aa r X i v : . [ m a t h . L O ] S e p PARAMETRIZED MEASURING AND CLUB GUESSING
DAVID ASPER ´O AND JOHN KRUEGER
Abstract.
We introduce
Strong Measuring , a maximal strengthening of J.T. Moore’s Measuring principle, which asserts that every collection of fewerthan continuum many closed bounded subsets of ω is measured by some clubsubset of ω . The consistency of Strong Measuring with the negation of CH is shown, solving an open problem from [2] about parametrized measuringprinciples. Specifically, we prove that Strong Measuring follows from MRP together with Martin’s Axiom for σ -centered forcings, as well as from BPFA .We also consider strong versions of Measuring in the absence of the Axiom ofChoice.
Club guessing principles at ω are well–studied natural weakenings of Jensen’s ♦ principle. Presented in a general form, they assert the existence of a sequence ~C = h c α : α ∈ ω ∩ Lim i , where each c α is a club of α , such that ~C guesses clubsof ω in some suitable sense. ~C guessing a club D of ω usually means that thereis some (equivalently, stationarily many) δ ∈ D such that c δ ∩ D is a suitably largesubset of c δ ; for example, we could require that c δ ⊆ D , in which case the resultingstatement is called club guessing , or that c δ ∩ D is cofinal in δ , in which case wecall the resulting statement very weak club guessing .Unlike the case of their versions at cardinals higher than ω , for which there arenon–trivial positive ZFC theorems (see, for example, [14]), club guessing principlesat ω are independent of ZFC . On the one hand, all of these principles obviouslyfollow from ♦ , and hence they hold in L , and they can always be forced by countablyclosed forcing. On the other hand, classical forcing axioms at the level of ω , suchas the Proper Forcing Axiom ( PFA ), imply the failure of even the weakest of theseprinciples. It should nevertheless be noted that Martin’s Axiom + ¬ CH is compa-tible with Club Guessing. This is because Martin’s Axiom can always be forced bya c.c.c. forcing, and the fact that every club of ω in a generic extension via a c.c.c.forcing contains a club of ω from the ground model implies that a club–guessingsequence from the ground model remains club–guessing in the extension. (On theother hand, this is of course not the case for ♦ since the negation of CH violates ♦ .) Measuring is a particularly strong failure of Club Guessing due to J. T. Moore([8]). Let X and Y be countable subsets of ω with the same supremum δ . We saythat X measures Y if there exists β < δ such that X \ β is either contained in,or disjoint from, Y . Measuring is the statement that for any sequence h c α : α ∈ Date : August 2018; revised July 2019.The first author acknowledges support of EPSRC Grant EP/N032160/1. The second authorwas partially supported by the National Science Foundation Grant No. DMS-1464859.2010
Mathematics Subject Classification:
Primary 03E05, 03E35; Secondary 03E57.
Key words and phrases.
Strong Measuring, Club Guessing, MRP, BPFA. ω ∩ Lim i , where each c α is a closed subset of α , there exists a club D ⊆ ω suchthat for all limit points δ ∈ D of D , D ∩ δ measures c δ .Measuring can be viewed as a strong negation of Club Guessing since, as is easyto see, it implies the failure of Very Weak Club Guessing. Measuring follows fromthe Mapping Reflection Principle ( MRP ), and therefore from
PFA , and it can beforced over any model of
ZFC .From Measuring as a vantage point, one can attempt to consider even strongerfailures of Club Guessing. In this vein, the following parametrized family of strength-enings of Measuring was considered in [2].
Definition.
For a cardinal κ , let Measuring <κ denote the statement that whenever ~ C = hC α : α ∈ ω ∩ Lim i is a sequence such that each C α is a family of fewer than κ many closed subsets of α , there exists a club D ⊆ ω with the property that forevery limit point δ of D and every c ∈ C δ , D ∩ δ measures c . For a cardinal λ , let Measuring λ denote Measuring <λ + . In the situation given by the above definition, we say that D measures ~ C . Wealso define Strong Measuring to be the statement
Measuring < ω .In the present article we contribute to the body of information on Measuringand related strong failures of Club Guessing (see also [8], [3], [5], [6], and [2]). Oneof the questions left unresolved in [2] is whether Measuring ω is consistent at all.Answering this question was the motivation for the work in the present article. Ourmain result is that Strong Measuring + ¬ CH is consistent. In fact, this statementfollows from MRP + Martin’s Axiom for the class of σ -centered posets, and alsofrom BPFA . We also show the failure, in
ZFC , of
Measuring κ , where κ is amongsome of the classical cardinal characteristics of the continuum. Finally, we considervery strong versions of Measuring in contexts in which the Axiom of Choice fails.1. Background
We review some background material and notation which is needed for under-standing the paper. Let c denote the cardinality of the continuum 2 ω . A set S ⊆ [ ω ] ω is a splitting family if for any infinite set x ⊆ ω , there exists A ∈ S suchthat A splits x in the sense that both x ∩ A and x \ A are infinite. The splitting num-ber s is the least cardinality of some splitting family. Given functions f, g : ω → ω ,we say that g dominates f if for all n < ω , f ( n ) < g ( n ). We say that g eventuallydominates f if there is some m < ω such that f ( n ) < g ( n ) for all n > m . A family B ⊆ ω ω is bounded if there exists a function g ∈ ω ω which eventually dominatesevery member of B , and otherwise it is unbounded . The bounding number b is theleast cardinality of some unbounded family. Both cardinal characteristics s and b are uncountable.Let P be a forcing poset. A set X ⊆ P is centered if every finite subset of X has a lower bound. We say that P is σ -centered if it is a union of countablymany centered sets. Martin’s Axiom for σ -centered forcings ( MA ( σ -centered)) isthe statement that for any σ -centered forcing P and any collection of fewer than c many dense subsets of P , there exists a filter on P which meets each dense set in thecollection. More generally, let m ( σ -centered) be the least cardinality of a collectionof dense subsets of some σ -centered forcing poset for which there does not exist a We can also prove the consistency of Strong Measuring with the continuum being arbitrarilylarge. This result will appear in a sequel to the present article.
