aa r X i v : . [ m a t h . L O ] J un DIAMONDS
SAHARON SHELAH
Abstract. If λ = χ + = 2 χ > ℵ then diamond on λ holds. Moreover, if λ = χ + = 2 χ and S ⊆ { δ < λ : cf( δ ) = cf( χ ) } is stationary then ♦ S holds.Earlier this was known only under additional assumptions on χ and/or S . Date : Oct.7, 2009.Research supported by the United-States-Israel Binational Science Foundation (Grant No.2002323), Publication No. 922. The author thanks Alice Leonhardt for the beautiful typing. Introduction
We prove in this paper several results about diamonds. Let us recall the basicdefinitions and sketch the (pretty long) history of the related questions.The diamond principle was formulated by Jensen, who proved that it holds in L for every regular uncountable cardinal κ and stationary S ⊆ κ . This is a predictionprinciple, which asserts the following: Definition 1.1. ♦ S (the set version).Assume κ = cf( κ ) > ℵ and S ⊆ κ is stationary, ♦ S holds when there is asequence h A α : α ∈ S i such that A α ⊆ α for every α ∈ S and the set { α ∈ S : A ∩ α = A α } is a stationary subset of κ for every A ⊆ κ .The diamond sequence h A α : α ∈ S i guesses enough (i.e., stationarily many)initial segments of every A ⊆ κ . Several variants of this principle were formulated,for example: Definition 1.2. ♦ ∗ S .Assume κ = cf( κ ) > ℵ and S is a stationary subset of κ . Now ♦ ∗ S holds whenthere is a sequence hA α : α ∈ S i such that each A α is a subfamily of P ( α ) , |A α | ≤ | α | and for every A ⊆ κ there exists a club C ⊆ κ such that A ∩ α ∈ A α for every α ∈ C ∩ S .We know that ♦ ∗ S holds in L for every regular uncountable κ and stationary S ⊆ κ . Kunen proved that ♦ ∗ S ⇒ ♦ S . Moreover, if S ⊆ S are stationary subsetsof κ then ♦ ∗ S ⇒ ♦ ∗ S (hence ♦ S ). But the assumption V = L is heavy. Tryingto avoid it, we can walk in several directions. On weaker relatives see [Sh:829] andreferences there. We can also use other methods, aiming to prove the diamondwithout assuming V = L .There is another formulation of the diamond principle, phrased via functions(instead of sets). Since we use this version in our proof, we introduce the following: Definition 1.3. ♦ S (the functional version).Assume λ = cf( λ ) > ℵ , S ⊆ λ, S is stationary. ♦ S holds if there exists adiamond sequence h g δ : δ ∈ S i which means that g δ ∈ δ δ for every δ ∈ S , and forevery g ∈ λ λ the set { δ ∈ S : g ↾ δ = g δ } is a stationary subset of λ .By Gregory [Gre76] and Shelah [Sh:108] we know that assuming λ = χ + = 2 χ and κ = cf( κ ) = cf( χ ) , κ < λ , and GCH holds (or actually just χ κ = χ or ( ∀ α <χ )( | α | κ < χ ) ∧ cf( χ ) < κ ), then ♦ ∗ S λκ holds (recall that S λκ = { δ < λ : cf( δ ) = κ } ).We have got also results which show that the failures of the diamond above astrong limit cardinal are limited. For instance, if λ = χ + = 2 χ > µ and µ > ℵ is strong limit, then (by [Sh:460]) the set { κ < µ : ♦ ∗ S λκ fails } is bounded in µ (recall that κ is regular). Note that the result here does not completely subsumethe earlier results when λ = 2 χ = χ + as we get “diamond on every stationary set S ⊆ λ \ S λ cf( χ ) ”, but not ♦ ∗ S ; this is inherent as noted in 3.4. In [Sh:829], a similar,stronger result is proved for ♦ S λκ : for every λ = χ + = 2 χ > µ, µ strong limit forsome finite d ⊆ Reg ∩ µ , for every regular κ < µ not from d we have ♦ S λκ , and even ♦ S for “most” stationary S ⊆ S λκ . In fact, for the relevant good stationary sets S ⊆ S λκ we get ♦ ∗ S . Also weaker related results are proved there for other regular λ (mainly λ = cf(2 χ )). IAMONDS 3
The present work does not resolve:
Problem 1.4.
