Cardinal Invariants of Strongly Porous Sets

In this work we study cardinal invariants of the ideal SP of strongly porous sets on 2. We prove that add(SP) = ω1 , cof(SP) = c and that it is consistent that non(SP) < add(N ), answering questions of Hrušák and Zindulka. We also find a connection between strongly porous sets on 2 and the Martin number for σ -linked partial orders, and we use this connection to construct a model where all the Martin numbers for σ -k-linked forcings are mutually different. 2010 Mathematics Subject Classification 03E17, 28A75 (primary); 03E35, 03E50 (secondary)


Introduction
In 1967, Dolženko [5] began the study of σ -porous sets 1 and since then, many applications have been found.One of these appears in [10], where Preiss and Zajíček proved that, given a Banach space X with a separable dual and a continuous convex function f on X , the set of points in which f is not Fréchet differentiable is σ -porous.Other applications can be found in Belna, Evans and Humke [2], Lindenstrauss and Preiss [7], Lindenstrauss, Preiss and Tišer [8], Reich and Zaslavski [11], Renfro [12] and Zaslavski [19].
We shall study the notion of strong porosity: given a metric space X, d , a subset A ⊆ X is strongly porous if there is a p > 0 such that for any x ∈ X and any 0 < r < 1, there is y ∈ X such that B p•r (y) ⊆ B r (x) \ A. In this paper we will refer to strongly porous sets as porous sets.We shall work mostly with porous sets in ω 2: we say that a set A ⊆ ω 2 is n-porous if for every s ∈ <ω 2 there is a t ∈ n 2 such that s t ∩ A = ∅.By s t we denote the concatenation of s followed by t, and by s we denote the cone of s, that is s = {f ∈ ω 2 : s ⊆ f }.It is easy to see (see Hrušák and Zindulka [6]) that 2 Osvaldo Guzmán, Michael Hrušák and Arturo Martínez-Celis a set A ⊆ ω 2 is porous if and only if there is an n ∈ ω such that A is n-porous.Zajíček [17] proved that a set A in a metric space X, d is σ -porous if and only if it is σ -lower porous, where A is lower porous if for every x ∈ X there exist ρ x > 0 and r x > 0 such that for any 0 ≤ r ≤ r x there is y ∈ X such that B ρx•r (y) ⊆ B r (x) \ A. Another classical notion of porosity is upper porosity: a set A in a metric space X, d is upper porous if for every x ∈ X there is ρ x > 0 and a sequence r n → 0 such that for every n ∈ ω there is y n ∈ X such that B ρx•rn (y n ) ⊆ B rn (x) \ A. It is easy to see that lower porosity implies upper porosity.
We will denote the σ -ideal2 generated by porous sets on ω 2 by SP , the σ -ideal generated by n-porous sets by SP n , and the σ -ideal generated by upper porous sets by UP.Observe that SP 1 is the ideal of countable sets of ω 2. Further research about different types of porosity can be found in Rojas-Rebolledo [16], Zajíček [17], Zapletal [18], Zelený [20] and Zelený and Zajíček [21].
Cardinal invariants of these σ -ideals have been studied in Brendle [3], Hrušák and Zindulka [6] and Repický [13,14,15].Recall that, given a σ -ideal I over a set X , the following are the standard cardinal invariants of I : In [6], Hrušák and Zindulka proved that the cardinal invariants of the σ -ideal of lower porous sets in the real line are the same as the cardinal invariants of SP .They proved that non(SP) < m σ-centered is consistent, that cov(SP) > cof(N ) is consistent, and that cov(SP) < c is consistent, where m σ-centered is the smallest cardinal where the Martin's axiom for σ -centered forcings fails and N is the ideal of sets of Lebesgue measure zero.
There are some analogous inequalities that hold for UP.In [15], M. Repický proved that non(UP) ≥ m σ-centered and cov(UP) ≤ cof(N ) holds.He proved in [13] that non(UP) ≥ add(N ) and in [3], J. Brendle proved that add(UP) = ω 1 and cof(UP) = c hold.In [6], Hrušák and Zindulka asked if the last three inequalities hold also for the SP ideal.In this work we show that add(SP) = ω 1 , cof(SP) = c and that it is consistent that non(SP) < add(N ).
Given k ∈ ω and a forcing notion P a subset A ⊆ P is k-linked if for every collection {a i : i ∈ k} of k elements of A, there is an a ∈ P such that for every i ∈ k, a ≤ a i .P is σ -k-linked if P is the countable union of k-linked subsets of P. We will denote by m k the Martin number for σ -k-linked forcings, that is, the smallest cardinal κ such that there is a σ -k-linked forcing P and κ many P-dense subsets of P such that no filter of P intersects them all.In [4], Brendle and Shelah constructed a model where all the Martin numbers of the form m 2 i are distinct.We will show a connection between σ -k-linked forcings and porous sets and we will use this connection to construct a model where all the Martin numbers m i are different all at once.If X, Y are sets, then Y X is the set of all functions from Y to X and <ω X = n∈ω n X .
If σ, s ∈ <ω X , then we will denote that σ is an initial segment of s by σ s.A set T ⊆ <ω X is a tree if it is closed under restrictions to initial segments.If T ⊆ <ω X is a tree, then by [T] we denote the set of branches of T , that is, With end(T) we will denote the end nodes of T , that is the nodes of T without extensions.In our forcing notation, the stronger conditions are the smaller ones.

