aa r X i v : . [ m a t h . L O ] A ug Splitting Localization and Prediction Numbers
Iv´an Ongay-Valverde ∗ Department of MathematicsUniversity of Wisconsin–MadisonEmail: [email protected]
First Draft: September 15, 2017Current Draft: August 7, 2019
Abstract
In this paper the work done by Newelski and Roslanowski in [9] isrevisited to solve a question posed by Blass about one of the possibleevasion and prediction numbers (see [3]). This led to define a variationof the k -localization property (the ( k + 1) ω -localization property) andthe use of a forcing notion with accelerating trees. In 1993 Newelski and Roslanowski defined the k -localization number, L k (see [9]), as the minimal cardinality of a family T of k -trees such that everyelement ( k + 1) ω is a branch of a tree in T . In their paper, they proved that L k +1 ≤ L k and that it is consistent tohave L k +1 < L k . In order to do this, they introduce the k -localization prop-erty that was later studied by Roslanowski [10] and Zapletal [12]. Theseproperties were also used by Geschke [6] to show that it is consistent tohave L i = f ( i ) for any non-increasing function from a natural number tothe cardinals with uncountable cofinality. ∗ Work done while being supported by CONACYT scholarship for Mexican studentstudying abroad. In their work, they originally studied ideals of unsymmetric games. The coveringnumbers of those ideals are the ones that we called k -localization numbers. The k -localization property says “all the reals in ω ω of the generic extension are abranch of a k -tree from the ground model.” k -localization property is not the minimum necessaryto have L Vk = L V [ G ] k . The minimum that we need to have would be theproperty “all the reals in ( k + 1) ω of the generic extension are a branchof a k -tree from the ground model” (we call this the ( k + 1) ω -localizationproperty). Does this mean that there is a cardinal characteristic that iscloser to the k -localization property than L k ?There is one. In his chapter of the Handbook of Set Theory [3], AndreasBlass talks about cardinal characteristics related to the concepts of evasionand prediction. At the end of that section, he introduces 36 variations ofthese cardinals and left as an open question to pin down 4 of them whoseidentity didn’t appear to be one of the known cardinal characteristic. Itturns out that the same proof of Newelski and Roslanowski shows that oneof these variations, specifically the prediction number for global adaptive k predictors, is not one of the known cardinal characteristics and, actually,gives countable many cardinal characteristics which, consistently, can takedifferent values (see Theorem 2.4).This triggers the following question: are the variation of prediction andthe k -localization number equal? This paper shows that they are not. It isconsistent to have all the prediction numbers mention above at value ℵ = c and all localization numbers at value ℵ (see Theorem 4.1).To do this we use a forcing that Noah Schweber and the author calledaccelerating tree forcing. We created it for a computability theory questionin a coauthor paper still in preparation and we will show in this paperthat countable support product of the accelerating tree forcing has the 3 ω -localization property.It has been pointed out to us that the accelerating tree forcing couldbe related to bushy tree forcing (as done in [8]) or other fast-growing treeforcing (as done in [4]). Although we got inspiration from them, the fact thatwe allowed long stretches with no split makes us believe that this forcing isconceptually of a different kind.It is also important to remark that countable support iteration and prod-uct of forcings with the ( k + 1) ω -localization property could also have the( k + 1) ω -localization property, as the k -localization property (see [12]). No-tice that these two properties are in the same line as the Sacks property. Itis unknown to the author if there is a bigger theory or theorem that handleall of them at once. This, we believe, is an interesting topic.About the structure of the paper, it has a first section with definition k +1) ω localization property are shown. Section 4 hasthe main theorem (Theorem 4.1) with some conclusions and open problems.The last section was added after the paper was first submitted, it includesresults that the author discover after sharing the work here presented.Finally, the author wants to thank Noah Schweber for his support andfor convincing me about publishing this work; Arnold Miller, for showingme a new way to order my thoughts, and Kenneth Kunen for all the adviceand guidance with this project and others.Also, special thanks to the referee who help refine the paper overall andgive important suggestions to improve the presentation of Theorem 3.6. These first definitions will be useful during the paper:
Definition 2.1.
1. We say that T ⊆ ω <ω is a tree if and only if given σ ∈ T we have that σ ↾ j ∈ T for all j < | σ | .2. A k -branching tree, is a tree such that every node has either 1 successoror k of them.3. A k -tree is a tree such that every node has at least 1 successor and nomore than k .Now, the following definition is due to [9] (they express it as the coveringnumber of an ideal): Definition 2.2.
The k -localization number, L k , is the smallest cardinalityof a family of k -branching trees that cover ( k + 1) ω .Notice that the definition is not trivial for k ≥
2. Furthermore, Newelskiand Roslanowski showed in [9] that, for k ≥ L k ≥ max { cov ( M ) , cov ( N ) } ,that L k +1 ≤ L k and that it is consistent that L k +1 < L k .On the other hand, in Blass’s chapter of the Handbook of Set Theory[3] he defines: Definition 2.3.
