simplied(;1)-morass.Inthepresentpaper,wewillgeneralizetheapproachtot - PDF document

simplied(;1)-morass.Inthepresentpaper,wewillgeneralizetheapproachtothree-dimensionalsystems,so-calledFSsystemsalongsimplied(;2)-morasses.WewillalsoobservethatunderaveryweakadditionalassumptiontheforcingobtainedfromaFSsystemalongasimpliedgap-1orgap-2morassisforcingequivalenttoasmallsubforcing.Animmediateconsequenceofthisand[11]is:Ifthereisasimplied(!1;1)-morass,thenthereexistsacccforcingofsize!1thataddsan!2-Suslintree.Thisimprovestheorem7.5.1.inTodorcevic'sbook[28]:Thereexistsconsistentlyacccforcingwhichaddsan!2-Suslintree.Themainresultis:Ifthereisasimplied(!1;2)-morass,thenthereexistsacccforcingofsize!1thataddsa0-dimensionalHausdortopologyon!3whichhasspreads()=!1.ThisforcingisobtainedbyaFSsystemalongasimplied(!1;2)-morass.Itsconditionsarenitefunctionsp:xp!2withxp!3!2.ByatheoremofHajnalandJuhasz[7],card(X)22s(X)=exp(exp(s(X))holdsforallHausdorspacesX.In[13],Juhaszexplicitlyraisesthequestionifthesecondexpisreallynecessary.BytheusualargumentusedforCohenforc-ing,acccforcingofsize!1preservesGCH.Henceourresultshowsthatitisconsistentthatthereexistsa0-dimensionalHausdorspaceXwiths(X)=!1suchthatcard(X)=22s(X).Sofar,theconsistencyofcard(X)=22s(X)hasonlybeenknownforthecases(X)=!.Theexampleisthe0-dimensional,hereditarilyseparable,hereditarilynormalspaceconstructedfrom}byFedor-cuk[5].TheauthorwouldliketothankProfessorJuhaszforpointingthisouttohim.WhilethegeneralmethodofFSsystemscanbegeneralizedstraightforwardlytohigherdimensions,wecannotexpectthattheconsistencystatementscannaivelybeextendedbyraisingthecardinalparameters.Inparticular,wecan-notexpecttobeaabletoconstructfroma(!1;3)-morassacccforcingofsize!1whichaddsaT2spaceofsize!4andspread!1.Ifthiswaspossible,wecouldndsuchaforcinginL.However,bytheusualargumentusedforCohenforcingitpreservesGCHwhichcontradictsthetheoremofHajnalandJuhasz.Thereasonwhythisgeneralizationdoesnotworkisthatthegap-3caseyieldsafour-dimensionalconstruction.Therefore,theniteconditionsofourforcinghavetottogetherappropriatelyinfourdirectionsinsteadofthreeandthatisimpossible.Soifandhowastatementgeneralizestohigher-gapsdependsheavilyontheconcreteconditions.Theauthorstartedtodevelopthemethodofforcingalongmorasses,becausehewasinterestedinsolvingconsistencyquestionslikethefollowingforhighercardinals:CanthereexistasuperatomicBooleanalgebrawithwidth!andheight!2(BaumgartnerandShelah[2],Martinez[16])?Isitpossiblethatthereisafunctionf:!2!2!!suchthatfisnotconstantonanyrectanglewithinnitesides(Todorcevic[26,28])?However,theexistenceofsuchaBooleanalgebraaswellastheexistenceofsuchafunctioncontradictsGCH.SotogettheconsistencieswehavetodestroyGCH.HenceasimpleapplicationofFSsystemswillnotworkbecauseofthepropertieswedescribedabove.Therefore,wewillintroduceso-calledlocalFSsystemsalongsimpliedmorasses.2 