Mathias and Silver forcing parametrized by density
Giorgio Laguzzi, Heike Mildenberger, Brendan Stuber-Rousselle
aa r X i v : . [ m a t h . L O ] F e b MATHIAS AND SILVER FORCING PARAMETRIZED BYDENSITY
GIORGIO LAGUZZI, HEIKE MILDENBERGER, BRENDAN STUBER-ROUSSELLE
Abstract.
We define and investigate versions of Silver and Mathiasforcing with respect to lower and upper density. We focus on proper-ness, Axiom A, chain conditions, preservation of cardinals and addingCohen reals. We find rough forcings that collapse 2 ω to ω , while othersare surprisingly gentle. We also study connections between regularityproperties induced by these parametrized forcing notions and the Baireproperty. Introduction
Forcings consisting of tree conditions are the dramatis personae exten-sively studied in descriptive set theory and set theory of the reals. Theoriginal interest on this type of forcings was mainly due to their applica-tions in resolving problems of independence risen from the study of cardinalcharacteristics, infinitary combinatorics and regularity properties. Throughthe years, the study of the combinatorial properties of tree-forcings has beenintensified and has been interesting on its own.From the forcing point of view, trees can be thought as conditions whosestem decides a finite fragment of the generic real, and the rest of the treeabove describes the possible paths of the generic real. In this sense, Co-hen forcing can be understood as the simplest tree-forcing notion, whoseconditions decide a finite fragment of the generic and then leave all pathsextending the stem possible. In general, the fatter the tree, the more free-dom the path of the generic real; the slimmer the tree, the more restrictivethe conditions on the path of the generic real. Under this point of view aSacks tree can be seen as the other extreme, since in this case the tree can beshrunk as much as desired, and the only requirement is to keep perfectness.Many other tree-forcings in between have been introduced and extensivelystudied, among them Silver and Mathias trees . In this paper we focus onstudying some variants where the set of nodes above the stem are governedby restrictions imposed on the density of the set of splitting nodes. Imposingfatness or slimness conditions though, may result in a non-proper forcing. Date : Feb. 9, 2021. The nodes of the Mathias (or Silver) tree p are all t ∈ <ω that are initial segments ofone of the possible generic branches that are compatible with a Mathias condition ( s, A )(or a Silver condition f p ). Each condition in Mathias forcing and in Silver forcing, when conceivedas a tree, comes with an infinite set A p such that any node t of p whoselength is in A p is a splitting node. Also other tree forcings, like Sacks orMiller, can be thinned out to the subsets of those conditions who come withsuch a set A p of overall splitting levels. In Bukovský-Namba forcing theconditions with uniform splitting levels are even dense, see e.g. [4, Theorem2.2].We let the A p range over prescribed families that are sets of positive loweror upper density in ω . There is some work on forcings of this type. Grigorieff[9] parametrised Silver forcing by having the A p range over a P -point andHalbeisen [10, Ch 24] generalised this to P -families. Mathias forcing with aRamsey ultrafilter [17] is a versatile notion of forcing. Farah and Zapletal[8, Ch. 9] used the coideal of the density zero ideal as a reservoir for theinfinite component of Mathias conditions.In Sections 2 to 5 we investigate whether the variants we introduce ofSilver and of Mathias forcing are proper at all. We show that Mathias withlower density ≥ ε is a disjoint union of σ -centered forcings (see Proposition3.6) and that Silver forcing with positive lower density collapses 2 ω to ω (seeTheorem 5.1). We also show that the lower density is far from the notion ofa measure (see Proposition 5.3). This result fits in the framework presentedin [15].In Section 6 and 7 we are concerned with regularity properties of theseforcings. Regularity properties of tree-forcings by themselves are a wellstudied field in mathematics. The following table illustrates which of themost popular regularity property of a subset of 2 ω correspond to whichtree-forcing P . Tree-forcing P P -measurableCohen C Baire propertyRandom B Lebesgue measurableSilver V completely doughnut [18]Sacks S Marczewski set [22]Mathias MA completely RamseyIn many cases adding a Cohen real is enough to establish a dependence be-tween Baire property and the measurability given by the tree-forcing. Thiswas made explicit in [14, Proposition 3.1.]. Proposition 7.1 is an improve-ment of this result.In Section 8 we construct a model in which all On ω -definable sets are V + ε -measurable but Σ ( C ) fails. In particular Σ ( V + ε ) ⇒ Σ ( C ) fails.In Section 9 we return to a question we asked in [13], namely: Does thesplitting forcing SP have the Sacks property? We give a partial negative an-swer in Theorem 9.13. Independently, Jonathan Schilhan found a completenegative answer to our question [21]. REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 3
In the remainder of this introduction, we set up our notation.
Definition 1.1.
Let X be non-empty.(a) We let X <ω = { s : ( ∃ n < ω )( s : n → X ) } . The set X <ω is par-tially ordered by the initial segment relation E , namely s E t if s = t ↾ dom( s ). We use ⊳ for the strict relation.(b) A set p ⊆ X <ω is called a tree if it is closed under initial segments,i.e. ( t ∈ p ∧ s E t ) → s ∈ p . The elements of p are called nodes .(c) A node is called a splitting node if it has at least two immediatesuccessors in p . We write Split( p ) for the set of splitting nodes of p .(d) A tree p is called perfect if for every s ∈ p there is a splitting node t D s .(e) For t ∈ p , we write splsuc( t ) for the shortest splitting node extending t . When t is splitting, then splsuc( t ) = t .(f) The stem of p , short stem( p ), is the E -least splitting node of p if itexists.(g) For n < ω we let Split n ( p ) consist of all splitting nodes t in p suchthat there are exactly n splitting nodes preceding t i.e., stem( p ) = t ⊳t ⊳ · · · ⊳t n − ⊳t , in particular Split ( p ) = { stem( p ) } . Analogously,we define Split ≤ n ( p ).(h) For n < ω , let Lev n ( p ) := { t ∈ p : | t | = n } .(i) For t ∈ p we let p ↾ t = { s ∈ p : s E t ∨ t ⊳ s } .(j) Let p ⊆ X <ω be a tree such that for any t ∈ p and any n there is s ∈ p , t E s such that | s | > n . The body or rump of a tree p , short[ p ], is the set { f ∈ X ω : ( ∀ n )( f ↾ n ∈ p ) } . Definition 1.2.
A partial ordering ( P , ≤ ) is called a tree-forcing , if there isa non-empty set X such that- All conditions p ∈ P are perfect trees on X <ω .- For all p ∈ P and t ∈ p the restriction p ↾ t is again a condition in P .- The partial order is the inclusion i.e., q ≤ p iff q ⊆ p .We are interested in tree-forcings P defined over 2 <ω or ω <ω with thefollowing additional property. Definition 1.3.
Let p be a perfect tree defined over 2 <ω or ω <ω . The tree p is called uniformly splitting , if there is an infinite set A p ∈ [ ω ] ω such that ∀ s ∈ p ( | s | ∈ A p ⇔ s ∈ Split( p )) . A tree-forcing P is called uniformly splitting tree-forcing , if all conditions p ∈ P are uniformly splitting.In this paper we study two well-known examples of uniformly splittingtree forcings: The Mathias forcing MA and the Silver forcing V . Example 1.4. (1) The Mathias forcing is given by conditions ( s, A ) ∈ MA if s ∈ [ ω ] <ω and A ∈ [ ω ] ω and max( s ) < min( A ). ( t, B ) ≤ ( s, A )if t \ s ⊆ A , t ⊇ s , and B ⊆ A . Now a Mathias tree p = p ( s, A ) GIORGIO LAGUZZI, HEIKE MILDENBERGER, BRENDAN STUBER-ROUSSELLE is given by p ⊆ <ω and t ∈ p if t is the characteristic functionof s ∪ s ′ for some finite s ′ ⊆ A . Of course, now A p = A , andstronger conditions correspond to subtrees. All trees are perfect andrestrictions p ↾ t are again conditions.(2) A condition f is a Silver condition if there is an infinite set A suchthat f : ω \ A →
2. So we get a Silver tree p ⊆ <ω by letting t ∈ p ↔ ( ∀ n ∈ | t | \ A )( t ( n ) = f ( n )) . Then A p = A . Stronger conditions are extensions and again corre-spond to subtrees. Definition 1.5.
For a set A ⊆ ω we define the upper density d + ( A ) andlower density d − ( A ) of A via:(1) d + ( A ) := lim sup n →∞ | A ∩ n | n ,(2) d − ( A ) := lim inf n →∞ | A ∩ n | n . Definition 1.6.
Let P be a uniformly splitting tree-forcing defined over 2 <ω or ω <ω . For ε ∈ (0 ,
1] we define two subforcings P + ε and P − ε , and we definethe upper and lower positive density versions:(1) p ∈ P + ε if p ∈ P and A p has upper density ≥ ε .(2) p ∈ P − ε if p ∈ P and A p has lower density ≥ ε .(3) p ∈ P + if p ∈ P and A p has upper density > p ∈ P − if p ∈ P and A p has lower density > q is stronger than p iff q is a subset of p .We focus our attention on MA + ε , MA − ε , MA + , MA − and the same forSilver. We order our investigation now in pairs, according to the densityrequirement. Some steps work also for general P .2. Upper density ≥ ε Definition 2.1.
A notion of forcing ( P , ≤ ) has Axiom A if there are partialorder relations h≤ n : n < ω i such that(a) q ≤ n +1 p implies q ≤ n p , q ≤ p implies q ≤ p ,(b) If h p n : n < ω i is a fusion sequence, i.e., a sequence such that for any n , p n +1 ≤ n p n , then there is a lower bound p ∈ P , p ≤ n p n .(c) For any maximal antichain A in P and and n ∈ ω and any p ∈ P thereis q ≤ n p such that only countably many elements of A are compatiblewith q . Equivalently, for any open dense set D and any n , p , thereis a countable set E p of conditions in D and q ≤ n p such that E p ispredense below q .A notion of forcing ( P , ≤ ) has strong Axiom A if the set of com-patible elements in (c) is even finite. REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 5
Axiom A entails properness and strong Axiom A implies ω ω -bounding(see, e.g., [20, Theorem 2.1.4, Cor 2.1.12]). Remark 2.2.