Assume χ is singular, λ = χ + = 2 χ , do we have ♦ S λ cf( χ ) ? (You mayeven assume G.C.H.).However, the full analog result for 1.4 consistently fails, see [Sh:186] or [Sh:667];that is: if G.C.H., χ > cf( χ ) = κ then we can force a non-reflecting stationary S ⊆ S χ + κ such that the diamond on S fails and cardinalities and cofinalities arepreserved; also G.C.H. continue to hold. But if χ is strong limit, λ = χ + = 2 χ , stillwe know something on guessing equalities for every stationary S ⊆ S λκ ; see [Sh:667].Note that this S (by [Sh:186], [Sh:667]) in some circumstances has to be “small”( ∗ ) if ( χ is singular, 2 χ = χ + = λ, κ = cf( χ ) and) we have the square (cid:3) χ (i.e.there exists a sequence h C δ : δ < λ, δ is a limit ordinal i , so that C δ is closedand unbounded in δ , cf( δ ) < χ ⇒ | C δ | < χ and if γ is a limit point of C δ then C γ = C δ ∩ γ ) then ♦ S λκ holds, moreover, if S ⊆ S λκ reflects in astationary set of δ < λ then ♦ S holds, see [Sh:186, § ♦ S for every stationary S ⊆ S ∗ ” forsuitable S ∗ ⊆ λ . Now usually this was deduced from the stronger statement ♦ ∗ S .However, the results on ♦ ∗ S cannot be improved, see 3.4.Also if χ is regular we cannot improve the result to ♦ S λχ , see [Sh:186] or [Sh:587],even assuming G.C.H. Furthermore the question on ♦ ℵ when 2 ℵ = ℵ = 2 ℵ wasraised. Concerning this we show in 3.2 that ♦ S ℵ ℵ may fail (this works in othercases, too). Question 1.5.
Can we deduce any ZFC result on λ strongly inaccessible?By Dˇzamonja-Shelah [DjSh:545] we know that failure of SNR helps (SNR standsfor strong non-reflection), a parallel here is 2.3(2).For λ = λ <λ = 2 µ weakly inaccessible we know less, still see [Sh:775], [Sh:898]getting a weaker relative of diamond, see Definition 3.5(2). Again failure of SNRhelps.On consistency results on SNR see Cummings-Dˇzamonja-Shelah [CDSh:571],Dˇzamonja-Shelah [DjSh:691].We thank the audience in the Jerusalem Logic Seminar for their comments in thelecture on this work in Fall 2006. We thank the referee for many helpful remarks.In particular, as he urged for details, the proof of 2.5(1),(2) just said “like 2.3”,instead we make the old proof of 2.3(1) prove 2.5(1),(2) by minor changes and makeexplicit 2.5(3) which earlier was proved by “repeat of the second half of the proofof 2.3(1)”. We thank Shimoni Garti for his help in the proofreading. Notation .
1) If κ = cf( κ ) < λ = cf( λ ) then we let S λκ := { δ < λ : cf( δ ) = κ } .2) D λ is the club filter on λ for λ a regular uncountable cardinal. SAHARON SHELAH Diamond on successor cardinals
Recall (needed only for part (2) of the Theorem 2.3):
Definition 2.1.
1) We say λ on S has κ -SNR or SNR( λ, S, κ ) or λ has strong non-reflection for S in κ when S ⊆ S λκ := { δ < λ : cf( δ ) = κ } so λ = cf( λ ) > κ = cf( κ )and there are h : λ → κ and a club E of λ such that for every δ ∈ S ∩ E for some club C of δ , the function h ↾ C is one-to-one and even increasing; (note that without lossof generality α ∈ nacc( E ) ⇒ α successor and without loss of generality E = λ , so µ ∈ Reg ∩ λ \ κ + ⇒ SNR( µ, S ∩ µ, κ )). If S = S λκ we may omit it. Remark . Note that by Fodor’s lemma if cf( δ ) = κ > ℵ and h is a functionfrom some set ⊇ δ and the range of h is ⊆ κ then the following conditions areequivalent:( a ) h is one-to-one on some club of δ ( b ) h is increasing on some club of δ ( c ) Rang( h ↾ S ) is unbounded in κ for every stationary subset S of δ .Our main theorem is: Claim 2.3.
Assume λ = 2 χ = χ + .1) If S ⊆ λ is stationary and δ ∈ S ⇒ cf( δ ) = cf( χ ) then ♦ S holds.2) If ℵ < κ = cf( χ ) < χ and ♦ S fails where S = S λκ (or just S ⊆ S λκ is a stationarysubset of λ ) then we have SNR( λ, κ ) or just λ has strong non-reflection for S ⊆ S λκ in κ . Definition 2.4.