Finally,weshouldalsomentionthatbesideshistoricizedforcingand-functionsthereisanother,quitedierentmethodtoproveconsistenciesintwo-cardinalcombinatorics.Thisisthemethodofforcingwithmodelsassideconditionsorwithsideconditionsinmorasses.ModelsassideconditionswereintroducedbyTodorcevic[25,27],whichwasfurtherdevelopedbyKoszmider[15]tosideconditionsinmorasses.Unliketheothermethods,itproducesproperforcingswhichareusuallynotccc.Thisissometimesnecessary.Forexample,KoszmiderprovedthatifCHholds,thenthereisnocccforcingthataddsasequenceof!2manyfunctionsf:!1!!1whichisorderedbystrictdominationmodnite.However,heisabletoproduceaproperforcingwhichaddssuchasequence[15].Moreonthemethod,includingadiscussionofitsrelationshipwiththatofusing-functions,canbefoundinMorgan'spaper[19].Inthecontextofourapproach,thisraisesthequestionifitispossibletodenesomethinglikeacountablesupportiterationalongamorass.2Simpliedgap-2morassesInthissection,wewillrecallthedenitionofsimpliedgap-2morassesandsummarizetheirpropertiestotheextentnecessaryforourapplications.Exceptfortheorem2.3(a)andlemma2.6(7),allresultsinthissectionareduetoVelleman[29,31].Nevertheless,wewillusuallyquotetheauthor'spaper[11]onFSsystemsalonggap-1morassesinsteadof[29],becausewehopethatinthiswaytheconnectiontoFSsystemsbecomesclearer.Asimplied(;1)-morassisastructureM=hhji;hFjiisatisfyingthefollowingconditions:(P0)(a)0=1,=+,80.(b)Fisasetoforder-preservingfunctionsf:!.(P1)jFjforall.(P2)If ,thenF =ffgjf2F ;g2Fg.(P3)If,thenF;+1=fid;hgwherehissuchthath=idandh()forsome.(P4)Ifisalimitordinal,1;2andf12F1,f22F2,thentherearea1;2 ,g2F andj12F1 ,j22F2 suchthatf1=gj1andf2=gj2.(P5)Forall&#x-349;0,=Sff[]j;f2Fg.Oursimplied(;1)-morassesarewhatarecalledneatsimplied(;1)-morassesin[29].Vellemanshowstherethatifthereisoneofhissimplied(;1)-morassesthereisaneatone.Note,moreover,thatitisequivalenttoreplace\h()forsome"in(P3)with\h(+)=+forsomeandallsuchthat+".Thisiseasilyseenusing(P5)and(P2).4 (6)Ifandb2G,thenfb=f(b)f.Assumeinthefollowingthat0,'0='forandG0=Gfor.Andletforthemomentbeingf=id,f=idforallandf=idforall.Letf2G00.Thenwecandeneanembeddingasfollows:Ifandb2G,thenf(b)=fb.Wecallsuchanembeddingfaleft-branchingembedding.Therearemanyleft-branchingembeddings,oneforeverychoiceoff.Anembeddingfisright-branchingifforsome,(1)f=id(2)f(+)=+if+(3)f=idfor(4)f=idfor(5)f2G(6)f[G]=G0f()f()if.Anamalgamationisafamilyofembeddingsthatcontainsallpossibleleft-branchingembeddings,exactlyoneright-branchingembeddingandnothingelse.Theright-branchingembeddingcorrespondstothemapshfrom(P3)inthegap-1case.Therefore,wewillusuallydenoteitbyh.Let!beregularandhh'j+i;hGj+iiasimplied(+;1)-morasssuchthat'forall.Lethjibeasequencesuchthat0and=+.LethFjibesuchthatFisafamilyofembeddingsfromhh'ji;hGjiitohh'ji;hGjii.Thisisasimplied(;2)-morassifithasthefollowingproperties:(1)jFjforall.(2)If ,thenF =ffgjf2F ;g2Fg.Herefgisthecompositionoftheembeddingsfandg,whicharedenedintheobviousway:(fg)=fg()gforand(fg)=fg()g()gfor.(3)If,thenF;+1isanamalgamation.(4)Ifisalimitordinal,1;2andf12F1,f22F2,thentherearea1;2 ,g2F andj12F1 ,j22F2 suchthatf1=gj1andf2=gj2.(5)Forall,2Lim:(a)=Sff[]j;f2Fg.(b)Forall,'=Sff[']j9(f2Fandf()=)g.(c)Forall,G=Sff[G]j9(f2F;f()=andf()=)g.Theorem2.3(a)IfV=L,thenthereisasimplied(;2)-morassforallregular&#x-408;!.6 (2)88b2Gf(b)=f#()f(b).Furthermore,thesefunctionshavethefollowingproperties:(3)Iff()andb2Gf(),then99c2G9d2Gf()b=f(c)d:(4)88b2Gfb=f(b)f.(5)If,b2Gandc2G,thenf(bc)=f(b)f(c).(6)If ,f2F ,g2Fand,then( fg)=fg()g(fg)#()=fg()g()(g#())f#(g())and( fg)=fg()g()gforall.Proof:SeeVelleman[31],lemma2.1.2Fromthepreviouslemma,wegetofcoursealsomaps( 0st)fors0tand(t).3FSsystemsalongmorassesInthissection,werecallthedenitionofFSsystemsalonggap-1morassesgivenin[11]andgeneralizeittothegap-2case,whichisstraightforward.LetPandQbepartialorders.Amap:P!Qiscalledacompleteembeddingif(1)8p;p02P(p0p!