Additivity and cofinality
The main goal of this section is to prove that add(SP) = ω 1 and cof(SP) = c.We will use the following notion.For each σ ∈ <ω 2, we will recursively define a finite tree T σ as follows: T ∅ = {∅}, and if T σ is defined then It is easy to see that, for each σ ∈ <ω 2, T σ is a finite 2-porous tree and that if σ τ , then T σ ⊆ T τ .For each f ∈ ω 2, define T f = n∈ω T f n .It follows easily that each T f is a 2-porous tree.
We will show that the family {T f : f ∈ ω 2} is the family we were looking for: Let X ∈ SP .Without loss of generality we will assume that X = i∈ω [T i ], where T i is an i + 1-porous tree.We must show that the set To finish the proof we will see that each B s,t,n has at most 2 n+1 − 1 elements: Suppose this is not the case and let s, t ∈ <ω 2, n ∈ ω and {f i } i<2 n+1 ⊆ B s,t,n .Extend s to σ such that σ ∈ end(T t ).Let j ∈ ω be such that the set a = {f i j : i < 2 n+1 } has 2 n+1 elements and let The tree , but this contradicts the fact that f k ∈ B s,t,n .This implies that each B s,t,n is finite, and therefore B is countable.
We can now prove the main result of this section.
Observe that this last proof can be used to show that add(SP n ) = ω 1 = add(SP) and cof(SP n ) = c = cof(SP).

Uniformity number
In this section we will study some properties concerning the uniformity number of porosity ideals.We will prove the consistency of non(SP) < add(N ).We will also develop some tools that we will use later in the paper.We will need the following concept, inspired by the concept of a k-Sacks tree in <ω k.
We will denote by AS k the σ -ideal generated by the branches of k-anti-Sacks trees.
This notion is the analogue of the notion of 1-porous tree in <ω k and it is closely related to the k-Sacks forcing.Recall that a k-Sacks tree T is a tree on <ω k such that for every s ∈ T , there is a t ∈ T such that, for every i < k, t i ∈ T .The k-Sacks forcing S k is the collection of all k-Sacks trees ordered by reverse inclusion.It is well-known that the k-Sacks forcing is equivalent to Borel( ω k)/AS k (see Newelski and Rosłanowski [9]).
Using a similar argument to the one we gave in the last section, it is possible to show that add(AS k ) = ω 1 and that cof(AS k ) = c.Alternatively, a proof of this fact can be found in [9].
The ideals SP k and AS 2 k share many properties.Many of the results in this work will concern properties of the ideals AS k that are also valid for the ideals SP k , and the proofs for both ideals are almost the same.Whenever we state a property about one of these ideals that is also valid for the other one, we will only give the proof for the ideal AS k .
We shall introduce a notion that we will use to keep the uniformity number small in a forcing extension.