Let p ∈ P + ε . We define an increasing sequence h k pn ∈ ω : n ∈ ω i as follows: n = 0 : We put k p := min( A p ) ,n > We let k pn := min { k ∈ ω : ( k > k pn − ∧ | A p ∩ k | ≥ k ( ε − − n )) } . Such a sequence has the following property for n ∈ ω : | A p ∩ k pn | k pn ≥ ε − − n , and therefore witnesses d + ( A p ) ≥ ε . We use the sequences h k pn ∈ ω : n ∈ ω i to define a stronger- n -relation ≤ n on P + ε . Definition 2.3.
We define a decreasing sequence of partial order relations h≤ n : n ∈ ω i on P + ε as follows: q ≤ n p ⇔ q ≤ p ∧ k qn = k pn ∧ A q ∩ k qn = A p ∩ k pn . Observe that given two conditions satisfying q ≤ n p we must have k qi = k pi for i ≤ n . Fact 2.4.
Let P ∈ { MA , V } . Let h q n : n ∈ ω i be a fusion sequence in P + ε .Then, fusions exist. Especially, q = T q n is a condition in P + ε . Silver forcing with upper density ε > . We quickly establish thatthe Silver forcing with positive upper density ε has strong Axiom A andthus is a proper forcing that does not add unbounded reals. The proof is astraightforward generalization of the standard case. Lemma 2.5.
The forcing V + ε has strong Axiom A.Proof. We take the partial order relations ≤ n as defined in Definition 2.3.It is easy to see that the requirements (a) and (b) from Definition 2.1 arefulfilled. We make sure that the strong version of (c) can be fulfilled as well.So, fix an open dense set D ⊆ V + ε , n ∈ ω and a condition p ∈ V + ε . We haveto find a finite set E p ⊆ D and a condition q ≤ n p such that E p is predensebelow q .To this end, let h k pm : m < ω i be the sequence associated to the condition p asgiven in Remark 2.2. Let k := | A p ∩ k pn | denote the number of splitting levelsup to k pn and { t i ∈ Split( p ) : i < k } enumerate all splitting nodes of lengthexactly k pn . We construct a decreasing sequence p =: q ≥ q ≥ · · · ≥ q k =: q such that q i +1 ↾ t i ∈ D, i < k . Fix i < k . Now, pick p i ≤ q i ↾ t i in D andcopy p i into q i above level k pn , more precisely: q i +1 := { s ∈ q i : ( | s | ≤ k pn ∨ ( | s | > k pn ∧∃ t ∈ p i ( k pn ≤ m < | s | → s ( m ) = t ( m )))) } . GIORGIO LAGUZZI, HEIKE MILDENBERGER, BRENDAN STUBER-ROUSSELLE
It is clear that the resulting tree q satisfies q ↾ t ∈ D , whenever t ∈ q is oflength at least k pn . Moreover, since we have pruned the tree q only above k pn we also made sure that k pn = k qn ∧ A p ∩ k pn = A q ∩ k pn , especially q ≤ n p . For the finite set E p we can simply put E p := { q ↾ t : t ∈ Lev k pn ( q ) } . This completes the proof. (cid:3) Corollary 2.6.
The forcing V + ε is proper and ω ω -bounding for each ε ∈ (0 , . Mathias forcing with upper density ε > . In the following, weinvestigate the differences between MA + ε and the classical Mathias forcing MA and Mathias forcing MA ( F ) with respect to a filter F . The decisivedifference to the classical forcing is that MA + ε adds Cohens. From MA ( F )it already differs by the fact that the set { A ⊆ ω : d + ( A ) ≥ ε } is not closedunder finite intersection and thus not a filter. Note that this does not yetimply that the two forcing are not forcing equivalent. Definition 2.7.
Let F be a filter over ω . The partial order MA ( F ) consistsof all p ∈ MA such that the set of splitting levels A p is an element of thefilter F . MA ( F ) is ordered by inclusion.Remember that a filter F over ω is called Canjar , if the correspondingMathias forcing MA ( F ) does not add dominating reals. See for instance [5],where Michael Canjar constructed an ultrafilter U under the assumption d = c such that MA ( U ) does not add dominating reals. The followinglemma shows that MA + ε cannot be equivalent to MA ( F ), for F Canjar.
Lemma 2.8. MA + ε adds dominating reals.Proof. Let G be MA + ε -generic over V and let x G = S { s : ∃ A (( s, A ) ∈ G ) } denote the the MA + ε -generic real. Observe that x G has upper density ε inthe generic extension V [ G ]. Therefore, the following function is well definedin V [ G ]: f ( n ) := min (cid:26) k : | x G ∩ k | k ≥ ε − − n (cid:27) . Remember that in Remark 2.2 we assigned to each condition p ∈ MA + ε asequence h k pn ∈ ω : n < ω i . That f is dominating now follows from thefollowing two facts:(i) ∀ p ∈ MA + ε , p (cid:13) ∀ n < ω ( f ( n ) ≥ k pn ),(ii) ∀ g ∈ ω ω ∀ p ∈ MA + ε ∃ q ≤ p, q (cid:13) ∀ ∞ n ( k qn ≥ g ( n )). (cid:3) There is a more general reason why MA + ε and MA ( F ) for any filter F cannot be forcing equivalent. Since F is a filter, any two conditions p, q ∈ MA ( F ) with the same stem are compatible and in particular MA + ε is σ -centered. REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 7
Lemma 2.9. MA + ε does not satisfy the countable chain condition.Proof. Without loss of generality let ε = 1. We construct an injective func-tion f : 2 ω → [ ω ] ω such that each x ∈ ω with | x − (1) | = ω get mapped to aset with upper density 1. To this end we define f together with an auxiliaryfunction g : ω → ωg (0) := 0 g ( i + 1) := 2 g ( i ) n ∈ f ( x ) : ⇔ there is i ∈ ω, x ( i ) = 1 and n ∈ [ g ( i ) , g ( i + 1))Now, choose an almost disjoint family { A α ∈ [ ω ] ω : α < c } of size continuumand let x α denote the corresponding characteristic function i.e., x α ( i ) = 1iff i ∈ A α . Then, { f ( x α ) : α < c } is an almost disjoint family consisting ofsets of upper density 1. In particular, the conditions { ( hi , f ( x α )) : α < c } form an antichain in MA +1 of size the continuum. (cid:3) Now we turn our attention to a comparison with the classical Mathiasforcing MA . Usually the first step in showing that a given forcing satisfies pure decision (compare [1, Lemma 7.4.5]) is to show that the forcing satisfies quasi pure decision (compare [1, Lemma 7.4.6]). We will see that although MA + ε fails to satisfy pure decision it still satisfies quasi pure decision. Lemma 2.10. MA + ε satisfies quasi pure decision i.e., given a condition p = ( s, A ) ∈ MA + ε and an open dense set D ⊆ MA + ε there is B ⊆ A suchthat the following holds:If there is ( t, C ) ≤ ( s, A ) and ( t, C ) ∈ D, then ( t, B \ (max( t ) + 1)) ∈ D. Proof.
The proof is a straightforward generalization of [1, Lemma 7.4.5.].To make sure that the final set B has upper density ≥ ε we have to usefinite sets b n instead of singletons.We construct B = S { b n ∈ [ ω ] <ω : n ∈ ω } together with a decreasingsequence { B n ⊆ A : n ∈ ω } . We start with B = A and b = min( A ).So, assume we have constructed all sets up to B n and b n . Let { t , . . . , t k − } enumerate all subsets of S { b , . . . , b n } . We construct B n +1 as a decreasingsequence B n =: B n +1 ⊇ · · · ⊇ B kn +1 =: B n +1 . Let j < k be given.Case 1: There exists C ⊆ B jn +1 \ (max( t j ) + 1) such that ( s ∪ t j , C ) ∈ D .Then put B j +1 n +1 := C .Case 2: Otherwise put B j +1 n +1 := B jn +1 .Finally put B n +1 := B kn +1 . Since B n +1 has upper density ≥ ε we can find k n +1 ∈ ω such that | ( S j ≤ n b j ∪ B n +1 ) ∩ k n +1 | k n +1 ≥ ε − − n − and set b n +1 := B n +1 ∩ k n +1 . (cid:3) Corollary 2.11.
The forcing MA + ε has Axiom A. GIORGIO LAGUZZI, HEIKE MILDENBERGER, BRENDAN STUBER-ROUSSELLE
Proof.