1. A k globally adaptive predictor is a sequence of func-tions π = h π n : n ∈ ω i with π n : ω n → [ ω ] k . We say that a function f ∈ ω ω is predicted by π if there is m ∈ ω such that for all n > m , f ( n ) ∈ π n ( f ↾ n ). 3. The k globally prediction number, v gk , is the minimal cardinality of aset of k globally adaptive predictors that predict all functions in ω ω .3. The k globally evasion number, e gk , is the minimal cardinality of a setof functions in ω ω such that the whole set is not predicted by a single k globally adaptive predictor.It is important to make some remarks about the last definition: • The ‘adaptive’ part refers to the fact that π n is not constant. Non-adaptive objects are closer to slaloms (or traces). • The ‘globally’ part of the definition refers to the fact that we have π n for all n ∈ ω . It is possible to define predictors using π n for n ∈ D ( ω . • Blass do not give a notation for this number, so the notation v gk and e gk is introduced here. • These definitions are not trivial for k ≥ v gk and e gk are duals between them and, by the work donein [3], we know that m σ − centered ≤ e gk ≤ add( N ). So, by duality, we knowthat cof( N ) ≤ v gk ≤ c . Also, from the definition, we have that v gk +1 ≤ v gk .Reading the definition more carefully we can notice that all the functionsthat are predicted by a k -globally adaptive predictor are covered by ℵ many k -branching trees, so v gk is also the minimum cardinal of a set of k -branchingtrees (or k -trees) that cover ω ω .Furthermore, in [9], we have the following result: Theorem 2.4.
Given k ≥ , it is consistent to have ZFC+ cof ( N ) = v gk +1 < v gk = c . This theorem is a corollary of the proof of:
Theorem 2.5 (Newelski, Roslanowski [9]) . Given k ≥ , it is consistent tohave ZFC+ cof ( N ) = L k +1 < L k = c . Specifically, it comes from two facts: first, that the forcings that wereused have the k -localization property. This is that “every real in ω ω is abranch of a k -tree from the ground model”, this keeps L k +1 and v gk +1 at ℵ ;and the forcing adds a function in ( k + 1) ω that is not the branch of any k -tree from the ground model. Notice that this function is also a functionin ω ω that is not the branch of any k -tree from the ground model. Once youtake a countable support product, this makes L k and v gk of size c .4he relation between these two cardinal characteristics is more evidentonce we realize, from the tree definition of L k and v gk , that L k ≤ v gk . So,a natural question arises of whether L k = v gk . The goal of this paper is toanswer the question in a negative way. In this setting it is better to understand some of the processes as combina-torial principles instead of parts of a forcing argument. Because of that, thefollowing lemmas come in pairs: one is a combinatorial statement and thefollowing one is the forcing result.
Lemma 3.1.
Given { f i : i ∈ I } ⊆ ω with | I | = 3 n you can find S ⊆ I with | S | = n such that { f i ↾ n : i ∈ S, n ∈ ω } is a -tree.Proof. We will do the proof by induction.For n = 0 and n = 1 it is trivially true.Now, assume that it is true for n , we will prove it for n + 1.Given { f i : i ∈ I } ⊆ ω with | I | = 3 n +1 if all of them are the samefunction then take the first n + 1 of them, they make trivially a 2-tree. Onthe other hand, if there are two of them that are different, find the firstnatural number m such that two of them differ. Notice that, using a pigeonhole principle, there is a value k ∈ J ⊆ I , with | J | ≥ n such that for all i ∈ J we have f i ( m ) = k .Now, take i ∈ I such that f i ( m ) = k and let S ′ ⊆ J be the indexset of size n given after using the induction hypothesis over J . Notice that { f i ↾ j : j ∈ ω } ∪ { f i ↾ j : i ∈ S ′ , j ∈ ω } forms a 2-tree and that S = S ′ ∪ { i } has size n + 1. Lemma 3.2.
There is a forcing notion that adds a function from ω to ω that is not predicted by any k -global adaptive predictor but such that all realsin ω are a branch of a -tree in the ground model.Proof. Definition 3.3.