Definition 3.2 Let P be a forcing notion and let
We say that P strongly preserves non(AS k ) in A (P strongly preserves non(SP k ) in A) if for every P-name .X of a k-anti-Sacks tree (k-porous tree) there is a . X]".We will say that P strongly preserves non(AS k ) (P strongly preserves non(SP k )) if P strongly preserves non(AS k ) in ω k (P strongly preserves non(SP k ) in ω 2).

Osvaldo Guzmán, Michael Hrušák and Arturo Martínez-Celis
It is easy to see that, if P strongly preserves non(AS k ) in A, then P "A / ∈ AS k " and if P strongly preserves non(AS k ), then P strongly preserves non(AS k ) in A for every A ⊆ ω k.The following lemma is easy to prove.Lemma 3.3 Suppose that a forcing notion P strongly preserves non(SP k ) for every k ∈ ω , then P " ω 2 ∩ V / ∈ SP".
Proof The proof is straightforward from the definitions.
The next lemma shows that there is a connection between anti-Sacks trees and σ -k-linked forcings.
Lemma 3.4 Let P be a forcing notion.If P is σ -k-linked, then P strongly preserves non(AS k ).
Proof Let {P i : i ∈ ω} be a sequence of k-linked subsets such that P = i∈ω P i .Let .
A be a P-name of a k-anti-Sacks tree.Define T n ⊆ ω k as follows: T n = {s ∈ <ω k : ∃p ∈ P n (p "s ∈ .

A")}
We claim that, for each n ∈ ω , T n is a k-anti-Sacks tree.Suppose this is not the case, so there is an s ∈ T n such that, for every i ∈ k, s i ∈ T n .For every i ∈ k, we can pick a condition p i ∈ P n such that p i "s i ∈ .
A". Let p ∈ P be such that, for every i ∈ k, p ≤ p i .Then p "∀i ∈ k(s i ∈ .A)".This contradicts the fact that .
A is a P-name of a k-anti-Sacks tree.
To conclude the proof, note that for every where n is such that p ∈ P n .
The lemma above is optimal in the sense that, for each k, there is a σ -(k − 1)-linked forcing P k such that P k " ω k ∩ V ∈ AS k " and therefore P k does not strongly preserve AS k .This will be proved in the next section.There is also a relation between porous sets in ω 2 and σ -k-linked forcings.Lemma 3.5 Let P be a forcing notion.If P is σ -2 k -linked, then P strongly preserves non(SP k ).
Proof The proof is similar to the previous lemma.
We shall show that strong preservation of non(AS k ) and non(SP k ) is preserved by finite support iterations.
(1) If P is a forcing notion such that P strongly preserves non(I) in A and .
Q is a P-name for a forcing such that P " .Q strongly preserves non(I) in A", then P * .Q strongly preserves non(I) in A.
Q α strongly preserves non(I) in A" for each α ∈ γ , then P γ strongly preserves non(I) in A.
Proof We will only show the case when I = AS k as the other cases are similar.The part ( 1) is easy, we will proceed with part (2) by induction on γ .It is easy to see that the lemma holds for successor ordinals, and if γ has uncountable cofinality we can use a standard reflection argument to show that P γ strongly preserves non(AS k ) in A, so it is enough to show that the lemma holds for γ = ω .Let .T be a P ω -name of a k-anti-Sacks tree.For each n ∈ ω , let .
T n be a P n -name for the following set: . T n = {s ∈ <ω k : P (n,ω) "s ∈ .

T"}
It is easy to see that each .
T n is a P n name for a k-anti-Sacks tree.Now we use that each P n strongly preserves non(AS k ) to find a family {T j i : i, j ∈ ω} such that, for each n ∈ ω , if x ∈ A and x / ∈ i∈ω [T n i ], then P n "x / ∈ [ .
T n ]".It is easy to see that the set Y = {[T j i ] : i, j ∈ ω} is the set we are looking for.
We will now prove the consistency of non(SP) < add(N ).For constructing the model we are looking for, we will use the amoeba forcing A in the following presentation: Here Borel( ω 2) represents the collection of Borel subsets of the Cantor space and µ is the standard Lebesgue measure over ω 2. The order is given by A ≤ B if and only if A ⊆ B. The following lemma is well-known (see eg Bartoszyński and Judah [1]).We include the simple proof for the sake of completeness.