1) For a filter D on a set I let Dom( D ) := I and S is called D -positive when S ⊆ I ∧ ( I \ S ) / ∈ D and D + = { S ⊆ Dom( D ) : S is D -positive } and we let D + A = { B ⊆ I : B ∪ ( I \ A ) ∈ D } (so if D = D λ , the club filter on theregular uncountable λ then D + is the family of stationary subsets of X ).2) For D a filter on a regular uncountable cardinal λ which extends the club filter,let ♦ D means: there is ¯ f = h f α : α ∈ S i which is a diamond sequence for D (or a D -diamond sequence) which means that S ∈ D + and for every g ∈ λ λ theset { α < λ : g ↾ α = f α } belongs to D + ; so ¯ f is also a diamond sequence for thefilter D + S , (clearly ♦ S means ♦ D λ + S for S a stationary subset of the regularuncountable λ ).A somewhat more general version of the theorem is Claim 2.5.
1) Assume λ = χ + = 2 χ and D is a λ -complete filter on λ whichextends the club filter. If S ∈ D + and δ ∈ S ⇒ cf( δ ) = cf( χ ) then we have ♦ D + S .2) We have ♦ D when : ( a ) λ = λ <λ ( b ) ¯ f = h f α : α < λ i lists ∪{ α λ : α < λ } ( c ) S ∈ D + ( d ) ¯ u = h u α : α ∈ S i and u α ⊆ α for every α ∈ S ( e ) χ = sup {| u α | + : α < λ } < λ ( f ) D is a χ + -complete filter on λ extending the club filter ( g ) ( ∀ g ∈ λ λ )( ∃ D + δ ∈ S )[ δ = sup { α ∈ u δ : g ↾ α ∈ { f β : β ∈ u δ }} ] . IAMONDS 5
3) Assume λ = χ + = 2 χ and ℵ < κ = cf( χ ) < χ, S ⊆ S λκ is stationary, and D is a λ -complete filter extending the club filter on λ to which S belongs. If ♦ D fails then SNR( λ, S, κ ) .Proof. Proof of 2.3 Part (1) follows from 2.5(1) for D the filter D λ + S . Part (2)follows from 2.5(3) for D the filter D λ + S . (cid:3) Proof.
Proof of 2.5Proof of part (1)Clearly we can assume ⊛ χ > ℵ as for χ = ℵ the statement is empty.Let ⊛ h f α : α < λ i list the set { f : f is a function from β to λ for some β < λ } .For each α < λ clearly | α | ≤ χ so let ⊛ ,α h u α,ε : ε < χ i be ⊆ -increasing continuous with union α such that ε < χ ⇒| u α,ε | ≤ ℵ + | ε | < χ .For g ∈ λ λ let h g ∈ λ λ be defined by ⊛ ,g h g ( α ) = Min { β < λ : g ↾ α = f β } .Let cd, h cd ε : ε < χ i be such that ⊛ ( a ) cd is a one-to-one function from χ λ onto λ such thatcd(¯ α ) ≥ sup { α ε : ε < χ } (when ¯ α = h α ε : ε < χ i )( b ) for ε < χ , cd ε is a function from λ to λ such that¯ α = h α ε : ε < χ i ∈ χ λ ⇒ cd ε (cd(¯ α )) = α ε (they exist as λ = λ χ , in the present case this holds as 2 χ = χ + = λ ).Now we let (for β < λ, ε < χ ): ⊛ f β,ε be the function from Dom( f β ) into λ defined by f β,ε ( α ) = cd ε ( f β ( α ))so Dom( f β,ε ) = Dom( f β ).Without loss of generality ⊛ α ∈ S ⇒ α is a limit ordinal.For g ∈ λ λ and ε < χ we let S εg = n δ ∈ S : δ = sup { α ∈ u δ,ε : for some β ∈ u δ,ε we have g ↾ α = f β,ε } o . Next we shall show ⊛ for some ε ( ∗ ) < χ for every g ∈ λ λ the set S ε ( ∗ ) g is a D -positive subset of λ . SAHARON SHELAH