We say that T ⊆ [ m ∈ ω m Y n ∈ m ( n + 1) is an accelerating tree ifand only if it is a subtree of [ m ∈ ω m Y n ∈ m ( n + 1), if every node has an extension5hat splits and given σ ∈ T such that there are k i ∈ ω , i < n , such that σ ↾ k i is a splitting node (i.e., σ has n splits before it) then σ has either 1successor or at least n + 2. Let P be the forcing notion whose conditions are of the form h τ, T i with τ ∈ [ m ∈ ω m Y n ∈ m ( n +1) and T an accelerating subtree of [ m ∈ ω m Y n ∈ m ( n +1) extending τ . We say that h τ ′ , T ′ i ≤ h τ, T i if and only if τ ⊆ τ ′ , T ′ ⊆ T and τ ′ ∈ T . For a node ρ ∈ T , let T ρ = { τ ∈ T : τ ⊆ ρ ∨ ρ ⊆ τ } Notice that given any k -tree U ⊆ ω <ω and a condition h τ, T i , there is ρ ∈ T that is not a node in U (for example, go to a split with k + 1 nodes,one of them is not in U ). Furthermore, if we take the condition h ρ, T ρ i ,none of the branches of T ρ are branches of U . This shows that forcing withaccelerating tree forcings adds a function from ω to ω that is not predictedby any k -global adaptive predictor.Now, we will only give a sketch on how to proof that all reals in 3 ω area branch of a 2-tree in the ground model. This part of the theorem is acorollary of Lemma 3.6 letting κ = 1. Also, the forcing is the set theoreticalversion of the forcing used in a paper to appear with Noah Schweber . Fromthat proof, translating from computability theory to set theory, we have thedesired result.Now, the sketch: the idea is that given a condition p ∈ P and a P -name˙ f such that p (cid:13) ˙ f ∈ ω we can define, in V, a 2-tree, A , and a condition q ≤ p such that q (cid:13) ˙ f ∈ A . To do this, you can prune the tree of p , call it T , in such a way that given σ ∈ T , an n -th split node of T , there is g σ ∈ n such that h σ, T σ i (cid:13) ˙ f ↾ n = g σ .We are looking for T ′ ⊆ T such that { g σ : σ ∈ T ′ n -split of T , n ∈ ω } is inside a ground model 2-tree we do the following process: if we alreadydecided that σ ∈ T ′ then we look for an extension of σ with 3 n successors,say σ i , i ∈ n . Now we look for M ∈ ω , ρ i extending σ i and g i ∈ M suchthat h ρ i , T ρ i i (cid:13) ˙ f ↾ M = g i . Now, running the above combinatorial Lemma During the conference ’Set Theory of the Reals’, BIRS-CMO Oaxaca, August 2019,it was brought to our attention that this forcing was originally defined by Geschke in [5]as Miller Lite Forcing. We decide to define the acceleration tree forcing using pairs to create a strongerresemblance to the effective analogue of accelerating trees of ω ω (paper in preparationwith Noah Schweber). Furthermore, this will allow us to easily define ( T ) in Lemma 3.6. S ⊆ n such that { g i ↾ j : i ∈ S, j < M } is a 2-tree (definable in V using the definability lemma of forcing). Finally, we ensure that there isno spliting between σ i and ρ i in T’ and we ask that σ i , ρ i ∈ T ′ if and onlyif i ∈ S . Lemma 3.4.
Given { f ji : i ∈ I, j ∈ k } ⊆ ω with k ∈ ω , | I | = N ( n, k ) abig enough number and m ∈ ω that makes { f ji ↾ l : i ∈ I, j ∈ k, l ∈ m + 1 } a -tree such that if f ji ↾ m = f ts ↾ m with t = j we have that f ji = f ts then youcan find S ⊆ I with | S | = n such that { f ji ↾ l : i ∈ S, j ∈ k, l ∈ ω } is a -tree.Proof. We will prove this by induction over k .At k = 1, we need N ( n, ≥ n so that we can use Lemma 3.1 to bedone.Now, assuming we have the case for k we will prove it for k + 1. We need N ( n, k + 1) ≥ N ( n,k ) , with this we can use Lemma 3.1 over { f ki : i ∈ I } to get J ⊆ I such that | J | = N ( n, k ) and { f ti ↾ l : i ∈ J, t = k, l ∈ ω } is a2-tree. Now, we can use our induction hypothesis over { f ti : i ∈ J, t ∈ k } toget S ⊆ J of size n such that { f ji ↾ l : i ∈ S, j ∈ k, l ∈ ω } is a 2-tree.We just need to show that { f ji ↾ l : i ∈ S, j ∈ k +1 , l ∈ ω } = { f ki ↾ l : i ∈ S, l ∈ ω }∪{ f ji ↾ l : i ∈ S, j ∈ k, l ∈ ω } is a 2-tree.Assume that we have a ∈ ω and h i, j i , h s, t i , h g, h i ∈ S × ( k + 1) differentbetween them such that f ji ↾ a = f ts ↾ a = f hg ↾ a . We have to show that |{ f ji ↾ ( a + 1) , f ts ↾ ( a + 1) , f hg ↾ ( a + 1) }| ≤ . Taking into account that { f ji ↾ l : i ∈ I, j ∈ k + 1 , l ∈ m + 1 } , { f ki ↾ l : i ∈ S, l ∈ ω } and { f ji ↾ l : i ∈ S, j ∈ k, l ∈ ω } are 2-trees, the only case left tocheck is when a ≥ m and j , t and h are not all the same but at least oneof them is equal to k . Without lost of generality, assume that h = k and j = k .Since a ≥ m we have that f kg ↾ m = f ji ↾ m . Using the fact that j = k , andour theorem’s hypothesis, we have that f kg = f ji , so {| f ji ↾ ( a + 1) , f ts ↾ ( a + 1) , f hg ↾ ( a + 1) }| ≤ . It is important to remark that in these combinatorial lemmas it is neverused that the domain of the functions is ω , so these lemmas are also truefor 3 n . 7 efinition 3.5. A forcing notion has the ( k + 1) ω localization property ifand only every function in ( k + 1) ω in the generic extension is a branch of a k -tree from the ground model. Lemma 3.6.
