3 All meager filters may be null 9 9 1 Tomek Bartoszynski∗ Martin Goldstern† n Boise State University Bar Ilan University a J Boise, Idaho Ramat Gan, Israel 5 and 1 Hebrew University ] Jerusalem O L Haim Judah† Saharon Shelah†‡ . Bar Ilan University Hebrew University h t Ramat Gan, Israel Jerusalem a m February 1, 2008 [ 1 v 6 Abstract 0 2 Weshow that it is consistent with ZFCthat all filterswhich havethe 1 BairepropertyareLebesguemeasurable. Wealsoshowthattheexistence 0 of a Sierpinski set implies that there exists a nonmeasurable filter which 3 has theBaire property. 9 / The goal of this paper is to show yet another example of nonduality between h t measure and category. a SupposethatF isanonprincipalfilteronω. IdentifyF withthesetofchar- m acteristicfunctionsofitselements. UnderthisconventionF becomesasubsetof v: 2ω andaquestionaboutitstopologicalormeasure-theoreticalpropertiesmakes i sense. X It has been proved by Sierpinski that every non-principal filter has either r Lebesgue measure zero or is nonmeasurable. Similarly it is either meager or a does not have the Baire property. In [T] Talagrand proved that Theorem 0.1 There exists a measurable filter which does not have the Baire property. ∗Theauthor thankstheLadyDavisFellowshipTrustforfullsupport †SupportedbytheIsraelAcademyofSciences (BasicResearch Fund) ‡Publication434 1 In fact we have an even stronger result. In [Ba] it is proved that Theorem 0.2 Every measurable filter can be extended to a measurable filter which does not have the Baire property. We show that the dual result is false. 1 A model where all meager filters are null In this section we prove the following theorem: Theorem 1.1 It is consistent with ZFC that every filter which has the Baire property is measurable. Proof We will use the following more general result: Theorem 1.2 Let V |= GCH and suppose that V[G] is a generic extension extension of V obtained adding ω2 Cohen reals. Then in V[G] for any two sets A,B ⊂ 2ω if A+B = {a+b : a ∈ A,b ∈ B} is a meager set then either A or B has measure zero. Proof Note that we apply this lemma only for the case A = B. Therefore to simplify the notation we assume that A = B. The proof of the general case is almost the same. We follow [Bu]. We will use the following notation. Let Fn(X,2)={s:dom(s)∈[X]<ω and range(s)⊂{0,1}} be the notion of forcing adding |X|-many Cohen reals. For s ∈ Fn(X,2) let [s]={f ∈2X :s⊂f}. Let V |= GCH be a model of ZFC and let Gω2 be a Fn(ω2,2)-generic filter over V. Clearly c = SGω2 is a generic sequence of ω2 Cohen reals and V[c]=V[G ]. ω2 Let {F : n ∈ ω} be a sequence of closed, nowhere dense sets such that n A+A ⊆ F . Without loss of generality we can assume that {F : n ∈ Sn∈ω n n ω}∈V. Let {aξ :ξ <ω2} be an enumeration of all elements of A. For every ξ <ω2 let a˙ξ be a name for aξ. In other words for every ξ < ω2 we have a countable set Iξ ⊂ ω2 such that a˙ξ is a Borel function from 2Iξ into 2ω. Moreover aξ is the value of of the function a˙ on Cohen real i.e. a˙ (c|I )=a . In addition