ARAMETRIZED MEASURING AND CLUB GUESSING 3 filter which meets each dense set in the collection. Note that MA ( σ -centered) isequivalent to the statement that m ( σ -centered) equals c .The Bounded Proper Forcing Axiom ( BPFA ) is the statement that whenever P is a proper forcing and h A i : i < ω i is a sequence of maximal antichains of P each of size at most ω , then there exists a filter on P which meets each A i ([9]).We note that BPFA implies c = ω ([12, Section 5]). It easily follows that BPFA implies Martin’s Axiom, and in particular, implies MA ( σ -centered). The forcingaxiom BPFA is equivalent to the statement that for any proper forcing poset P andany Σ statement Φ with a parameter from H ( ω ), if Φ holds in a generic extensionby P , then Φ holds in the ground model ([7]).An open stationary set mapping for an uncountable set X and regular cardinal θ > ω is a function Σ whose domain is the collection of all countable elementarysubstructures M of H ( θ ) with X ∈ M , such that for all such M , Σ( M ) is an open, M -stationary subset of [ X ] ω . By open we mean in the Ellentuck topology on [ X ] ω ,and M -stationary means meeting every club subset of [ X ] ω which is a member of M (see [12] for the complete details). In this article, we are only concerned with theseideas in the simplest case that X = ω and for each M ∈ dom(Σ), Σ( M ) ⊆ ω . Inthis case, being open is equivalent to being open in the topology on ω with basisthe collection of all open intervals of ordinals, and being M -stationary is equivalentto meeting every club subset of ω in M .For an open stationary set mapping Σ for X and θ , a Σ -reflecting sequence is an ∈ -increasing and continuous sequence h M i : i < ω i of countable elementarysubstructures of H ( θ ) containing X as a member satisfying that for all limit ordinals δ < ω , there exists β < δ so that for all β ≤ ξ < δ , M ξ ∩ X ∈ Σ( M δ ). The Mapping Reflection Principle ( MRP ) is the statement that for any open stationaryset mapping Σ, there exists a Σ-reflecting sequence. We will use the fact that forany open stationary set mapping Σ, there exists a proper forcing which adds aΣ-reflecting sequence ([12, Section 3]). Consequently,
MRP follows from
PFA .2.
Parametrized Measuring and Club Guessing
Let X and Y be countable subsets of ω with the same supremum δ . We say that X measures Y if there exists β < δ such that X \ β is either contained in, or disjointfrom, Y . Measuring is the statement that for any sequence h c α : α ∈ ω ∩ Lim i ,where each c α is a closed and cofinal subset of α , there exists a club D ⊆ ω suchthat for all limit points α of D , D ∩ α measures c α .The next two results are due to J. T. Moore ([8]). Theorem 2.1.
MRP implies Measuring.
Theorem 2.2.
BPFA implies Measuring.
We now describe parametrized forms of measuring which were introduced in [2].Let ~ C = hC α : α ∈ ω ∩ Lim i be a sequence such that each C α is a collection of closedand cofinal subsets of α . A club D ⊆ ω is said to measure ~ C if for all α ∈ lim( D )and all c ∈ C α , D ∩ α measures c . Definition 2.3.
For a cardinal κ , let Measuring <κ denote the statement that when-ever ~ C = hC α : α ∈ ω ∩ Lim i is a sequence such that each C α is a collection of fewerthan κ many closed and cofinal subsets of α , then there exists a club D ⊆ ω whichmeasures ~ C . For a cardinal λ , let Measuring λ denote Measuring <λ + . DAVID ASPER ´O AND JOHN KRUEGER
Observe that the principle Measuring is the same as
Measuring . If κ < λ , thenclearly Measuring <λ implies Measuring <κ . It is easy to see that Measuring c is false. Definition 2.4.