(p0)(p))(2)8p;p02P(pandp0areincompatible$(p)and(p0)areincompatible)(3)8q2Q9p2P8p02P(p0p!((p0)andqarecompatibleinQ)).In(3),wecallpareductionofqtoPwithrespectto.Ifonly(1)and(2)hold,wesaythatisanembedding.IfPQsuchthattheidentityisanembedding,thenwewriteP?Q.WesaythatPQiscompletelycontainedinQifidP:P!Qisacompleteembedding.Lethh'j+i;hGj+iibeasimplied(+;1)-morass.Wewantto"iterate"alongit.Thisleadstothefollowingdenition.WesaythathhPj++i;hstjsti;hej+iiisaFSsystemalonghh'j+i;hGj+iiifthefollowingconditionshold:(FS1)hPj++iisasequenceofpartialorderssuchthatP?PifandP=SfPjgfor2Lim.(FS2)hstjstiisacommutativesystemofinjectiveembeddingsst:P(s)+1!P(t)+1suchthatiftisalimitpointin,thenP(t)+1=[fst[P(s)+1]jstg:8 equalityholdsby(FS5).Itfollowsfromthepreviousobservationthath n(p)jn2!iisdecreasing.Sotherecursivedenitionabovebreaksdownatsomepoint,i.e. n(p)=0forsomen2!.Hencesupp(p)=f n(p)jn2!gisnite.Lemma3.1Ifp()andq()arecompatiblefor=max(supp(p)\supp(q)),thenpandqarecompatible.Proof:See[11],lemma4.1.2Theorem3.2Let;�!becardinals,regular.LethhPj+i;hstjsti;hejiibeaFSsystemalonga(;1)-morassM.AssumethatallPwithsatisfythe-cc.ThenP+alsodoes.Proof:See[11],lemma4.2.2Now,letMbeasimplied(;2)-morass.WesaythathhPj++i;hstjsti;h0stjs0ti;hej+i;he0jiiisaFSsystemalongMifthefollowingconditionshold:(FS21)hhPj++i;hstjsti;hej+iiisaFSsystemalonghh'j+i;hGj+ii.LetQ=fpsupp(p)jp2P++g.DeneapartialorderonQbysettingpqidom(q)dom(p)andp()q()forall2dom(q).SetQ :=fp2Qjdom(p) g.(FS22)h0stjs0tiisacommutativesystemofinjectiveembeddings0st:Q(s)+1!Q(t)+1suchthatiftisalimitpointin0,thenQ(t)+1=Sf0st[Q(s)+1]js0tg.(FS23)e0:Q+1!Q.(FS24)Lets0tand=0st.If(0)=0,s0=h(s);0iandt0=h(t);0i,then0st:Q(s)+1!Q(t)+1extends0s0t0:Q0+1!Q0+1.Henceforf2F,wemaydenef=Sfstjs=h;i;t=h;f()ig.(FS25)If0st(s)+1=id(s)+1,then0st=idQ(s)+1.(FS26)(a)If,thenQiscompletelycontainedinQ+1insuchawaythate0(p)isareductionofp2Q+1.10 withspread!1.Theconstructionhastwoimportantprecursors.Thoseare,rstly,theconstructionofacccforcingthataddsan!2-Suslintreein[11]and,secondly,Velleman'sproof[31]thatthemodeltheoreticgap-3theoremholdsinL.Inthefollowing,wewillreferto[11]and[31]fromtimetotimetopointoutsimilaritiesbetweentheconstructions.Wehopethatthismakesthewholeproofmorecomprehensible.Let:!beanorder-preservingmap.Then:!inducesmaps:!2!!2and:(!2)2!(!2)2intheobviousway::!2!!2;h ;i7!h( );i:(!2)2!(!2)2;hx;i7!h(x);i:Basically,wewilldenethemapsoftheFSsystembysetting(p)=[p].Now,westartourconstructionofP.Intherststep,wedenepartialordersP()for!3andQ()for!2.Inthesecondstep,wethinoutP()andQ()tothePandQwhichformtheFSsystemalongthegap-2morass.Assumethatasimplied(!1;2)-morassasintheprevioussectionisgiven.WedeneP()byinductiononthelevelsofhh'j!2i;hGj!2iiwhichweenumerateby!2.BaseCase:=0ThenweonlyneedtodeneP(1).LetP(1):=fp2Pjxp1!g.SuccessorCase:=+1WerstdeneP(').Letitbethesetofallp2Psuchthat(1)xp'!(2)p('!),h�1[p('!)]2P(')(3)p('!)andh�1[p('!)]arecompatibleinPwherehisasin(P3)inthedenitionofasimpliedgap-1morass.Forall'P()isalreadydened.For''setP()=fp2P(')jxp!g:Setst:P((s)+1)!P((t)+1);p7!st[p]:Itremainstodenee.Ifp2rng(),thensete(p)=�1(p).Ifp2P('),thensete(p)=p.Andifp=2rng()[P('),thensete(p)=p('!)