aa r X i v : . [ m a t h . L O ] A p r MORE ON TREES AND COHEN REALS
GIORGIO LAGUZZI, BRENDAN STUBER-ROUSSELLE
Abstract.
In this paper we analyse some questions concerning trees on κ ,both for the countable and the uncountable case, and the connections withCohen reals. In particular, we provide a proof for one of the implications leftopen in [6, Question 5.2] about the diagram for regularity properties. Introduction
Throughout the paper we deal with trees on η <κ , with κ ≥ ω being any regularcardinal and η ≥ η is infinite then η regular too.A tree-forcing P is a poset whose conditions are perfect trees p ⊆ η <κ withthe property that for every p ∈ P and every t ∈ p one has p ↾ t := { t ′ ∈ p : t ′ ⊆ t ∨ t ⊆ t ′ } ∈ P ; the ordering is q ≤ p ⇔ q ⊆ p . In case κ = ω and η ∈ { , ω } some of the most popular tree-forcings are for instance: the Hechler forcing D ([1,Def. 3.1.9, p.104]), eventually different forcing E ([1, Def. 7.4.8, p.366]), Sacksforcing (see [2, p.3]), Silver forcing V (see [2, p.4]), Miller forcing M (see [2, p.3]),Laver forcing (see [2, p.3]), Mathias forcing R (see [2, p.4]), random forcing B (see[1, p. 99]). The relation between tree-forcings and Cohen reals has been ratherextensively developed in the literature. The reason to study such connections fordifferent types of tree-forcing notions was mainly to “separate” different kinds ofcardinal characteristics, in particular from cov ( M ). We can associate a tree-forcing P in a standard way with a notion of P -nowhere dense sets, P -meager sets and P -measurable sets. { meager } Definition 1.1.
Given P a tree-forcing notion and X ⊆ η κ a set of κ -reals, we say that: • X is P -nowhere dense if ∀ p ∈ P ∃ q ≤ p ([ q ] ∩ X = ∅ ) , and we put N P := { X : X is P -nowhere dense } . • X is P -meager if there are A i ∈ N P such that X ⊆ S i ∈ κ A i , and we put I P = { X : X is P -meager } . • X is P -measurable if ∀ p ∈ P ∃ q ≤ p ([ q ] ∩ X ∈ I P ∨ [ q ] \ X ∈ I P ) . • A family Γ of subsets of κ -reals is called well-sorted if it is closed undercontinuous pre-images. We abbreviate the sentence “every set in Γ is P -measurable” by Γ( P ).For example when P is the Cohen forcing C , then C -meagerness coincides withtopological meagerness and C -measurability coincides with the Baire Property.When P is the Random forcing B , then B -meagerness coincides with Lebesguemeasure zero and B -measurability coincides with Lebesgue measurability. The presence of Cohen reals added by a tree-forcing P has an impact both on thestructure of I P and on the corresponding notion of P -measurability, as specified inthe tables introduced below. More specifically, if P adds a Cohen real then the wayof coding the P -generic into a Cohen real often induces a construction providingΓ( P ) ⇒ Γ( C ) (e.g., see [5, Theorem 3.1] where such a connection is shown in case of P = D ). Moreover the presence of a coded Cohen real often implies that N P and I P do not coincide. For instance, this holds for the Hechler forcing D and for the even-tually different forcing E . Both these forcings are ccc, and indeed σ -centered. So,a natural question that arises is whether one can find a non-ccc tree-forcing notion P for which Γ( P ) ⇒ Γ( C ) and I P = N P . In this paper we give a positive answer, bydefining and analysing a variant of Mathias forcing in the space 3 ω instead of 2 ω .As a more general question, for a tree-forcing P , one can consider the four proper-ties mentioned so far, namely: 1) P adds Cohen reals; 2) Γ( P ) ⇒ Γ( C ); 3) I P = N P ;4) P is ccc. So for instance, if we consider the most popular tree-forcings we get thefollowing table, where T stands for the variant of Mathias forcing defined in Section2, and M full is the variant of Miller forcing where we require that every splittingnode splits into the whole ω . The results in Table 1 without an explicit referenceare deemed as folklore. Table 1