Strong Measuring is the statement that
Measuring < c holds. Since the intersection of countably many clubs in ω is club, Measuring easilyimplies Measuring ω . In particular, Measuring together with CH implies StrongMeasuring. We will prove in Section 3 the consistency of Strong Measuring togetherwith ¬ CH . We also observe at the end of that section that Measuring does not imply Measuring ω . Proposition 2.5 ([2]) . Measuring s is false.Proof. Fix a splitting family S of cardinality s . For each limit ordinal α < ω , fixa function f α : ω → α which is increasing and cofinal in α . For each A ∈ S , let c α,A = S { ( f α ( n ) , f α ( n + 1)] : n ∈ A } , which is clearly closed and cofinal in α . Let C α := { c α,A : A ∈ S } . Then ~ C := hC α : α ∈ ω ∩ Lim i is a sequence such that foreach α , C α is a collection of at most s many closed and cofinal subsets of α .Let D ⊆ ω be a club. Fix α ∈ lim( D ). We will show that there exists a memberof C α which D ∩ α does not measure. Define x := { n < ω : D ∩ ( f α ( n ) , f α ( n + 1)] = ∅} . Since α ∈ lim( D ), x is infinite. As S is a splitting family, we can fix A ∈ S which splits x . So both x ∩ A and x \ A are infinite. We claim that D ∩ α does notmeasure c α,A .Suppose for a contradiction that for some β < α , ( D ∩ α ) \ β is either a subsetof, or disjoint from, c α,A . Since A ∩ x is infinite, we can fix n ∈ A ∩ x such that f α ( n ) > β . Then n ∈ x implies that D ∩ ( f α ( n ) , f α ( n + 1)] = ∅ , and n ∈ A impliesthat ( f α ( n ) , f α ( n + 1)] ⊆ c α,A . It follows that ( D ∩ α ) \ β meets c α,A . By thechoice of β , this implies that ( D ∩ α ) \ β is a subset of c α,A . But x \ A is alsoinfinite, so we can fix m ∈ x \ A such that f α ( m ) > β . Then m ∈ x implies that D ∩ ( f α ( m ) , f α ( m + 1)] = ∅ , and m / ∈ A implies that ( f α ( m ) , f α ( m + 1)] is disjointfrom c α,A . Thus, there is a member of ( D ∩ α ) \ β which is not in c α,A , which is acontradiction. (cid:3) We will prove later in this section that
Measuring b is also false.We now turn to parametrized club guessing. We recall some standard defini-tions. Consider a sequence ~L = h L α : α ∈ ω ∩ Lim i , where each L α is a cofinalsubset of α with order type ω (that is, a ladder system ). We say that ~L is a clubguessing sequence , weak club guessing sequence , or very weak club guessing sequence ,respectively, if for every club D ⊆ ω , there exists a limit ordinal α < ω such that:(1) L α ⊆ D ,(2) L α \ D is finite, or(3) L α ∩ D is infinite, respectively.We say that Club Guessing , Weak Club Guessing , or
Very Weak Club Guessing holds, respectively, if there exists a club guessing sequence, a weak club guessingsequence, or a very weak club guessing sequence, respectively. It is well knownthat Measuring implies the failure of Very Weak Club Guessing (see Proposition2.8 below).
Definition 2.6.
Let ~ L = hL α : α ∈ ω ∩ Lim i be a sequence where each L α is anon–empty collection of cofinal subsets of α with order type ω . The sequence ~ L issaid to be a club guessing sequence , weak club guessing sequence , or very weak ARAMETRIZED MEASURING AND CLUB GUESSING 5 club guessing sequence , respectively, if for every club D ⊆ ω , there exists a limitordinal α < ω and some L ∈ L α such that: (1) L ⊆ D , (2) L \ D is finite, or (3) L ∩ D is infinite, respectively. Definition 2.7.
For a cardinal κ , let CG <κ , WCG <κ , and VWCG <κ , respectively,be the statements that there exists a club guessing sequence, weak club guessingsequence, or very weak club guessing sequence hL α : α ∈ ω ∩ Lim i , respectively,such that for each α , |L α | < κ . Let CG κ , WCG κ , and VWCG κ denote the statements CG <κ + , WCG <κ + , and VWCG <κ + , respectively. Clearly, if κ < λ , then CG <κ implies CG <λ , and similarly with WCG and
VWCG .Observe that Club Guessing, Weak Club Guessing, and Very Weak Club Guessingare equivalent to CG , WCG , and VWCG , respectively. Obviously, CG c is true.The weakest forms of club guessing principles which are not provable in ZFC arewhen the index is < c . Proposition 2.8.
