Effective Symbolic Dynamics and Complexity

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Effective Symbolic Dynamics and Complexity
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Toska, Ferit
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Doctorate ( Ph.D.)
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Cenzer, Douglas A
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Larson, Jean A
Jury, Michael Thomas
Zapletal, Jindrich
Ray, Gregory Brian

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complexity -- computability -- dynamics -- machine -- process -- randomness -- strict
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In the second chapter we investigate the computability of countable subshifts in one dimension, and their members. Subshifts of Cantor-Bendixson rank two contain only eventually periodic elements. Any rank two subshift of bi-infinite sequences is decidable. Subshifts of rank three may contain members of arbitrary Turing degree. In contrast, effectively closed subshifts of rank three contain only computable elements, but effectively closed subshifts of rank four may contain members of higher degree. There is no subshift of rank omega + 1. In the third chapter we introduce a notion of description for infinite sequences and their sets. Strict process machine complexity of a computable sequence is zero whereas there are sequences of arbitrary Turing degree with zero complexity. SPM complexity of a sequence or of a subset of Cantor space is equal to its effective Hausdorff dimension. In the fourth chapter we investigate the compressibility of countable subsets of Cantor space. 1-decidable sets of Cantor-Bendixson rank two describe each other where the rate of description is determined by the limit points. If P and Q are strongly homeomorphic then P 1-compresses Q and vice versa.
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Thesis (Ph.D.)--University of Florida, 2013.
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EFFECTIVESYMBOLICDYNAMICSANDCOMPLEXITYByFERITTOSKAADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOLOFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENTOFTHEREQUIREMENTSFORTHEDEGREEOFDOCTOROFPHILOSOPHYUNIVERSITYOFFLORIDA2013

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c2013FeritToska 2

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Idedicatethistoexistence. 3

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ACKNOWLEDGMENTS IwouldliketorecognizemyadvisorDr.DouglasCenzerforhissupport.Faculty,staff,andgraduatestudentsoftheUniversityofFloridaDepartmentofMathematics,specicallytheLogicgroup,andSebastianWymanandThanosGentimishavecontributedtothecompletionofthisprojectaswell.Ithankthemall.IwouldalsoliketoacknowledgetheToskafamilyinTurkeyaswellastheToska'sinGainesvilleforbeingwithmeduringthisodyssey. 4

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TABLEOFCONTENTS page ACKNOWLEDGMENTS .................................. 4 ABSTRACT ......................................... 6 CHAPTER 1INTRODUCTION ................................... 7 1.1Summary .................................... 7 1.1.1CountableSubshifts .......................... 7 1.1.2AlgorithmicRandomness ....................... 9 1.2Preliminaries .................................. 10 2COUNTABLESUBSHIFTS ............................. 24 2.1Preliminaries .................................. 24 2.2SubshiftsofRankTwo ............................. 25 2.3SubshiftsofRankThreeandFour ...................... 37 3SPMCOMPLEXITYOFINFINITESTRINGS ................... 45 3.1Preliminaries .................................. 45 3.2EffectiveHausdorffDimensionandSPMComplexity ............ 49 3.3SPMComplexityofSubsetsofCantorSpace ................ 53 4COMPRESSIBILITYOFCOUNTABLESUBSETSOFCANTORSPACE .... 59 4.1Preliminaries .................................. 59 4.2RankTwoSubsetsof2N ............................ 63 REFERENCES ....................................... 77 BIOGRAPHICALSKETCH ................................ 81 5

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AbstractofDissertationPresentedtotheGraduateSchooloftheUniversityofFloridainPartialFulllmentoftheRequirementsfortheDegreeofDoctorofPhilosophyEFFECTIVESYMBOLICDYNAMICSANDCOMPLEXITYByFeritToskaMay2013Chair:DouglasCenzerMajor:MathematicsInthesecondchapterweinvestigatethecomputabilityofcountablesubshiftsinonedimension,andtheirmembers.SubshiftsofCantor-Bendixsonranktwocontainonlyeventuallyperiodicelements.Anyranktwosubshiftin2Zisdecidable.SubshiftsofrankthreemaycontainmembersofarbitraryTuringdegree.Incontrast,effectivelyclosed(01)subshiftsofrankthreecontainonlycomputableelements,but01subshiftsofrankfourmaycontainmembersofarbitrary02degree.Thereisnosubshiftofrank!+1.Inthethirdchapterweintroduceanotionofdescriptionforinnitesequencesandtheirsets.StrictprocessmachinecomplexityofacomputablesequenceiszerowhereastherearesequencesofarbitraryTuringdegreewithzerocomplexity.SPMcomplexityofasequenceorofasubsetofCantorspaceisequaltoitseffectiveHausdorffdimension.InthefourthchapterweinvestigatethecompressibilityofcountablesubsetsofCantorspace.1-decidablesetsofCantor-Bendixsonranktwodescribeeachotherwheretherateofdescriptionisdeterminedbythelimitpoints.IfPandQarestronglyhomeomorphicthenP1-compressesQandviceversa. 6