We take the partial order relations ≤ n as defined in Definition 2.3and also recall the sequences h k pn : n < ω i from Remark 2.2. The crucialpart is to make sure that the requirement (c) from Axiom A (Definition 2.1)is fulfilled. So, fix an open dense set D ⊆ MA + ε , n < ω and a condition p = ( s, A p ) ∈ MA + ε . Let N be big enough and { t i ∈ [ ω ] <ω : i < N } enumerate all subsets of A p ∩ k pn . We define a decreasing sequence A p ⊇ B ⊇ B ⊇ , . . . , ⊇ B N such that for i < N : ∀ ( t, C ) ∈ D (cid:0) ( t, C ) ≤ ( s ∪ t i , B i ) → ( t, B i +1 \ (max( t ) + 1)) ∈ D (cid:1) . We start with B := A p \ (max( t ) + 1). To get the sets B i for i >
0, simplyapply Lemma 2.10 to the condition ( s ∪ t i − , B i − ) and the open dense set D .Finally, we set B := ( A ∩ k pn ) ∪ B N and E p := { ( t, B \ (max( t ) + 1)) : ∃ C ∈ [ ω ] ω (( t, C ) ≤ ( s, B ) ∧ ( t, C ) ∈ D ) } .Then, ( s, B ) ≤ n ( s, A ) and E p is a countable predense set below ( s, B ). (cid:3) Proposition 2.12. MA + ε adds a Cohen real.Proof. Take N ∈ ω such that 1 /N < ε . Divide ω into N + 1 disjoint sets a i ⊆ ω, i < N + 1 of density 1 / ( N + 1) i.e., d + ( a i ) = d − ( a i ) = 1 / ( N +1) , i < ( N + 1) . Then each set A ⊆ ω with upper density ≥ /N cannotbe completely contained in a single set a i . Furthermore it must intersectat least two of them infinitely often. Let x G be the canonical name for the MA + ε -generic real and h n k : k ∈ ω i enumerate all integers n such that x G ( n ) = 1. We define a MA + ε -name ˙ c via:˙ c ( k ) := ( , ∃ i < ( N + 1) { n k , n k +1 } ⊆ a i , else.We claim that ˙ c is Cohen. For this purpose, fix t ∈ <ω , ( s, A ) ∈ MA + ε .W.l.o.g. we can assume that | s − ( { } ) | is even. Let r ∈ <ω maximal suchthat ( s, A ) (cid:13) r E ˙ c . We have to find ( s ′ , A ′ ) ≤ ( s, A ) with the property( s ′ , A ′ ) (cid:13) r a t E ˙ c . We construct ( s ′ , A ′ ) as a decreasing sequence ( s, A ) =:( s , A ) ≥ ( s , A ) ≥ · · · ≥ ( s | t | , A | t | ) = ( s ′ , A ′ ) such that | s i − ( { } ) | + 2 = | s i +1 − ( { } ) | , i < | t | and ( s i , A i ) (cid:13) r a t ↾ i E ˙ c . We only carry out the firststep of the construction. Take i < ( N + 1) such that A ∩ a i is infinite. Put m := min( A ∩ a i ). There are two cases: t (0) = 0 : Define M := min(( A ∩ a i ) \ ( m + 1)) and put s := s a m a M,A := A \ ( M + 1) . Then ( s , A ) (cid:13) r a t (0) E ˙ c.t (0) = 1 : Define M := min( A \ (( m + 1) ∪ a i )) and put s := s a m a M,A := A \ ( M + 1) . Then ( s , A ) (cid:13) r a t (0) E ˙ c. The rest of the construction is carried out analogously. (cid:3)
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Thus, in contrast to MA we get. Corollary 2.13. MA + ε does not satisfy pure decision. Lower density ≥ ε In this section we investigate the Mathias forcing with lower density ≥ ε .The first observation is. Observation 3.1.
The family of sets F := { A ⊆ ω : d − ( A ) = 1 } is afilter. So, MA − is in fact equivalent to Mathias forcing MA ( F ) with respect tothe filter F .Thus, we get: Fact 3.2. MA − satisfies the countable chain condition. Now, we will see that the forcing MA − ε is a disjoint union of σ -centredforcings. Definition 3.3.
Let
A, B ⊆ ω . The lower density of B with respect to A isdefined by: d − A ( B ) := lim inf n →∞ | A ∩ B ∩ n || A ∩ n | . Lemma 3.4.
Let B ⊆ A ⊆ ω, d − ( A ) > and d − A ( B ) < .Then, d − ( B ) < d − ( A ) . Proof.
Let B ⊆ A ⊆ ω be as in the lemma. Then, d − A ( B ) < ∃ n < ω )( ∀ k < ω )( ∃ m ≥ k ) (cid:18) | B ∩ m || A ∩ m | < − − n (cid:19) . So, we in fact get( ∃ ∞ m ) (cid:0) | B ∩ m | < (1 − − n ) · | A ∩ m | (cid:1) . (cid:3) Corollary 3.5. If d − ( A ) = ε , then F ( A ) := { B ⊆ A : d − ( B ) = ε } is afilter. Proposition 3.6.
Let ε ∈ (0 , .The forcing notion MA − ε is equivalent to a disjoint union of σ -centredforcings.Proof. First, note that D := { p = ( s, A p ) ∈ MA − ε : d − ( A p ) = ε } is an opendense subset of MA − ε . Fix a maximal antichain A ⊆ D . For each p ∈ A therestriction of MA − ε to p is denoted by MA − ε ↾ p = { q ∈ MA − ε : q ≤ p } . Then,each MA − ε ↾ p has the countable chain condition. Indeed, fix p = ( s, A p ) ∈ A .By Corollary 3.5 we know that the set { B ⊆ A p : d − ( B ) = ε } is closedunder finite intersections. Thus, any two conditions q , q ≤ p which havethe same stem, are compatible.Finally, we can set MA − ε = S p ∈A MA − ε ↾ p. (cid:3) Corollary 3.7. MA − ε is proper for each ε ∈ (0 , . Next, we show that MA − ε has antichains of size the continuum, whenever ε < Lemma 3.8.
Let ε ∈ (0 , . There is a family of sets { A f : f ∈ I } suchthat(1) I ⊆ ω has size continuum,(2) d − ( A f ) ≥ ε, for all f ∈ I ,(3) d − ( A f ∩ A g ) < ε , for all f = g in I .Proof. Let ε ∈ (0 , I ⊆ ω . To thisend, let I be any family of functions of size continuum with the property ∀ f, g ∈ I ( f = g → ∃ ∞ i, j ( f ( i ) = 1 > g ( i ) ∧ f ( j ) = 0 < g ( j ))) . (3.1)Next, we define for n ∈ ω the interval I n := [2 n +1 , n +2 ) ⊆ ω . Then, allintervals are pairwise disjoint and each interval I n has size 2 n +1 . Now, wedefine for each f ∈ I a function F f ∈ ω ω via F f ( n ) := n X i =0 f ( i ) · n − i . Observe, that the functions F f have the property that for n ∈ ω ( F f ( n ) ∈ [0 , n +1 )), in other words there are 2 n +1 possible values for F f ( n ).We define the sets A f of lower density ≥ ε recursively over n as a disjointunion A f = S n A nf , such that A nf ⊆ I n . To this end, fix n and let k be theclosest natural number ≤ ε · n +1 , so k := max { j ∈ ω : j ≤ ε · n +1 } . Wedifferentiate two cases:If 2 n +1 + F f ( n ) + k − < n +2 : We set A nf := { n +1 + F f ( n ) , n +1 + F f ( n ) + 1 , . . . , n +1 + F f ( n ) + k − } Otherwise, there is j ≤ k − n +1 + F f ( n ) + j = 2 n +2 − A nf := { n +1 , . . . , n +1 + k − − j } ∪ { n +1 + F f ( n ) , . . . , n +1 + F f ( n ) + j } Finally we set A f := S A nf . It follows directly from the definition that eachset A f has density ε and for n ∈ ω we get ( A f ∩ I n = A nf ).It is left to show that for f = g the set A f ∩ A g has lower density < ε . So,fix f, g ∈ I and let i be minimal such that f ( i ) = g ( i ). W.l.o.g assume f ( i ) = 1 and g ( i ) = 0, so in particular F f ( n ) > F g ( n ) for all n ≥ i . Byproperty (3.1) of I there is i > i such that f ( i ) = 1 and g ( i ) = 0. Then,for all n > i we have | F f ( n ) − F g ( n ) | ≥ n − i + 2 n − i − ( i − X i>i n − i + 2 n − i − > n − i +1 . REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 11
Claim 3.9.
There is a strictly positive constant c ∈ (0 , such that for all n ∈ ω | A f ∩ A g ∩ I n || I n | ≤ ε − c. Clearly, the claim above implies d − ( A f ∩ A g ) < ε . So, fix n ∈ ω . Sincewe are only interested in the size of A f ∩ A g ∩ I n , we might assume, bya transformation argument, that g ( i ) = 0 , i < n and f (0) = 0. Thus, A ng consists of the first k elements of I n i.e., { n +1 , . . . , n +1 + k − } and F f ( n ) − F g ( n ) = F f ( n ) < n . Again, we have to differentiate two cases:If F f ( n ) + 2 n +1 + k − < n +2 : Then, | A nf ∩ A ng ∩ I n | n +1 ≤ k − n − i +1 n +1 ≤ ε − i . Otherwise F f ( n ) + 2 n +1 + k − ≥ n +2 . In this case we must have ε > / | A nf ∩ A ng ∩ I n | n +1 ≤ n +1 − n +1 − k )2 n +1 ≤ · ε − . We can set c := min { − i , (1 − ε ) } . (cid:3) Corollary 3.10. MA − ε has antichains of size continuum for ε ∈ (0 , .Proof. Take the family of sets { A f : f ∈ I } as in the lemma above. Then, { ( hi , A f ) : f ∈ I } ⊆ MA − ε is an antichain of size continuum. (cid:3) Corollary 3.11.
There is no filter F such that MA ( F ) and MA − ε are forcingequivalent.Proof. This follows from the previous corollary, together with the fact, that MA ( F ) is a σ -centered forcing for each filter. (cid:3) Analog to the upper density case MA + ε we get that MA − ε adds Cohenreals. Proposition 3.12. MA − ε adds Cohen reals.Proof. We can repeat the proof of Proposition 2.12. Again we divide ω into N + 1 disjoint sets a i ⊆ ω, i < N + 1 of density 1 / ( N + 1), where N is suchthat 1 /N < ε . Now to carry out the rest of the construction it enough to seethat any set of lower density ε intersects at least two of the sets a i infinitelyoften. (cid:3) Let x G be the generic real added by MA − ε . Then, x G has lower density0 and upper density ε in the generic extension. So one might try to usethe same recipe as in Lemma 2.8, where it was proven that MA + ε addsdominating reals. However, condition (ii) from the proof is not satisfied andthe following question remains open. Question 3.13.