Proof of ⊛ Assume this fails, so for every ε < χ there is g ε ∈ λ λ such that S εg ε is not D -positive and let E ε be a member of D disjoint to S εg ε . Define g ∈ λ λ by g ( α ) := cd( h g ε ( α ) : ε < χ i ) and let h g ∈ λ λ be as in ⊛ ,g , i.e. h g ( α ) = Min { β : g ↾ α = f β } .Let E ∗ = { δ < λ : δ is a limit ordinal such that α < δ ⇒ h g ( α ) < δ } , clearly itis a club of λ hence it belongs to D and so E = ∩{ E ε : ε < χ } ∩ E ∗ belongs to D as D is λ -complete and χ + 1 < λ .As S is a D -positive subset of λ there is δ ∗ ∈ E ∩ S . For each α < δ ∗ as δ ∗ ∈ E ⊆ E ∗ clearly h g ( α ) < δ ∗ and α as well as h g ( α ) belong to ∪{ u δ ∗ ,ε : ε <χ } = δ ∗ , but h u δ ∗ ,ε : ε < χ i is ⊆ -increasing hence ε δ ∗ ,α = min { ε : α ∈ u δ ∗ ,ε and h g ( α ) ∈ u δ ∗ ,ε } is not just well defined but also ε ∈ [ ε δ ∗ ,α , χ ) ⇒ { α, h g ( α ) } ⊆ u δ ∗ ,ε .As cf( δ ∗ ) = cf( χ ), by an assumption on S , it follows that for some ε ( ∗ ) < χ theset B := { α < δ ∗ : ε δ ∗ ,α < ε ( ∗ ) } is unbounded below δ ∗ .So( a ) α ∈ B ⇒ { α, h g ( α ) } ⊆ u δ ∗ ,ε ( ∗ ) and( b ) α ∈ B ⇒ g ↾ α = f h g ( α ) ⇒ V ε<χ [ g ε ↾ α = f h g ( α ) ,ε ] ⇒ g ε ( ∗ ) ↾ α = f h g ( α ) ,ε ( ∗ ) .But δ ∗ ∈ E ⊆ E ε ( ∗ ) hence δ ∗ / ∈ S ε ( ∗ ) g ε ( ∗ ) by the choice of E ε ( ∗ ) , but by (a) + (b) andthe definition of S ε ( ∗ ) g ε ( ∗ ) recalling δ ∗ ∈ S we have sup( B ) = δ ∗ ⇒ δ ∗ ∈ S ε ( ∗ ) g ε ( ∗ ) , (where h g ( α ) plays the role of β in the definition of S εg above), contradiction. So the proofof ⊛ is finished.Let χ ∗ = ( | ε ( ∗ ) | + ℵ ) hence δ ∈ S ⇒ | u δ,ε ( ∗ ) | ≤ χ ∗ and χ + ∗ < λ as χ ∗ < χ <λ because ℵ , ε ( ∗ ) < χ < λ . Now we apply 2.5(2) which is proved below with λ, S, D, χ + ∗ , h f β,ε ( ∗ ) : β < λ i , h u δ,ε ( ∗ ) : δ ∈ S i here standing for λ, S, D, χ, ¯ f , ¯ u there.The conditions there are satisfied hence also the conclusion which says that ♦ D holds. (cid:3) (1) Proof.
Proof of 2.5(2)Let ⊠ h cd ε : ε < χ i and cd be as in ⊛ in the proof of part (1), possible as we areassuming χ < λ = λ <λ ⊠ for β < λ and ζ < χ let f β,ζ be the function with domain Dom( f β ) suchthat f β,ζ ( α ) = cd ζ ( f β ( α )) ⊠ for g ∈ λ λ define h g ∈ λ λ as in ⊛ in the proof of part (1), i.e. h g ( α ) =Min { β : g ↾ α = f β } .If 2 <χ < λ our life is easier but we do not assume this. For δ ∈ S let ξ ∗ δ be a cardinal,and let h ( α δ,ξ , α δ,ξ ) : ξ < ξ ∗ δ i list the set { ( α , α ) ∈ u δ × u δ : Dom( f α ) = α } ,note that ξ ∗ δ < χ , recalling | u δ | < χ by clause (e) of the assumption. We now try tochoose (¯ v ε , g ε , E ε ) by induction on ε < χ , (note that ¯ v ε is defined from h g ζ : ζ < ε i (see clause (e) of ⊠ below) so we choose just ( g ε , E ε )), such that: ⊠ ( a ) E ε is a member of D and h E ζ : ζ ≤ ε i is ⊆ -decreasing with ζ ( b ) ¯ v ε = h v εδ : δ ∈ S ∩ E ′ ε i when E ′ ε = ∩{ E ζ : ζ < ε } ∩ λ so is λ if ε = 