we ξ ξ ξ ξ can find a dense Gδ set Hξ ⊆2Iξ such that a˙ξ|Hξ is a continuous function. For α,ξ,η <ω2 define ξ ≃α η if 1. I and I are order isomorphic, ξ η 2. theorder-isomorphismbetweenI andI transfersa˙ ontoa˙ andH onto ξ η ξ η ξ H , η 2 3. I ∩α=I ∩α. ξ η Notice that for every α<ω2 the relation ≃α is an equivalence relation with ω1 many equivalence classes. Lemma 1.3 There exists α⋆ <ω such that 2 ∀ξ,β ∃η (ξ ≃ η & I ∩(β−α⋆)=∅) . α⋆ η Proof For every α < ω2 let Eα be the set {[ξ]α : ξ < ω2} of ≃α-equivalence classes. Let Eα0 ={E ∈Eα : sup(min(Iη −α))<ω2} and η∈E E1 =E −E0 . α α α Let γ(α)= sup(sup(min(I −α))) . η E∈E0 η∈E α Note that γ(α)<ω2 since |Eα|<ℵ1. Findα⋆ <ω2 suchthatγ(α)<α⋆ forallα<α⋆ andcf(α⋆)=ω1. Weclaim that α⋆ satisfies the statement of the lemma. Takeanyξ <ω2 andanyβ. Ifβ <α⋆ orIξ ⊆α⋆,thenwecanchooseη =ξ. So assume β >α⋆ andI −α⋆ 6=∅. There is α<α⋆ suchthat I ∩α=I ∩α⋆. ξ ξ ξ Let E =[ξ] . α Case 1 E ∈E0. Then α sup(min(I −α))≤γ(α)<α⋆ η η∈E which is a contradiction since min(I −α)≥α⋆ and ξ ∈E. ξ Case 2 E 6∈E0. So α sup(min(Iη −α))=ω2 η∈E hence there is η ∈E with min(I −α)≥β i.e. I ∩(β−α)=∅. η η SoI ∩α⋆ =I ∩α=I ∩α=I ∩α⋆, wherethe lastequalityholdsbecause ξ ξ η η I ∩(α⋆−α)⊆I ∩(β−α)=∅. Also I ∩(β−α⋆)⊆I ∩(β−α)=∅. η η η η Let α⋆ be the ordinal from the above lemma. Work in V′ =V[c|α⋆]. For every ξ <ω2 define Dξ ={s∈Fn(ω2−α⋆,2):cl(a˙ξ([s])) has measure zero } . Lemma 1.4 Dξ is dense in Fn(ω2−α⋆,2) for every ξ <ω2. 3 Proof Notice that it is enough to show that D ∩Fn(I −α⋆,2) is dense in ξ ξ Fn(Iξ−α⋆,2) for ξ <ω2. Suppose that this fails. Find ξ < ω2 and s0 ∈ Fn(Iξ −α⋆,2) such that for all s⊇s0 the set cl(a˙ξ([s])) has positive measure. Using the lemma with β > sup(Iξ) we can find η < ω2 such that ξ ≃α⋆ η and (Iξ −α⋆)∩(Iη −α⋆) = ∅. Notice that there exists t0 ∈ Fn(Iη −α⋆,2) (the image of s0 under the isomorphismbetween Iξ and Iη) such that for every t⊇t0 the set cl(a˙η([t])) has positive measure. Since s0 and t0 have disjoint domains, s0∪t0 ∈Fn(ω2−α⋆,2). Find n∈ω and a condition u ∈ Fn(ω2 −α⋆,2) extending s0 ∪t0 such that u ⊢ a˙ξ(c˙)+ a˙η(c˙) ∈ Fn. u can be written as u1∪u2∪u3 where s0 ⊆ u1 ∈ Fn(Iξ −α⋆,2), t0 ⊆u2 ∈Fn(Iη−α⋆,2)andu3 ∈Fn(ω2−(Iξ∪Iη∪α⋆),2). Bytheassumption the setscl(a˙ξ([u1])),cl(a˙η([u2])) havepositivemeasure. By well-knowntheorem of Steinhaus the set cl(a˙ξ([u1]))+cl(a˙η([u2])) contains an open set (hence also (cl(a˙ξ([u1]))+cl(a˙η([u2])))−Fn contains an open set). Using the fact that a˙ξ and a˙η are continuous functions we can find u1 ⊆ s1 ∈ Fn(Iξ − α⋆,2) and u2 ⊆ t1 ∈ Fn(Iη −α⋆,2) such that (cl(a˙ξ([s1]))+cl(a˙η([t1])))∩Fn = ∅. But this is a contradiction since s1∪t1∪u3 ⊢a˙ξ(c˙)+a˙η(c˙)6∈Fn . Notice that for ξ <ω2 D ={s∈Fn(I ): there exists a closed measure zero set F ∈V′ ξ ξ such that s ⊢a˙ (c˙)∈F} . ξ Therefore by the above lemma A⊆[{F :F is a closed measure zero set coded in V′} . Since V contains Cohen reals over V′, the union of all closedmeasure zero sets coded in V′ has measure zero in V. We conclude that A has measure zero. Let F be a non-principal