Adding Cohen I P = N P Γ( P ) ⇒ Γ( C ) c.c.c D , E ✓ ✓ ✓ ([5, Theorem 3.1] ) ✓ B ✗ ✗ ✗ ([14]) ✓ V , M , R ✗ ✗ ✗ ✗ T ✓ (Lemma 2.4) ✓ (Lemma 2.5) ✓ (Proposition 3.3) ✗ M full ✓ ✗ ✓ ([8, Theorem 3.4]) ✗ Note that the table above refers to the tree-forcings in the ω -case, and so definedon spaces like 2 ω , ω ω or [ ω ] ω .For κ > ω we could consider the same table, but then the situation changes andwe can get several different developments. We always assume κ <κ = κ .(1) For D κ (and similarly for E κ ), the constructions done for the ω -case (e.g.,the proof of [5, Theorem 3.1]) easily generalises;(2) for the κ -Silver forcing, the situation seems to depend on whether κ isinaccessible or not; but it is rather independent of whether we considerclub splitting or other version of < κ -closure;(3) for κ -Mathias forcing, the situation is drastically different from the ω -case,as we can prove a strict connection with the Baire property and Cohenreals;The table for κ uncountable then appears as follows, where κ denotes any car-dinal, λ any inaccessible cardinal and γ any not inaccessible cardinal: ORE ON TREES AND COHEN REALSApril 24, 2020 3
Table 2
Adding Cohen I P = N P Γ( P ) ⇒ Γ( C ) κ + -c.c D κ , E κ ✓ (Definition 48 [4]) ✓ ✓ (Reamrk 4.7) ✓ M Club κ ✓ (Proposition 77 [4]) ✗ (Lemma 3.8. [6]) ✓ ✗ V Club λ ✗ ✗ ✗ (Theorem 4.11. [6]) ✗ V Club γ ? ? ? ✗ R Club κ ✓ (Remark 30 [11]) ✓ (Lemma 4.1. [6]) ✓ ✓ R κ ✓ (Remark 30 [11]) ✓ (Lemma 4.6) ✓ (Proposition 31 [11]) ✗ Basic notions and definitions. The elements in η κ are called κ -reals or κ -sequences,where η is also a regular cardinal, usually η = 2 or η = κ . Given s, t ∈ η <κ wewrite s ⊥ t iff neither s ⊆ t nor t ⊆ s (and we say s and t are incompatible). Thefollowing notations are also used. • A tree p ⊆ η <κ is a subset closed under initial segments and its elementsare called nodes . We consider < κ -closed trees p , i.e., for every ⊆ -increasingsequence of length < κ of nodes in p , the supremum (i.e., union) of thesenodes is still in p . Moreover, we abuse of notation denoting by | t | the ordinaldom( t ). • We say that a < κ -closed tree p is perfect iff for every s ∈ p there exists t ⊇ s and α, β ∈ η , α = β , such that t a α ∈ p and t a β ∈ p ; we call such t a splitting node (or splitnode ) and set Split ( p ) := { t ∈ p : t is splitting } . • We say that a splitnode t ∈ p has order type α (and we write t ∈ Split α ( p ))iff ot( { s ∈ p : s ( t ∧ s ∈ Split ( p ) } , ( ) = α . • stem ( p ) is the longest node in p which is compatible with every node in p ; p ↾ t := { s ∈ p : s is compatible with t } . • [ p ] := { x ∈ η κ : ∀ α < κ ( x ↾ α ∈ p ) } is called the set of branches (or body ) of p . • succ ( t, p ) := { α ∈ η : t a α ∈ p } , for t ∈ p . • A poset P is called tree-forcing if its conditions are perfect trees and forevery p ∈ P , and every t ∈ p , one has p ↾ t ∈ P too. Remark . When comparing different notions of P -measurablity, i.e., investigatingthe relationship between Γ( P ) and Γ( Q ) for different tree-forcings P and Q , we oftenrefer to different topological spaces. As Brendle pointed out explicitly in [2] theidea is to consider the analogue versions in the space of strictly increasing sequences ω ↑ ω which can be seen to be almost isomorphic to the spaces we deal with (for thedetails see paragraph 1.2 in [2]). The only case that is not covered in [2] is 3 ω . Inthis paper we need to implement this case as well, as we are going to work with itin the coming section. Actually in trying to describe a suitable isomorphism, weneed to consider a special subspace, in the same fashion as we do when we consideronly the subspace of 2 ω consisting of binary sequences that are not eventually 0.Analogously we consider H := { x ∈ ω : ∃ ∞ n ( x ( n ) = 2) } and we define theappropriate map ϕ : H → ω ↑ ω as follows: we fix the lexicographic enumeration b : 2 <ω → ω . So b ( s ) ≤ b ( t ), whenever s ⊆ t and in particular b ( hi ) = 0. For every x ∈ H let { n k : k ∈ ω } enumerate the set of all inputs n such that x ( n ) = 2. Thendefine σ x := h x ( i ) : 0 ≤ i < n i and for every j ∈ ω , σ xj +1 := h x ( i ) : n j < i < n j +1 i . GIORGIO LAGUZZI, BRENDAN STUBER-ROUSSELLE
Finally put ϕ ( x ) := h b ( σ x ) , b ( σ x )+ b ( σ x )+1 , b ( σ x )+ b ( σ x )+ b ( σ x )+2 , . . . i = h X i ≤ n b ( σ xi )+ n : n ∈ ω i . One can easily check that ϕ is an isomorphism.2. A variant of Mathias forcing { section2 }{ variant of mathias forcing } Definition 2.1.