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CHAPTER1INTRODUCTION 1.1SummaryWewillpresentasummaryofthefollowingchaptersalongwithahistoricalsurveyonthebackgroundtheories. 1.1.1CountableSubshiftsTheresultsofChapter2appearedin[ 9 ].Aninitialversionofthispaperwaspresentedat[ 8 ].Thereisalonghistoryofinteractionbetweencomputabilityanddynamicalsystems.ATuringmachinemaybeviewedasadynamicalsystemwhichproducesasequenceofcongurationsorwordsbeforepossiblyhalting.Thereversenotionofusinganarbitrarydynamicalsystemforgeneralcomputationhasgeneratedmuchinterestingwork.Seeforexample[ 1 16 ].InChapter2wewillconsidercomputableaspectsofcertaindynamicalsystemsovertheCantorspace2Nandtherelatedspace2Z.ThestudyofcomputabledynamicalsystemsispartoftheNerodeprogramtostudytheeffectivecontentoftheoremsandconstructionsinanalysis.Weihrauch[ 44 ]hasprovidedacomprehensivefoundationforcomputabilitytheoryonvariousspaces,includingthespaceofcompactsetsandthespaceofcontinuousrealfunctions.ThecomputabilityofJuliasetsintherealshasbeenstudiedbyCenzer[ 5 ]andKo[ 21 ];thecomputabilityofcomplexdynamicalsystemshasbeeninvestigatedbyRettingerandWeihrauch[ 37 ]andbyBravermanandYampolski[ 2 ].ComputablesubshiftsandtheconnectionwitheffectivesymbolicdynamicswereinvestigatedbyCenzer,DashtiandKing[ 10 ].Atotal,TuringcomputablefunctionalF:2N!2Nisalwayscontinuousandthuswillbetermedcomputablycontinuousorjustcomputable.Effectivelyclosedsets(alsoknownas01classes)areacentraltopicincomputabilitytheory;see[ 4 ].ItwasshownforanycomputablycontinuousfunctionF:2N!2N,It[F]isadecidable01classand,conversely,anydecidable01subshift 7

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PisIt[F]forsomecomputablemapF.InChapter2,01subshiftsareconstructedin2Nandin2Zwhichhavenocomputableelementsandarenotdecidable.Thusthereisa01subshiftwithnon-trivialMedvedevdegree.J.Miller[ 34 ]showedthatevery01Medvedevdegreecontainsa01subshift.Simpson[ 41 ]studied01subshiftsintwodimensionsandobtainedasimilarresultthere.InChapter2,wewillconsidermorecloselythestructureofcountablesubshifts,usingtheCantor-Bendixson(CB)derivative.WewillcompareandcontrastcountablesubshiftsofniteCBrankwith01subshiftsofniteCBrankaswellaswitharbitrary01classesofniterank.TheoutlineofChapter2isasfollows.Section2.1containsdenitionsandpreliminaries.Section2.2focusesonsubshiftsofranktwoandhassomegeneralresultsaboutperiodicandeventuallyperiodicmembersofsubshifts.WeshowthatifQisasubshiftofranktwo,theneverymemberofQiseventuallyperiodic(andthereforecomputable)andfurthermoreifQ2Z,thenthemembersofranktwoareperiodicandQisadecidableclosedset.However,thereareranktwosubshiftsin2NofarbitraryTuringdegreeandranktwo01subshiftsofarbitraryc.e.degree,sothatranktwoundecidable01subshiftsexistin2N.Wegiveconditionsunderwhicharanktwosubshiftin2Nmustbedecidable.Weshowthatthereisnosubshiftofrank!+1.Insection2.3,westudysubshiftsofrankthreeandfour.WeshowthatsubshiftsofrankthreemaycontainmembersofarbitraryTuringdegreeandthatsubshiftsofrankthreein2ZmayhavearbitraryTuringdegree.Incontrast,weshowthat01subshiftsofrankthreecontainonlycomputableelements,but01subshiftsofrankfourmaycontainmembersofarbitraryc.e.degree.Moregenerally,weshowthatforanygiven01classPofranktwo,thereisasubshiftQofrankfoursuchthatthedegreesofthemembersofPandthedegreesofthemembersofQareidentical. 8