Does MA − ε add dominating reals? Note, a positive answer to this question for ε = 1 would also be a positiveanswer to [19][Question 38] from Raghavan. Upper density > MA + i.e., the forcing consisting of Mathiasconditions p ⊆ <ω such that the corresponding set of splitting levels A p hasstrictly positive upper density. Definition 4.1.
The density zero ideal Z is defined by: Z := { A ⊆ ω : d + ( A ) = 0) } , and the corresponding coideal is denoted by Z + = { A ⊆ : d + ( A ) > } .Observe that for p ∈ MA we have p ∈ MA + iff A p ∈ Z + .We show that the forcing MA + is proper. In order to do this, we need thefollowing Lemma (compare [8, Lemma 9.6.]). Lemma 4.2. MA + is forcing equivalent to the two step iteration of P ( ω ) / Z + ∗ MA ( ˙ F ) , where ˙ F is a P ( ω ) / Z + -generic filter. For sake of completeness we sketch a proof.
Proof.
We define a map i : MA + −→ P ( ω ) / Z + ∗ MA ( ˙ F ) via i (( s, A )) :=([ A ] Z + , ( s, A )), where [ A ] Z + = { B ⊆ ω : A ∆ B ∈ Z} is the correspondingequivalence class. It is not hard to see that i preserves being stronger andbeing incompatible. We show that the range of i is dense. To this end, fix acondition ([ A ] Z + , ( s, ˙ C )). Then ˙ C is a P ( ω ) / Z + -name for an infinite subsetof ω such that [ A ] Z + (cid:13) ˙ C ∈ ˙ F . This implies [ A ] Z + (cid:13) A \ ˙ C ∈ Z . So we get i (( s, A )) = ([ A ] Z + , ( s, A )) ≤ ([ A ] Z + , ( s, ˙ C )) . (cid:3) Corollary 4.3. MA + is proper.Proof. Let
Fin denote the ideal of finite subsets of ω . In [7, Theorem 1.3.]Farah proved that P ( ω ) / Z + is forcing equivalent to the two step iterationof P ( ω ) / Fin and a measure algebra of Maharam character c and therefore isproper. So, MA + is a finite iteration of proper forcings. (cid:3) Theorem 4.4. MA + adds Cohen reals. The Theorem follows from [8, Lemma 9.8.] and the fact that the coideal Z + is not semiselective (compare [6, Definition 2.1.]). However, since wewill make use of the explicit construction of the Cohen real, we also give aproof. We freely identify sequences x ∈ ≤ ω with their corresponding setsof natural numbers x ∈ [ ω ] ≤ ω via x
7→ { n : x ( n ) = 1 } . Proof.
We define maximal antichains A n in ( Z + , ⊆ ∗ ) as follows: For n ∈ ω let A in := { k ∈ ω : k = i mod 2 ( n +1) } and put A n := { A in : i < ( n +1) } ,e.g. A consists of the even and odd numbers. Let x G ∈ ω be a MA + -generic real. Then, in the generic extension V [ x G ], there is for each n ∈ ω exactly one i n < ( n +1) such that x G ⊆ ∗ A i n n . To simplify notations wedenote with A n this unique A i n n . We define two sequences h n i : i ∈ ω i and h m i : i ∈ ω \ { }i as follows: We start with n := min( x G ). When n i isknown, we put m i +1 := min( x G \ ( n i + 1)). To define n i +1 we differentiate REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 13 two cases: If x G \ ( m i +1 + 1) ⊆ A m i +1 we put n i +1 := m i +1 . Otherwise weput n i +1 := min( x G \ (( m i +1 + 1) ∪ A m i +1 )) . Observe that by definition wehave m i ≤ n i . We put c ( i ) := ( , if x G \ ( m i +1 + 1) ⊆ A m i +1 , else.We show that c is Cohen. For this purpose, fix a condition ( s, A ) ∈ MA + .By density we can assume that there is i < ω such that m i , n i and c ↾ i aredecided by ( s, A ) but none of the values of m i +1 , n i +1 nor c ( i ). We musthave max( s ) = n i . Fix j ∈
2. It is enough to find ( t, B ) ≤ ( s, A ) such that( t, B ) (cid:13) c ( i ) = j and in addition ( t, B ) does not decide m i +2 , n i +2 or c ( i +1).Depending on the value of j we distinguish two cases: j = 0: Let m := min( A ) , t := s ∪ { m } and B := A m ∩ A \ ( m + 1). j = 1: Find m ∈ A such that A A m (such an m always exists since A has positive upper density and A m has density 2 − ( m +1) ). Next, pick n ∈ A \ A m and put t := s ∪{ m }∪{ n } . Finally, define B := A \ ( n +1).In both cases we have max( t ) = n i +1 , ( t, B ) decides the value of c ( i ) to be j but neither does it decide m i +2 , n i +2 and nor c ( i + 1). (cid:3) Next, we show that MA + adds dominating reals. Theorem 4.5. MA + adds dominating reals.Proof. Let Z ∗ = { A ⊆ ω : ω \ A ∈ Z} be the dual filter of the density zeroideal. In [11, Corollary 3] Hrušak and Minami showed that MA ( F ) addsdominating reals, whenever F is a filter extending Z ∗ . We use Lemma 4.2.Let ˙ F denote the P ( ω ) / Z + -generic filter. Then, in the generic extension V [ ˙ F ] the filter ˙ F extends ( Z ∗ ) V and therefore MA ( ˙ F ) adds dominatingreals over V . Specifically, MA + adds dominating reals over V . (cid:3) Positive lower density
In this section we show that V − collapses the continuum to ω . We alsoconstruct an uncountable antichain in the partial order consisting of setswith strictly positive lower density ordered by inclusion. Theorem 5.1.
The forcing V − collapses the continuum to ω . We first prove a lemma:
Lemma 5.2.
There is a strictly increasing function ℓ : ω × ω × ω ∪{− } → ω ,a small natural number r and a V − -name ˙ F for a function F : ω × ω × ω → in V [ G ] with the following property: ( ∀ ̺ ∈ ω )( ∀ p ∈ V − )( ∀ k ∈ ω \ { } )( ∀ k ∈ ω ) (cid:16)(cid:0) d − ( A p ) ≥ k ∧ ∀ n ≥ k | A p ∩ [ ℓ ( k, k , n ) , ℓ ( k, k , n + 1)) | ℓ ( k, k , n + 1) − ℓ ( k, k , n ) ≥ k (cid:1) → ( ∃ q ̺ ≤ p ) (cid:0) d − ( A q ̺ ) ≥ rk ∧ q ̺ (cid:13) ( ∀ n ∈ ω ) ˙ F ( k, k , n ) = ̺ ( n ) (cid:1)(cid:17) (5.1) Proof. r , z , z and m will be arranged later. The function ℓ is definedrecursively by ℓ ( k, k , n ) = , if n = − ,z k ( k + 1) , if n = 0 ,mk · ℓ ( k, k , n − , if n > . Although the function ℓ in fact depends on three variables it makes sense toabbreviate it when the values k and k are clear from the context. In sucha situation we also write ℓ n instead of ℓ ( k, k , n ).We let x G be the generic branch and define f ( k, k , n ) = the closest natural number to | x − G [ { } ] ∩ ℓ n | ℓ n · z kz . We define in V [ G ] the following function: F ( k, k , n ) = ( , if f ( k, k , n ) is even;1 , else.So, assume that we are given a condition p ∈ V − that meets the premise ofthe implication (5.1) for k and k i.e., d − ( A p ) ≥ k and ( ∀ n ≥ k ) | A p ∩ [ ℓ n ,ℓ n +1 ) | ℓ n +1 − ℓ n ≥ k . Let ̺ ∈ ω be given. Recall any Silver condition p is uniquely describedby a function f p : ω \ A p →
2. By induction on n < ω we define an increasingsequence of partial functions f p = f q − ⊆ f q ⊆ f q ⊆ . . . such that for all n ≥ f q n ↾ ℓ n − = f q n − ↾ ℓ n − ,(2) f q n ↾ [ ℓ n − , ℓ n ) ⊇ f q n − ↾ [ ℓ n − , ℓ n ) , (3) f q n ↾ [ ℓ n , ∞ ) = f p ↾ [ ℓ n , ∞ ) , From the three conditions above it already follows that each correspond-ing tree q n will be a member of V − . Additionally, we make sure that thefollowing holds as well for n ≥ | A qn ∩ [ ℓ n − ,ℓ n ) | ℓ n − ℓ n − ≥ rk (5) q n (cid:13) ∀ m ≤ nF ( k, k , m ) = ̺ ( m ) . The conditions (1) − (4) together make sure that the decreasing sequence h q n : n < ω i has a lower bound in V − , namely q := T n q n . Condition (5)ensures that q (cid:13) ∀ n ∈ ωF ( k, k , n ) = ̺ ( n ). Thus, we can set q ̺ := q andare done.Now for the step from n − n : We have | A p ∩ [ ℓ n − , ℓ n ) | ℓ n − ℓ n − ≥ k . This means we can add at most ( r − r · k · ( ℓ n − ℓ n − ) elements of [ ℓ n − , ℓ n ) ∩ A p to dom( f q n ) and still make sure that condition (4) is met. We will laterchoose which of them are mapped to 0 and which are mapped to 1 by f q n . REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 15
We define two approximations to f and F respectively, which do notdepend on the generic element x G . We let f ( k, k , n, p ) = the closest natural number to | f − p [ { } ] ∩ ℓ n | ℓ n · z kz , and F ( k, k , n, p ) = ( , if f ( k, k , n, p ) is even;1 , else.Now at most z z · ℓ n k many new values are needed to change f ( k , k, n, p )to f ( k , k, n, q ) such that the quotient | f − p [ { } ] ∩ ℓ n | ℓ n · z kz ∈ ω and such that F ( k , k, n, q ) coincides with ̺ ( n ). However, we have to becareful not to contradict condition (4). Especially, the following must hold: z z · ℓ n k ≤ ( r − r · k · ( ℓ n − ℓ n − ) = ( r − r · k · ( mk − mk · ℓ n On the other hand, we need to ensure that q n decides F ( k, k , n ) to be F ( k , k, n, q n ). By construction and in particular condition (4), we get thatthe amount of digits that are in ℓ n \ dom( f q n ) is ℓ n rk . Hence we need z z · ℓ n k > · ℓ n rk Both inequalities are true if we have:2 · r < z z · ℓ n k ≤ ( r − r · ( mk − mk . We can take, e.g., r = m = 4, z = 3 and z = 2 and get < ≤ · . (cid:3) Proof of the Theorem.