We define T as the tree-forcing consisting of perfect trees p ⊆ <ω with A p ⊆ ω such that: • for every t ∈ p ( | t | ∈ A p ⇔ t ∈ Split ( p )), we refer to A p as the set ofsplitting levels of p ; • if t ∈ Split ( p ), then t is fully splitting (i.e., for every i ∈ t a i ∈ p ); • for every s ⊇ stem ( p ), if s / ∈ Split ( p ) then s a / ∈ p ; • for every s, t ∈ p , | s | = | t | , s, t / ∈ Split ( p ), one has ∀ i ∈ s a i ∈ p ⇔ t a i ∈ p ) . Intuitively, any condition p ∈ T is a perfect tree in 3 <ω such that at any level n ∈ ω either p uniformly splits, or uniformly takes the same value.Note that T is not c.c.c. . To show that let E ⊆ ω be the set of even numbersand O = ω \ E . For each a ⊆ O we define a condition p a ∈ T in the following way:on even levels we uniformly split and on odd levels n we uniformly choose the value1 whenever n ∈ a and 0 otherwise, so p a := { t ∈ <ω : ∀ n ∈ O ∩ | t | (( n ∈ a → t ( n ) = 1) ∧ ( n a → t ( n ) = 0)) } . We claim that { p a : a ⊆ O } is an antichain. In fact, let a, b ⊆ O be two differentsubsets and fix n ∈ O such that n ∈ a \ b or n ∈ b \ a . W.l.o.g. assume n ∈ a \ b .Then each branch x through p a must satisfy x ( n ) = 1, whereas each branch y through b satisfies y ( n ) = 0. Thus [ p a ] ∩ [ p b ] = ∅ and in particular p a ⊥ p b .Under a certain point of view T seems to behave like the original Mathias forcing R . For instance, the following proof showing that T satisfies Axiom A follows thesame line as for R . However, going more deeply one has to be careful, as even if T still satisfies quasi pure decision (Lemma 2.3), it fails to satisfy pure decision(Lemma 2.4). Thus, we examine these proofs in closer detail to better understandthe main differences between T and R . Proposition 2.2. T satisfies Axiom A.Proof. We define the partial orderings h≤ n : n ∈ ω i in the expected way: For p, q ∈ T we put q ≤ n p if and only if q ≤ p and the two sets of splitting levels A q and A p coincide on the first n + 1 elements. So, in particular q ≤ p implies stem ( q ) = stem ( p ). It is easy to check that fusion sequences exist. Let p ∈ T , k ∈ ω and D ⊆ T a dense subset be given. We show that there is a stronger condition q ≤ k p and a finite set E ⊆ D pre dense below q . This proves that T satisfiesAxiom A. Let A p = { n i : i < ω } be an increasing enumeration of the splittinglevels of p . Observe that there are exactly 3 k nodes t ∈ p of length n k . Each ofthose nodes is splitting, so that there are exactly 3 k +1 immediate successor-nodes.Let { t i : i < k +1 } enumerate all nodes t ∈ p of length n k + 1. We construct q ≤ k p together with a decreasing sequence p = q ≥ q ≥ ... ≥ q k +1 = q . Assumewe want to construct q j +1 . Find p j ∈ D so that p j ≤ q j ↾ t j (this is always possible ORE ON TREES AND COHEN REALSApril 24, 2020 5 since D is dense). We define q j +1 to be the condition which is obtained from q j , bycopying p j above each node in q j of length n k + 1. More precisely: q j +1 := { t ∈ q j : ( | t | ≤ n k + 1 ∨ ( | t | > n k + 1 ∧ ∃ s ∈ p j ∀ n ∈ ω ( n k < n < | t | → s ( n ) = t ( n )))) } . It follows from the construction that for q := q k +1 and j < k +1 we must have q ↾ t j ≤ p j . In particular, we have that q ≤ k p . Put E := { p j : j < k +1 } . Wewant to check that E is pre dense below q . Therefore, let r ≤ q be given. Thenthere is j < k +1 such that r ↾ t j ≤ q ↾ t j . But also q ↾ t j ≤ p j ∈ E and so r and p j arecompatible via r ↾ t j . (cid:3) { quasi-pure } Lemma 2.3. T satisfies quasi pure decision, i.e., for every open dense D ⊆ T , p ∈ T , there is q ≤ p satisfying what follows: if there exists q ′ ≤ q such that q ′ ∈ D , then q ↾ stem ( q ′ ) ∈ D as well.Proof. Let p ∈ T and D ⊆ T open dense be given. We construct a fusion sequence p = q ≥ q ≥ ... such that the fusion q = T k q k witnesses quasi pure decision.Assume we are at step k + 1 of the construction i.e. we have already constructed q k . Let A q k = { n i : i ∈ ω } be the corresponding set of splitting levels. Let { t j ∈ q k : j ∈ k } enumerate all nodes in q k of length n k . Similar to above weconstruct a decreasing sequence q k = q k ≥ q k ≥ ... ≥ q k k . Assume we are at step j < k . There are two cases: Case 1 : There is no stronger condition p ′ ≤ q jk in D with stem ( p ′ ) = t j . Then donothing and put q j +1 k := q jk . Case 2 : Otherwise there is a p ′ ≤ q jk in D with stem ( p ′ ) = t j . As in the proofabove we define q j +1 k := { t ∈ q jk : ( | t | ≤ n k + 1 ∨ ( | t | > n k + 1 ∧ ∃ s ∈ p ′ ∀ n ∈ ω ( n k < n < | t | → s ( n ) = t ( n )))) } ;specifically q j +1 k ↾ t j = p ′ . Finally defining q k +1 := q k k , we get that the correspondingtwo sets of splitting levels A q k and A q k +1 coincide on the first k + 1 elements andtherefore q k +1 ≤ k q k . This completes the construction.Before showing that the fusion q := T k q k witnesses quasi pure decision we makethe following observation: Since in the ( k + 1)-th step in the construction of thefusion the k -th splitting level is fixed, we know for each k ∈ ω and l > k that q ≤ k q l . Therefore the two sets of splitting levels A q and A q l coincide on the first l elements.Now let q ′ ≤ q in D be given. Put t := stem ( q ′ ). Again we denote the splittinglevels of q by A q = { n k : k ∈ ω } and take n k such that | t | = n k . We look at theconstruction of q k +1 . Then there is j < k with t j = t . Since q ′ ≤ q ≤ q jk and q ′ ∈ D we know that in the construction of q j +1 k case 2 was applied i.e. q j +1 k ↾ t = p ′ forsome p ′ ∈ D . Thus, using openness of D and q ↾ t ≤ q j +1 k ↾ t , we also get q ↾ t ∈ D . (cid:3) { t adds cohen reals } Lemma 2.4. (1) T does not satisfy pure decision.(2) T adds Cohen reals.Proof. (1). We have to find a condition p ∈ T and a sentence ϕ such that no q ≤ p decides ϕ . We prove something slightly stronger: Given any p ∈ T we can find a GIORGIO LAGUZZI, BRENDAN STUBER-ROUSSELLE sentence ϕ p such that there is no q ≤ p deciding ϕ p .So let p ∈ T and q ≤ p be given (i.e. q ≤ p ∧ stem ( p ) = stem ( q )). Let ˙ z be the T -name for the generic real. It is clear that (cid:13) T ∃ ∞ n ˙ z ( n ) = 2. We can define aname ˙ σ z ∈ ω ω ∩ V T such that (cid:13) T ˙ σ z ( k ) = k -th 2 occurring in ˙ z. This means that in any generic extension V [ z ] the evaluation of ˙ σ z enumerates theset { k ∈ ω : z ( k ) = 2 } ∈ V [ z ]. For k ∈ ω we define ϕ k := “there are even many 1’s occuring in ˙ z between ˙ σ z ( k ) and ˙ σ z ( k + 1)” . Put k := |{ n < | stem ( q ) | : stem ( q )( n ) = 2 }| and let n q < n q denote the first twosplitting levels of q . Take q , q ≤ q such that(1) stem ( q )( n q ) = 0 and stem ( q )( n q ) = 2,(2) stem ( q )( n q ) = 1 and stem ( q )( n q ) = 2.Then there are at least k + 1 many 2’s occurring in stem ( q i ), therefore ϕ k is decidedby q i , i ∈ q (cid:13) ϕ k ⇔ q (cid:13) ¬ ϕ k . This proves that q does not decide ϕ k .