0( c ) h v ζδ : ζ ≤ ε i is ⊆ -decreasing with ζ for each δ ∈ S ∩ E ′ ε ( d ) g ε ∈ λ λ IAMONDS 7 ( e ) v εδ = { ξ < ξ ∗ δ : if ζ < ε then g ζ ↾ α δ,ξ = f α δ,ξ ,ζ } (so if ε is a limit ordinal then v εδ = T ζ<ε v ζδ and ε = 0 ⇒ v εδ = ξ ∗ δ )( f ) if δ ∈ E ′ ε ∩ S then v ε +1 δ $ v εδ or δ > sup { α δ,ξ : ξ ∈ v ε +1 δ } .Next ⊕ we cannot carry the induction, that is for all ε < χ .Why? Assume toward contradiction that h (¯ v ε , g ε , E ε ) : ε < χ i is well defined. Let E := ∩{ E ε : ε < χ } , it is a member of D as D is χ + -complete. Define g ∈ λ λ by g ( α ) := cd( h g ε ( α ) : ε < χ i ). Let E ∗ = { δ < λ : δ a limit ordinal such that h g ( α ) < δ and δ > sup(Dom( f α ) ∪ Rang( f α )) for every α < δ } , so E ∗ is a club of λ hence it belongs to D . By assumption (g) of the claim the set S g := { δ ∈ S : δ = sup { α ∈ u δ : ( ∃ β ∈ u δ )( f β = g ↾ α ) }} is D -positive, so we can choose δ ∈ E ∩ E ∗ ∩ S g . Hence B := { α ∈ u δ : ( ∃ β ∈ u δ )( f β = g ↾ α ) } is an unbounded subset of u δ and let h : B → u δ be h ( α ) =min { β ∈ u δ : f β = g ↾ α } , clearly h is a function from B into u δ . Now α ∈ B ∧ ζ <χ ⇒ f h ( α ) = g ↾ α ∧ ζ < χ ⇒ f h ( α ) ,ζ = g ζ ↾ α , so for α ∈ B the pair ( α, h ( α ))belongs to { ( α δ,ξ , α δ,ξ ) : ξ ∈ v εδ } for every ε < χ . Hence for any ε < χ we have B ⊆ { α δ,ξ : ξ ∈ v εδ } so δ = sup { α δ,ξ : ξ ∈ v εδ } .So for the present δ , in clause (f) of ⊠ the second possibility never occurs.So clearly h v εδ ∗ : ε < χ i is strictly ⊆ -decreasing, i.e. is ⊂ -decreasing which isimpossible as | v δ ∗ | = ξ ∗ δ ∗ < χ . So we have proved ⊕ hence we can assume ⊕ there is ε < χ such that we have defined our triple for every ζ < ε but wecannot define for ε . So we have h (¯ v ζ , g ζ , E ζ ) : ζ < ε i .As in ⊞ ( e ), let ⊙ E ′ ε be λ if ε = 0 and ∩{ E ζ : ζ < ε } if ε > S ∗ := S ∩ E ′ ε .Clearly ¯ v ε is well defined, see clauses (b),(e) of ⊠ , and for δ ∈ S ∗ let F δ = { f α δ,ξ ,ε : ξ ∈ v εδ } , so each member is a function from some α ∈ u δ ⊆ δ into some ordinal < δ .Let ⊙ S ∗ := { δ ∈ S ∗ : there are f ′ , f ′′ ∈ F δ which are incompatible as functions }⊙ S ∗ := { δ ∈ S ∗ : δ / ∈ S ∗ but the function ∪ { f : f ∈ F δ } has domain = δ }⊙ S ∗ = S ∗ \ ( S ∗ ∪ S ∗ ).For δ ∈ S ∗ let g ∗ δ = ∪{ f : f ∈ F δ } , so by the definition of h S ∗ ℓ : ℓ = 1 , , i clearly g ∗ δ ∈ δ δ . Now if h g ∗ δ : δ ∈ S ∗ i is a diamond sequence for D then we are done.So assume that this fails, so for some g ∈ λ λ and member E of D we have δ ∈ S ∗ ∩ E ⇒ g ∗ δ = g ↾ δ . Without loss of generality E is included in E ′ ε . But thenwe could have chosen ( g, E ) as ( g ε , E ε ), recalling ¯ v ε was already chosen. Easily thetriple ( g ε , E ε , ¯ v ε ) is as required in ⊕ , contradicting the choice of ε in ⊕ so we aredone proving part (2) of Theorem 2.3 hence also part (1). (cid:3) (2) SAHARON SHELAH Proof.