For any cardinal κ ≥ , Measuring <κ implies the failure of VWCG <κ .Proof. Suppose for a contradiction that
Measuring <κ and VWCG <κ both hold. Fixa very weak club guessing sequence ~ L = hL α : α ∈ ω ∩ Lim i such that each L α hascardinality less than κ . Observe that for each α , every member of L α is vacuouslya closed subset of α since it has order type ω .By Measuring <κ , there exists a club D ⊆ ω which measures ~ L . Let E be theclub set of indecomposable limit ordinals α > ω in lim( D ) such that ot( D ∩ α ) = α .Since ~ L is a very weak club guessing sequence, there exists a limit ordinal α and L ∈ L α such that L ∩ E is infinite. In particular, α is a limit point of E , and henceof D .Since D measures ~ L and L ∈ L α , D ∩ α measures L . So we can fix β < α suchthat ( D ∩ α ) \ β is either a subset of, or disjoint from, L . Now L ∩ E , and hence L ∩ D , is infinite. As L has order type ω , this implies that L ∩ D is cofinal in α . Bythe choice of β , ( D ∩ α ) \ β must be a subset of L . But since α ∈ E , ot( D ∩ α ) = α and α is indecomposable, which implies that ot(( D ∩ α ) \ β ) = α . As α > ω , thisis impossible since ( D ∩ α ) \ β is a subset of L and L has order type ω . (cid:3) In particular, since Strong Measuring is consistent, so is the failure of
VWCG < c .(The consistency of ¬ VWCG < c together with c arbitrarily large was previouslyshown in [4].) Proposition 2.9 (Hruˇs´ak [5]) . VWCG b is true.Proof. Fix an unbounded family { r α : α < b } in ω ω . For each limit ordinal δ < ω ,fix a cofinal subset C δ of δ with order type ω and a bijection h δ : ω → δ . Let C δ ( n )denote the n -th member of C δ for all n < ω . For all limit ordinals δ < ω and α < b , define A αδ := C δ ∪ [ { h δ [ r α ( n )] \ C δ ( n ) : n < ω } . It is easy to check that for all δ and α , A αδ has order type ω and sup( A αδ ) = δ .Given a club C ⊆ ω , let δ be a limit point of C and let g C,δ : ω → ω be thefunction given by g C,δ ( n ) = min { m < ω : h δ ( m ) ∈ C \ C δ ( n ) } . DAVID ASPER ´O AND JOHN KRUEGER
Now let α < b be such that r α ( n ) > g C,δ ( n ) for infinitely many n . It then followsthat | A αδ ∩ C | = ω . (cid:3) By Propositions 2.8 and 2.9, the following is immediate.
Corollary 2.10.
Measuring b is false. An obvious question is whether the parametrized versions of club guessing areactually the same as the usual ones. We conclude this section by showing that theyare not.Recall that a forcing poset P is ω ω -bounding if every function in ω ω ∩ V P isdominated by a function in ω ω ∩ V . Lemma 2.11 (Hruˇs´ak) . Assume that
VWCG fails. Let P be any ω -c.c., ω ω -bounding forcing. Then P forces that VWCG fails.Proof.
Since P is ω -c.c. and ω ω -bounding, a standard argument shows that when-ever p ∈ P and p forces that ˙ b ∈ ω ω , then there exists a function b ∗ ∈ ω ω such that p forces that b ∗ dominates ˙ b .Let us show that whenever p ∈ P , δ < ω , and p forces that ˙ X is a cofinal subsetof δ of order type ω , then there exists a set Y with order type ω such that p forcesthat ˙ X ⊆ Y . To see this, fix a bijection f : ω → δ and a strictly increasing sequence h α n : n < ω i cofinal in α with α = 0. We claim that there exists a P -name ˙ b for afunction from ω to ω such that p forces that for all n < ω , ˙ b ( n ) is the least m < ω such that ˙ X ∩ [ α n , α n +1 ) ⊆ f [ m ]. This is true since p forces that ˙ X has order type ω and hence that ˙ X ∩ [ α n , α n +1 ) is finite for all n < ω . Fix a function b ∗ : ω → ω such that p forces that b ∗ dominates ˙ b . Now let Y := [ { f [ b ∗ ( n )] ∩ [ α n , α n +1 ) : n < ω } . It is easy to check that Y has order type ω and p forces that ˙ X ⊆ Y .Now we are ready to prove the proposition. So suppose that p ∈ P forces that h ˙ X α : α ∈ ω ∩ Lim i is a very weak club guessing sequence. By the previousparagraph, for each limit ordinal α < ω we can fix a cofinal subset Y α of α withorder type ω such that p forces that ˙ X α ⊆ Y α . We claim that h Y α : α ∈ ω ∩ Lim i is a very weak club guessing sequence in the ground model, which completes theproof. So consider a club C ⊆ ω . Then C is still a club in V P . Fix q ≤ p anda limit ordinal α < ω such that q forces that ˙ X α ∩ C is infinite. Then clearly q forces that Y α ∩ C is infinite, so in fact, Y α ∩ C is infinite. (cid:3) Proposition 2.12.