(2). We now show with a similar idea that T adds Cohen reals. Again let ˙ z bethe T -name for the generic real and let ˙ σ z be as above. For every k ∈ ω , • c ( k ) = 0 iff |{ i ∈ ω : ˙ σ z ( k ) ≤ i < ˙ σ z ( k + 1) ∧ ˙ z ( i ) = 1 }| is even • c ( k ) = 1 iff |{ i ∈ ω : ˙ σ z ( k ) ≤ i < ˙ σ z ( k + 1) ∧ ˙ z ( i ) = 1 }| is odd.Then (cid:13) T c ∈ ω . We want to show that c is Cohen. So fix p ∈ T , σ ∈ <ω and let c p ⊆ c be the part of c decided by p . We aim to find q ≤ p such that q (cid:13) c p a σ ⊆ c .This is sufficient to show that c is Cohen.Let k = | c p | , i.e. k is minimal such that c ( k ) is not decided by p . Define p = q ≥ q ≥ · · · ≥ q | σ | by recursion as follows.Assume we have constructed q j , j < | σ | . Let n j < n j be the first two splittinglevels of q j . For i ∈ t i ∈ q j of length n j + 1 so that t i ( n j ) = i and t i ( n j ) = 2.Put q ij := q j ↾ t i . Then we must have |{ m ∈ ω : n j ≤ m < n j ∧ stem ( q ij )( m ) = 1 }| = mod 2 σ ( j )(2.1) { gl1 }{ gl1 } for exactly one i ∈
2. Let q j +1 = q ij such that (2.1) holds.Then by construction, for every j < | σ | , q | σ | (cid:13) c ( | c p | + j ) = σ ( j ), i.e., q | σ | (cid:13) c p a σ ⊆ c . (cid:3) Before moving to the issue concerning the ideals I T and N T , we have to clarifythe space that we are interesting in working with. To understand the point letus consider the standard Mathias forcing R . If we work in the Cantor space 2 ω literally, then we end up with a trivial example to show that N R = I R , namely theset of “rational numbers”, i.e., the set Q := { x ∈ ω : ∃ n ∀ m ≥ n ( x ( m ) = 0) } . Ina similar fashion one can check that the sets N n := { x ∈ ω : x ( i ) = 2 ∀ i ≥ n } are T -nowhere dense, but the union S n ∈ ω N n is not. We leave the straightforwardproof to the reader.For the same argument we specified in Remark 2, indeed the space we re-ally refer to when we work with the standard Mathias forcing is not literally ORE ON TREES AND COHEN REALSApril 24, 2020 7 ω , but is the subspace obtained via the identification of [ ω ] ω and 2 ω , i.e., theset { x ∈ ω : ∃ ∞ n ( x ( n ) = 1) } . In such a space the counterexample disappearsand indeed we get I R = N R . The main difference we want to make is that T behaves completely differently. In fact even when we take the “proper” space H := { x ∈ ω : ∃ ∞ n ( x ( n ) = 2) } we cannot show that N T = I T , as the follow-ing result highlights (where the ideals are considered in the space H ). { tˆ0 Tˆ0 } Lemma 2.5. N T = I T .Proof. Given z ∈ H consider σ z ∈ ω ω as in the proof of the previous Lemma andalso remind c z ∈ ω be as follows: • c z ( k ) = 0 iff |{ i ∈ ω : σ z ( k ) ≤ i < σ z ( k + 1) ∧ z ( i ) = 1 }| is even • c z ( k ) = 1 iff |{ i ∈ ω : σ z ( k ) ≤ i < σ z ( k + 1) ∧ z ( i ) = 1 }| is odd.Then define M n := { z ∈ H : ∀ k ≥ n ( c z ( k ) = 0) } . We claim each M n is T -nowhere dense, but S n ∈ ω M n is not. In fact given n ∈ ω and p ∈ T we can lengthen the stem of p to get a stronger condition p ′ ≤ p suchthat { k < | stem ( p ′ ) | : p ′ ( k ) = 2 } has size > n . Let A p ′ := { n i : i ∈ ω } . Now wetake t ∈ Split ( p ′ ) extending stem ( p ′ ) a t ( n ) = 2 such that t ( n ) = 2 and theset of { k > | stem ( p ′ ) | : t ( k ) = 1 } is odd. Then q := p ′ ↾ t a M n . On the other hand there is always a branch z ∈ [ p ] ∩ H such that for all k > stem ( p ), c z ( k ) = 0. (cid:3)
3. Γ( P ) ⇒ Γ( C )We now prove a rather general result, showing how the “Cohen coding” allowsus to prove a classwise connection between P -measurability and Baire property.Beyond its own interest, the technique used will also permit us to apply it in otherspecific cases that we will summarize along the paper, in particular to answer aquestion connected to the diagram of regularity properties at uncountable inves-tigated in [6]. Recall that a family of sets Γ is well-sorted if it is closed undercontinuous pre-images and Γ( P ) stands for “every set in Γ is P -measurable”. { gamma p implies gamma c } Proposition 3.1.