Now it is easy to see that (cid:13) V − ( ∀ ̺ ∈ V ∩ ω )( ∃ k > ∃ k )( ∀ nF ( k, k , n ) = ̺ ( n )) . Simply fix p and ̺ . Compute k and k for p such that the prerequirement ofthe implication (5.1) are fulfilled. This is always possible since d − ( A p ) > q ̺ ≤ p as in the lemma above.Hence, (cid:13) V − ( k, k ) F ( k, k , · ) is a surjection from ω × ω onto 2 ω ∩ V . (cid:3) MA − has large antichains. The following lemma establishes thatbelow any condition p ∈ MA − there is an antichain of size continuum. Proposition 5.3.
There is a family of sets { A f ⊆ ω : f ∈ I } such that(1) I ⊆ ω has size continuum,(2) d − ( A f ) ≥ / , for all f ∈ I .(3) d − ( A f ∩ A g ) = 0 , for all f = g in I . Proof.
First, we fix a suitable set of indices I ⊆ ω . Therefore pick anyfamily of functions I of size continuum with the property, that for any twodifferent functions f, g ∈ I there are infinitely many n ∈ ω ( f ( n ) = g ( n )). Inorder to define the sets A f we will define three auxiliary sets B , B , B ⊆ ω and a sequence h k nj : n ∈ ω, j ∈ ∪ {− }i such thati) k n − < k n < k n < k n = k n +1 − , for n < ω ,ii) k nj − k nj < − n , for j < , n < ω ,iii) d − ( B i ∪ B j ) ≥ /
2, for i = j ,iv) d − ( B i ) = 0, for i < B i recursively over n ∈ ω as a disjoint union of sets { B ni } n such that each set B ni is a subset of [ k n − , k n ). Start by defining B i := ∅ , i < k j := j + 1 , j ∈ ∪ {− } . Now assume we haveconstructed B mi and k mj for m < n, i < j ∈ ∪ {− } . We start with k nj . Let k n − := k n − and choose k nj , j < i ) and ii ) are fulfilled and additionally the difference k nj − k nj − is an evennatural number for each j <
3. We perform three construction steps todefine the sets B ni :a) Divide the interval [ k n − , k n ) evenly between the two sets B n and B n and avoid the set B n entirely i.e.,[ k n − , k n ) ∩ B n := ∅ , [ k n − , k n ) ∩ B n := { k n − , k n − + 2 , . . . , k n − } , [ k n − , k n ) ∩ B n := { k n − + 1 , k n − + 3 , . . . , k n − } . b) Divide the interval [ k n , k n ) evenly between the two sets B n and B n and avoid the set B n entirely i.e.,[ k n , k n ) ∩ B n := ∅ , [ k n , k n ) ∩ B n := { k n , k n + 2 , . . . , k n − } , [ k n , k n ) ∩ B n := { k n + 1 , k n + 3 , . . . , k n − } . c) Divide the interval [ k n , k n ) evenly between the two sets B n and B n and avoid the set B n entirely i.e.,[ k n , k n ) ∩ B n := ∅ , [ k n , k n ) ∩ B n := { k n , k n + 2 , . . . , k n − } , [ k n , k n ) ∩ B n := { k n + 1 , k n + 3 , . . . , k n − } . This completes the construction of the sets B ni and we can put B i := S n B ni .We check that conditions iii) and iv) are fulfilled. Let i = j be given. Byconstruction steps a ) − c ), we know that B i ∪ B j selects at least each sec-ond natural number of each interval [ k n − , k n ). Since the intervals [ k n − , k n )partition ω we get iii). Condition iv) follows from | B i ∩ k ni | k ni ≤ k ni − k ni < − n . REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 17
Now, we are in a position to define the sets A f , f ∈ I . For n ∈ ω and i ∈
3, welet A ni := B ni ∪ B ni , where i and i are chosen such that { i, i , i } = 3. Then,we set A f := S n A nf ( n ) . We have to verify that the sets A f satisfy conditions(2) and (3). That each set A f has lower density ≥ / B i . So let f, g ∈ I be two different functions andtake n such that f ( n ) = g ( n ). W.l.o.g. assume f ( n ) = 0 and g ( n ) = 1.Then, from construction step c) it follows that B ∩ B ∩ [ k n , k n ) = ∅ andthus | A f ∩ A g ∩ k n | k n ≤ k n k n < − n . Since f and g differ on infinitely many n we get d − ( A f ∩ A g ) = 0. (cid:3) The results of the previous sections are summarized in the table below.
P MA + ε V + ε MA − MA − ε V − ε MA + V + MA − V − proper ✓ ✓ ✓ ✓ ✓ ✗ c.c.c ✗ ✗ ✓ ✗ ✗ ✗ ✗ ✗ ✗ Cohen ✓ ✗ ✓ ✓ ✓ ✓ dominating ✓ ✗ ✓ Measurability
In this section we compare different notions of measurability. We firstestablish some notations.
Definition 6.1.
Let X be a non-empty set and P be a tree-forcing definedover X ω .(1) A subset X ⊆ X ω is called P -nowhere dense, if( ∀ p ∈ P )( ∃ q ≤ p )([ q ] ∩ X = ∅ ) . We denote the ideal of P -nowhere dense sets with N P .(2) A subset X ⊆ X ω is called P -meager if it is included in a countableunion of P -nowhere dense sets. We denote the σ -ideal of P -meagersets with I P .(3) A subset X ⊆ X ω is called P -measurable if( ∀ p ∈ P )( ∃ q ≤ p )([ q ] \ X ∈ I P ∨ [ q ] ∩ X ∈ I P ) . (4) A family Γ ⊆ P ( X ω ) is called well-sorted if it is closed under con-tinuous pre-images. We abbreviate the sentence “every set in Γ is P -measurable” by Γ( P ).We make two useful observations concerning the measurability of a set X . Observation 6.2.