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1.1.2AlgorithmicRandomnessInChapter3andChapter4,wewillturnourattentiontoalgorithmicrandomness,anareaofmathematicstryingtodenethenotionofrandomnessusingthetoolsofcomputabilitytheory.Earlyin20thcenturyvonMises[ 35 ]assertedthatanyrandomsequenceshouldfollowthelawoflargenumbers.HissearchforotherrulesarandomsequenceshouldobeywasaddressedbyChurch[ 13 ].Hewastherstmakingtheconnectionbetweenrandomnessandcomputability.Theseruleshadtobecomputable.Later,MartinLof[ 32 ]gaveacharacterizationofrandomnumbersusingmeasuretheoreticaltests.Therearetwootherapproachesleadingtothesamenotionofrandomness.Kolmogorov[ 22 ]andSolomonoff[ 43 ]developedthenotionofKolmogorovcomplexity,followingtheintuitionthatarandomsequenceshouldbeincompressible.Thethirdnotionismotivatedbytheideathatarandomsequenceshouldbeunpredictable.FollowingthisideaLevy[ 26 ]denedmartingalesandSchnorr[ 38 40 ]effectivizedit.TheideaofnitestringsdescribingothernitestringsisthecentralaspectofKolmogorovcomplexity.Differentfamiliesofmachinesusedforthedescriptionsystemsgiverisetodifferentnotionsofcomplexity.Levin[ 24 25 ]usedmonotonemachinestocharacterizerandomnessandSchnorr[ 39 40 ]introducedprocesscomplexity.InChapter3wewilluseanothertypeofmachines,strictprocessmachineswhichcanbefoundintheworksofZvonkinandLevin[ 45 ],andinSolomonoff[ 43 ].TheinterplaybetweenprocessmachinesandothernotionsofcomplexitiesisstudiedbyDay[ 15 ].InSection3.1,wewillusestrictprocessmachinestoformalizeanotionofdescriptionforinnitesequences.Wewilldenestrictprocessmachine(SPM)complexityofinnitestrings.WeshowthattheSPMcomplexityofcomputablesequencesarezero.Ontheotherhand,therearesequenceswithzerocomplexityofanyTuringdegree. 9

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Insection3.2weprovethattheSPMcomplexityofaninnitestringisequaltoitseffectiveHausdorffdimension.TheclassicaldenitionofHausdorffdimensiongoesbacktoworksofLebesgue[ 23 ],Caratheodory[ 3 ]andHausdorff[ 18 ].EffectiveversionofHausdorffdimensionwasgivenbyLutz[ 29 31 ]usingmartingales.Mayordomo[ 33 ]providedacharacterizationofeffectiveHausdorffdimensionintermsofKolmogorovcomplexity.Insection3.3weextendthenotionofdescriptiontothesetsofinnitestrings.Thenotionsofcompression,descriptionspaceandrepresentationspaceareintroduced.WeprovideadenitionofSPMcomplexityforsubsetsofCantorspaceandprovethatitisequaltoeffectiveHausdorffdimension.InChapter4westudythecompressibilityofcountablesubsetsofCantorspace.Insection4.1wedenen-decidabilityofaclosedsetwhichisthebackboneofstrictprocessmachinesconstructedinthefollowingsection.Insection4.2wefocusonranktwosubsets.Givenanytwo1-decidablesetPandQ,weconstructamachineMsothatPisanM-descriptionofQ.Weseethattherateofdescriptiondoesonlydependonthelimitpointsandthattheisolatedpointscanbecompressedwithanyratewedesire.Weobservethathomeomorphismisnotsufcienttoentail1-compressibility.Thenotionofstronghomeomorphismisintroducedtocapturethisidea. 1.2PreliminariesInthissectionourgoalistointroducebackgroundtheoriesthatneededtopresentthisdissertation.Ourworkcanbeconsideredasapartofmathematicscontainedinsymbolicdynamics,computabilitytheoryandalgorithmicrandomness.Whiletheseareascoveraverybroadrangeofinterestourscopewillbelimited.Wewillalsointroducesomenotationalconventionsinthissection.DynamicalSystems:Thetheoryofdynamicalsystemsaimstounderstandtheasymptoticbehaviorofaniterativeprocess.Theseprocessescanbecategorizedascontinuousdynamical 10