Proof of part (3)We use cd,cd ε (for ε < χ ), hh u α,ε : ε < χ i : α < λ i , h f α : α < λ i , h f α,ε : α <λ, ε < χ i and S εg for ε < κ as in the proof of part (1).Recall κ , a regular uncountable cardinal, is the cofinality of the singular cardinal χ and let h χ γ : γ < κ i be increasing with limit χ . For every γ < κ we ask:The γ -Question: Do we have: for every g ∈ λ λ , the following is a D -positive subsetof λ : { δ ∈ S : S γ [ g ] ∩ δ is a stationary subset of δ } where S γ [ g ] := { ζ < λ : cf( ζ ) ∈ [ ℵ , κ ),sup( u ζ,χ γ ) = ζ and for arbitrarily large α ∈ u ζ,χ γ for some β ∈ u ζ,χ γ and ε < χ γ ,we have Dom( f β ) = α and g ↾ α = f β,ε } .Case 1: For some γ < κ , the answer is yes.Choose h C δ : δ ∈ S i such that C δ is a club of δ of order type cf( δ ) = κ .For δ ∈ S ⊆ S λκ let u δ := ∪{ u α,χ γ : α ∈ C δ } .Clearly ⊞ | u δ | ≤ κ + χ γ < χ ⊞ for every g ∈ λ λ for D -positively many δ ∈ S , we have δ = sup { α ∈ u δ : g ↾ α ∈ { f β,ε : ε < χ γ and β ∈ u δ }} .Why ⊞ holds? Given g ∈ λ λ , let h g ∈ λ λ be defined by h g ( α ) = min { β < λ : g ↾ α = f β } , so h g ( α ) ≥ α (but is less than λ ). Let E g = { δ < λ : δ is a limitordinal such that ( ∀ α < δ ) h g ( α ) < δ } , so E g is a club of λ and let E ′ g be the setof accumulation points of E g , so E ′ g , too, is a club of λ . By the assumption of thiscase, the set S ′ := { δ ∈ S : δ ∩ S γ [ g ] is a stationary subset of λ } is D -positive,hence S ′′ := S ′ ∩ E ′ g is a D -positive subset of λ . Let δ ∈ S ′′ , by E ′ g ’s definition, wecan find B δ ⊆ E g ∩ δ unbounded in δ , so without loss of generality B δ is closed.But S γ [ g ] ∩ δ is a stationary subset of δ , recalling δ ∈ S ′′ , so B δ = B δ ∩ S γ [ g ] ∩ C δ is a stationary subset of δ as B δ , C δ are closed unbounded subsets of δ .Clearly ζ ∈ B δ ⇒ ζ ∈ C δ ⇒ u ζ,χ γ ⊆ u δ by the definitions of B δ and u δ . Also ζ ∈ B δ ⇒ ζ ∈ S γ [ g ] ⇒ ( ζ is a limit ordinal) ∧ ζ = sup( u ζ,χ γ ) = sup { α ∈ u ζ,χ γ :( ∃ β ∈ u ζ,χ γ )( ∃ ε < χ γ )( g ↾ α = f β,ε ) } ⇒ (( ζ is a limit ordinal) ∧ ζ = sup { α ∈ u δ ∩ ζ :( ∃ β ∈ u δ \ α )( ∃ ε < χ γ )( g ↾ α = f β,ε ) } ).As B δ is unbounded in δ being stationary we are done proving ⊞ .Now without loss of generality every δ ∈ S is divisible by χ hence δ = χ γ δ andlet u ′ δ = u δ ∪ { χ γ α + ε : α ∈ u δ , ε < χ γ } , so u δ is an unbounded subset of δ , andlet f ′ β = f α,ε when β = χ γ α + ε, ε < χ γ . So translating what we have is: ⊞ ( a ) h f ′ α : α < λ i is a sequence of members of ∪{ β λ : β < λ } ( b ) for δ ∈ S, u ′ δ is an unbounded subset of δ of cardinality ≤ χ γ × χ γ = χ γ ( < χ )( c ) for every g ∈ λ λ for D -positively many δ ∈ S we have δ = sup { α ∈ u ′ δ : ( ∃ β ∈ u ′ δ )( g ↾ α = f ′ β ) } .Now we can apply part (2) with h f ′ α : α < λ i , h u ′ δ : δ ∈ S i replacing ¯ f , h u δ : δ ∈ S i .So as there we can prove ♦ S , hence we are done.Case 2: For every γ < κ the answer is no. IAMONDS 9