It is consistent that ¬ VWCG and CG ω both hold.Proof. Let V be a model in which CH holds and VWCG fails. Such a model wasshown to exist by Shelah [13]. Let P be an ω -c.c., ω ω -bounding forcing posetwhich adds at least ω many reals; for example, random real forcing with productmeasure is such a forcing. We claim that in V P , CG ω holds but VWCG fails. ByLemma 2.11,
VWCG is false in V P . In V , define ~ L = hL α : α ∈ ω ∩ Lim i by letting L α be the collection of all cofinal subsets of α with order type ω . Since CH holds,the cardinality of each L α is ω . If C is a club subset of ω in V P , then since P is ω -c.c., there is a club D ⊆ ω in V such that D ⊆ C . In V , fix d ⊆ D with ordertype ω , and let α := sup( d ). Then d ∈ L α and d ⊆ C . Thus, ~ L witnesses that CG ω holds in V P . (cid:3) ARAMETRIZED MEASURING AND CLUB GUESSING 7 The Consistency of Strong Measuring and ¬ CH As we previously mentioned, Measuring is equivalent to
Measuring ω , and there-fore under CH , Measuring is equivalent to Strong Measuring. In this section weestablish the consistency of Strong Measuring with the negation of CH . More pre-cisely, we will prove that MRP together with MA ( σ -centered) implies Strong Mea-suring, and BPFA implies Strong Measuring. Recall that both
MRP and
BPFA imply that c = ω ([12]).A set M is suitable if for some regular cardinal θ > ω , M is a countableelementary substructure of H ( θ ). We will follow the conventions introduced inSection 1 that the properties “open” and “ M -stationary” refer to open and M -stationary subsets of ω (where ω is considered as a subspace of [ ω ] ω ). Proposition 3.1.
Assume that M is suitable. Let δ := M ∩ ω . Suppose that Y is a collection of open subsets of δ such that for any finite set a ⊆ Y , T a is M -stationary. Then there exists a σ -centered forcing P and a collection D of densesubsets of P of size at most |Y| + ω such that whenever G is a filter on P in someouter model W of V with ω V = ω W which meets each member of D , then thereexists a set z ⊆ δ in W which is open, M -stationary, and satisfies that for all X ∈ Y , z \ X is bounded in δ .Proof. Define a forcing poset P to consist of conditions which are pairs ( x, a ), where x is an open and bounded subset of δ in M and a is a finite subset of Y . Let( y, b ) ≤ ( x, a ) if y is an end-extension of x , a ⊆ b , and y \ x ⊆ T a .Since M is countable, there are only countably many possibilities for the firstcomponent of a condition. If ( x, a ) , . . . , ( x, a n ) are finitely many conditions withthe same first component, then easily ( x, a ∪ . . . ∪ a n ) is a condition in P which isbelow each of the conditions ( x, a ) , . . . , ( x, a n ). It follows that P is σ -centered.For each X ∈ Y , let D X denote the set of conditions ( x, a ) such that X ∈ a .Observe that D X is dense. For every club C of ω which is a member of M , let E C denote the set of conditions ( x, a ) such that x ∩ C is non–empty. We claim that E C isdense. Let ( x, a ) be a condition. Since T a is M -stationary and lim( C ) \ (sup( x )+1)is a club subset of ω in M , we can find a limit ordinal α in C ∩ ( T a ) which is inthe interval (sup( x ) , δ ). Since α ∈ T a and T a is open, we can find β < γ < δ suchthat α ∈ ( β, γ ) ⊆ T a . As sup( x ) + 1 < α , without loss of generality sup( x ) < β .By elementarity, the interval b := ( β, γ ) is in M . It follows that ( x ∪ b, a ) is acondition, x ∪ b end-extends x , and ( x ∪ b ) \ x = b ⊆ T a . Thus, ( x ∪ b, a ) ≤ ( x, a ),and since α ∈ C , ( x ∪ b, a ) ∈ E C .Let D denote the collection of all dense sets of the form D X where X ∈ Y , or E C where C is a club subset of ω belonging to M . Then |D| ≤ |Y| + ω . Let G bea filter on P in some outer model W with ω V = ω W which meets each dense set in D . Define z := S { x : ∃ a ( x, a ) ∈ G } . Note that since z is a union of open sets, it isopen (using the fact that being open is absolute between V and W ). For each club C ⊆ ω which lies in M , there exists a condition ( x, a ) which belongs to G ∩ E C ,and thus x ∩ C = ∅ . Therefore, z ∩ C = ∅ . Hence, z is M -stationary.It remains to show that for all X ∈ Y , z \ X is bounded in δ . Consider X ∈ Y .Then we can fix ( x, a ) ∈ G ∩ D X , which means that X ∈ a . Now the definitionof the ordering on P together with the fact that G is a filter easily implies that z \ x ⊆ X . Therefore, z \ X ⊆ x , and hence z \ X is bounded in δ . (cid:3) DAVID ASPER ´O AND JOHN KRUEGER
Corollary 3.2.
Assume that M is suitable. Let δ := M ∩ ω . Suppose that Y isa collection of less than m ( σ -centered) many open subsets of δ such that for anyfinite set a ⊆ Y , T a is M -stationary. Then there exists a set z ⊆ δ which is open, M -stationary, and satisfies that for all X ∈ Y , z \ X is bounded in δ .Proof. Fix a σ -centered forcing P and a collection D of dense subsets of P of size atmost |Y| + ω as described in Proposition 3.1. Since m ( σ -centered) is uncountable, |D| < m ( σ -centered). Hence, there exists a filter G on P which meets each denseset in D . By Proposition 3.1, there exists a set z ⊆ δ which is open, M -stationary,and satisfies that for all X ∈ Y , z \ X is bounded in δ . (cid:3) Proposition 3.3.
