , but • for every perfect set P ⊆ H there are x, y ∈ P with ( B + x ) ∩ ( B + y )countable?Similarly if “uncountable / countable” are replaced with “infinite / finite”, respec-tively.Let us also remind two other questions related to our results. The first one callsfor a “dual” results. Problem 7.7. Is it consistent to have a Borel set B ⊆ H such that • B has uncountably many pairwise disjoint translations, but OREL SETS WITHOUT PERFECTLY MANY OVERLAPPING TRANSLATIONS, III 45 • there is no perfect of pairwise disjoint translations of B ?Assumptions of Conclusion 7.1 and Conclusion 7.2 bring the question what isthe value of the first cardinal λ = λ ω such that Pr ε ( λ ) holds true every ε < ω . Problem 7.8. Is λ ω = ℵ ω ? Does Pr ε ( ℵ ω ) hold true for all ε < ω ? References [1] Marek Balcerzak, Andrzej Roslanowski, and Saharon Shelah. Ideals without ccc. Journal ofSymbolic Logic , 63:128–147, 1998. arxiv:math/9610219.[2] Tomek Bartoszy´nski and Haim Judah. Set Theory: On the Structure of the Real Line . A KPeters, Wellesley, Massachusetts, 1995.[3] Udayan B. Darji and Tam´as Keleti. Covering R with translates of a compact set. Proc. Amer.Math. Soc , 131:2598–2596, 2003.[4] M´arton Elekes and Tam´as Keleti. Decomposing the real line into Borel sets closed underaddition. 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Reports on Mathematical Logic , 54:3–43, 2019. arXiv:1806.06283.[12] Saharon Shelah. Disjoint translations. F1927[13] Saharon Shelah. Polish groups and AA ⊆ BB . F1926[14] Saharon Shelah. Borel sets with large squares. Fundamenta Mathematicae , 159:1–50, 1999.arxiv:math/9802134.[15] Piotr Zakrzewski. On Borel sets belonging to every invariant ccc σ –ideal on 2 N . Proc. Amer.Math. Soc. , 141:1055–1065, 2013. Department of Mathematics, University of Nebraska at Omaha, Omaha, NE 68182-0243, USA E-mail address : [email protected] Institute of Mathematics, The Hebrew University of Jerusalem, 91904 Jerusalem,Israel, and Department of Mathematics, Rutgers University, New Brunswick, NJ 08854,USA E-mail address : [email protected] URL ::