filter. Denote by Fc = {X ⊆ ω : ω −X ∈ F}. Fc is an ideal and it is very easy to see that F is measurable (has the Baire property) iff Fc is measurable (has the Baire property). Lemma 1.5 F +F =Fc. Proof Suppose that X,Y ∈ F. Then {n : X(n)+Y(n) = 0} ⊇ X−1(1)∩ Y−1(1) ∈ F. In general F +···+F is equal to F or Fc depending whether there is an even or odd number of F’s. Let V|=GCH and suppose that V[G] is a generic extension of V obtained by adding ω2 Cohen reals. By the above lemma if F is a meager filter then Fc =F +F is meager. So by 1.2 F has measure zero. 4 2 Filters which are meager and nonmeasurable Theorem 1.1 shows that in order to construct a filter which is meager and nonmeasurable we need some extra assumptions. In [T] Talagrand showed that Theorem 2.1 Suppose that the real line is not the union of <2ℵ0 many mea- sure zero sets. Then there exists a nonmeasurable filter which is meager. Letκbearegularuncountablecardinal. RecallthatS isageneralizedSierpinski set ofsize κ if |S∩H <κ for everynull setH. It is clear that allS′ ⊆S of size κ are also nonmeasurable. Theorem 2.2 AssumethatthereexistsageneralizedSierpinskiset. Thenthere exists a nonmeasurable meager filter. Proof Let S be a generalized Sierpinski set of size κ. Build a sequence {x : ξ ξ <κ}⊂S andanelementary chainofmodels {M :ξ <κ}of size κ suchthat ξ 1. {x :ξ <α}⊂M for α<κ, ξ α 2. x is a random real over M for β >α. β α Suppose that M ,x are already constructed for β <α. Since S is a Sierpinski β β set [{S∩H :H is a null set coded in Mβ for β <α} has size <κ. Let x be any element of S avoiding this set. α Let X = x−1(1) for ξ < κ. Let F be the filter generated by the family ξ ξ {X :ξ <κ}. We will show that F has the required properties. ξ For X ⊂ω let |X ∩n| d(X)= lim n→∞ n if the above limit exists. By easy induction we show that for ξ1,...,ξn < κ we have d(Xξ1 ∩···∩ X )=2−n. This shows that ξn |X∩n| F ⊆{X ⊂ω :liminf >0} n→∞ n whichisameagerset. TocheckthatF isnonmeasurablenoticethatF contains the nonmeasurable set {x :ξ <κ} . ξ It is an open problem whether one can construct a meager nonmeasurable filter assuming the existence of a nonmeasurable set of size ℵ1. We only have some partial results. Let b be the size of the smallest unbounded family in ωω and let unif be the size of the smallest nonmeasurable set. For X ⊆ ω let f ∈ ωω be an increasing function enumerating X. For a X filter F let F⋆ ={f :X ∈F}. In [J] it is proved that X 5 Theorem 2.3 For every filter F, F has the Baire property iff F⋆ is bounded. Theorem 2.4 Supposethatunif <b. Then thereexistsanonmeasurablefilter which is meager. Proof Let X ⊆2ω be a nonmeasurable set of size unif. Let M be a model of thesamesizecontainingX asasubset. ThenM∩2ωdoesnothavemeasurezero, soitisnonmeasurable. ConsideranyfilterF suchthatM |=F is an ultrafilter. F generates a filter in V and this filter is meager by 2.3 and the fact that it is generated by unif < b many elements. On the other hand M |= 2ω = F ∪Fc andwe knowthat M∩2ω is a nonmeasurableset. Hence