Let ( g γ , E γ ) exemplify that the answer for γ is no; so g γ ∈ λ λ and E γ ∈ D . Let E = T γ<κ E γ , so E is a member of D . Let g ∈ λ λ be defined by g ( α ) = cd( h g γ ( α ) : γ < κ i ˆ(0) χ ), i.e. cd ε ( g ( α )) is g γ ( α ) if γ < κ and is 0 if ε ∈ [ κ, χ ).Let E g := { δ < λ : δ a limit ordinal such that if α < λ then h g ( α ) < δ and δ > sup(Dom( f α ) ∪ Rang( f α )) } . We now define h : λ → κ as follows ⊞ for β < λ ( a ) if cf( β ) / ∈ [ ℵ , κ ) or β / ∈ E g then h ( β ) = 0( b ) otherwise h ( β ) = min { γ < κ : β = sup { α ∈ u β,χ γ : for some α ∈ u β,χ γ and ε < χ γ we have g ↾ α = f α ,ε }} . Now ⊞ h : λ → κ is well defined.Why ⊞ holds? Let β < λ . If cf( β ) / ∈ [ ℵ , κ ) or β / ∈ E g then h ( α ) = 0 < κ byclause (a) of ⊞ . So assume cf( β ) ∈ [ ℵ , κ ) and β ∈ E g . Let h γ β,ε : ε < cf( β ) i be increasing with limit β and let γ β,ε = min { γ : g ↾ γ β,ε = f γ } , so ε < cf( β ) ⇒ γ β,ε < β as β ∈ E g . But h u β,χ ζ : ζ < cf( χ ) i is ⊆ -increasing with union β sofor each ε < cf( β ) there is ζ = ζ β,ε < cf( χ ) such that { γ β,ε , γ β,ε } ⊆ u β,χ ζ . Ascf( β ) < κ = cf( χ ) for some ζ < κ the set { ε < cf( β ) : ζ β,ε < ζ } is unbounded incf( β ). So ζ can serve as γ in clause (b) of ⊞ so h ( β ) is well defined, in particularis less than κ so we have proved ⊞ . ⊞ if δ ∈ S ∩ E γ then for some club C of δ the function h ↾ C is increasing.Why ⊞ holds? If not, then by Fodor’s lemma for some γ < κ the set { δ ′ ∈ δ ∩ S : h ( δ ′ ) ≤ γ } is a stationary subset of δ , and we get contradiction to the choice of E γ so ⊞ holds indeed.So h is as promised in the claim. (cid:3) Note
Observation 2.6. If κ ∗ < λ are regular, S λκ ∗ strongly does not reflect in λ forevery κ ∈ Reg ∩ κ ∗ and Π(Reg ∩ κ ∗ ) < λ , then :( a ) S λ<κ ∗ can be divided to ≤ Π(Reg ∩ κ ∗ ) sets, each not reflecting in any δ ∈ S λ<κ ∗ in particular( b ) S λ ℵ can be divided to ≤ Π(Reg ∩ κ ∗ ) sets each not reflecting in any δ ∈ S λ<κ ∗ . Remark .
1) Of course if λ has κ -SNR then this holds for every regular λ ′ ∈ ( κ, λ ).2) We may state the results, using λ ∗ κ (see below). Definition 2.8.
For each regular κ let λ ∗ κ = Min { λ : λ regular fails to have κ − SNR } , and let λ ∗ κ be ∞ (or not defined) if there is no such λ . IAMONDS 11 Consistent failure on S A known question was:
Question 3.1.
For θ ∈ {ℵ , ℵ } do we have (2 ℵ = 2 ℵ = ℵ ⇒ ♦ S ℵ θ )?So for θ = ℵ the answer is yes (by 2.3(1)), but what about θ = ℵ ? We notedsome years ago that easily: Claim 3.2.
Assume V | = GCH or even just ℵ ℓ = ℵ ℓ +1 for ℓ = 0 , , . Then someforcing notion P satisfies ( a ) P is of cardinality ℵ ( b ) forcing with P preserves cardinals and cofinalities ( c ) in V P , ℵ = 2 ℵ = ℵ , ℵ = ℵ ( d ) in V P , ♦ S fails where S = { δ < ℵ : cf( δ ) = ℵ } , moreover ( ∗ ) there is a sequence ¯ A = h A δ : δ ∈ S i where A δ an unbounded subset of δ of order type ω satisfying ( ∗∗ ) if ¯ f = h f δ : δ ∈ S i , f δ ∈ ( A δ ) ( ω ) , then there is f ∈ ( ω ) ( ω ) such that δ ∈ S ⇒ δ > sup( { α ∈ A δ : f ( α ) ≤ f δ ( α ) } ) .Remark . Similarly for other cardinals.
Proof.
