433 revision1998-05-01 modified1998-05-27 LENGTH OF BOOLEAN ALGEBRAS AND ULTRAPRODUCTS

r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-27LENGTH OF BOOLEAN ALGEBRAS ANDULTRAPRODUCTSMENACHEM MAGIDOR AND SAHARON SHELAHAbstract.We prove the consistency with ZFC of “the length of an ultraproduct of Boolean algebras is smaller than the ultraproduct of the lengths”.Similarly for some other cardinal invariants of Boolean algebras.0.IntroductionOn the length of Boolean algebras (the cardinality of linearly ordered subsets)see Monk [M1],[M2](and Definition 1.1below).In Shelah [Sh 345,§1]it is said that Koppelberg and Shelah noted that by the L o´s theorem for an ultrafilter D on κand Boolean algebras B i (i <κ)we have(∗)| i<κLength(B i )/D|≤Length(i<κB i /D ),andµi <Length(B i )⇒| i<κµi /D|<Length( i<κB i /D ).D.Peterson noted that the indicated proof fails,but holds for regular ultra-filters (see [Pe97]).Now the intention in [Sh 345]was for Length +,i.e.(∗)+| i<κLength +(B i )/D|≤Length +(i<κB i /D ),where Length +(B )is the first cardinal not represented as the cardinality ofa linearly ordered subset of the Boolean Algebra (the only difference being the case the supremum is not attained).Here we prove that the statement (∗)may fail (see Theorem 1.3and Proposition 1.6).The situation is similar for many cardinal invariants.Of course,if (∗)fails then (using ultraproducts of (H (χ),B i ):i <κ or see e.g.Ros l anowski,Shelah [RoSh 534,§1])we have {i <κ:Length +(B )is a limit cardinal }∈D ,and i<κLength +(B i )/D is λ–like for some successor cardinal λ.Hence{i <κ:Length +(B i )is a regular cardinal }∈Dr e v i s i o n :1998-05-01 m o d i f i e d :1998-05-272MENACHEM MAGIDOR AND SAHARON SHELAHhence {i :Length +(B i )is an inaccessible cardinal }∈D ,so the example we produce is in some respect the only one possible.(Note that our convention is that “inaccessible”means regular limit (>ℵ0),not necessarily strong limit.)More results on cardinal invariants of ultraproducts of Boolean algebras can be found in [Sh 462],[Sh 479],[RoSh 534]and [Sh 620],[RoSh 651].This paper is continued for other cardinal invariants (in particular spread)in [ShSi 677].We thank Otmar Spinas and Todd Eisworth for corrections and com-ments.1.The main resultDefinition 1.1.(1)For a Boolean algebra B ,let its length,Length(B ),be sup {|X |:X ⊆B ,and X is linearly ordered (in B )}.(2)For a Boolean algebra B ,let its strict length,Length +(B ),besup {|X |+:X ⊆B,and X is linearly ordered (in B )}.Remark 1.2.(1)In Definition 1.1,Length +(B )is (equivalently)the firstλsuch that for every linearly ordered X ⊆B we have |X |<λ.