F is nonmeasurable. The previous theorem depended on the implication: If F has measure zero then M ∩2ω has measure zero. This implicationisnottrue ingeneralforanysetX ∈M havingoutermeasure 1 in M as is showed by the following example. ExampleItis consistentwithZFCthattherearemodelsM ⊂V suchthat only some sets which have outer measure 1 in M have measure 0 in V. Let V=L[c][hrξ :ξ <ω1i] where c is a Cohen real over L and hrξ :ξ <ω1i is a sequence of random reals over L[c] (added side by side). Let M = L[hr : ξ ξ <ω1i]. Consider the set X =L∩2ω. It is known that X is a nonmeasurable set in M but X has measure 0 in V. On the other hand the set {rξ : ξ < ω1} is nonmeasurable in V. We conclude the paper with a canonical example of a filter which does not generate an ultrafilter. In other words we have the following: Theorem 2.5 Let M be a model for ZFC and let r be a real which does not belong to M. Then there exists a filter F such that M |=F is an ultrafilter but M[r]|={X ⊆ω :∃Y ∈F Y ⊆X} is not an ultrafilter . Proof Let {k :n∈ω} be a fast increasing sequence of natural numbers. Let n T be a tree on 2<ω such that: 1. For s∈T we have |s|=k iff s⌢0∈T and s⌢1∈T, n 2. let {s1,...,s2n} be the list of T ∩ 2kn in lexicographical order. Then for every w ⊆ P(2n)−{∅,2n} there exists m ∈ [kn+1,kn+1) such that s (m)=0 iff l∈w, l 3. there is no m∈ω such that for all s∈T ∩2m+1 we have s(m)=0 or for all s∈T ∩2m+1 we have s(m)=1. 6 Let S ⊆T be a subtree of T. Define A0 ={m:∀s∈S∩2m+1 s(m)=0} and S A1 ={m:∀s∈S∩2m+1 s(m)=1} . S Let J be the ideal generated by sets {A0,A1 :S is a perfect subtree of T}. S S One can easily verify that all finite subsets of ω belong to J. Lemma 2.6 J is a proper ideal. Proof Let S1,...,Sm be perfect subtrees of T. Find n sufficiently big so that |Sj∩2kn|>mforj ≤m. Lets1,...,s2n bethelistofT∩2kn inlexicographical ordering. Let w1,...,wm be such that Sj ∩2kn = {si :i∈ wj} for j ≤ m. Let w ={min(w1),...,min(wm)}. Then for all j, wj 6⊆w and wj ∩w 6=∅. By the definition of T there is k < k such that w = {l : s (k) = 0}. By the property n l of w for every j ≤ m there exist s0,s1 ∈ Sj ∩2kn such that s0(k) = 0 and s1(k)=1. Therefore k 6∈A0 ∪A1 ∪···∪A0 ∪A1 . S1 S1 Sm Sm LetF beanyultrafilterinM extendingthefilter{ω−X :X ∈J}. Letr be a real which does not belong to M. Without loss of generality we can assume that r is a branch through T. Assume that F generates an ultrafilter and let X = {n : r(n) = 1}. We r can assume that there exists an element X ∈ F such that X ⊆ X . Let r S = {s ∈ T : ∀k ∈ X (|s| > k → s(k)= 1)}. Clearly r is a branch through S. But in that case S contains a perfect subtree S1 ⊆ S (since it contains a new branch). Therefore X ⊆A1 ∈J. Contradiction. S1 References [Ba] T. Bartoszynski On the structure of the filters on a countable set to appear [Bu] M. Burke notes of June 17, 1989 [J] H. Judah Unbounded filters on ω, in Logic Colloquium 1987 [T] M. Talagrand Compacts de fonctions mesurables et filtres non- mesurables, Studia Mathematica, T.LXVII, 1980. 7