Lemma 3.7
The amoeba forcing is σ -n-linked for every n ∈ ω .
Therefore A ∈ A C .Now we must show that, for every clopen set C ⊆ ω 2, the intersection K of an arbitrary family {A j : j ∈ n} ⊆ A C is an element of A. This is a consequence of the following calculations: As a consequence, 1 2 < µ(K).Therefore K ∈ A.
We are ready to prove the main result of this section.The method of the proof was suggested to us by J. Brendle.
Proof Start with a model V such that V |= CH .Let P = {P α , .
Q α : α < ω 2 } be a finite support iteration of the amoeba forcing.It follows from the lemmas above that P strongly preserves non(SP k ) for every k ∈ ω and therefore P " ω 2 ∩ V / ∈ SP".As a consequence, we have that

Martin numbers of σ-k-linked forcings
It is easy to see that m 2 ≤ m 3 ≤ . . .and, for each k > 1, it is possible to get the consistency of m k < m k+1 by forcing with a finite support iteration of σ -(k + 1)-linked forcings over a model of CH.In [4], Brendle and Shelah constructed a model where all the cardinals of the form m 2 k are different.In this section we will construct a model where all the Martin numbers m i are different at the same time.In this model, the cardinals non(AS i ) will be different all at once (so will be the cardinals non(SP i )).We will use the following forcing notions.Given k > 2 let Osvaldo Guzmán, Michael Hrušák and Arturo Martínez-Celis Proof Again, we will only check that P k+1 is σ -k-linked, the other part is done in a similar way: For any two finite (k + 1)-anti-Sacks trees s, t of height ht(s), ht(t) respectively, define It is easy to see that P k+1 = {P(s, t) : s, t are finite (k + 1)-anti-Sacks trees}.We will show that every P(s, t) is k-linked.Let { s, F i : i < k} ⊆ P(s, t) and let F = i<k F i .We must show that s, F ∈ P k+1 .The properties (a) and (c) are easily verified, so the only thing left to do is to show that , then, for every i < k, σ only has (at most) one immediate successor in F i and therefore it is always possible to find a From these last two propositions we get the following result.
Proof This follows easily from the proof of the Proposition 4.1 and the last proposition.
For the proof of the main theorem we will need the following notion.
Definition 4.4 Given a regular cardinal κ and Observe that the existence of a κ, I -Luzin set implies that non(I ) ≤ κ.Recall that Cohen reals are added at every limit step of countable cofinality of a finite support iteration of arbitrary length.One common application of Cohen reals is that they are used to construct Luzin sets with special properties.The following lemma is one of those applications.
Q α : α ∈ κ be a finite support iteration of length κ such that L α " . Q α = P i * P k ", then L "There is a κ, AS i -Luzin set and there is a κ, SP k -Luzin set".
Proof Working in V[G], let L = {f α : α ∈ κ ∧ α has countable cofinality} be a family of Cohen reals such that each f α is added at the α-th stage of the iteration.Using the Proposition 4.1, it is easy to show that V <κ .The κ, SP k -Luzin set is found in a similar way.
In the lemma above, it is clear that if we replace L α " .Q α = P i * P k " by L α " .Q α = P i ", then, in the extension, we still have a κ, AS i -Luzin set.The following theorem is the main tool we will use to prove the main result of this section.
Q α : α ∈ ω ω be a finite support iteration of length ω ω such that, for each i > 1 and each α ∈ [ω i , ω i+1 ), L α " .Q α = P i+1 * Q i+1 ", where Q i+1 = P i+1 when i + 1 is a number of the form 2 k + 1 and Q i+1 = {∅} in all the other cases (for α < ω 2 , L α " .Q α = {∅}").We will show that the extension is the model we are looking for.As usual in this work, we will only show that there are ℵ i , AS i -Luzin sets for every i > 2 (the rest can be done in a similar way).Using the lemma above, for each i > 2, in V[G ωi ] there is a ℵ i , AS i -Luzin set L i .The only thing left to do is to show that . First, using Lemma 3.4, Lemma 3.6 and Proposition 4.2, we observe that L [ωi,ωω] strongly preserves non The actual value of c in the model above may depend on V .For example, if V |= GCH, then it is easy to see that V[G] |= c = ℵ ω+1 .Lemma 4.7 Let κ be a regular cardinal, let I ∈ {SP n , AS k : n > 0, k > 1} and let L be a κ, I -Luzin set.If P is a forcing notion such that |P| < κ, then P strongly preserves non(I ) in L.
Proof We will do the case when I ∈ {AS i : i > 1}; the other cases are similar.Let We are ready to prove the main result of this section.
Proof Start with a model V like the one constructed in Theorem 4.6 and V |= c = ℵ ω+1 .Using a standard bookkeeping argument, it is possible to construct a finite support iteration P of length ω ω+1 of σ -k-linked forcings of size smaller than ℵ k+1 (for every k > 1), such that any partial order which appears in an intermediate model is listed cofinally along the iteration.Now, using Lemmas 3.4, 3.6 and 4.7, it is possible to show that, for every k > 2, P strongly preserves non We note that, as each small σ -k-linked forcing appears in an intermediate model in the iteration, we have V To finish the proof, use the fact that non(SP) does not have countable cofinality and that, for every It follows from SP 1 ⊆ SP 2 ⊆ SP 3 ⊆ . . .that ω 1 = non(SP 1 ) ≤ non(SP 2 ) ≤ non(SP 3 ) ≤ . . .≤ non(SP) and we proved in the theorem above that each inequality can be consistently strict.It is important to remark that none of these numbers is comparable with m σ-centered .An argument for this can be found in Hrušák and Zindulka [6].