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systemsanddiscretedynamicalsystems.Ourinterest,symbolicdynamics,fallsinthediscretecategory.Aprocessdeterminedbytheiterationofafunctionisanexampleofadiscreteprocess.Thetheoryinvestigatestheeventualbehaviorofthepointsx,f(x),f2(x),...,fn(x),....Formoreinformationondiscretedynamicalsystemssee[ 17 ]or[ 14 ].SymbolicDynamics:Symbolicdynamicscanbestudiedasasubeldofformallanguagetheory,independentfromdynamicalsystems.Fromtheperspectiveofcomputabilitytheory,bothapproacheshavetheirownmerits.Hence,weshalladoptanduseboth.Weneedtointroducesomebasicdenitionsandnotations:ForanysetA,weletcard(A)denotethecardinalityofthesetA.ThesetN=f0,1,2,...gdenotesthesetofnaturalnumbers,andZ=f...,-2,-1,0,1,...gdenotesthesetofintegers.Analphabetisasetofletters.Awordisanitestringofletters.Foranysetandanyi2N,idenotesthestringsoflengthifrom,denotesthesetofallnitestringsfrom.Ndenotesthesetofcountablyinnitesequencesfrom,thatis,thesetoffunctionsmappingNinto,andZdenotesthesetoffunctionsmappingZinto.Wewrite
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Giventwostringsvandw,theconcatenationv_wisdenedbyv_w=(v(0),v(1),...,v(m)]TJ /F3 11.955 Tf 11.96 0 Td[(1),w(0),w(1),...,w(n)]TJ /F3 11.955 Tf 11.95 0 Td[(1)),wherejvj=mandjwj=n.Fora2,wewritew_a(orjustwa)forw_(a)andwewritea_w(orjustaw)for(a)_w.Wesaywisaninitialsegmentorprexofv(writtenwv)ifv=w_xforsomex2.Thisisequivalenttosayingthatw=vmforsomem2N.Wealsosaywisasufxofvifv=x_wforsomex2;wisafactorofvifv=x_w_yforsomex,y2.Wealsowritewvifwvandw6=v.ForanyX2Nandn2N,theinitialsegmentXnis(X(0),...,X(n)]TJ /F3 11.955 Tf 11.96 0 Td[(1)).ForZ2ZandijfromZ,Z[i,j]denotesthenitestring(Z(i),Z(i+1),...,Z(j));Z[i,j),Z(i,j],andZ(i,j)aresimilarlydened.ThesedenitionsalsoapplytoZ2Naswellastonitestrings.WesaythatawordvisafactorofX2NorofX2Zifv=X[i,j]forsomeiandj.WesaythatXavoidsvifvisnotafactorofX.Forastringw2andanyX2N,wewritewXifw=Xnforsomen.Foranyw2nandanyX2N,weletw_X=(w(0),...,w(n)]TJ /F3 11.955 Tf 11.95 0 Td[(1),X(0),X(1),...).ForX2N,letX)]TJ /F3 11.955 Tf 11.35 -4.34 Td[(=(,X(2),X(1),X(0));thatis,foreachi2N,letX)]TJ /F3 11.955 Tf 7.09 -4.34 Td[(()]TJ /F7 11.955 Tf 9.3 0 Td[(i)]TJ /F3 11.955 Tf -420.91 -14.45 Td[(1)=X(i),sothatX)]TJ /F2 11.955 Tf 11.06 -4.34 Td[(2Z)]TJ /F1 11.955 Tf 6.75 -7.15 Td[(,whereZ)]TJ /F3 11.955 Tf 11.06 -4.34 Td[(=fi2Z:i<0g.ForX,Y2N,letZ=X)]TJ /F3 11.955 Tf 7.08 -4.34 Td[(.Y2Zbedenedsothat,foralln2N,Z(n)=Y(n)andZ()]TJ /F7 11.955 Tf 9.3 0 Td[(n)]TJ /F3 11.955 Tf 11.96 0 Td[(1)=X(n).Wedenotebyv!theinniteconcatenationv_v_;similarlyv)]TJ /F14 7.97 Tf 6.58 0 Td[(!=v_vandv1=v)]TJ /F14 7.97 Tf 6.59 0 Td[(!.v!.Next,weintroducethenotionofformallanguageandlanguagecharacterizationofsubshifts. Denition1.1. 1.Ifisanitealphabet,thenthefull-shiftisthecollectionofallinnitesequences(orbi-innitesequences)ofsymbolsfrom.2.Theshiftmaponisdenedby(w)=(w(1),...,w(jwj)]TJ /F3 11.955 Tf 18.99 0 Td[(1)).ForX22N,(X)=(X(1),X(2),...).ForZ22Z,Y=(Z)isdenedsothatY(i)=Z(i+1). 12