Let ~ C = hC α : α ∈ ω ∩ Lim i be a sequence such that each C α isa collection of less than m ( σ -centered) many closed and cofinal subsets of α . Thenthere exists an open stationary set mapping Σ such that, if W is any outer modelwith the same ω in which there exists a Σ -reflecting sequence, then there exists in W a club subset of ω which measures ~ C .Proof. For each limit ordinal α < ω , let D α := { α \ c : c ∈ C α } . Observe that each D α is a collection of fewer than m ( σ -centered) many open subsets of α .We will define Σ to have domain the collection of all countable elementary sub-structures M of H ( ω ). Consider such an M and we define Σ( M ). Note that M issuitable. Let δ := M ∩ ω . We consider two cases. In the first case, there does notexist a member of D δ which is M -stationary. Define Σ( M ) = δ , which is clearlyopen and M -stationary.In the second case, there exists some member of D δ which is M -stationary.A straightforward application of Zorn’s lemma implies that there exists a non–empty set Y M ⊆ D δ such that for any a ∈ [ Y M ] <ω , T a is M -stationary, andmoreover, Y M is a maximal subset of D δ with this property. Since Y M ⊆ D δ , |Y M | < m ( σ -centered). So the collection Y M satisfies the assumptions of Corollary3.2. It follows that there exists a set z M ⊆ δ which is open, M -stationary, andsatisfies that for all X ∈ Y M , z M \ X is bounded in δ . Now define Σ( M ) := z M .This completes the definition of Σ. Consider an outer model W of V with thesame ω , and assume that in W there exists a Σ-reflecting sequence h M δ : δ < ω i .Let α δ := M δ ∩ ω for all δ < ω . Let D be the club set of δ < ω such that α δ = δ .We claim that D measures ~ C .Consider δ ∈ lim( D ). Then δ = α δ = M δ ∩ ω . Let M := M δ . We first claimthat if c ∈ C δ and δ \ c is not M -stationary, then for some β < δ , ( D ∩ δ ) \ β ⊆ c .Fix a club subset E of ω in M which is disjoint from δ \ c . By the continuity ofthe Σ-reflecting sequence, there exists β < δ such that E ∈ M β . We claim that( D ∩ δ ) \ β ⊆ c . Let ξ ∈ ( D ∩ δ ) \ β . Then E ∈ M ξ , and hence by elementarity, ξ = M ξ ∩ ω ∈ E . Since E is disjoint from δ \ c , ξ ∈ c .We split the argument according to the two cases in the definition of Σ( M ). Inthe first case, there does not exist a member of D δ which is M -stationary. Consider c ∈ C δ . Then δ \ c is not M -stationary. By the previous paragraph, there exists β < δ such that ( D ∩ δ ) \ β ⊆ c .In the second case, there exists a member of D δ which is M -stationary. Consider c ∈ C δ . Then X := δ \ c ∈ D δ . We consider two possibilities. First, assume that X is in Y M . By the choice of Y M and z M , we know that z M \ X is bounded in δ . So fix β < δ so that z M \ β ⊆ X . By the definition of being a Σ-reflectingsequence, there exists β < δ so that for all β ≤ ξ < δ , M ξ ∩ ω ∈ Σ( M ) = z M . ARAMETRIZED MEASURING AND CLUB GUESSING 9
Let β := max { β , β } . Consider ξ ∈ ( D ∩ δ ) \ β . Then ξ ≥ β implies that ξ = M ξ ∩ ω ∈ z M . So ξ ∈ z M \ β ⊆ X = δ \ c .Secondly, assume that X is not in Y M . By the maximality of Y M , there exists aset a ∈ [ Y M ] <ω such that X ∩ T a is not M -stationary. Fix a club E in M which isdisjoint from X ∩ T a . By the continuity of the Σ-reflecting sequence, there exists β < δ such that E ∈ M β . Consider ξ ∈ ( D ∩ δ ) \ β . Then E ∈ M ξ , which impliesthat ξ = M ξ ∩ ω ∈ E . Thus, ξ is not in X ∩ T a . On the other hand, letting a = { X , . . . , X n } , for each i ≤ n the previous paragraph implies that there exists β i < δ such that ( D ∩ δ ) \ β i ⊆ X i . Let β ∗ be an ordinal in δ which is larger than β and β i for all i ≤ n . Consider ξ ∈ ( D ∩ δ ) \ β ∗ . Then by the choice of β , ξ / ∈ X ∩ T a .By the choice of the β i ’s, ξ ∈ T a . Therefore, ξ / ∈ X = δ \ c , which means that ξ ∈ c . Thus, ( D ∩ δ ) \ β ∗ ⊆ c . (cid:3) Corollary 3.4.