[h�1[p('!)]:LimitCase:2Lim12 Thereasonwhyweusefinsteadoff2Fisthatfdoesnotmapthesupportofaconditioncorrectly.Foranexample,considerthecase=+1andletf2Fberight-branching.Letbethesplittingpointoff,i.e.f()=.Assumethatp2Q(),2dom(p)anddom(p())'!.Letf[p]bedenedbydom(f[p])=f[dom(p)]andf[p](f()):=f f[p()]forall2dom(p).Wewillshowthatf[p]=2Q().Todoso,noticerstthatf=f#()fbylemma2.6(1).However,f=id',becausefisright-branchingwithsplittingpoint.Sof=f#().Hencef[p]()=f f[p()]=f#()[p()]becausedom(p())'!andf=id.However,thiscontradictsthefactthatallq2Q()areoftheformq=rsupp(r)forsomer2P(!3)becauseinthiscaseq()6=g[q]forallg2G ,q2P(' )and bythedenitionofthesupportofacondition.Thisproblemdoesobviouslynotoccur,ifweconsiderf[p].Lemma4.3(a)Iff2Fandp2Q(),thenf[p]2Q().(b)Ifs0tandp2Q((s)+1),then 0st[p]2Q((t)+1).Proof:Setq:=f[p].Letdom(p)=f1:::nganddom(q)=f1:::ng:=ff(1):::f(n)g.Bythedenitionofthesupportofacondition,alliaresuccessorordinals.Andf(i�1)=f(i)�1bythedenitionoff.Setq(i�1)=ei�1(q(i)).Thenitsucestoprovethattherearefunctionsgi2Gi;i+1�1suchthat(1)q(i+1�1)=gi[q(i)](2)q(i)=2rng(i�1),q(i)=2P('i�1):Sincepisacondition,therearefunctionsji2Gi;i+1�1suchthatp(i+1�1)=ji[p(i)]:Sowecansetgi=fi;i+1�1(ji)f#(i):Weneedtocheck(1).Werstprovethatfi+1�1 f[ei+1�1(p(i+1))]=ei+1�1(q(i+1)):Toseethis,weuselemma2.6(4)whichsays88b2Gfb=f(b)f:Applyingitfor=i+1�1,=i+1andb=id'i+1�1,wegetq(i+1)('i+1�1!(i+1�1))=fi+1 f[p(i+1)]('i+1�1!(i+1�1))==fi+1�1 f[p(i+1)('i+1�1!(i+1�1))]wheretherstequalityholdsbythedenitionofq=f[p].14 InthefollowingwethinoutQ( )toQ toobtainaFSsystemalongourgap-2morass.WedeneQ byinductiononthelevelsofhhj!1i;hF0j!1ii.BaseCase:=0ThenweonlyneedtodeneQ1.LetQ1=Q(1).SuccessorCase:=+1WerstdeneQ.Todoso,letP'bethesetofallp2P(')suchthat(1)(h h)�1[p]2P'(2)g�1[p(')]and(h h)�1[p]arecompatibleforallg2Gwherehistheuniqueright-branchingembeddingofF.SetQ=fp(supp(p)\)jp2P'g:Fort2T0setQ(t)+1=fp2Qjdom(p)(t)+1gandQ=SfQjgfor2Lim.Set0st:Q(s)+1!Q(t)+1;p7! 0st[p]:Itremainstodenee0.Ifp2rng(0),thensete0(p)=0�1(p).Ifp2Q,thensete0(p)=p.Andifp=2rng(0)[Q,thenchoosear2P'withp=rsupp(r)andsetq:=[fg�1[r(')]jg2Gg[(h h)�1[r]=r()[(h h)�1[r]:Sete0(p)=q(supp(q)\).LimitCase:2LimFort2T0setQ(t)+1=Sf0st[Q(s)+1]js0tgandQ=SfQjgfor2Limwhere0st:Q(s)+1!Q(t)+1;p7! 0st[p].Finally,setP=fp2P()jpsupp(p)2Q!2gandP:=P!3.Wethinkthatsomeexplanationsareappropriate.LetusrstcompareourdenitiontoVelleman'sconstructionin[31].Hisproofofthegap-3theoremistheorem5.3of[31].HehastoconstructastructureA.Assumethathis+=!1.ThenheconstructsAbyconstructingforevery!1astructureAandtakingadirectlimit.However,thesystemofelementaryembeddingsheusestotake16 ThenitwillturnoutthatFisatleastnotyetdeterminedon!3f!+njn2!�D;2!2g:Thiswillbeusedinlemma4.6,whichisthecrucialstepforprovingthatPaddsaHausdorspace.Remark3:Assumethat=+1andthathistheright-branchingembeddingofF.Letp1;p22P'becompatibleandg2G.Thenalsog[p1]andh h[p2]arecompatible,i.e.g[p1]andh h[p2]agreeonthecommonpartoftheirdomains.Toprovethis,leth ;i2dom(g[p1])\dom(h h[p2])g(h 1;1i)=h ;ih h(h 2;2i)=h ;i:Sincehisright-branching,h=h.Letbethecriticalpointoff.Then!andtherefore=1=2.By(6)inthedenitionofright-branching,thereexistsab2Gsuchthatf(b)=g.Hence,by(6)inthedenitionofembedding,hb=gh:Sothereexistsh ;i2'!suchthathb(h ;i)=gh(h ;i)=h ;ih(h ;i)=h 1;ib(h ;i)=h 2;i:By(5)inthedenitionofright-branchingembedding,h2G.Hencep1( 1;)=p1()( ;).Moreover,p2( 2;)=p2()( ;)becauseb2G.However,p1andp2arecompatible.Therefore,alsop1()andp2()arecom-patible.Sop1()( ;)=p2()( ;).Thisinturnimpliesp1( 1;)=p2( 2;).Henceg[p1]( ;)=h h[p2]( ;).That'swhatwewantedtoshow.Thesameargumentshowsforallp2P'andallg2Gthatg[p]2P',h h[p]2P'andg[p][(h h)[p]2P'.Forarbitrary!1andf2Fdenef f:'!!'