Countable product of the accelerating tree forcing has the ω localization property. Newelski and Roslanowski, in [9], define the k -localization property asthe fact that all branches of ω ω are cover by a k -tree of the ground model.This property was deeply study later by Roslanowski, in [10], and by Zaple-tal, in [12]. They found that the k -localization property is preserved undermost of the used countable support product and iteration of proper forcings.Our forcing does not have the 2-localization property, it will have aversion of that for 3 ω : the 3 ω localization property. Our proof will resemblethe one did by Newelski and Roslanowski, nevertheless, it is possible thatthere are results in the lines of the other two papers. Proof.
First, given a tree and n >
0, we let ( T ) n be the set of all nodessuch that they are the successors of the n -th split. As a convention, given p = h s, T i a forcing condition, we have that ( T ) = { s } . Now, given elementsof the accelerating tree forcing we will define for n ≥ p = h s, T i ≤ n p ′ = h s ′ , T ′ i if and only if h s, T i ≤ h s ′ , T ′ i and ( T ′ ) k = ( T ) k , for all 1 ≤ k ≤ n ,and p ≤ p ′ if and only if p ≤ p ′ . Notice that, since these are subtrees of [ m ∈ ω m Y n ∈ m ( n + 1), these orders have the fusion property and satisfy Axiom A(as in [1]).Assume that we have a countable support product of the acceleratingtree forcing of length κ . Call the final partial order P κ , as notation we willexpress q ∈ P κ as q = h r, T i and q ( α ) = h r ( α ) , T ( α ) i .Given F ∈ [ κ ] <ω and η : F → ω , we define p ≤ F,η q if and only if p ≤ q and for all α ∈ F we have that p ( α ) ≤ η ( α ) q ( α ). Furthermore,given σ ∈ Q α ∈ F T ( α ) and p ∈ P κ we define p ∗ σ to be p ( β ) if β / ∈ F and p ( β ) ∗ σ ( β ) = h σ ( β ) , T σ ( β ) i if β ∈ F (following the notation of Lemma 3.2).The orders ≤ F,η have the fusion property under the following conditions:given p n +1 ≤ F n ,η n p n with S n ∈ ω F n = S n ∈ ω supp ( p n ) and lim n →∞ η n ( α ) = ∞ for all α ∈ S n ∈ ω F n we have that there exist q ∈ P κ such that q ≤ F n ,η n p n for all n ∈ ω .In order to complete the proof, it is enough to define the following con-cept and show the following claim: Definition 3.7.
Given (cid:13) P “ ˙ f ∈ ω ”. We say that the 5-tuple h q, F, η, m, A i consolidates ˙ f if and only if the following is satisfied:8. q = h r, T i ∈ P κ , F ∈ [ κ ] <ω , η : F → ω , m ∈ ω .2. A ⊆ Working in V , suppose that (cid:13) P “ ˙ f ∈ ω ” and h q, F, η, m, A i that consolidates ˙ f . Then there are M ′ > m , A ′ ⊂ Suppose M ∈ ω and σ ∈ Q α ∈ F ( T ( α )) ν ( α ) . Then thereexists q ∗ ∈ P κ such that q ∗ ≤ F,ν p and q ∗ ∗ σ forces a value to ˙ f ↾ M .Proof. Standard, see Lemma 1.7 of [2] Observation B For M ∈ ω there exists there exists q ∗ ∈ P κ , q ∗ ≤ F,ν p ,such that for every σ ∈ Q α ∈ F ( T ( α )) ν ( α ) we have that q ∗ ∗ σ forces a valueto ˙ f ↾ M .Proof. Standard, see Corollary 1.10 of [2]Now, given p ≤ F,ν p and M > m we define z p,M = |{ a ∈ M : ∃ σ ∈ Y α ∈ F ( T ( α )) ν ( α ) ( p ∗ σ (cid:13) ˙ f ↾ M = a ) }| . Notice that z p,M ≤ k · N ( n, k ). Therefore, we can find p + = h r + , T + i ≤ F,η p and M ′ > m such that z p + ,M ′ has maximum value.Passing to a ≤ F,ν condition we may also demand that for every σ ∈ Q α ∈ F ( T + ( α )) ν ( α ) = Q α ∈ F ( T ( α )) ν ( α ) , the condition p + ∗ σ forces a valueto ˙ f ↾ M ′ . Observation C Suppose that σ , σ ∈ Q α ∈ F ( T + ( α )) ν ( α ) , M ′′ ≥ M ′ , p ≤ F,ν p + and a , a ∈ M ′′ . If a = a and p ∗ σ (cid:13) ˙ f ↾ M ′′ = a and p ∗ σ (cid:13) ˙ f ↾ M ′′ = a then a ↾ M ′ = a ↾ M ′ . roof. Suppose towards a contradiction that a ↾ M ′ = a ↾ M ′ . We can find p ′′ ≤ F,ν p ≤ F,ν p + ≤ F,ν p such that for every σ ∈ Q α ∈ F ( T + ( α )) ν ( α ) ,the condition p ′′ ∗ σ forces a value to ˙ f ↾ M ′′ . Then, z p ′′ ,M ′′ > z p + ,M ′ , acontradiction.For every σ ∈ Q α ∈ F ( T + ( α )) η ( α ) = Q α ∈ F ( T ( α )) η ( α ) there are N ( n, k )many ρ σ ∈ Q α ∈ F ( T + ( α )) ν ( α ) = Q α ∈ F ( T ( α )) ν ( α ) such that for all α ∈ F \ { β } we have that σ ( α ) = ρ σ ( α ). Fix an enumeration of these ρ anddefine σ ⌢ i = ρ σi (if we have σ ( β ) = σ ( β ) then ρ σ i = ρ σ i