Let X be a set of size ≤ κ endowed with the discrete topology, X κ the topological product space equipped with the bounded topology (i.e., the topologygenerated by [ t ] := { x ∈ X κ : x ⊇ t } with t ∈ X <κ ), P be a < κ -closed tree-forcing notion defined on X <κ . Assume there exist two maps ϕ : X κ → κ and ϕ ∗ : X <κ → <κ such that:a) ϕ is continuous,b) ∀ i < κ ϕ ( x ) ↾ i = ϕ ∗ ( x ↾ i ) ,c) ∀ q ∈ P ∀ s ∈ <κ ∃ σ ∈ q such that ϕ ∗ ( σ ) ⊇ ϕ ∗ ( stem ( q )) a s. Then Γ( P ) implies Γ( C ) . We note that the second condition implies ϕ [[ p ]] ⊆ [ ϕ ∗ ( stem ( p ))] for each p ∈ P .The third condition intuitively means that the map ϕ ∗ is below any condition almostsurjective. The key step for the proof is the following lemma. Lemma 3.2.
Let P , ϕ, ϕ ∗ be as in the Proposition and X ⊆ κ . Define Y := ϕ − [ X ] . Assume there is q ∈ P such that Y ∩ [ q ] is P -comeager in [ q ] . Then X ∩ [ ϕ ∗ ( stem ( q ))] is comeager in [ ϕ ∗ ( stem ( q ))] . GIORGIO LAGUZZI, BRENDAN STUBER-ROUSSELLE
Proof.
We are assuming Y ∩ [ q ] is P -comeager, for some q ∈ P . This implies thatthere is a collection { A α : α < κ ∧ A α is P -open dense in [ q ] } such that T α A α ⊆ [ q ] ∩ Y . W.l.o.g. assume A α ⊇ A β , whenever α < β < κ . Let t = ϕ ∗ ( stem ( q )).We want to show that ϕ [ Y ] ∩ t = X ∩ t is comeager in [ t ] i.e., we want to find { B α : α < κ } open dense sets in [ t ] such that T α B α ⊆ X ∩ [ t ]. Given σ ∈ κ <κ werecursively define on the length of σ a set { q σ : σ ∈ κ <κ } ⊆ P with the followingproperties: q hi = q , ∀ σ ∈ κ <κ S i [ ϕ ∗ ( stem ( q σ a i ))] is open dense in [ ϕ ∗ ( stem ( q σ ))], ∀ σ ∈ κ <κ ∀ i ∈ κ ([ q σ a i ] ⊆ A | σ | ∧ q σ a i ≤ q σ ).Assume we are at step α = | σ | . Fix σ ∈ κ α arbitrarily and then put t σ = ϕ ∗ ( stem ( q σ )). We first make sure that 2 . holds. Therefore let { s i : i < κ } enumerate 2 <κ . By condition c ) from Proposition 3.1 we can find p i ≤ q σ suchthat ϕ ∗ ( stem ( p i )) ⊇ t σ a s i . Since each A α is P -open dense in [ q ] we can find foreach i < κ an extension q i ≤ p i such that [ q i ] ⊆ T α ≤| σ | A α . This ensures thatalso 3 . holds and we put q σ a i := q i . At limit steps λ , we put for every σ ∈ κ λ , q σ := T β< | σ | q β . Finally we put B α := S { ϕ [[ q σ ]] : σ ∈ κ α } . We have to check that T α B α ⊆ X ∩ [ t ]. Since t = stem ( q ) and q σ ≤ q we get ϕ [[ q σ ]] ⊆ ϕ [[ q ]] ⊆ [ ϕ ∗ ( t )]and therefore B α ⊆ [ t ] for each α ∈ κ . On the other hand by construction of B α +1 we know ϕ − [ B α +1 ] ⊆ A α and hence ϕ − [ T α B α ] ⊆ T α A α which implies T α B α ⊆ X . (cid:3) Proof of the proposition.
Let X ∈ Γ be given and put Y := ϕ − [ X ]. Then also Y ∈ Γ, since Γ is well-sorted and ϕ is continuous. We now use the lemma to showthat for every t ∈ <κ there exists t ′ ⊇ t such that X ∩ [ t ′ ] is meager or X ∩ [ t ′ ] iscomeager.Fix t ∈ <κ arbitrarily and pick p ∈ P such that ϕ ∗ ( stem ( p )) ⊇ t . By assumption Y is P -measurable, and so: • in case there exists q ≤ p such that Y ∩ [ q ] is P -comeager; put t ′ := ϕ ∗ ( stem ( q )). By the lemma above, X ∩ [ t ′ ] is comeager in [ t ′ ]; • in case there exists q ≤ p such that Y ∩ [ q ] is P -meager, then apply thelemma above to the complement of Y , in order to get X ∩ [ t ′ ] be meager in[ t ′ ], with t ′ := ϕ ∗ ( stem ( q )).By the remark directly after Definition 1.1 this suffices to complete the proof. (cid:3) { prop-mathias-baire } Proposition 3.3.
Let Γ be a well-sorted family of sets. Then Γ( T ) ⇒ Γ( C ) . Proof.