Assume
MRP and MA ( σ -centered). Then Strong Measuring holds.Proof. Let ~ C = hC α : α ∈ ω ∩ Lim i be a sequence such that each C α is a collectionof fewer than c many closed and cofinal subsets of α . We claim that there exists aclub subset of ω which measures ~ C . By MA ( σ -centered), m ( σ -centered) equals c .So each C α has size less than m ( σ -centered).By Proposition 3.3, there exists an open stationary set mapping Σ such that,if W is any outer model with the same ω in which there exists a Σ-reflectingsequence, then there exists in W a club subset of ω which measures ~ C . Applying MRP , there exists a Σ-reflecting sequence in V . Thus, in V there exists a clubsubset of ω which measures ~ C . (cid:3) Corollary 3.5.
Assume
BPFA . Then Strong Measuring holds.Proof.
Let ~ C = hC α : α ∈ ω ∩ Lim i be a sequence such that each C α is a collectionof fewer than c = ω many closed and cofinal subsets of α . We claim that thereexists a club subset of ω which measures ~ C . Since c = ω , ~ C is a member of H ( ω ).Thus, the existence of a club subset of ω which measures ~ C is expressible as a Σ statement involving a parameter in H ( ω ). By BPFA , it suffices to show that thereexists a proper forcing which forces that such a club exists.Now
BPFA implies Martin’s Axiom, and in particular, that m ( σ -centered) isequal to c . So each C α has size less than m ( σ -centered). By Proposition 3.3, thereexists an open stationary set mapping Σ such that, if W is any outer model withthe same ω in which there exists a Σ-reflecting sequence, then there exists in W a club subset of ω which measures ~ C . By [12, Section 3], there exists a properforcing P which adds a Σ-reflecting sequence, so in V P there is a club subset of ω which measures ~ C . (cid:3) We now sketch a proof that
MRP alone does not imply Strong Measuring. Inparticular, Measuring does not imply Strong Measuring. Start with a model of CH in which there exists a supercompact cardinal κ . Construct a forcing iteration P inthe standard way to obtain a model of MRP . To do this, fix a Laver function f : κ → V κ . Then define a countable support forcing iteration h P α , ˙ Q β : α ≤ κ, β < κ i as follows. Given P α , consider f ( α ). If f ( α ) happens to be a P α -name for someopen stationary set mapping, then let ˙ Q α be a P α -name for a proper forcing whichadds an f ( α )-reflecting sequence. Otherwise let ˙ Q α be a P α -name for Col ( ω , ω ).Now define P := P κ . Arguments similar to those in the standard construction of amodel of PFA can be used to show that P forces MRP . The forcing for adding a Σ-reflecting sequence for a given open stationary setmapping does not add reals ([12, Section 3]). In particular, it is vacuously ω ω -bounding. The property of being proper and ω ω -bounding is preserved undercountable support forcing iterations ([1, Theorem 3.5]), so P is also ω ω -bounding.In particular, V ∩ ω ω is an unbounded family in V P , and it has size ω since CH holds in V . It follows that the bounding number b is equal to ω . But by Corollary2.9, Measuring b is false. So P forces that Measuring ω is false. As c = ω in V P ,Strong Measuring fails in V P .We also note that Strong Measuring plus c = ω is consistent with the existenceof an ω -Suslin tree. Namely, both the forcing for adding a Σ-reflecting sequence fora given open stationary set mapping Σ, as well as any σ -centered forcing, preserveSuslin trees ([11]). And the property of being proper and preserving a given Suslintree is preserved under countable support forcings iterations ([10]). So starting witha model in which there exists an ω -Suslin tree S and a supercompact cardinal κ , wecan iterate forcing similar to the argument in the preceding paragraphs to producea model of MA ( σ -centered) plus MRP in which S is an ω -Suslin tree. By Corollary3.4, Strong Measuring holds in that model.4. Measuring Without the Axiom of Choice
Another natural way to strengthen
Measuring is to allow, in the sequence to bemeasured, not just closed sets, but also sets of higher Borel complexity. This lineof strengthenings of
Measuring was also considered in [2]. For completeness, we areincluding here the corresponding observations.The version of
Measuring where one considers sequences ~X = h X α : α ∈ ω ∩ Lim i , with each X α an open subset of α in the order topology, is of course equivalentto Measuring . A natural next step would therefore be to consider sequences in whicheach X α is a countable union of closed sets. This is obviously the same as allowingeach X α to be an arbitrary subset of α . Let us call the corresponding statement Measuring ∗ : Definition 4.1.