for all i ).Given σ ∈ Q α ∈ F ( T + ( α )) η ( α ) and i ∈ N ( n, k ), define f σi ∈ M ′ to besuch that p + ∗ σ ⌢ i (cid:13) “ ˙ f ↾ M ′ = f σi ”. Since h q, F, η, m, A i consolidates ˙ f , wehave that { f σi ↾ l : i ∈ N ( n, k ) , t ∈ Q α ∈ F ( T + ( α )) η ( α ) , l ∈ m + 1 } ⊆ A is a2-tree such that if f σ i ↾ m = f σ j ↾ m with σ = σ we have that f σ i = f σ s .Furthermore, using observation C, we have that given σ , σ elements of Q α ∈ F ( T + ( α )) η ( α ) , i, j ∈ N ( n, k ) with σ = σ we have that if f σ i ↾ M ′ = f σ j ↾ M ′ with σ = σ we have that f σ i = f σ s . With this, any 2-tree thatcomes from { f σi : i ∈ N ( n, k ) , σ ∈ Q α ∈ F ( T + ( α )) η ( α ) } will satisfy the re-quirements of the claim.Now we can use Lemma 3.4 on { f σi : i ∈ N ( n, k ) , σ ∈ Q α ∈ F ( T + ( α )) η ( α ) } so we can find S ⊆ N ( n, k ) of size n such that { f σi : i ∈ S, σ ∈ Y α ∈ F ( T + ( α )) η ( α ) } is a 2-tree.To complete the claim, we use: • M ′ , • A ′ = { f ti : i ∈ S, t ∈ k } ∪ A and • q ′ ≤ F,η q define as q ′ ( α ) = p + ( α ) for all α = β and q ′ ( β ) = h r ′ ( β ) , T ′ ( β ) i where r ′ ( β ) = r + ( β ) and T ′ ( β ) is an accelerating subtree of T + ( β ) suchthat ( T ′ ( β )) ν ( β ) = { ρ σi ( β ) : σ ∈ Y α ∈ F ( T + ( α )) η ( α ) , i ∈ S } . Corollary 3.9. Countable support product of accelerating tree forcing hasthe ( k + 1) ω localization property for all k ≥ . roof. To prove this, it is enough to show that the ( k + 1) ω localizationproperty is implied by the ( s + 1) ω localization property for k ≥ s ≥ 2, then,the result is a corollary of Lemma 3.6.Fix a surjective function f : ( k + 1) → ( s + 1). Notice that this functioninduces a surjective function f ∗ : ( k + 1) ω → ( s + 1) ω . Now, working in ageneric extension given a s -tree T from the ground model, ( f ∗ ) − [ T ] is a k -tree from the ground model.Therefore, if in the generic extension ( s + 1) ω is covered by s -trees fromthe ground model, then ( k +1) ω is covered by k -trees from the ground model.Now, the following definition can let us expand our last result a littlemore. Definition 3.10. Forcing with k -branching trees of k <ω is the forcing notionthat uses subtrees of k <ω such that every node has either 1 or k successors.This forcing is used in [9] where Newelski and Roslanowski showed thatthis forcing has the k -localization property, i.e., that every function of ω ω in the generic extension is the branch of a k -tree from the ground model.Notice that this property implies the ( k + 1) ω localization property. A firststep in order to investigate if the countable support products of forcingswith the ( k + 1) ω localization property still has the ( k + 1) ω localization istrue for a bigger spectrum of forcings than the accelerating tree forcing isto show the following lemmas, that are analogues of Lemma 3.4 and 3.6: Lemma 3.11. Given { f ji : i ∈ I, j ∈ a } ⊆ ( k +1) ω with a ∈ ω , | I | = N ( n, a ) a big enough number and m ∈ ω that makes { f ji ↾ l : i ∈ I, j ∈ a, l ∈ m + 1 } a k -tree such that if f ji ↾ m = f ts ↾ m with t = j we have that f ji = f ts then youcan find S ⊆ I with | S | = n such that { f ji : i ∈ S, j ∈ l } is a k -tree.Proof. This follows from the proofs of Lemma 3.1 and Lemma 3.4, in thoselemmas we had k = 2. The same reasoning will give us this lemma. Lemma 3.12. Countable support product of alternating accelerating treeforcing and forcing with k -branching trees of k <ω has the ( k +1) ω localizationproperty.Proof. Notice that the orders ≤ n also make sense when forcing with k -branching trees of k ω .The proof in full detail will have the same extension as the proof ofLemma 3.6. Nevertheless, here we give a sketch of how to combine thetechnique used in [9] and the proof of 3.6.12verything works the same changing 2 for k and 3 for k + 1. Now, toshow the analogue of Claim 3.8 we will have two cases:1. If you are extending a node that comes from an accelerating tree, thenuse Lemma 3.12 instead of Lemma 3.6. Everything else works thesame.2. If you are extending a node that comes from a k -branching tree insteadof using Lemma 3.12, it is enough to find a condition like p + . Sincethe next split only has k successors, they naturally form