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3.ForanysubsetSof,letQZS=fX22Z:Xavoidswforallw2SgandletQNS=fX22N:Xavoidswforallw2Sg.WewillwriteQSwhenthereisnoconfusion.ThesetSisalsocalledthesetofforbiddenwordsover.Usingthesetofforbiddenwordswedeneshiftspacesasfollows: Denition1.2. AshiftspaceisasubsetofthefullshiftsuchthatQ=QSforsomecollectionofforbiddenwordsover.Wecanalsodescribeashiftspaceusingthewordsthatareallowedratherthanforbidden. Denition1.3. 1.AlanguageLisasubsetof.2.LetPbeasubsetofafullshift.ThelanguageofP,denotedbyL(P)isfu2:uoccursinXforsomeX2Pg Proposition1.1. LetLbealanguage.Then,L=L(P)forsomeshiftspacePifandonlyifLsatisesthefollowingconditions:1.ifw2Landuisafactorofvthenu2L.2.ifw2Lthentherearenonemptywordsu,v2Lsuchthatuwv2L3.LetPandQbetwoshiftspaces.Then,P=QifandonlyifL(P)=L(Q).4.PisashiftspaceifandonlyifwheneverX[i,j]2L(P)foralli,jthenX2P.Formoreinformationonsymbolicdynamicssee[ 27 ]andonformallanguagessee[ 42 ].CantorSpace:Fortherestofthisdissertationwewillset=f0,1g,andthefullshiftwillbedenotedby2N.AshiftspaceP2Nwillbecalledasubshift.CantorSpace,thesetofinnitebinarysequences,isourmainobjectofstudy.Welistsomebasicfactsabout2N:Itstopologygeneratedbyabasisofintervals,oftheform[w]=fX:wXg. 13

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Intervalsareclopen,thatisbothopenandclosed.2Niscompact.Thesametopologyisgeneratedbythemetricddenedasfollows:ForallX,Y22N,letd(X,Y)=1=2nifnisthelargestksuchthatXk=Yk.LetUf0,1g.TheopensetgeneratedbyUis[U]=fX22N:uXforsomeu2UgAsubsetPof2NisclopenifandonlyifP=[U]forsomenitesubsetUoff0,1g.Lebesguemeasureofaninterval[w],denotedby([w])is2jwj.Treesoff0,1gand01classes: Denition1.4. AtreeTisasubsetoff0,1gwhichcontainstheemptystringandclosedunderinitialsegments.Thatis,ifv2Tanduv,thenu2T.Wesaythatw2Tisanimmediatesuccessorofv2Tifw=vaforsomea2f0,1g.Theelementsofatreeareoftenreferredasnodes. Denition1.5. 1.ForanytreeT,X22NissaidtobeaninnitepaththroughTifXn2Tforalln.Welet[T]denotethesetofinnitepathsthroughT.2.ForanytreeT,wesaythatanodew2TisextendibleifthereexistsX2[T]suchthatwX.ThesetofextendiblenodesofTisdenotedbyExt(T).3.Anodeusuchthatv=2T,butu2Tforalluv,iscalledadeadendofT.WewilllistsomeusefulfactsabouttheinteractionoftreesandsubsetsofCantorSpace: Proposition1.2. 1.(Konig'sLemma)IfTisaninnitetreethen[T]6=;.2.AsubsetQof2NisclosedifandonlyifQ=[T]forsometreeT.3.ForanyclosedsetP,TP=fw2f0,1g:P\[w]6=;gisatreesuchthatP=[TP]. 14