(2)If Length +(B )is a limit cardinal then Length +(B )=Length(B );and if Length +(B )is a successor cardinal then Length +(B )=(Length(B ))+.Theorem 1.3.Suppose V satisfies GCH above µ(for simplicity),κis measurable,κ<µ,µis λ+-hypermeasurable (somewhat less will suffice),F is the function such that F (θ)=the first inaccessible >θ,and λ=F (µ)iswell defined,and χ<µ,χ>22κ.Then for some forcing notion P not collapsing cardinals,except those in the interval (µ+,λ)[so in V P we have µ++=λ=F V (µ)],and not adding subsets to χ,in V P ,we have:(α)in V P the cardinal µis a strong limit of cofinality κ,(β)for some strictly increasing continuous sequence µi :i <κ of(strong limit)singular cardinals >χwith limit µ,each λi =:F V (µi )is still inaccessible and for any normal ultrafilter D ∈V on κwe have:i<κλi /D has order type µ++=F V (µ).Definition 1.4.A forcing notion Q is directed µ-complete if:for a directed quasi-order I (so (∀s 0,s 1∈I )(∃t ∈I )(s 0≤I t &s 1≤t ))of cardinality <µ,and ¯p = p t :t ∈I such that p t ∈Q and s ≤I t ⇒p s ≤Q p t ,there is p ∈Q such that t ∈I ⇒p t ≤Q p .r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-27LENGTH OF BOOLEAN ALGEBRAS AND ULTRAPRODUCTS 3Proof Without loss of generality for every directed µ–complete forcing notion Q of cardinality at most λsatisfying the λ-c.c.,in V Q the cardinal µis still λ-hypermeasurable.[Why?If µsupercompact,use Laver [L],if µis just λ-hypermeasurable see more in Gitik Shelah [GiSh 344].]Let Q be the forcing notion adding λCohen subsets to µ,i.e.,{f :f a partial function from λto {0,1},|Dom(f )|<µ}.In V ,let R =Levy(µ+,<λ)={f :f a partial two place function such that [f (α,i )defined ⇒0<α<λ&i <µ+&f (α,i )<α]and |Dom(f )|<µ+}(so R collapses all cardinals in (µ+,λ)and no others,so in V R the ordinal λbecomes µ++).Clearly R is µ+–complete and hence adds no sequence of length ≤µof members of V .In V Q ,there is a sequence ¯D= D i :i <κ of normal ultrafilters on µas in [Mg4]and,¯g = g i,j :i <j ≤κ ,g i,j ∈µH (µ)witness this (that is D i ∈(V Q )κ/D j ,in fact D i is equal to g i,j /D j in the Mostowski collapse of(V Q )κ/D j ).Let ¯D ˜= D ˜i :i ≤κ ,¯g ˜= g ˜i,j :i <j ≤κ be Q –names of such sequences.Notethat a Q –name A ˜of a subset of µis an object of size ≤µ,i.e.,it consists of a µ–sequence ofµ–sequences of members of Q ,say p i,j :j <µ :i <µ ,and function f :µ×µ−→{0,1}such that each {p i,j :j <µ}is a maximal antichain of Q and p i,j Q “i ∈A ⇔f (i,j )=1”.So the set of members of Q and the set of Q –names of subsets of µare thesame in V and in V R .So in V R ×Q the sequence ¯Dstill gives a sequence of normal ultrafilters as required in [Mg4]as witnessed by ¯g = g i,j :i <j ≤κ .Also the Magidor forcing P (¯D ,¯g )(from there)for changing the cofinality ofµto κ(not collapsing cardinals not adding subsets to χ,the last is just by fixing the first element in the sequence)is the same in V Q and V R ×Q andhas the same set of names of subsets of µ.We now use the fact that P (¯D ,¯g )satisfies the µ+-c.c.