Let P be a tree-forcing and X ⊆ X ω a set of reals.(1) If H is P -comeager, then X is P -measurable if and only if H ∩ X is P -measurable. (2) If I P = N P , then X is P -measurable if and only if ( ∀ p ∈ P )( ∃ q ≤ p )([ q ] ⊆ X ∨ [ q ] ∩ X = ∅ ) . For Sacks, Laver, Miller, Silver and Mathias forcing we have N P = I P . Incase of the Silver forcing the usual proof for I V = N V also works for V + ε ,since it only makes use of fusion sequences. But it is unclear whether wecan expect the same for the other versions of Silver forcing.In case of the Mathias forcing however, we don’t have an equality in noneof the four versions of the forcing. Theorem 6.3. (1) I V + ε = N V + ε ,(2) I P = N P for P ∈ { MA + ε , MA − ε , MA + , MA − } .Proof. (1) As we mentioned above, the first part of the Theorem is a straight-forward generalization of the proof for the usual Silver forcing. The onlydifference being that one has to use the partial orderings defined as in Def-inition 2.3 to ensure that fusions exist in V + ε . (2) We divide the proof into cases.( ≥ ε ) Let ε ∈ (0 ,
1] be given. The proof is closely intertwined with the factthat the forcings MA + ε and MA − ε add Cohen reals. We quickly recallthe construction of the Cohen real from Lemma 2.12 and define afunction ϕ such that the image of the generic real is a Cohen real.Fix ε > , N ∈ ω such that 1 /N < ε and a partition { a i } i 6∈ N P .( > 0) In this case the meager set which is not nowhere dense lies directlyat hand. Claim 6.4. The set Z + is P -meager but not P -nowhere dense, for P ∈ { MA + , MA − } . The proof works for both forcings analogously. We only check itfor MA + explicitly. For n ∈ ω we define the sets N n := { x ∈ ω : d + ( x ) ≥ /n } . Let n ∈ ω and p ∈ MA + be fixed. We can easilyfind q ≤ p such that d + ( A q ) < /n . Such a condition q satisfies theproperty ∀ x ∈ [ q ]( d + ( x ) < /n ) and in particular [ q ] ∩ N n = ∅ . Thisproves that each set N n is MA + -nowhere dense and so Z + = S n N n is MA + -meager. To see that Z + cannot be MA + -nowhere dense itis enough to check that each condition p = ( s, A p ) ∈ MA + contains REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 19 a branch x of positive upper density. Clearly, the rightmost branch(i.e. x ( i ) = 1 ⇔ i ∈ s ∪ A p ) fulfills this requirement. (cid:3) Since subsets of P -meager sets are P -meager as well, P -meager and P -comeager sets are P -measurable we get the following. Corollary 6.5. The sets Z , Z ∗ and Z + are P -measurable, P ∈ { MA + , MA − } . Our next goal is to compare different notions of measurability. Let P beany tree-forcing. The statement“Γ( P ) ⇒ Γ( C ), for each well-sorted family Γ”is true for various tree-forcings adding Cohen reals e.g. Hechler forcing D ,Eventually different forcing E , a Silver like version of Mathias forcing T introduced in [14, Definition 2.1.] and the full-splitting Miller forcing FM [12, Definition 1.1.]. In fact, it appears reasonable enough to ask. Question 6.6. Does each tree-forcing P adding a Cohen real, necessarilysatisfy Γ( P ) ⇒ Γ( C ) , for each well sorted family Γ ? A partial answer to this question was given in [14, Proposition 3.1.]. Werestate the proposition in a slightly modified version, that will allow us togeneralize the result later on. Proposition 6.7. Let X be a set of size ≤ ω and P be tree-forcing defined on X <ω . Equip X with the discrete topology and X ω with the product topology.Let ϕ ∗ : X <ω → <ω and ϕ : X ω → ω be two mappings that satisfy thefollowing conditions:(1) ϕ ∗ is order preserving and ϕ ∗ ( hi ) = hi ,(2) ϕ is continuous,(3) ( ∀ p ∈ P ) ϕ [ p ] is open dense in [ ϕ ∗ (stem( p ))] ,(4) ( ∀ p ∈ P ) ∀ t D ϕ ∗ (stem( p )) ∃ p ′ ≤ p ( ϕ ∗ (stem( p ′ )) D t ) . Then Γ( P ) ⇒ Γ( C ) , for each well-sorted family. To understand why Proposition 6.7 is a partial answer to Question 6.6,it should be noted that given P , ϕ as in the proposition and x G a P -genericreal, we have that ϕ ( x G ) is Cohen. Theorem 6.8. Γ( MA + ε ) ⇒ Γ( C ) , for each well sorted family Γ .Proof. Our aim is to apply Proposition 6.7. To this end let H ∗ := { s ∈ <ω : | s − ( { } ) | is even } denote the set of all finite sequences with an evennumber of 1 ′ s . For s ∈ H ∗ let h n k : k < | s − ( { } ) |i enumerate s − ( { } ).Observe that H ∗ is dense in 2 <ω . Now fix N ∈ ω and a partition { a i } i< ( N +1) of ω such that ε > / ( N +1) and such that each set a i has density 1 / ( N +1).We define the function ϕ ∗ as expected. Let k ∈ ω and s ∈ H ∗ be such that2 k < | s − ( { } ) | . We define: ϕ ∗ ( s )( k ) := ( , ∃ i < ( N + 1)( { n k , n k +1 } ⊆ a i )1 , else. Let H and ϕ : H → <ω be defined as in the proof of Theorem 6.3 (2).Now, given x ∈ H let h n xk : k < ω i enumerate x − { } . Then, ϕ ∗ ( x ↾ n x k ) isdefined for each k and ϕ ( x ) = S k ϕ ∗ ( x ↾ n x k ).Note that the same argument used in the proof to show that MA + ε addsCohen reals, gives us ϕ [ p ] = [ ϕ ∗ (stem( p ))]. Especially, condition (3) fromthe proposition is fulfilled. It is straightforward to check that the other threeconditions are satisfied as well and since the set H is MA + ε -comeager, wecan apply the proposition. (cid:3) MA + -measurability In this section we examine MA + -measurability and give a generalizationof Proposition 6.7.By Theorem 4.4 we know that MA + adds Cohen reals. So, it seems reason-able enough to try the same method used in Section 6 to prove Γ( MA + ε ) ⇒ Γ( C ), with the forcing MA + . However, in doing so one encounters the fol-lowing problem. The coding from Theorem 4.4 used to generate the Cohenreal, uses information of the whole condition and does not depend solely onthe stem. To make it clear what we mean, we quickly explain how using theproof of Theorem 4.4 one gets a coding function defined on a MA + -comeagerset.7.1. Construction of the coding function ϕ . Recall that we definedin the proof of Theorem 4.4 families A n := { A in : i < ( n +1) } , where A in := { k ∈ ω : k = i mod 2 ( n +1) } . Each A n is a maximal antichain in( Z + , ⊆ ∗ ) and each set A in has density 2 − ( n +1) . Now we define D n := { A ∈ [ ω ] ω : ∃ i < − ( n +1) ( A ⊆ ∗ A in ) } . Then for each n ∈ ω and ( s, A ) ∈ MA ( Z + )there is B ∈ D n such that ( s, B ) ≤ ( s, A ). This already implies that the set D := T D n is MA ( Z + )-comeager and thus it is enough to find a suitablecoding function ϕ defined on D instead of 2 ω . First, note that for x ∈ D we also must have the property ( ∀ n ∃ ! i n ( x ⊆ ∗ A i n n )) and we abbreviate A i n n with A n . So, we can define for x ∈ D two sequences h n i : i < ω i and h m i : i < ω \ { }i as in the proof of Theorem 4.4. Finally, ϕ : D → ω isdefined as ϕ ( x )( i ) := ( , if x \ ( m i +1 + 1) ⊆ A m i +1 , else.This definition of the coding function ϕ seems promising. However, when onetakes a close look at possible candidates for a corresponding ϕ ∗ , it becomesclear that ϕ ∗ cannot depend solely on the stem of a condition. Especially,we cannot apply Proposition 6.7 as it is stated in Section 6. We have to finda generalization like the following. Proposition 7.1. Let X , Y be sets of size ≤ ω , P , Q be tree-forcings definedon X <ω , Y <ω respectively. Equip X , Y with the discrete topology and X ω , Y ω with the product topology. Let ϕ ∗ : P → Q and ϕ : X ω → Y ω be two mappingsthat satisfy the following conditions: REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 21 (1) ϕ ∗ is order preserving and ϕ ∗ (1 P ) = 1 Q ,(2) ϕ is continuous,(3) ∀ p ∈ P ϕ [ p ] is Q -open dense in [ ϕ ∗ ( p )] ,(4) ∀ p ∈ P ∀ q ≤ ϕ ∗ ( p ) ∃ p ′ ≤ p ( ϕ ∗ ( p ′ ) ≤ q ) . Then Γ( P ) ⇒ Γ( Q ) , for each well-sorted family Γ . The key difference is that ϕ ∗ is a map from P to Q , instead of beingdefined for finite sequences.Before we turn to the proof of the proposition we investigate further if wemight apply it to MA + and C . To this end we already have defined a MA + -comeager set D and a coding function ϕ : D → ω satisfying ϕ ( x G ) is Cohen,where x G is MA + -generic. We want to define ϕ ∗ : MA + → C . Let ˙ x G bethe canonical name for the MA + -generic real. For p ∈ MA + let r p ∈ <ω bemaximal such that p (cid:13) ϕ ( ˙ x G ) D r p and put ϕ ∗ ( p ) := r p . Observe that wehave the following properties: • ϕ ∗ is order preserving and ϕ ∗ (( hi , ω )) = hi , • ϕ is not continuous, • It follows from the proof that ϕ ( ˙ x G ) is Cohen, that conditions (3)and (4) of Proposition 7.1 are satisfied.This shows that we almost get Γ( MA + ) ⇒ Γ( C ), for any well-sorted familyΓ. In fact, the only time we need ϕ to be continuous is to ensure that thepre-image of a regular set Y ∈ Γ is again regular. So, if we change therequirement of Γ of being well-sorted and instead assume that the family ofsets Γ is closed under pre-images of the ϕ constructed above we get: Corollary 7.2. Let ϕ be defined as in the beginning of this section and Γ bea family of sets closed under pre-images of ϕ . Then Γ( MA + ) ⇒ Γ( C ) holds. Now we prove Proposition 7.1. The key step is the following lemma. Lemma 7.3. Let P , Q , ϕ, ϕ ∗ be as in the Proposition and Y ⊆ Y ω . Define X := ϕ − [ Y ] . Assume there is p ∈ P such that X ∩ [ p ] is P -comeager in [ p ] .Then Y ∩ [ ϕ ∗ ( p )] is Q -comeager in [ ϕ ∗ ( p )] .Proof. We are assuming X ∩ [ p ] is P -comeager, for some p ∈ P . This impliesthat there is a collection { A n : n < ω ∧ A n is P -open dense in [ p ] } such that T n A n ⊆ [ p ] ∩ X . W.l.o.g. assume A n ⊇ A n +1 , for all n . Let q = ϕ ∗ ( p ). Wewant to show that ϕ [ X ] ∩ [ q ] = Y ∩ [ q ] is Q -comeager in [ q ] i.e., we want tofind { B n : n < ω } Q -open dense sets in [ q ] such that T n B n ⊆ Y ∩ [ q ]. Given σ ∈ c <ω we recursively define on the length of σ a set { p σ : σ ∈ c <ω } ⊆ P with the following properties: p hi = p , ∀ σ ∈ c <ω S i [ ϕ ∗ ( p σ a i )] is Q -open dense in [ ϕ ∗ ( p σ )], ∀ σ ∈ c <ω ∀ i ∈ ω ([ p σ a i ] ⊆ T k ≤| σ | A k ∧ p σ a i ≤ p σ ).Assume we are at step n . Fix σ ∈ c n arbitrarily and then put q σ = ϕ ∗ ( p σ ).We first make sure that 2 . holds. For this purpose, let { q i : i < c } enumerateall conditions in Q below q σ . By condition (4) from Proposition 7.1 we can find p i ≤ p σ such that ϕ ∗ ( p i ) ≤ q i . Since each A k is P -open dense in [ p ] wecan find for each i < c an extension p σ a i ≤ p i such that [ p σ a i ] ⊆ T k ≤ n A k .This ensures that also 3 . holds. Finally, we put B n := S { ϕ [[ p σ ]] : σ ∈ c n } .We have to check that each set B n is Q -open dense in [ q ] and T n B n ⊆ Y ∩ [ q ].So fix n ∈ ω and q ′ ≤ q = ϕ ∗ ( p ). In the first construction step this q ′ wasenumerated, say by i < c so q ′ = q i and ϕ ∗ ( p h i i ) ≤ q ′ . Especially, ϕ ∗ ( p σ ) ≤ q ′ , whenever σ ∈ c n , σ (0) = i . By condition (3) from the Proposition ϕ [[ p σ ]]is Q -open dense in [ ϕ ∗ ( p σ )]. This proves that B n is Q -open dense in [ q ] . By construction of B n +1 we know B n ⊆ ϕ [ T k ≤ n +1 A k ] and hence \ B n ⊆ ϕ [ \ n A n ] ⊆ ϕ [[ p ] ∩ X ] ⊆ ϕ [ p ] ∩ Y ⊆ [ q ] ∩ Y. (cid:3) Proof of the proposition. Let Y ∈ Γ be given and put X := ϕ − [ Y ]. Thenalso X ∈ Γ, since Γ is well-sorted and ϕ is continuous. We now use thelemma to show that for every q ∈ Q there exists q ′ ≤ q such that Y ∩ [ q ′ ] is Q -meager or Y ∩ [ q ′ ] is Q -comeager.Observe that by conditions (1) and (4) we get ϕ ∗ [ P ] is dense in Q . Nowfix q ∈ Q arbitrarily and pick p ∈ P such that ϕ ∗ ( p ) ≤ q . By assumption X is P -measurable, and so: • in case there exists p ′ ≤ p such that X ∩ [ p ′ ] is P -comeager; put q ′ := ϕ ∗ ( p ′ ). By the lemma above, Y ∩ [ q ′ ] is Q -comeager in [ q ′ ]; • in case there exists p ′ ≤ p such that X ∩ [ p ′ ] is P -meager, then applythe lemma above to the complement of X , in order to get Y ∩ [ q ′ ]be meager in [ q ′ ], with q ′ := ϕ ∗ ( p ′ ). (cid:3) Corollary 7.4. Let P , Q , ϕ ∗ and ϕ be as in proposition. Then ϕ ( x G ) is Q -generic, where x G is a P -generic real.Proof of the Corollary. Fix an Q -open dense set D ⊆ Q . We want to showthat the conditions p ∈ P such that p (cid:13) ∃ q ∈ D ( ϕ ( x G ) ∈ [ q ]) is dense in P . To this end, fix a condition p ∈ P . Then, since D is dense in Q there is q ′ ∈ D below ϕ ∗ ( p ). By condition (4) of Proposition 7.1 there is p ′ ≤ p suchthat ϕ ∗ ( p ′ ) ≤ q ′ . By condition (3) we know that each condition r ∈ P forces ϕ ( x G ) into [ ϕ ∗ ( r )] and hence p ′ (cid:13) ϕ ( x G ) ∈ [ ϕ ∗ ( p ′ )] ⊆ [ q ′ ]. (cid:3) In light of this one might also generalize Question 6.6 to the following. Question 7.5. Let P , Q be two tree-forcings and assume P adds a Q -generic.Does Γ( P ) ⇒ Γ( Q ) hold, for each well-sorted family Γ ? A model for Σ ( V + ε ) ∧ ¬ Σ ( C )We construct a model in which the implication Γ( V + ε ) ⇒ Γ( C ) fails forΓ = Σ . REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 23 We recall that the shortest splitting node extending s ∈ <ω is denoted bysplsuc( s ) (see Definition 1.1 (e)). Lemma 8.1. Let ε ∈ (0 , and p ∈ V + ε . Let ¯ ϕ : Split( p ) → <ω such that ¯ ϕ (stem( p )) := hi and for every t ∈ Split( p ) and j ∈ { , } , ¯ ϕ (splsuc( t a h j i )) := ¯ ϕ ( t ) a h j i . Let ϕ : [ p ] → ω be the expansion of ¯ ϕ , i.e. for every x ∈ [ p ] , ϕ ( x ) := S n ∈ ω ¯ ϕ ( t n ) , where h t n : n ∈ ω i is a E -increasing sequence of splitting nodesin p such that x = S n ∈ ω t n .If c is Cohen generic over V , then V [ c ] | = ∃ p ′ ∈ V + ε ∧ p ′ ⊆ p ∧ ∀ x ∈ [ p ′ ]( ϕ ( x ) is Cohen over V ) . Proof. For a finite tree T ⊆ <ω we define the set of terminal nodes Term( T ) := { s ∈ T : ¬∃ t ∈ T ( s ⊳ t ) } . Consider the following forcing P consisting offinite trees T ⊆ <ω such that for all s, t ∈ T the following holds:(1) If s, t ∈ Term( T ), then | s | = | t | .(2) If s, t Term( T ) and | s | = | t | , then s a i ∈ T iff t a i ∈ T , i ∈ P is ordered by end-extension: T ′ ≤ T iff T ′ ⊇ T and ∀ t ∈ T ′ \ T ∃ s ∈ Term( T )( s E t ).Note P is countable and non-trivial, thus it is equivalent to Cohen forcing C . Let p G := S G , where G is P -generic over V . We claim that p ′ := ¯ ϕ − ” p G satisfies the required properties. It is left to show that:(1) for every x ∈ [ p ′ ] one has ϕ ( x ) is Cohen generic, i.e., every y ∈ [ p G ]is Cohen generic;(2) p ′ ∈ V + ε .For proving (1), let D be an open dense subset of C and T ∈ P . It isenough to find T ′ ≤ T such that every t ∈ Term( T ′ ) is a member of D .Let { t j : j < N } enumerate all terminal nodes in T and pick r N , sothat for every j < N , ( t j a r N ∈ D ). Then put T ′ := { t ∈ <ω : ∃ t j ∈ Term( T )( t E t j a r N } . Hence T ′ ≤ T and T ′ (cid:13) ∀ y ∈ [ p G ] ∃ t ∈ ( T ′ ∩ D )( t ⊳ y ).Hence we have proven that (cid:13) C ∀ y ∈ [ p G ] ∃ t ∈ D ( t ⊳ y ) , which means every y ∈ [ p G ] is Cohen generic over V .For proving (2), one has to verify that the resulting set of splitting levels A p ′ has upper density ≥ ε . It is easy to see that given any condition T ∈ P and n ∈ ω one can always find an end-extension T n ≤ T such that T n (cid:13) ∃ k < ω (cid:18) | A p ′ ∩ k | k ≥ ε − − n (cid:19) . (cid:3) Proposition 8.2. Let C ω be an ω -product with finite support and let G be C ω -generic over the constructible universe L . Then, for every ε ∈ (0 , L ( R ) L [ G ] | = “All On ω -definable sets are ( V + ε ) -measurable” ∧ ¬ Σ ( C ) . Proof. The argument is the same as in the proof of [2, Proposition 3.7]. Fix ε ∈ (0 , α ≤ ω C α denote the forcing adding α Cohen reals. Let G be C ω -generic over L .Let X be an On ω -definable set of reals, i.e. X := { x ∈ ω : ψ ( x, v ) } for aformula ψ with a parameter v ∈ On ω , and let p ∈ V + ε . We aim to find q ≤ p such that [ q ] ⊆ X or [ q ] ∩ X = ∅ .We can find α < ω such that v, p ∈ L [ G ↾ α ]. Let ϕ : [ p ] → ω be asin Lemma 8.1. Let c = G ( α ) be the next Cohen real and write C for the α -component of C ω . We let b = qJ ( ψ ( ϕ − ( c ) , v )) K C α = y C and b = qJ ψ ( ϕ − ( c ) , v )) K C α = y C . Then, by C -homogeneity, b ∧ b = and b ∨ b = . Hence, by applyingLemma 8.1, one can then find q ≤ p such that q ⊆ b or q ⊆ b and for every x ∈ [ q ], ϕ ( x ) is Cohen over L [ G ↾ α ]. We claim that q satisfies the requiredproperty. • Case q ⊆ b : note for every x ∈ [ q ], ϕ ( x ) is Cohen over L [ G ↾ α ],and so L [ G ↾ α ][ ϕ ( x )] | = J ψ ( ϕ − ( ϕ ( x )) , v ) K C α +1 . Hence L [ G ] | = ∀ x ∈ [ q ]( ψ ( x, v )), which means L [ G ] | = [ q ] ⊆ X . • Case q ⊆ b : we argue analogously and get L [ G ] | = ∀ x ∈ [ q ]( ¬ ψ ( x, v )),which means L [ G ] | = [ q ] ∩ X = ∅ .Moreover in L [ G ] it is well-known that Σ ( C ) fails (see [3, Theorem 5.8] and[1, 6.5.3, p. 313]). Hence in L [ G ] all On ω -definable sets are V + ε -measurable,but there is a Σ set not satisying the Baire property. As a consequence, inparticular we obtain L ( R ) L [ G ] | = Σ ( V + ε ) ∧ ¬ Σ ( C ). (cid:3) We conclude this section by summarizing our results from Sections 6,7and 8. P MA + ε MA + V + ε I P = N P ✗ ✗ ✓ Γ( P ) ⇒ Γ( C ) for all Γ Γ = P ( ω ) Σ ( V + ε ) Σ ( C )9. SP does not have the Sacks property In this section we return to one of our questions from [13]. First, we recallthe definition of splitting tree. Definition 9.1. A tree p ⊆ <ω is called splitting tree , short p ∈ SP , if forevery t ∈ p there is k ∈ ω such that for every n ≥ k and every i ∈ { , } there is t ′ ∈ p , t E t ′ such that t ′ ( n ) = i . We denote the smallest such k by K p ( t ). The set SP is partially ordered by q ≤ SP p iff q ⊆ p .On SP we can define a stronger n relation. Definition 9.2. For two conditions p, q ∈ SP we let q ≤ n p if: q ≤ p ∧ Split ≤ n ( p ) = Split ≤ n ( q ) ∧ ∀ t ∈ Split ≤ n ( p )( K p ( t ) = K q ( t )) . REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 25 Assume we are given a condition p ∈ SP and a node t ∈ p . Then, foreach q ≤ p ↾ t and n > K p ( t ) there are two nodes t , t ∈ Lev n ( q ) such that t i ( n − 1) = i, ( i ∈ Definition 9.3. Let p ∈ SP . For each t ∈ p and each n > K p ( t ) we define(9.1) K ( p, t, n ) = min {| Lev n ( q ) | : q ≤ p ↾ t ) } The key observation is Lemma 9.4. For q ≤ p , t ∈ q , and n > K q ( t ) = K p ( t ) we have (9.2) K ( q, t, n ) ≥ K ( p, t, n ) . Proof. Fix r ≤ q ↾ t such that | Lev n ( r ) | = K ( q, t, n ). Then, also r ≤ p ↾ t since K q ( t ) = K p ( t ) and thus | Lev n ( r ) | ≥ K ( p, t, n ). (cid:3) However, the observed property is not yet strong enough, since it needsthe proviso K q ( t ) = K p ( t ). Theorem 9.5. There is p ∈ SP such that for any q ≤ p , t ∈ q , ( ∀ ∞ ℓ )( ∀ t ∈ Lev ℓ ( q ))( | t − { }| < log ( ℓ ) )(9.3)The proof of the theorem consists of the next three lemmata. Definition 9.6. To improve readability we fix the following sequence h l n : n < ω i given by l n = 2 n − . Lemma 9.7. There is a perfect tree T such that(1) K T ( ∅ ) = 0 ,(2) ( ∀ ℓ )( | Lev ℓ ( T ) | = ℓ + 1) ,(3) ( ∀ n )( ∀ ℓ ∈ [ ℓ n , ℓ n +1 )( ∀ t ∈ Lev ℓ ( T )) | t − { }| ≤ n ,(4) ( ∀ t ∈ T )( ∀ n ∈ ω )( t a h , . . . , | {z } n i ∈ T ) .Proof. We let Lev ( T ) = {h i , h i} . We go by induction on n . SupposeLev ℓ n ( T ) is ≤ lex - increasingly enumerated by s ℓ n , , . . . , s ℓ n ,ℓ n . Now for 1 ≤ j < ℓ n we let Lev ℓ n + j ( T ) be increasingly enumerated by s ℓ n + j,i , i ≤ ℓ n + j + 1, as follows s ℓ n + j,i = s ℓ n + j − ,i a , for i < j − ,s ℓ n + j, j − = s ℓ n + j − , j − a ,s ℓ n + j, j − = s ℓ n + j − , j − a ,s ℓ n + j,i +1 = s ℓ n + j − ,i a , for i ∈ [2 j − , ℓ n + j ) . Then each t ∈ Lev ℓ n +1 ( T ) has exactly one i ∈ [ ℓ n , ℓ n +1 ) such that t ( i ) = 1.Items (1), (2) and (4) follow immediately from the construction. (cid:3) In many cases a picture of the object helps to understand the cumbersomeindices. Here is the specific construction of T up to level l = 7: ∅ 10 1000 10000 10000 100 10000000 10000000 Definition 9.8. For T ⊆ <ω and s ∈ <ω we let s a T = { s a t : t ∈ T } . Definition 9.9. By induction on m we define T m . T is as in Lemma 9.7.Now suppose that T m is defined. For each t ∈ Split( T m ) \ T m − (for m = 0,we let T − = {∅} ) we choose the minimal n such that | t | ∈ [ ℓ n , ℓ n +1 ). Nowwe let f ( t ) = t a h , , , . . . , i such that the string of zeros is so long that | f ( t ) | = ℓ n +1 . Note that f ( t ) ∈ T m . Now we let T m +1 = [ { f ( t ) a T : t ∈ Split( T m ) \ T m − } . Note that the T is on purpose. We let p = S { T m : m < ω } . Lemma 9.10. Let p be as in Definition 9.9. Then p ∈ SP .Proof. Let t ∈ Split( p ) and let m be minimal such that t ∈ T m . If m = − t = stem( T ) = ∅ and already K T ( t ) = 0. Now assume that m ≥ f ( t ) ∈ T m and f ( t ) is the stem of f ( t ) a T ⊆ T m +1 . So K T m +1 ( t ) = | f ( t ) | . In particular K p ( t ) is defined and at most | f ( t ) | . (cid:3) Lemma 9.11. Let p be as in the definition and ℓ ∈ ω . For any t ∈ Lev ℓ ( p ) ,there are at most log ( ℓ ) elements i < | t | such that t ( i ) = 1 .Proof. For each t ∈ p , | t | ≤ ℓ n +1 , there is some j ≤ n ≤ log ( | t | ) and are t i ∈ T i , i ≤ j , such that t ⊳ f ( t ) E t ⊳ f ( t ) E · · · E t j = t and for i ≤ j there are n i ≤ n such that n i < n i +1 for 0 ≤ i ≤ j and and | t i | ∈ [ ℓ n i , ℓ n i +1 ). We let t − = ∅ . Now by the properties of T , the number REES WITH POSITIVE UPPER DENSITY, COMPILED February 12, 2021 27 of k ∈ [ | t i − | , | t i | ) such that t i ( k ) = 1 is bounded by n i +1 ≤ n + 1 ≤ log ( ℓ ).Since also j ≤ log ( ℓ ), we have | t − { }| ≤ log ( ℓ ) . (cid:3) Now we turn our attention to the Sacks property and the question whether SP satisfies this property. Definition 9.12. (1) h S n : n < ω i is called an f -slalom if S n ⊆ [ ω ] f ( n ) .(2) A forcing P has the Sacks property if for any f : ω → ω \ {∅} suchthat lim n f ( n ) = ∞ and any P -name τ for a real and any condition p there is an f -slalom h S n : n < ω i and there is q ≤ p such that q (cid:13) ( ∀ ∞ n )( τ ( n ) ∈ S n ) . Theorem 9.13. At least under a condition p as in Theorem 9.5 the forcing SP does not satisfy the Sacks property. The proof consists of the following two lemmata. Lemma 9.14. Let p be as in the Theorem 9.5. Then ( ∀ q ≤ p )( ∀ t ∈ Split( q ))( ∀ ∞ n ) (cid:16) K ( q, t, ℓ n ) ≥ ℓ n · log ( ℓ n ) (cid:17) . Proof. Suppose not and fix a q ≤ p and a node t ∈ Split( q ) such that K q ( t ) = k and ( ∃ ∞ n ) K ( q, t, ℓ n ) < ℓ n · log ( ℓ n ) =: M n . We fix such an n such that ℓ n > k . We fix a slimmest K q ( t ) witnessingtree W ⊆ Lev ≤ ℓ n ( q ↾ t ) of width K ( q, t, ℓ n ) < M n at level ℓ n . That means | Lev ℓ n ( W ) | = K ( q, t, ℓ n ) and on each height i ∈ [ k, ℓ n ) there is at least one s ∈ W such that s ( i ) = 1. Hence there is s ∈ Lev ℓ n ( W ) such that s hasat least M n · ( ℓ n − k ) many i ∈ [ k, ℓ n ) such that s ( i ) = 1. We pick such an s ∈ W ∩ Lev ℓ n ( q ). However now s shows1 M n · ( ℓ n − k ) > ℓ n · M n = log ( ℓ n ) . This contradicts the property (9.3) that says that there are at most log ( ℓ n ) many i < | s | such that s ( i ) = 1. (cid:3) Now we apply the next lemma with g ( n ) = n · log ( n ) . Lemma 9.15. Let g : ω → ω be such that lim n →∞ g ( n ) = ∞ . Let p be suchthat ( ∀ q ≤ p )( ∀ t ∈ Split( q ))( ∀ ∞ n )( K ( q, t, ℓ n ) ≥ g ( ℓ n )) . For any function f : ω → ω \ {∅} there is a SP -name τ such that for any f -slalom h S n : n <ω i ∈ V p (cid:13) SP ( ∃ ∞ n )( τ ( n ) S n ) . Proof. Fix f ∈ ω ω ∩ V . We give a name ˙ h f ∈ ω ω ∩ V SP so that for everyslalom S ∈ ([ ω ] <ω ) ω ∩ V , with | S ( n ) | ≤ f ( n ), we have that p forces that ˙ h f is not captured by S , i.e., p (cid:13) ( ∃ n ∈ ω ) ˙ h f ( n ) / ∈ S ( n ) . We let ˙ x be the name for the SP -generic real, i.e. the union S { stem( p ) : p ∈ G } .Let code : 2 <ω → ω be an injective function.We fix an increasing subsequence h m f,i : i < ω i ⊆ h ℓ n : n < ω i such that( ∀ i )( g ( m f,i ) > f ( i )) . We define τ = ˙ h f := h code( ˙ x ↾ m f,n ) : n ∈ ω i . We aim to show that any q ≤ p forces that ˙ h f cannot be captured by any f -slalom in the ground model. So fix an f -slalom S ∈ V and q ≤ p .Let t = stem( q ). It is enough to find n ∈ ω , r ≤ q such that r (cid:13) ˙ h f ( n ) / ∈ S ( n ). Pick n ∈ ω such that for any n ′ ≥ n K ( q, t, m f,n ′ ) ≥ g ( m f,n ′ ) > f ( n ′ ) . Hence | Lev m f,n ( q ↾ t ) | ≥ g ( m f,n ) > f ( n ). Let { t k ∈ q ↾ t : | t | = m f,n , k References [1] Tomek Bartoszyński and Haim Judah. 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