Consider H := { x ∈ ω : ∃ ∞ n x ( n ) = 2 } . As we remarked right aboveLemma 2.5, H is T -comeager. Thus we have for each set X ⊆ ω : X is T -measurable ⇔ X ∩ H is T -measurable.Since we are only concerned with T -measurability we can work with the set H instead of the whole space 3 ω . We want to apply Proposition 3.1. For an element x ∈ H let A x = { n i : i < ω } be an increasing enumeration of all n ∈ ω such that x ( n ) = 2. This is by definition of H an infinite set. Using this notation we define ORE ON TREES AND COHEN REALSApril 24, 2020 9 a function ϕ : H → ω via: ϕ ( x )( i ) = ( |{ j < ω : n i < j < n i +1 ∧ x ( j ) = 1 }| is even1 else.Note that ϕ is surjective but not injective and observe that ϕ induces a map ϕ ∗ :3 <ω → <ω such that for each x ∈ H and i < ω we have ϕ ( x ) ↾ i = ϕ ∗ ( x ↾ n i ). Wehave to check that a ) , b ) and c ) from Proposition 3.1 are satisfied. Condition b ) isclear. For condition a ) we have to show that the pre-image of a basic open set in 2 ω is open in H (regarding the induced topology of 3 ω on H ). Therefore let s ∈ <ω be given. It follows ϕ − [[ s ]] = [ t ∈ <ω ,ϕ ∗ ( t )= s [ t ] ∩ H which is a union of basic open sets in H .So we are left to show that c ) holds as well. Therefore fix q ∈ T and s ∈ <ω .Let A q = { n i : i < ω } be the corresponding set of splitting levels and s =( i , . . . , i k ). Then we can lenghten stem ( q ) in order to have the parity of 1s betweentwo subsequent 2 according to the corresponding i j , that means we find t ∈ q suchthat ϕ ∗ ( t ) ⊇ ϕ ∗ ( stem ( q )) a s .So we are able to apply Proposition 3.1 and get Γ( T ) ⇒ Γ( C ). (cid:3) Some results for the uncountable case
In this section we investigate some issues concerning Table 2. We will alwaysassume that κ is an uncountable regular cardinal such that κ = 2 <κ . Definition 4.1 (Club κ -Miller forcing M Club κ ) . A tree p ⊆ κ <κ is called κ -Millertree if it is pruned, < κ -closed and(a) for every s ∈ p there is an extension t ⊇ s in p such that succ ( t, p ) ⊆ κ isclub. Such a splitting node t is called club-splitting .(b) for every x ∈ [ p ] the set { α < κ : x ↾ α is club-splitting } is club. Remark:
Both (a) and (b) ensure that M Club κ is a < κ -closed forcing. The set oftrees that consist of nodes that are either club-splitting or not splitting is a densesubset of M Club κ .The following result highlights the connection with κ -Cohen reals. We remarkthat a similar result (though in a different context, dealing with a version of M κ satisfying (a) but not (b)) has been proven by Mildenberger and Shelah in [12]. { prop:miller-baire } Proposition 4.2.
Let Γ be a well-sorted family of subsets of κ -reals. Then Γ( M Club κ ) ⇒ Γ( C ) .Proof. We introduce a coding function ϕ ∗ : κ <κ → <κ . Therefore fix a κ sizedfamily { S t ⊆ κ : t ∈ <κ } of pairwise disjoint stationary sets such that the unionof all S t ’s covers κ (this is possible since we assume κ = 2 <κ ). Let σ ∈ κ <κ . Wedefine ϕ ∗ ( σ ) = t i a t i a . . . a t i α a . . . , with σ ( α ) ∈ S t iα for all α < | σ | . Then ϕ ∗ induces a function ϕ : κ κ → κ via ϕ ( x ) ↾ α := ϕ ∗ ( x ↾ α ).It is easy to see that such maps ϕ and ϕ ∗ satisfy the three conditions in Propo-sition 3.1; a ) and b ) are clear, so we only check condition c ). So fix q ∈ M Club κ and t ∈ <κ . Let τ = stem ( q ). Since succ ( τ, q ) ⊆ κ is club and S t is stationary, we can pick β ∈ S t ∩ succ ( τ, q ). Then τ a β ∈ q and ϕ ∗ ( τ a β ) = ϕ ∗ ( τ ) a t . Using Proposition3.1 we obtain Γ( M Club κ ) ⇒ Γ( C ) as desired. (cid:3) { miller-cohen } Remark . The map ϕ we used in Proposition 4.2 allows us to read off a Cohen κ -real from the M Club κ -generic. Indeed, let { S t ⊆ κ : t ∈ <κ } , ϕ ∗ and ϕ be asabove. Let ˙ z be the M Club κ -name for the generic κ -real and ˙ c the M Club κ -name suchthat (cid:13) M Club κ ˙ c = ϕ ( ˙ z ) ∈ κ . We claim that ˙ c is κ -Cohen in every generic extension.Therefore fix p ∈ M Club κ and let c p ∈ <κ be the initial part of ˙ c decided by p so c p = ϕ ∗ ( stem ( p )). Let t ∈ <κ be given. We want to find q ≤ p such that q (cid:13) c p a t ⊆ ˙ c .Since stem ( p ) is club-splitting we can find an α ∈ S t ∩ { α < κ : stem ( p ) a α ∈ p } and take q to be p ↾ stem ( p ) a α i.e. stem ( q ) extends stem ( p ) a α . This implies that ϕ ∗ ( stem ( q )) ⊇ c p a t and therefore q (cid:13) c p a t ⊆ ˙ c as demanded.We also remark that the fact that M Club κ adds Cohen κ -reals is not new and itwas proven in [4], even if the authors use a different coding map.Differently from T , the Cohen-like behaviour of the M Club κ -generic does not havean impact on the ideals, as shown in the next result. Lemma 4.4. N M Club κ = I M Club κ Proof.