There is an ℵ -complete ℵ -c.c. forcing notion P not collapsing cardinals,not changing cofinalities, preserving 2 ℵ ℓ = ℵ ℓ for ℓ = 0 , , | P | = ℵ such thatin V P , we have ( ∗ ), in fact more than ( ∗ ) holds - see [Sh:587]. Let Q be the forcingof adding ℵ Cohen or just any c.c.c. forcing notion of cardinality ℵ adding ℵ reals (can be Q ˜ , a P -name). Now we shall show that P ∗ Q , equivalently P × Q isas required:Clause (a): | P ∗ Q | = ℵ ; trivial.Clause (b):Preserving cardinals and cofinalities; obvious as both P and Q do this.Clause (c): Easy.Clause (d): In V P we have ( ∗ ) as exemplified by say ¯ A = h A δ : δ ∈ S i . We shallshow that V P ∗ Q | = “ ¯ A satisfies ( ∗∗ )”. Otherwise in V P ∗ Q we have ¯ f = h f δ : δ ∈ S i say in V [ G P , G Q ] a counterexample then in V [ G P ] for some q ∈ Q and ¯ f ˜ we have V [ G P ] | = ( q (cid:13) Q “ ¯ f ˜ = h f ˜ δ : δ ∈ S i where f ˜ δ : A δ → ω for each δ ∈ S form a counterexample to ( ∗ )”) . Now in V [ G P ] we can define ¯ g = h g δ : δ ∈ S i ∈ V [ G P ] where g δ a function withdomain A δ , by g δ ( α ) = { i : q f ˜ δ ( α ) = i } . I.e. there is ¯ A = h A δ : δ ∈ S i where A δ is an unbounded subset of δ of order type ω satisfying: ⊕ if ¯ f = h f δ : δ ∈ S i , f δ ∈ ( A δ ) ω then there is f ∈ ( ω ) ω such that for every δ ∈ S ℵ ℵ forevery α ∈ A δ large enough we have f ( α ) = f δ ( α ). So in V [ G P ] we have q (cid:13) Q “ V δ ∈ S ( ∀ α ∈ A δ ) f ˜ δ ( α ) ∈ g δ ( α ) } ”. Also g δ ( α ) is a countablesubset of ω as Q satisfies the c.c.c.For δ ∈ S we define a function g δ : A δ → ω by letting g δ ( α ) = (sup( g δ ( α )) + 1hence g δ ( α ) < ω so h g δ : δ ∈ S i is as required on ¯ f in ( ∗∗ ) in V [ G P ], of course.Apply clause ( ∗∗ ) in V [ G P ] to h g δ : δ ∈ S i so we can find g : ω → ω such that V δ ∈ S δ > sup { α ∈ A δ , g δ ( α ) > g ( α ) } . Now g is as required also in V [ G P ][ G Q ]. (cid:3) ♦ ∗ S (of course thedemand in clause (e) and (f) in claim 3.4 below are necessary, i.e. otherwise ♦ ∗ S holds). The answer is not as: (the restriction in (e) and in (f) are best possible). Observation 3.4.
Assume λ = λ <λ , S ⊆ S λκ .Then for some P ( a ) P is a forcing notion( b ) P is of cardinality λ + satisfying the λ + -c.c.( c ) forcing with P does not collapse cardinals and does not change cofinality( d ) forcing with P adds no new η ∈ λ> Ord( e ) ♦ ∗ S fails for every stationary subset S of λ such that( α ) S ⊆ S λκ when ( ∃ µ < λ )[ µ <κ> tr = λ ]or just( β ) α ∈ S ⇒ | α | < cf( α ) > tr > | α | ( f ) ( Dℓ ) S , see below, fails for every S ⊆ S λκ when α ∈ S ⇒ | α | < cf( α ) > tr = λ .Recalling Definition 3.5.
1) For µ ≥ κ = cf( κ ) let µ <κ> tr = {|T | : T ⊆ κ ≥ µ is closed underinitial segments (i.e. a subtree) such that |T ∩ κ> µ | ≤ µ } .2) For λ regular uncountable and stationary S ⊆ λ let ( Dℓ ) S mean that there is asequence ¯ P = hP δ : δ ∈ S i witnessing it which means:( ∗ ) ¯ P ( a ) P δ ⊆ δ δ has cardinality < λ ( b ) for every f ∈ λ λ the set { δ ∈ S : f ↾ δ ∈ P δ } is stationary(for λ successor it is equivalent to ♦ S ; for λ strong inaccessible it is trivial). Proof.
Proof of 3.4 Use P = adding λ + , λ -Cohen subsets.The proof is straight. (cid:3) Remark . The consistency results in 3.4 are best possible, see [Sh:829].
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