The covering number
It follows from the fact that AS 2 ⊆ AS 3 ⊆ . . .that c = cov(AS 2 ) ≥ cov(AS 3 ) ≥ . ... We can show that every pair of these numbers can be consistently different.
Let P be a finite support iteration of length ω 1 of the P k+1 forcing defined above and let G ⊆ P be a generic filter over V .It follows that P is an iteration of σ -k-linked forcing notions and therefore P strongly preserves non(AS k ).In T α : α ∈ ω 1 } is a collection of P-names for k-anti-Sacks trees, then we can use the fact that P strongly preserves non(AS k ) to show that there is a collection {C α : α ∈ ω 1 } ⊆ AS k such that if x ∈ ω k and x / ∈ {C α : α ∈ ω 1 }, then P "x / ∈ α∈ω 1 [ .
An alternative proof of this proposition follows from the results proven in Newelski and Rosłanowski [9].If k > 1 then a tree T ⊆ <ω ω is a k-tree if every s ∈ T has at most k immediate successors.A forcing notion P has the k-localization property if P "∀f ∈ ω ω (∃T ∈ V (T is a k-tree and f ∈ [T]))".It is easy to see that if P has the k-localization property, then P " (AS k+1 ∩ V) = ω (k + 1)".Let S k = {T ⊆ <ω k : ∀s ∈ T(∃t ∈ T(∀i ∈ k(s t ∧ t i ∈ T)))} be the k-Sacks forcing ordered by inclusion.It turns out that S k is forcing equivalent to Borel( ω k)/AS k and that if P is the countable support iteration or the countable support product of length ω 2 of the forcing S k , then P has the k-localization property (see [9]).As a consequence, in the extension cov(AS k+1 ) = ω 1 and cov(AS k ) = ω 2 .
Obviously it is impossible to separate infinitely many of the cov(SP n ) at the same time.This suggests the following: Question 5.2 How many of the cov(SP n ) can be separated at the same time?
We do not even know how to separate three of them.Another question we have is the following: Question 5.3 Is it possible to get the consistency of ZFC + ∀k ∈ ω (cov(SP) < cov(SP k ))?
We are also interested in the relationship between non(SP) and cov(SP).It follows from the fact that the Cohen forcing is σ -centered that, in the Cohen model, non(SP) < cov(SP).However, we do not know if it is possible to construct a model where non(SP) > cov(SP).
Definition 2.1 Let k ∈ ω .A tree T ⊆ <ω 2 is a k-porous tree if for every s ∈ <ω 2 there is t ∈ k 2 such that s t / ∈ T .There is a family {T f : f ∈ ω 2} of 2-porous trees such that for every X ∈ SP , the set {f ∈ ω 2 : [T f ] ⊆ X} is countable.