The proof is rather standard. We report a sketch of it here just for com-pleteness. Given { D i : i < κ } a family of M Club κ -open dense sets and p ∈ M Club κ wesimply construct a fusion sequence { q i : i < κ } so that q := T i<κ q i ≤ p , for every i < κ , [ q i ] ⊆ D i , and for every j < i , q i ≤ j q j , i.e., q i ≤ q j and for every j ≤ i , Split j ( q i ) = Split j ( q j ). This can be done via an easy recursive construction: at limitsteps i , simply put q i := T j
We first clarify what is meant with N R κ : X ⊆ [ κ ] κ is called R κ -nowheredense if for each ( s, A ) ∈ R κ there is a stronger condition ( t, B ) ≤ ( s, A ) such that ∀ x ∈ X ∀ y ∈ [ B ] κ ( x = t ∪ y ) . (4.1) { equa }{ equa } We define an equivalence relation on the set of countably infinite subsets of κ . For a, b ∈ [ κ ] ω let a ∼ b : ⇔ | a △ b | < ω . We fix a system of representatives. For a ∈ [ κ ] ω we denote the representative of { b ∈ [ κ ] ω : b ∼ a } with ˜ a . Then we define acoloring function C : [ κ ] ω → { , } as follows: C ( a ) = ( | a △ ˜ a | is even1 else.We can identify x ∈ [ κ ] κ with it’s increasing enumeration χ : κ → κ given by χ ( ξ ) := min { x \ S α<ξ χ ( α ) } . Let { α i : i < κ } enumerate the limit ordinals < κ .For x ∈ [ κ ] κ and i < κ we define the countable set b xi := { x ( ξ ) : α i < ξ < α i +1 } ⊆ κ . ORE ON TREES AND COHEN REALSApril 24, 2020 11
Claim:
The set X i := { x ∈ [ κ ] κ : ∀ j > i C ( b xj ) = 0 } is R κ -nowhere dense for all i < κ , but their union is not. Proof of the claim . Let ( s, A ) be a κ -Mathias condition and i < κ be given. Fix j > i . Then A ⊆ κ is of size κ . By removing at most one element of A , we find A ′ ⊆ A such that C ( b A ′ j ) = 1. We extend s with the first α j +1 elements of A ′ to get t := s ∪{ A ′ ( ξ ) : ξ ≤ α j +1 } ∈ κ <κ . Now we can shrink A ′ to B := A ′ \ ( A ′ ( α j +1 )+1)in order to obtain a κ -Mathias condition ( t, B ) ≤ ( s, A ) fulfilling the requirement(4.1). This proves the claim.However the union X := S i<κ X i can not be R κ -nowhere dense. In fact, let( s, A ) be a κ -Mathias condition. We can always find for i > otp( s ) a subset B ⊆ A of size κ such that C ( b Bj ) = 0, for all j > i and hence (4.1) is false for X i and( s, B ). (cid:3) (The coloring introduced above requires AC. However the result needs not AC, aswe can also consider another kind of coloring, as noted by Wohofsky and Koelbingduring the writing of [7]: fix S ∈ [ κ ] κ stationary and co-stationary and define thecoloring C : [ κ ] ω → { , } by C ( a ) := 0 iff sup a ∈ S .) { reamrk-kappa-hechler } Remark . Proposition 3.1 also applies for P ∈ { D κ , E κ } . The coding function ϕ : κ κ → κ we need in this case is given by ϕ ( x )( i ) = x ( i ) mod 2, similarly to the ω -case. It is straightforward to prove that such a ϕ (and the natural corresponding ϕ ∗ ) satisfies the required properties of Proposition 3.1. st References [1] T. Bartoszy´nski and H. Judah,
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