a k -tree.Everything else works the same as the proof of Claim 3.8 or you canuse the technique used in [9]. Theorem 4.1. It is consistent with ZF C that ∀ k ≥ L k < v gk = c ) .Proof. Starting with a model of ZF C + GCH we can make a countablesupport product of the accelerating tree forcing describe in Lemma 3.2.Using Axiom A, as in in [1], we know that the product preserves cardinalsand that c = ℵ . Also, by Lemma 3.6, the resulting model will have L k = L = ℵ . We just need to show that in the extension v gk = ℵ = c .Let P ω = Q α ∈ ω Q α be the countable support product of acceleratingtree forcings. Let G = { c α : α ∈ ω } be generic over P ω . Now, for all β < ω let T β ⊆ ω ω be a k ( β )-tree, with k ( β ) ∈ ω , in V [ G ].Now, in V , we can find ˙ T ( β ) a P α ( β ) -name for some α ( β ) ∈ ω . So,there is γ ∈ ω such that α ( β ) < γ for all β . Therefore, we have that T β ∈ V [ { c α : α < γ } ] for all β ∈ ω .Notice that if T is an accelerating tree in V , then at the split k ( β ) + 1it has a branch that is not in T β in V [ { c α : α < γ } ]. Then avoiding T β is adense condition (in V [ { c α : α < γ } ]) for accelerating trees from V .Since c γ is a V -accelerating forcing generic over V [ { c α : α < γ } ], then c γ is not a branch of any k -tree in V [ { c α : α < γ } ], k ∈ ω . Therefore, c γ isnot a branch of any T β .This shows that, in V [ G ], ω ω is not cover by { T β : β ∈ ω } . Since thiswas an arbitrary collection we have that v gk = ℵ for all k ∈ ω .13his theorem proves that it is consistent that v gk = L k and answers thequestion from Blass about the identity of v gk : they indeed are a differentcardinal characteristic from the ones that are known.Furthermore, we can see that there are more ways to do this split: Theorem 4.2. For all s ≥ it is consistent with ZF C that ∀ k ≥ L s +1 < L s = v gk = c ) .Proof. Following the same strategy as above, starting with a model of ZF C + GCH we can make a countable support product of the accelerating tree forc-ing alternated with forcing with s + 1-branching trees of ( s + 1) ω . Just asbefore, we know that the product preserves cardinals and that c = ℵ . Also,by Lemma 3.12, the resulting model will have L s +1 = ℵ . We just need toshow that, in the extension, L s = v gk = ℵ = c .Let P ω = Q α ∈ ω Q α be the countable support product of acceleratingtree forcings, when α is even and forcing with s + 1 subtrees of ( s + 1) ω when α is odd. Let G = { c α : α ∈ ω } be generic over P ω .To see that v gk = ℵ = c , we can do the same as above. Now, showingthat L s = ℵ = c can be found in [9]. Nevertheless, for convenience to thereader, we give an argument here:For all β < ω let T β ⊆ ( s + 1) ω be a s -tree in V [ G ]. In V , we canfind ˙ T ( β ) a P α ( β ) -name for some α ( β ) ∈ ω . So, there is γ ∈ ω such that α ( β ) < · γ + 1 for all β . Therefore, we have that T β ∈ V [ { c α : α < · γ + 1 } ]for all β ∈ ω .Since c · γ +1 is a generic for the forcing using s + 1-branching trees (from V ) of ( s + 1) ω over V [ { c α : α < γ } ], then c · γ +1 is not a branch of any s -treein V [ { c α : α < γ } ] (same reasoning as in Theorem 4.1). Therefore, c · γ +1 isnot a branch of any T β .This shows that, in V [ G ], ( s + 1) ω is not cover by { T β : β ∈ ω } . Sincethis was an arbitrary collection we have that v gk = ℵ for all k ∈ ω . Question 4.3. What is the value of cof ( N ) in the above models? Theorem 4.2 shows that in order to have different values for v gk and L k it is not necessary that every L s have the same value. In this same venue,we can wonder if it is necessary that all v gs have the same value. In otherwords: Question 4.4. Can we have L k = v g < v g for all k ≥ ? ω , which are the two waysused in this paper to make v g = c . Another approach will be to use treesof ω ω that branches more than 3 times at each split. Nevertheless, I do notsee any good reason for that forcing to have the 3 ω -localization property.Maybe a modification of it can do the trick.Now, during the paper, the ( k + 1) ω localization property played a reallyimportant role. In order to show that it was preserved the proofs showedabove are really case specific. This is useful for our purposes, but a