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4.IfP=[T],thenTPwillequalthesetofextendiblenodesofT. Denition1.6. 1.AsubsetPof2Nisa01class(oreffectivelyclosedset)ifP=[T]forsomecomputabletreeT.2.A01classPisdecidableifTPiscomputable.Wenotethefollowingfactsabouta01classP:ThereisacomputabletreeTwithP=[T]suchthatExt(T)iscomputable.ThereisacomputabletreeTwithnodeadendssuchthatP=[T].WealsonotethatifTiscomputable,thenthesetofextendiblenodesisatreewhichisaco-c.e.subsetoff0,1gbutisnotingeneralcomputable.Formoreinformationon01classessee[ 6 ].Subshiftsof2ZwillbestudiedinChapter2.Welistheresomedenitionsandnotationsneeded:Thetopologyon2ZhasabasisofclopensetsoftheformfZ22Z:Z[)]TJ /F7 11.955 Tf 9.3 0 Td[(n,n]=wg,wherewhasoddlength.Abi-treeTisasetofnitestringsofoddlengthwhichisclosedundercentralsegments,thatis,(x()]TJ /F7 11.955 Tf 9.3 0 Td[(n)]TJ /F3 11.955 Tf 12.21 0 Td[(1),x()]TJ /F7 11.955 Tf 9.3 0 Td[(n),...,x(0),x(1),...,x(n),x(n+1))2Timpliesthat(x()]TJ /F7 11.955 Tf 9.3 0 Td[(n),x()]TJ /F7 11.955 Tf 9.3 0 Td[(n+1),...,x(0),x(1),...,x(n)]TJ /F3 11.955 Tf 11.96 0 Td[(1),x(n))2T.Forabi-treeT,Z22Zisabi-innitepaththroughTifZ[)]TJ /F7 11.955 Tf 9.3 0 Td[(n,n]2Tforalln.Herealso[T]isthesetofbi-innitepathsthroughTQ2ZisclosedifandonlyifQ=[T]forsomebi-treeT.Thedenitionsofeffectivelyclosedsets,extendiblenodes,anddecidableclosedisanalogoustothosegivenabovefor2N. Denition1.7. 1.AtreeT2Nissaidtobesubsimilarifforeveryvandw,vw2Timpliesw2T. 15

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2.Abi-treeTissaidtobesubsimilarifforeveryvofevenlengthandeveryw,vw2Torwv2Timpliesthatw2T.Nextwewillpointoutsomefactsconnectingformallanguagecharacterizationofsubshiftsandsubsimilartrees: Proposition1.3. 1.AclosedsetPisasubshiftifandonlyifPisshift-invariant,thatis(X)2PwheneverX2P.2.TheclosedsetPisasubshiftifandonlyifTPissubsmilar.3.(Cenzer,Dashti,King[ 10 ])ForanySf0,1g,QZSisasubshiftin2ZandQNSisasubshiftin2N.Furthermore,ifSisac.e.set,thenthesetsQZSandQNSare01classes.4.IfthesetQin2N(or2Z)isasubshift,thenthereisasetSofwordssuchthatQ=QNS(orQ=QZS).Furthermore,ifQisa01class,thenSmaybetakentobeac.e.set. Denition1.8. LetX22N.TheorbitofX,denotedbyO(X)isfY:Y=n(X)forsomen2NgFromthedenitionitfollowsthatforanyX22NtheclosureofO(X)isasubshift.Wewilloftenuseorbitstoconstructsubshifts.Forinstance,letA2N.Then,thesmallestclosedsubsetof2NcontainingO(X)foreachX2Aisasubshift.Cantor-BendixsonDerivative:InChapter2andChapter4wewillfocusoncountablesubshifts.Cantor-Bendixsonderivativewillbeanimportanttooltocharacterizecomplexityofclosedsets.WewillrstgiveageneraldenitionforCantor-Bendixsonderivative.AseparablecompletelymetrizablespaceiscalledPolish.ApointxofatopologicalspaceiscalledalimitpointifforeveryopenneighborhoodUofxthereisapointy2Usuchthaty6=x.Anisolatedpointofatopologicalspaceisapointthatisnotalimitpoint.AsubsetPofaspaceXiscalledperfectinXifitisclosedandallofitspointsarelimitpoints. 16