(see [Mg4]).Let P =(Q ×R )∗P (¯D ˜,¯g ˜),so again everyQ ∗P (¯D ˜,¯g ˜)-name involves only µdecisions so also V Q ∗P (¯D ˜,¯g ˜),V (Q ×R )∗P (¯D ˜,¯g ˜)have the same subsets of µ.So our only problem is to check conclusion (β)of Theorem 1.3.Let D ∈V be any normal ultrafilter on κ(so this holds also in V R ,V R ×Q ,(V R ×Q )P (¯D ˜)).Claim 1.4.1.In V Q ∗P (¯D ˜,¯g ˜),the linear orderi<κF (µi )/D has true cofinalityF (µ)=λ.Proof of the Claim:Clearly Q “for i <κwe have µµ/D ˜i is wellordered”(as D ˜i is ℵ1–complete)andQ“for i <κwe have µµ/D ˜i has cardinality 2µand even i<µ2|i |/D ˜i hascardinality 2µ”r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-274MENACHEM MAGIDOR AND SAHARON SHELAH[Why?As µ=µ<µand D i is a uniform ultrafilter on µ.In details,let h :µ>2→µbe one-to-one,and for each η∈µ2define g η∈µµby g η(i )=h (η i ).Thenη=ν∈µ2⇒{i <µ:g η(i )=g ν(i )}is a bounded subset of µand hence its complement belongs to D i but |{g η(i ):η∈µ2}|=2|i |].Consequently,for some F ∗∈µµwe haveQ “ ζ<µF ∗(ζ)/D ˜i is isomorphic to λ”.If we look at the proof in [L](or [GiSh 344])which we use above,we seethat w.l.o.g.F ∗is the F above (and so does not depend on i ).So let f ˜i,αbe Q -names such thatQ “for i <κand α<λ,f ˜i,α∈ζ<µF (ζ)andf i,α/D ˜i is the α-th function inζ<µF (ζ)/D ˜i .”In V Q let D κ= i<κD i and let B i ∈D i \j<iD j be as in [Mg4](you can alsoproduce them straightforwardly),so B i :i <κ is a sequence of pairwise disjoint subsets of µ.Define f α∈µµfor α<λas follows:f α B i =f i,α B i and f α (µ\i<κB i )is constantly zero.So f α:α<µ is <D κ–increasing and cofinal inζ<µF (ζ).Let D ˜κ,B ˜i ,f ˜αbe Q –names forced to be as above.Then Q“for α<β<λthe setA ˜α,β={ζ<µ:f α(ζ)<f β(ζ)}belongs to D ˜κ”.Now in V Q ,one of the properties of Magidor forcing P (¯D ,¯g )is thatP (¯D ,¯g )“for every A ∈D κ=i<κD i for every i <κlarge enough we have µ˜i ∈A ”(where µ˜i :i <κ is the increasing continuous κ–sequence cofinal in µwhich P (¯D ,¯g )adds).Since for every p ∈P (¯D ,¯g ),for some q ≥p we have (recall from [Mg4]that F q (i )is the set which q “says”µ˜i belongs to (when q does not forces avalue to µ˜i ))[F q (i )⊆A α,βfor every i <κlarge enough],hence in V Q ∗P (¯D ˜,¯g ˜),for α<β<λthe set {i <κ:¬(f α(µi )<f β(µi ))}is bounded,i.e., f α(µi ):i <κ :α<λ is <J bd κ–increasing ini<κF (µi ).r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-27LENGTH OF BOOLEAN ALGEBRAS AND ULTRAPRODUCTS 5On the other hand,in V Q ,assume p P (¯D ,¯g )“f ˜∈i<κF (µ˜i )”.W.l.o.g.p forces that F ∗(µ˜i )<µ˜i +1,and so by [Mg4](possibly increasing p ),we have,for some function h ,Dom(h )=κ,h (i )∈[i ]<ℵ0,that above p ,we know:f ˜(i )depends on the value of µ˜j for j ∈{i }∪h (i ).So we can define a functionf ∗∈µµ:f ∗(ζ)=sup {γ:for some i <κ,possibly µ˜i is ζandγis a possible value (abovep )of f ˜(i )}.So f ∗(ζ)<F (ζ),hence (in V Q )for some α,f ∗<D κf αand consequentlyp “f ∗<J bd κf α”.So f α:α<λ is,in V Q ∗P (¯D ˜,¯g ˜),<J bd κ–increasing and cofinal in i<κF (µi )which is more than enough for 1.4.1.Note that 1.4.1holds in V (R ×Q )∗P (¯D ˜,¯g ˜)too (remember that any sequenceαi <F (µi ):i <κ in V (Q ×R )∗P (¯D ˜,¯g ˜)is bounded by a function f ∈ θ∈Reg ∩µF (θ) V (see [Mg4])and also is itself in V Q ∗P (¯D ˜,¯g ˜)).But why,if αi <F (µi )andαi :i <κ ∈V (R ×Q )∗P (¯D ˜,¯g ˜),do we have that i<καi /D has cardinality ≤µ+(this means <λ)?It suffices to prove this inequality in the universe V 1=V Q ∗P (¯D ˜,¯g ˜).Now V κ1/D is well founded,hence there is an isomorphism from V κ1/D onto a transitive class which we now call M and let j be the isomorphism(=the Mostowski collapse).As µin V Q ∗P (¯D ˜,¯g ˜)is strong limit >κ,clearly α<µ⇒j ( α:i <κ /D )<µ;and as D is normal,and µi :i <κ is increasing continuous,we have j ( µi :i <κ /D )=µ.As αi <F (µi )we have (by the L o´s theorem):M |=“j ( αi :i <κ /D )is an ordinal smaller thanthe first inaccessible >µ”.But the property “not weakly inaccessible”is preserved by extending the universe (from M to V 1).So we finish.r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-276MENACHEM MAGIDOR AND SAHARON SHELAHis an ultrafilter on κextending D then (in (V )P )we have:theultraproducti<κλi /D ∗is λ–like.Proposition 1.6.Suppose λi :i <κ is a sequence of (weakly)inaccessiblecardinals >κ,D an ultrafilter on κ,and the linear order i<κ(λi ,<)/D is λ–like,λregular.(1)There are Boolean algebras B i (for i <κ)such that:(a)Length(B i )=Length +(B i )=λi ,(b)Length +( i<κB i /D )=λ,(c)if λ=µ+then Length( i<κB i /D )=µ.(2)Assume λi is regular >κ,|B i |=λi ,B i =α<λB i,α,B i,αincreasingcontinuous in α(the B i ,B i,αare Boolean algebras of course)andwe have(∀α<λi )(|B i,α|<λi )and(∗)0if i <κ,δ<λi ,cf(δ)=κ+then:B i,δ<◦B i (<◦is complete subalgebra sign)i.e.(∗)1B i ,B i,δif x ∈B i \B i,δ,x =0then for some y ∗we have:(a)y ∗∈B i,δand y ∗=0(b)for any y ∈B i,αsuch that B i |=“y ≤y ∗&y =0”:B i |=“y ∩x =0&x −y =0”Then Length +(i<κB i /D )≤λ.Proof (1)Let for i <κand α<λi ,B i αbe the Boolean algebra generated freely by {x i ζ:ζ<α}except⊗x i,αζ≤x i,αξfor ζ<ξ<α.Let B i be the free product of {B i α:α<λi }so B i is freely generated by {x i,αζ:α<λi ,ζ<α}except for ⊗.Let B i,βbe the subalgebra of B i generated by {x i,αζ:α<β,ζ<α}.Now clause (a)holds immediately,and the inequality ≥in clause (b)holds by the L o´s theorem,and the other inequality follows by part (2)of the stly clause (c)follows.(2)W.l.o.g.the set of members of B i is λi ,and the set of elements of B i,αis an initial segment.Let S i ={δ:δ<λi ,the set of members of B i,δis δand cf(δ)=κ+}r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-27LENGTH OF BOOLEAN ALGEBRAS AND ULTRAPRODUCTS 7Let (B,<∗,S )=i<κ(B i ,<i ,S i )/D ,with <i the order on the ordinals <λi(≤is reserved for the order in the Boolean algebra).So (|B |,<∗)is λ–like (where |B |is the set of elements of B ).Let y i :i <λ be an <∗increasing sequence of members of B .Let S def={δ<λ:cf(δ)=κ+and {y i :i <δ}has in (|B |,<∗)a least upper bound which we call y δand it belongs to S }.Now clearly (read [Sh 420,§1]if you fail to see;or assume 2κ<µand [Sh 111,2.3p.269]):⊕S ⊆λis stationary.Note:⊕is enough,as if X ⊆B is linearly ordered by <B ,let y i ∈X for i <λbe pairwise distinct;as <∗is λ–like w.l.o.g. y i :i <λ is <∗–increasing,and let S be as above.For each δ∈S apply (∗)1B,B {y :y<∗y δ}from the assumption to y δ,y δ(holds by L o´s theorem)and get y ∗i.Then apply Fodor lemma,and get a stationary subset S 2of S and an element y ∗such that for every i ∈S 2we have y ∗i =y ∗.Now the set of y ifor i ∈S 2is independent (check or see [Sh:92,4.1]).r e v i s i o n :1998-05-01 m o d i f i e d :1998-05-278MENACHEM MAGIDOR AND SAHARON SHELAHthen c (B i )=λi =c +(B i ),c (B )=µ+,λ=c +(B ).