!;h ;!+ni7!hf( );!f()+niforalln2!andf f:('!)2!('!)2;hx;i7!hf f(x);i:If=+1,thenFisanamalgamationby(3)inthedenitionofasimpliedgap-2morass.Hencef2Fiseitherleft-branchingorright-branching.Letp2P'andassumethatfisright-branching.Thenf f[p]=f f[p]18 andforP()weknow(FS1)already.By(),onehastoprovefor(FS2),(FS3)and(FS6)thatcertainconditionsareelementsofP.Inthecaseof(FS2),forexample,onehastoshowthatst(p)2P(t)+1forallp2P(s)+1.Inallthreecasesthat'snotdicult.2Thenexttwolemmascorrespondtolemma5.2andlemma5.3of[11].Lemma4.6willensurethatthegenerictopologicalspaceisHausdor.Lemma4.7willguaranteethatthespacehasspread!1.Lemma4.6Letp2Pand 6=2!3.ThenthereisqpinPand2!2suchthatq( ;)6=q(;).Proof:Weprovebyinductionoverthelevelsofthegap-2morass,whichweenumerateby!1,thefollowingClaim:Letp2P'and 6=2'.ThenthereisqpinP'and2!suchthatq( ;)6=q(;).BaseCase:=0Trivial.SuccessorCase:=+1Lethbetheright-branchingembeddingofF.Weconsiderfourcases.Case1: ;2rng(h)Letp2P'begiven,h( )= andh()=.Setp=(h h)�1[p][p().Bytheinductionhypothesis,thereexistsaq2P'anda=!+n2!(n2!)suchthatqpandq( ;)6=q(;).Setq=p[(h h)[q]and=!h()+n.Thenq2P'byremark3,qpandq( ;)=q( ;)6=q(;)=q(;).Case2: ;=2rng(h)Weconsidertwosubcases.Assumerstthat=2Lim.Thenchoosesome2[!(�1);![suchthat=2f2j91h1;2i2dom(p)g.Setq=p[fhh ;i;0i;hh;i;1ig:Bythechoiceof,q2P(').Accordingtothecasewhichwearein,q()=p()and(h h)�1[q]=(h h)�1[p].Henceqand(h h)�1[q]arecompatiblebecauseqand(h h)�1[q]arecompatible.Soq2P'anditisobviouslyaswanted.Now,supposethat2Lim.Assumew.l.o.g.that .Sett=h;i.20 q(;).Setq:=f f[q].Thenqisaswanted.2Lemma4.7Lethpiji2!2ibeasequenceofconditionspi2Psuchthatpi6=pjifi6=j.Lethiji2!2ibeasequenceofordinalsi2!3suchthati2dom(xpi)foralli2!2.Thenthereexisti6=jandp2Psuchthatppi;pj,hi;i;hj;i2xpandp(i;)=p(j;)forall2rng(xpj).Proof:Byrstextendingtheconditions,wemayassumethatxpi=dom(xpi)rng(xpi)foralli2!2.Hencehj;i2xpwillholdforall2rng(xpj)automatically.Moreover,wecanassumebythe
-systemlemmathatallxpiareisomorphicrelativetotheorderoftheordinals,thatpi=pjforalli;j2!2,that(i)=jif:dom(xpi)=dom(xpj),thatfrng(xpi)ji2!2gformsa
-systemwithroot
,andthat
=id
if:rng(xpi)=rng(xpj).Toprovethelemma,weconsidertwocases.Case1:rng(xpi)=
foralli2!2Thenweset=max(
).Sincethereare!2-manypiwhileP'+1hasonly!1-manyelements,thereexistpiandpjwithi6=jsuchthatpi(+1)=pj(+1).Hencebytheusualargumentspiandpjarecompatible.Setp=pi[pj.Thenpisaswanted,becausepi=pjand(i)=jif:dom(xpi)=dom(xpj).Case2:rng(xpi)6=
foralli2!2Thenfmin(rng(xpi)�
)ji2!2gisunboundedin!2.Foreveryi2!2choosei!1,fi2Fi!1,i2'iandpi2P'isuchthatpi=(fi)i fi[pi]andi=(fi)i(i):Sincethereare!2-manyiandpibutonly!1-manypossibleiandpi,wecanassumethati=j,i=jandpi=pjforalli;j2!2.Setp=pi,=iand=i.Let2!3besuchthatpi2Pforalli2!2.Lett=h!2;i.Letstsuchthatpi2rng(st)for!1-manyi2!2.Lets2T.Pickpisuchthatmin(rng(xpi)�
)&#x-278;!.Leti=min(rng(xpi)�
).Thenbythechoiceoffi,i2rng(fi).Letutbesuchthatu2Ti.Letfi(i)=i.Sincethereare!1-manyj2!2suchthatpj2rng(st),therearealso!1-manyj2!2suchthatpj2rng(ut).Ontheotherhand,rng((fi)i)iscountable.Sowecanpickaj2!2suchthat=2rng((fi)i),ut()=jandpj2rng(ut).Inthefollowingwewillshowthatthereexistsppi;pjsuchthathj;i2xpandp(i;)=p(j;)forall2rng(xpi).For!1,letfi=gijiwheregi2Fandji2F!1.Letgi(i)=iand beminimalsuchthat2rng((g i) i).For !1,let(gi)i()=,pi=(ji) ji[p],gi[
]=
andi=(ji)().Weprovebyinductionover !1thefollowingClaim1:Ifhi;ih;0i,thenthereexistsppisuchthath0;i2xp22 Hencebytheinductionhypothesis,thereexistsppisuchthath0;i2xpandp(i;)=p(0;)forall2rng(xpi)�
.Setp=h h[p]:Thenpisaswanted.Subcase2:0=2rng(h)andiExactlylikethebasecaseoftheinduction.Subcase3:0=2rng(h)andi.Thiscaseisacombinationofthebasecaseoftheinductionandofcase1.Lethi;ih;00ih;0i.Let2Gsuchthat(00)=0.Thenbytheinductionhypothesis,thereexistsppisuchthath0;i2xpandp(i;)=p(00;)forall2rng(xpi)�
.Setp=[p][(h h)[p][fhh0;i;pi(i;)ij2rng(xpi)�g:Byremark3,p2P.Weclaimthatpisaswanted.For2rng(xpi)�,p(i;)=p(0;)holdsbydenition.For2rng(xpi)\=rng(xpi)\,wehavep(0;)=p(00;)=p(i;)=pi(i;):Thisnishestheproofofthesuccessorstep.Limitcase:2LimBylemma4.4andby(4)and(5)inthedenitionofasimpliedgap-2morass,wecanpickaandaf2Fsuchthat02rng(f)andf f[pi]=pi.Letf(0)=0.Thenby(6)inthedenitionofembedding,hi;ih;0i.Hencewecanpickbytheinductionhypothesisappisuchthath0;i2xpandp(i;)=p(0;)forall2rng(xpi)�
.Setp=f f[p]:Thenpisobviouslyaswanted.Thisnishestheproofofclaim1.Finally,wecanprovebyinductionover!124 Lemma4.8(a)i:P!3!Q!2;p7!psupp(p)isadenseembedding.(b)Thereisaccc-forcingPofsize!1suchthatQ!2embeddsdenselyintoP.Proof:(a)Wehavei[P!3]=Q!2.Soitisclear,thati[P!3]isdenseinQ!2.Itremainstocheck(1)and(2)ofthedenitionofembedding.Itfollowsfromlemma4.2,that(1)holds.For(2)assumerstthatp;p02P!3arecompatible.Sothereisrp;p0inP!3.Hencei(r)i(p);i(p0)bylemma4.2.Soi(p);i(p0)2Q!2arecompatible.Converselyassumethati(p);i(p0)2Q!2arecompatible.Thenp;p02P!3arecompatiblebylemma3.1.(b)Note,thathhQj!3i;h0stjs0ti;he0j!1iiisanFSsystemalonghhj!1i;hF0j!1ii.HencewecandenePfromQ!2likewedenedQ!2fromP!3.ThatQ!2embeddsdenselyintoPisprovedlikebefore.2Beforeweprovethemaintheorem,letusrecallthedenitionofthespreadofatopologicalspace.Let(X;)beatopologicalspacewithtopology.AsubsetDXiscalleddiscreteifforeveryx2DthereexistsanU2suchthatU\D=fxg.Thespreads(X)ofXisdenedass(X)=!
supfcard(D)jDisadiscretesubsetofXg.Theorem4.9Ifthereisan(!1;2)-morass,thenthereisaccc-forcingPofsize!1thataddsa0-dimensionalT2topologyon!3whichhasspread!1.Proof:Bylemma4.8,P!3embeddsdenselyintoP.HenceP!3andPyieldthesamegenericextensions.SoitsucestoprovethatP:=P!3addsa0-dimensionalT2topologyon!3whichhasspread!1.Bylemma4.5,Pisccc.Therefore,itpreservescardinals.LetGbeP-generic.WesetF=Sfpjp2Gg.ThenF:!3!2!2byasimpledensityargument.Letbethetopologyon!3generatedbythesetsAi:=f2!3jF(;)=ig.ThusabaseforisformedbythesetsB":=TfA"()j2dom(")gwhere":dom(")!2isniteanddom(")!2.Henceis0-dimensional.Weclaimthatisaswanted.WerstshowthatitisT2.Wehavetoprovethatfor 6=thereissome2!2suchthatF( ;)6=F(;).ThisisclearbythegenericityofGandlemma4.6.Itremainstoprovethathasspread!1.Assumenot.Let_X,_hand_Bbenamesandp2Paconditionsuchthatp (_X!3,_h:!2!_Xisbijective,_B:!2!V,8i2!2_B(i)isabasicopenset,8i6=j2!2_h(i)2_B(i)^_h(j)=2_B(i)).Foreveryi2!2letpipandi,"ibesuchthatpi _h(i)=i^_B(i)=B"i.Bythepreviouslemma,therearei6=jandr2Psuchthatrpi;pj,hi;i;hj;i2xrandr(i;)=r(j;)forall2rng(xpj).Hencer _h(j)=j2_B(i)whichcontradictsthedenitionofp.226 sequencesp2P+suchthatP=fpjp2P+gandP+1=P_Q(where_QisaP-namesuchthatP (_Qisaforcing)).Thenp:+!V2P+iP p()2_Qforall2+andsupp(p):=f2+jP6 p()=1_Qgisnite.Fornite