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Theorem1.1. (Cantor-Bendixson)LetXbeaPollishspace.ThenXcanbeuniquelywrittenasX=P[C,wherePisaperfectsetandCiscountableopen. Denition1.9. ForanyPolishspaceX,ifX=P[C,wherePisperfectandCiscountablewithP\C=;,wecallPtheperfectkernelofX.NowwedenetheCantor-Bendixsonderivative: Denition1.10. ForanytopologicalspaceXdenethetheCantor-Bendixsonderivative,denotedbyD(X),tobethesetofnon-isolatedelementsofX.UsingtransniterecursionwedenetheiteratedCantor-BendixsonderivativesD(X)foranyordinalasfollows:D0(X)=XD+1(X)=D(D(X))D(X)=T
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Denition1.12. LetPbeacountableclosedsubsetof2N(or2Z)andX2P.ThentheCantor-BendixsonrankrkP(X)ofXinPisdenedtobetheleastintegernsuchthatX=2Dn+1(P).ForcountablePandX2Pwenotethefollowing:rkP(X)=0ifandonlyifXisisolatedinP.rk(P)isthesupremumoffrkP(X)+1:X2PgWenotethatinseveralarticlesoneffectivelyclosedsets[ 4 7 8 11 ],adifferentdenitionisgivenfortheCantor-Bendixsonrankofacountableclosedset,namelythetheleastordinalsuchthatD+1(P)=;.Thiswillalwaysbeonelessthanrk(P)asdenedabove.Thisalternativedenitionallowstherk(P)forP2Ntobeanycountableordinalandmakesrk(P)thesupremumoffrkP(X):X2Pg.Wewillusetheclassicaldenition.FormorebackgroundoncomputabilityandontheCantor-Bendixsonderivative,see[ 4 ],whichincludesthefollowing: Lemma1. LetFbeacontinuousmapfrom2Ninto2NandletPandQbeclosedsetssuchthatF[P]=Q.ThenforanyY2Q,rkQ(Y)maxfrkP(X):X2P&F(X)=Yg.Notethatthelemmaalsoholdsforthecontinuousfunctionsfrom2Zinto2Zsince2Zishomeomorphicto2N. Notation1.3. FortwoclosedsetsPandQin2N,letPQ=fX)]TJ /F3 11.955 Tf 7.09 -4.33 Td[(.Y:X2P&Y2Qg,whichwillbeaclosedsetin2Z. Remark1.4. NotethatourversionofPQiscomputablyhomeomorphictotheusualdenition(see[ 6 ])asthesetofsequencesXY=(X(0),Y(0),X(1),Y(1),...)forX2PandY2Q.Wewillalsoneedthefollowing(essentiallyTheorem4.1of[ 6 ]).LetdenotetheHessenbergsumofordinalsand.Forourpurposes,itsufcestonotethatfornaturalnumbersmandn,mn=m+nforniteordinalsand!n=n!=!+n. 18

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Lemma2. ForanyclosedsetsPandQin2N,anyX2PandanyY2Q,theCantor-BendixsonrankofX)]TJ /F3 11.955 Tf 7.09 -4.34 Td[(.YinPQequalsrkP(X)rkQ(Y).Henceifrk(P)andrk(Q)arenite,rk(PQ)=rk(P)+rk(Q))]TJ /F3 11.955 Tf 11.95 0 Td[(1.KolmogorovComplexity:Next,wewillintroduceKolmogorovcomplexityofnitestrings.Althoughsamenotionscanbestudiedforotherniteobjects,wewillrestrictourattentiontobinarystrings,thatiselementsoff0,1g.ThenotionsmachineanddescriptionplayacentralroleinstudyingKolmogorovcomplexity.Amachineisnothingbutapartialcomputablefunctionthatmapsnitestringstonitestrings.Usingthewordmachinemaynotbeneededtodevelopthetheoryofpartialcomputablefunctionsbutitprovidesusaveryusefulheuristicdevice.Weexpectfromourmachinestoperformatask.Weliketofeedinsomeinputtoourmachinesandwaitforanoutput.Asimilarbreakthroughhappenswhenweimagineourmachinesasdescriptionsystems.IfamachineMoninputuoutputsv,thatisMisapartialcomputablefunctiondenedonu2f0,1gwhereitsimageisv,wesaythatuisanM-descriptionofv.Anaturalquestionarises:Howgoodisadescription?.TheattemptsaddressingthisquestionleadstothetheoryofKolmogorovComplexityandalgorithmicrandomness.Intuitively,adescriptionisgoodifitisshort.Forexample,consideramachinethatoutputs`1n'wheren=max(k,l)sothatkisthenumberofoccurrencesof`0'andlisthenumberofoccurrencesof'1'inagiveninput.Then,both`0000'and`10101010'describesthesameoutput,namely`1111'.Actually,`0000'isthebetterone,beingoneofthetwoshortestM-descriptionsof`1111'.Intuitively,astringisconsideredcomplexifitishardtodescribe,inotherwordsifithasnoshortdescriptions.So,thequestionwhatisthelengthoftheshortestdescription?playsacentralrole.Wehavethefollowingdenition. Denition1.13. LetMbeamachineandv2f0,1g. 19