(Same proof.)(5)We can also get the parallel to 1.6for the independence number.LetB i be the Boolean algebra generated by {x i,αζ:ζ<αand α<λi }freely except(⊗5)x i,αζ∩x i,βξ=0if α<β<κ,ζ<α,ξ<β.Let I i be the ideal of B i which {x i,αζ:ζ<α,α<λi }generates.Clearly it is a maximal ideal.Let B i,αbe the ideal of B i generated by {x i,βζ:ζ<αand β<α}.Again w.l.o.g.the universe of B i is λi and letC i ={δ<λi :for x ∈B i we have:x <δiffx ∈B i,δ∨−x ∈B i,α}.It is a club of λi .The B i,βare not Boolean subalgebras of B i ,just Boolean subrings ;now (∗)0in proposition 1.6is changed somewhat.We will haveP B i =I i and (B,<∗,P ∗)= i<κ(B i ,<∗i ,P B i)/D .We know:(α)P ∗is a maximal ideal of B (by L o´s Theorem)(β)if i <κ,δ<λi is a limit ordinal and δ∈C i then for anyx ∈P B i there are x 0<δ,x 0∈P B i and x 1∈P B i disjoint to all members of P B i which are <i δand x =x 0∪x 1.Similarly for B add if you like Q i =C i ⊆λi .References[GiSh 344]Moti Gitik and Saharon Shelah.On certain indestructibility of strong cardinalsand a question of Hajnal.Archive for Mathematical Logic ,28:35–42,1989.[L]Richard Laver.Making supercompact indestructible under κ-directed forcing.Israel J.of Math.,29:385–388,1978.[Mg4]Menachem Magidor.Changing cofinality of cardinals.Fund.Math.,XCIX :61–71,1978.[M1]J.Donald Monk.Cardinal Invariants of Boolean Algebras .Lectures in Math-ematics.ETH Zurich,Birkhauser Verlag,Basel Boston Berlin,1990.[M2]J.Donald Monk.Cardinal Invariants of Boolean Algebras ,volume 142ofProgress in Mathematics .Birkh¨a user Verlag,Basel–Boston–Berlin,1996.[Pe97]Douglas Peterson.Cardinal functions on ultraproducts of Boolean algebras.Journal of Symbolic Logic ,62:43–59,1997.[RoSh 534]Andrzej Ros l anowski and Saharon Shelah.Cardinal invariants of ultrapod-ucts of 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Mathematicae,153:199–275,1997.math.LO/9609216.[Sh620]Saharon Shelah.Special Subsets of cf(µ)µ,Boolean Algebras and Maharammeasure Algebras.Topology and its Applications,99:135–235,1999.8thPrague Topological Symposium on General Topology and its Relations toModern Analysis and Algebra,Part II(1996).math.LO/9804156.[ShSi677]Saharon Shelah and Otmar Spinas.On incomparability and related cardi-nal functions on ultraproducts of Boolean algebras.Mathematica Japonica,52:345–358,2000.math.LO/9903116.Institute of Mathematics,The Hebrew University,Jerusalem91904,IsraelE-mail address:menachem@math.huji.ac.ilInstitute of Mathematics,The Hebrew University,Jerusalem91904,Israel, and Rutgers University,Mathematics Department,New Brunswick,NJ08854,USAE-mail address:shelah@math.huji.ac.il72-5-8991:deifidom1-5-8991:noisiver。