questionarises: Question 4.5. Can we show that the ( k + 1) ω -localization property is pre-served under countable support iteration and products? This is likely to be possible. In [12], Zapletal showed that the n -localizationproperty is preserved under countable support product and iteration of abroad variety of forcings (some kind of definable proper forcings).Finally, notice that v gk is a cardinal characteristic that is usually reallyclosed to c . This is not true in cardinal arithmetic, but it is true in the ChiconDiagram: all of these numbers are above cof ( N ). So, in order to work withthem, it is important to use forcing notions that are tame somehow (theycannot add Cohen or random reals, for example). In this case, we used aforcing notion with the ( k + 1) ω localization property but, in the literature,there are examples of properties like the Sacks property, the n -localizationproperty and, most recently, the shrink wrapping property (see [7]) that arealso tame with reals. It is important to notice that most of these ‘tameness’properties relates to the idea of keeping the new reals inside a tree of somesort. Question 4.6. Is there an underlying theorem (or meta theorem) that re-lates all (of some) of this tameness properties? One possible result could be that all of them are preserved under count-able support product of a variety of forcings, but I do not have any goodguess of whether this is possible or not. Time after this paper was sent for review, we were able to proved a coupleof results answering the questions that appear above. Since this are smallresults, we decided to include them here.15his first results answer question 4.3, they are the result of a conversationwith Corey Switzer during the XIX Graduate Student Conference in Logicin Madison, Wisconsin, April 2018. Lemma 5.1. The accelerating tree forcing has the Sacks property.Proof. Definition 5.2. Given a function f : ω → ω \ { } an slalom of growth f is a function s : ω → [ ω ] <ω such that | s ( n ) | ≤ f ( n ) for all n . We say that g ∈ ω ω goes thorugh s if and only if g ( n ) ∈ s ( n ) for all n . Definition 5.3. [11] We say that a forcing has the Sacks property if andonly if there is g ∈ V such that g : ω → ω \ { } and diverges to infinity suchthat for all f ∈ ω ω ∩ V [ G ] there is a tree T ∈ V such that f is a branch of T and the n -th level of T has size g ( n ).Notice that, given an slalom of growth f , we can generate a tree T suchthat its n -th level has size Q ni =0 f ( i ).To show that the accelerating tree forcing has the Sacks property we willshow that every real in ω ω ∩ V [ G ] goes through an slalom s ∈ V such that | s ( n ) | ≤ n !.Let P be the accelerating tree forcing. From Lemma 3.6 we know thatgiven a name ˙ f such that (cid:13) P ˙ f ∈ ω ω then there is a condition h p, T i ∈ P such that given σ ∈ T n (notation defined in Lemma 3.6) we have that thereis τ σ ∈ ω n such that h σ, T σ i (cid:13) ˙ f ↾ n = τ σ .In V define s : ω → [ ω ] <ω such that s ( n ) = { τ σ ( n ) : σ ∈ T n } . Since T is accelerating, we have that | s ( n ) | ≤ n !.It is important to mention that in August 2019 it was brought to ourattention that, in [5], Geschke showed indirectly that the accelerating treeforcing has the Sacks property . We hope that this more direct proof is bothmore convinient for the reader and, maybe, useful for future work. Theorem 5.4. In the forcing extension generated after forcing with count-able support product of accelerating tree forcing cof ( N ) = ℵ .Proof. He forced the Dual Coloring Axiom and showed that this axiom implies cov ( N ) = ℵ . heorem 5.5. (from [3]) cof ( L ) is the cardinality of the smallest family F of slaloms of growth f (for f : ω → ω \ { } increasing and diverging toinfinity) such that all reals in ω ω go through a slalom in F . Theorem 5.6. (from [11]) The countable support product of forcings thathave the Sacks property have the Sacks property. Notice that, if a forcing has the Sacks property then cof ( N ) V = cof ( N ) V [ G ] .Since the accelerating tree forcing has the Sacks property, this showsthat the model generated in Theorem 4.1 satisfies cof ( N ) V [ G ] = ℵ .In Theorem 4.2 we get the same using the fact that