Measuring ∗ holds if and only if for every sequence ~X = h X α : α ∈ ω ∩ Lim i , if X α ⊆ α for each α , then there is some club D ⊆ ω such that forevery limit point δ ∈ D of D , D ∩ δ measures X δ . It is easy to see that
Measuring ∗ is false in ZFC . In fact, given a stationary andco-stationary S ⊆ ω , there is no club of ω measuring ~X = h S ∩ α : α ∈ ω ∩ Lim i .The reason is that if D is any club of ω , then both D ∩ S ∩ δ and ( D ∩ δ ) \ S arecofinal subsets of δ for each δ in the club of limit points in ω of both D ∩ S and D \ S .The status of Measuring ∗ is more interesting in the absence of the Axiom ofChoice. Let C ω = { X ⊆ ω : C ⊆ X for some club C of ω } . Observation 4.2. ( ZF + C ω is a normal filter on ω ) Suppose ~X = h X δ : δ ∈ ω ∩ Lim i is such that (1) X δ ⊆ δ for each δ . (2) For each club C ⊆ ω , (a) there is some δ ∈ C such that C ∩ X δ = ∅ , and (b) there is some δ ∈ C such that ( C ∩ δ ) \ X δ = ∅ .Then there is a stationary and co-stationary subset of ω definable from ~X . ARAMETRIZED MEASURING AND CLUB GUESSING 11
Proof.
We have two possible cases. The first case is that in which for all α < ω ,either • W α = { δ < ω : α / ∈ X δ } is in C ω , or • W α = { δ < ω : α ∈ X δ } is in C ω .For each α < ω , let W α be W ǫα for the unique ǫ ∈ { , } such that W ǫα ∈ C ω ,and let W ∗ = ∆ α<ω W α ∈ C ω . Then X δ = X δ ∩ δ for all δ < δ in W ∗ . Itthen follows, by (2), that S = S δ ∈ W ∗ X δ , which of course is definable from ~C , is astationary and co-stationary subset of ω . Indeed, suppose C ⊆ ω is a club, andlet us fix a club D ⊆ W ∗ . There is then some δ ∈ C ∩ D and some α ∈ C ∩ D ∩ X δ .But then α ∈ S since δ ∈ W ∗ and α ∈ W ∗ ∩ X δ . There is also some δ ∈ C ∩ D and some α ∈ C ∩ D such that α / ∈ X δ , which implies that α / ∈ S by a symmetricalargument, using the fact that X δ = X δ ∩ δ for all δ < δ in W ∗ .The second possible case is that there is some α < ω with the property thatboth W α and W α are stationary subsets of ω . But now we can let S be W α , where α is first such that W α is stationary and co-stationary. (cid:3) It is worth comparing the above observation with Solovay’s classic result thatan ω –sequence of pairwise disjoint stationary subsets of ω is definable from anygiven ladder system on ω (working in the same theory). Corollary 4.3. ( ZF + C ω is a normal filter on ω ) The following are equivalent. (1) C ω is an ultrafilter on ω ; (2) Measuring ∗ ; (3) For every sequence h X α : α ∈ ω ∩ Lim i , if X α ⊆ α for each α , then thereis a club C ⊆ ω such that either • C ∩ δ ⊆ X δ for every δ ∈ C , or • C ∩ X δ = ∅ for every δ ∈ C .Proof. (3) trivially implies (2), and by the observation (1) implies (3). Finally,to see that (2) implies (1), note that the argument right after the definition of Measuring ∗ uses only ZF together with the regularity of ω and the negation of(1). (cid:3) In particular, the strong form of
Measuring ∗ given by (3) in the above observationfollows from ZF together with the Axiom of Determinacy.We finish this digression into set theory without the Axiom of Choice by observ-ing that any attempt to parametrize Measuring ∗ , in the same vein as we did with Measuring , gives rise to principles vacuously equivalent to
Measuring ∗ itself, at leastwhen the parametrization is done with the alephs. Specifically, given an aleph κ , let us define Measuring ∗ κ as the statement thatfor every sequence hX α : α ∈ ω ∩ Lim i , if each X α is a set of cardinality atmost κ consisting of subsets of α , then there is a club D ⊆ ω such that for everylimit point δ ∈ D of D , D ∩ δ measures X for all X ∈ X δ . Then Measuring ∗ ω isclearly equivalent to Measuring ∗ under ZF together with the normality of C ω andthe Axiom of Choice for countable families of subsets of ω (which of course followsfrom the Axiom of Choice for countable families of sets of reals, and therefore alsofrom ZF + AD ). On the other hand, working in ZF + C ω is a normal filter on ω , we have that Measuring ∗ ω follows vacuously from Measuring ∗ simply because This was pointed out by Asaf Karagila. under
Measuring ∗ there is no sequence hX α : α ∈ ω ∩ Lim i as in the definition of Measuring ∗ ω and such that |X α | = ω for some α ; indeed, Measuring ∗ implies, overthis base theory, that C ω is an ultrafilter (Corollary 4.3), and if C ω is an ultrafilterthen there is no ω -sequence of distinct reals, whereas the existence of a family ofsize ω consisting of subsets of some fixed countable ordinal clearly implies thatthere is such a sequence.We conclude the article with two natural questions. Question 4.4. Is Measuring p false? Question 4.5.
Are Measuring and Strong Measuring equivalent statements assu-ming Martin’s Axiom?
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David Asper´o, School of Mathematics, University of East Anglia, Norwich NR4 7TJ,UK
E-mail address : [email protected] John Krueger, Department of Mathematics, University of North Texas, 1155 UnionCircle
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