+andp2P+denep
2P+bysettingp
()=p()if2
p
()=_1Qif=2
where_1QisaP-namesuchthatP _1Q=1_Q.ForAP+andnite
+deneA
=fp
jp2Ag:If!1isregularandP
satisesthe-ccforallnite
+,thenP+alsosatisesthe-cc,asfollowsbythestandard
-systemargument.Theideaisnowtoensurethe-ccofeveryP
byconstructingitbyaFSsystemalongamorass.Thismotivatesthefollowingdenition:WesaythataFSiterationhPj+ilikeaboveisalocalFSsystemalonga(simplied)(;1)-morassMiforeverynite
+thereisaFSsystemhhQ
j+i;h
stjsti;he
jiialongMsuchthatP
?Q
+.Sofar,allthisisofcourseonlytheory.AsasimpleexampleletmeconsidertheforcingtoaddachainhXj!2isuchthatX!1,X�XisniteandX�Xhassize!1forall!2.ThenaturalforcingtodothiswouldbeP:=fp:apbp!2japbp!2!1nitegwherewesetpqiqpand8122aq82bp�bqp(1;)p(2;):Obviously,wewillsetX=f2!1jp(;)=1forsomep2GgforaP-genericG.ItiseasilyseenthathPj+iwithP=fp2PjapgcanbewrittenasFSiterationsuchthatP
=fp2Pjap
g.Ontheotherhand,itisnotsimplyaproduct.Unfortunately,italsodoesnotsatisfyccc.Toseethis,considerforevery!1thefunctionp:f0;1gfg!2wherep(0;)=1andp(1;)=0.ThenA=fpj2!1gisanantichainofsize!1.Therefore,weneedtothinouttheforcinginanappropriateway.Todothis,lethhj!1i;hFj!1iibeasimplied(!1;1)-morass.WewilldeneasystemhhPj!2i;hstjstiiwhichsatisesproperties(FS1)-(FS5)inthedenitionofFSsystemalongagap-1morass.Let:!beaorder-preservingmap.Then:!inducesmaps:!1!!1and:(!1)2!(!1)2intheobviousway::!1!!1;h ;i7!h( );i28 Claim:p2P ip2P,ap ,bp andforall andallf2F+1; f�1[p](+1fg)ismonotone:Basecase: =0Thenthereisnothingtoprove.Successorcase: =+1Assumerstthatp2P .Then,by(2)inthesuccessorstepofthedenitionofP!2,f�1[p];(id)�1[p]2P.Nowassumef2F+1; and.Thenf=ff0orf=f0forsomef02F+1;by(P2)and(P3).Sobytheinductionhypothesisf�1[p](+1fg)ismonotoneforallf2F+1; andall.Moreover,if=thentheidentityistheonlyf2F+1; .Inthiscasef�1[p](+1fg)ismonotoneby(3)inthesuccessorcaseofthedenitionofP.Nowsupposethatf�1[p](+1fg)ismonotoneforall andallf2F+1; .Wehavetoprovethat(2)and(3)inthesuccessorstepofthedenitionofPhold.(3)obviouslyholdsbytheassumptionbecausetheidentityistheonlyfunctioninF =F+1; .For(2),itsucesbytheinductionhypothesistoshowthatf�1[h�1[p]](+1fg)ismonotoneandf�1[(id)�1[p]](+1fg)ismonotoneforallf2F+1;.This,however,holdsby(P2)andtheassumption.Limitcase: 2LimAssumerstthatp2P.Let andf2F+1; .Wehavetoprovethatf�1[p](+1fg)ismonotone:BythelimitstepofthedenitionofP,thereare ,g2F andp2Psuchthatp=g[p].By(P4)thereare; ,g02F,f02Fandj2F suchthatg=jg0andf=jf0.Letp0:=g0[p].Then,bytheinductionhypothesis(f0)�1[p0](+1fg)ismonotone:However,(f0)�1[p0]=(f0)�1[j�1[p]]=f�1andwearedone.30 technical.InsteadwewilldirectlyshowthefollowingLemma5.2Ifp;q2P
andp();q()arecompatibleinPfor=max(supp(p)\supp(q)),thenpandqarecompatibleinP