forcing with k -branching trees also has the Sacks property.Finally, these are answers for different variations of question 4.4.In the flavor of the contructibility degrees , we have that: Lemma 5.7. If a forcing has the k + 1 localization property but it does nothave the k -localization then it do not have the ( k + 1) ω localization property.Proof. In V [ G ] let b ∈ ω ω be such that it is not in any k tree from the groundmodel. Now, let T be a k + 1-branching tree from V such that b ∈ T . Noticethat, in V , there is a bijection f : T → ( k + 1) <ω so, in V [ G ], this inducesa function f ∗ : [ T ] → ( k + 1) ω , where [ T ] are the branches of T .Notice that f ∗ ( b ) is also not in any k tree from the ground model. If itwere, say in A , we will have that b ∈ f − ( A ), but f − ( A ) is a k tree from V . Now, the following results is in the direction of the work done with NoahSchweber: Lemma 5.8. If a Turing degree has the property that every total functionit computes is a branch of a k + 1 branching computable tree but it computesa function that escapes every k branching computable tree then it computesa function in ( k + 1) ω that escapes every k branching tree.Proof. The same proof of the above theorem works, since given that T iscomputable, there is a computable function from it to ( k + 1) <ω .With this technique, we can answer in a negative way question 4.4. As in the paper The Cichon Diagram for Degrees of Constructibility by Corey Switzerthat can be at https://coreyswitzer.files.wordpress.com/2018/09/the-cichon-diagram-for-degrees-of-relative-constructibility.pdf heorem 5.9. The following equality is true v gk = max { v gk +1 , L k } . Fur-thermore, if v gk +1 < v gk then L k +1 < L k .Proof. Let v gk +1 = κ . This means that ω ω can be covered by κ k + 1-trees.Now, if L k = λ , this means that ( k + 1) ω can be covered by λ k -trees.Using a function like the one in the lemma above, given a cover of ω ω with k + 1 trees, we can cover each one of them with λ k -trees creating acover of κ · λ k -trees of ω ω . This means that v gk ≤ κ · λ = max { v gk +1 , L k } .Since v gk ≥ v gk +1 and v gk ≥ L k , we have that v gk = max { v gk +1 , L k } .For the furthermore, if v gk +1 < v gk then L k = v gk and L k +1 ≤ v gk +1 < v gk = L k This creates a new question: Question 5.10. Can L k be express as the maximum or the minimum ofother cardinal characteristics? Finally, the fact that v gk = max { v gk +1 , L k } also translate to computabilitytheory, but it becomes a trivial result: Lemma 5.11. Any Turing degree that computes a function f ∈ ω ω thatescapes all k computable trees either computes a function that escapes all k + 1 computable trees (that same f ) or a function g ∈ ( k + 1) ω that escapesall k -computable trees (the image of f under a certain computable function). eferences [1] J. E. Baumgartner. Iterated forcing. In Surveys in set theory , volume 87of London Math. Soc. Lecture Note Ser. , pages 1–59. Cambridge Univ.Press, Cambridge, 1983.[2] J. E. Baumgartner. Sacks forcing and the total failure of Martin’saxiom. Topology Appl. , 19(3):211–225, 1985.[3] A. Blass. Combinatorial cardinal characteristics of the continuum. In Handbook of set theory. Vols. 1, 2, 3 , pages 395–489. Springer, Dor-drecht, 2010.[4] K. Ciesielski and S. Shelah. A model with no magic set. J. SymbolicLogic , 64(4):1467–1490, 1999.[5] S. Geschke. A dual open coloring axiom. Ann. Pure Appl. Logic , 140(1-3):40–51, 2006.[6] S. Geschke and M. Kojman. Convexity numbers of closed sets in R n . Proc. Amer. Math. Soc. , 130(10):2871–2881, 2002.[7] O. Guzm´an Gonz´alez and D. Hathaway. Sacks forcing and the shrinkwrapping property. arXiv preprint arXiv:1708.07250 , 2017.[8] M. Khan and J. S. Miller. Forcing with bushy trees. Bull. Symb. Log. ,23(2):160–180, 2017.[9] L. Newelski and A. Ros l anowski. The ideal determined by the unsym-metric game. Proc. Amer. Math. Soc. , 117(3):823–831, 1993.[10] A. Ros l anowski. n -localization property. J. Symbolic Logic , 71(3):881–902, 2006.[11] S. Shelah. Proper and improper forcing . Perspectives in MathematicalLogic. Springer-Verlag, Berlin, second edition, 1998.[12] J. Zapletal. Applications of the ergodic iteration theorem.