.Proof:Theproofisasimpliedversionoftheproofoflemma3.1.Supposepandqareasinthelemma,butincompatible.Let(supp(p)[supp(q))�=f n::: 1g.Weprovebyinductionon1in,thatp( i)andq( i)areincompatibleforall1in.Since n=,thisyieldsthedesiredcontra-diction.Noterst,thatp( 1)andq( 1)areincompatiblebecauseotherwisep=st[p( 1)]andq=st[q( 1)]wereincompatible(fors2T 1,st).If 1=,wearedone.Soassumethat 16=.Theneitherp( 1)=ss[p( 1�1)]orq( 1)=ss[q( 1�1)]wheresst,s2T 1�1ands2T 1.Weassumeinthefollowingthatp( 1)=ss[p( 1�1)].Mutatismutandis,theothercaseworksthesame.Claim:p( 1�1)andq( 1�1)areincompatibleinP 1�1Assumenot.Thenthereisrp( 1�1);q( 1�1)inP' 1�1suchthatar=ap( 1�1)[aq( 1�1).Letr0:=ss[r].Thenr0[p( 1�1)]=p( 1)andr0[q( 1�1)]=q( 1)( 1 1).Inthefollowingwewillconstructanrp( 1);q( 1)whichyieldsthecontradictionwewerelookingfor.By(2)inthedenitionofP 1,q(; 1)q(; 1)forall2aqwhereq:=q( 1).Let~=maxf2aqjq(; 1)=0gifthesetisnotempty.Otherwise,set~=0.Setr=r0[fhh; 1i;0ij~;2ar0g[fhh; 1i;1ij~;2ar0g:Thenrisaswanted.Thisprovestheclaim.Itfollowsfromtheclaim,thatp( 2)andq( 2)areincompatible.Hencewecanprovethelemmabyrepeatingthisargumentinductivelynitelymanytimes.2Lemma5.3P:=P!2satisesccc.Proof:LetAPbeasetofsize!1.Bythe
-lemma,wemayassumethatfbpjp2Agformsa
-systemwithrootD.Wemaymoreoverassumethatforall2D,allf2F+1;!1andallp;q2Af�1[p](+1fg)f�1[q](+1fg)orf�1[p](+1fg)f�1[q](+1fg):ToseethisassumethatX=fapjp2Ag!2formsa
-systemwithroot
1.Fix2D.BythinningoutA,wecanensurethatwhenevera6=b2X,2a�b,2b�a,,t=h!1;i,st,s2T+1,then=2rng(st).This32 =((ap�(ap1[ap2))fg)\f[(+1fg)]:Toprovethatpf[(+1fg)]ismonotone;weconsidertherstcaserst.Let 2f[+1].If ;2ap1,thenp( ;)=p1( ;)p1(;)=p(;)becausep12P.Otherwisep( ;)p(;)bythedenitionofp.Thesecondcaseisprovedinthesamewaywherep1isreplacedbyp2.2Theorem5.4Ifthereisasimplied(!1;1)-morass,thenthereisaccc-forcingPwhichaddsachainhXj!2isuchthatX!1,X�XisniteandX�Xhassize!1forall!2.Proof:Bylemma5.3,Psatisesccc.Henceitpreservescardinals.Itiseasilyseenbyinductionalongthemorass,thatforevery2!2andevery2!1thesetsD=fp2Pj2apgandD0=fp2Pj2bpgaredenseinP.SoifGisP-generic,thenF=Sfpjp2GgisafunctionF:!2!1!2.SetX=f2!1jF(;)=1g.BythedenitionofonP,X�Xisniteforall.Finally,againbyaneasyinductionalongthemorasswecanprovethatforall2!1,2!2thesetD00;;=fp2Pj9 p(; )=0;p(; )=1gisdenseinP.ThisyieldsthatX�Xisuncountableforall!2.2References[1]JamesE.Baumgartner.Almost-disjointsets,thedensesetproblemandthepartitioncalculus.Ann.Math.Logic,9(4):401{439,1976.[2]JamesE.BaumgartnerandSaharonShelah.RemarksonsuperatomicBooleanalgebras.Ann.PureAppl.Logic,33(2):109{129,1987.[3]KeithJ.Devlin.Constructibility.PerspectivesinMathematicalLogic.Springer-Verlag,Berlin,1984.[4]Hans-DieterDonder.Anotherlookatgap-1morasses.InRecursiontheory(Ithaca,N.Y.,1982),volume42ofProc.Sympos.PureMath.,pages223{236.Amer.Math.Soc.,Providence,RI,1985.[5]V.V.Fedorcuk.Thecardinalityofhereditarilyseparablebicompacta.Dokl.Akad.NaukSSSR,222(2):302{305,1975.[6]SyD.Friedman.Finestructureandclassforcing,volume3ofdeGruyterSeriesinLogicanditsApplications.WalterdeGruyter&Co.,Berlin,2000.[7]A.HajnalandI.Juhasz.Discretesubspacesoftopologicalspaces.Nederl.Akad.Wetensch.Proc.Ser.A70=Indag.Math.,29:343{356,1967.[8]BernhardIrrgang.Constructing(!1;)-morassesfor!1.Unpublished.[9]BernhardIrrgang.Proposing(!1;)-morassesfor!1.Unpublished.34 [27]StevoTodorcevic.Partitionproblemsintopology,volume84ofContempo-raryMathematics.AmericanMathematicalSociety,Providence,RI,1989.[28]StevoTodorcevic.Walksonordinalsandtheircharacteristics,volume263ofProgressinMathematics.BirkhauserVerlag,Basel,2007.[29]DanVelleman.Simpliedmorasses.J.SymbolicLogic,49(1):257{271,1984.[30]DanVelleman.Gap-2morassesofheight!.J.SymbolicLogic,52(4):928{938,1987.[31]DanVelleman.Simpliedgap-2morasses.Ann.PureAppl.Logic,34(2):171{208,1987.36