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1.TheKolmogorovComplexityofvwithrespecttoMisCM(v)=minfjuj:M(u)=vgwheremin;=1.2.ThestringvissaidtoberandomrelativetoMifCM(v)jvj.Thedenitionaboveisnotsufcienttocaptureourintuitionaboutcomplexityofastring,foritdependsonamachine.Inparticular,ifastringvhasnoM-description,thenitsKolmogorovcomplexitywouldbe1.Thisdifcultyisaddressedbyusinguniversalmachinesdenedbelow. Denition1.14. AmachineMiscalledauniversalmachineifforanypartialcomputablefunctionN,thereisastringuN,calledcodingstring,suchthat(8v)M(uNv)=N(v).Theexistenceofauniversalmachineisgivenbytheenumerationtheorem.WewillxauniversalmachineanddenoteitbyV.TheKolmogorovcomplexityisdenedtobeCV(v),anddenotedbyC(v).ThechoiceofauniversalmachinedoesnotplayanimportantrolesinceforanymachineM,thereisaconstantc=juMj,thelengthofthecodingconstant,suchthatC(v)CM(v)+cforanyv2f0,1g.Asseeninthefollowingtheorem,Kolmogorovcomplexityhasadrawback.Thisisanimportantproblem,especiallyforthestudyofcomplexityoftheinnitestrings.For,itimpliesthatanyinnitestringhasnon-complexinitialsegmentsofanysize. Theorem1.5. Foranyd2N,thereexistsw2f0,1gsuchthatw=uvandC(w)>C(u)+C(v)+d.Onewayofavoidingthisdifcultyisusingprex-freemachines. Denition1.15. 1.AsetAf0,1giscalledprex-free,if8u,v2A(uv!u=v).2.AmachineMiscalledprex-freeifitsdomainisaprex-freeset. Theorem1.6. Thereexistsaprex-freeuniversalmachine.Wewillxauniversalprex-freemachineanddenoteitbyU.Theprex-freeKolmogorovcomplexityofastringvisdenedtobeCU(v)anditisdenotedbyK(v). 20

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ThefollowingresultrelatesKolmogorovcomplexityandprex-freeKolmogorovcomplexity. Theorem1.7. Thereexistsc1,c22Nsuchthatforallv2f0,1g,C(v)K(v)+c1andK(v)C(v)+2log(jvj)+c2.Next,wewilldeneanotherclassofdescriptionsystems. Denition1.16. LetM:f0,1g!f0,1gbeapartialcomputablemachine.1.Miscalledaprocessmachineifforallu,v2f0,1gsuchthatuvandM(u)#andM(v)#,thenM(u)M(v).2.Miscalledastrictprocessmachineifforallu,v2f0,1gsuchthatuvandM(v)#,thenM(u)#andM(u)M(v).Ourmaininterestwillbeonstrictprocessmachines(SPM).Wehavethefollowing: Theorem1.8. Thereexistsauniversalstrictprocessmachine.Oncewexauniversalstrictprocessmachine,saySwecandenethestrictprocesscomplexityofastringutobeCS(u)anddenoteitbyKs(u).Inchapter3,wewilluseuniversalstrictprocessmachinestodenestrictprocesscomplexityofaninnitestring.EffectiveHausdorffDimension:Inchapter3,weshowthatthestrictprocesscomplexityofaninnitestringisequaltoitseffectiveHausdorfdimension.Here,wepresentbasicdenitionsandresultsneeded.WestartwiththeclassicalHausdorffdimensiondenedonCantorspace. Denition1.17. LetP2NandAf0,1g.1.For0s1,thes-measureofaclopenset[u]iss([u])=2)]TJ /F11 7.97 Tf 6.59 0 Td[(sjuj.2.Aisann-coverofPifP[A]andAf0,1gn.3.Hsn(P)=inffPu2As(juj):Aisann-coverofPg4.Thes-dimensionalouterHausdorffmeasureofPisHs(P)=limnHsn(P).ThefollowingresultenablesustodeneHausdorffdimension. 21

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Theorem1.9. LetP2N.Then,thereiss2[0,1]suchthat1.Ht(P)=0forallt>sand2.Hr(P)=1forall0r
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ThefollowingresultwhichwillbeusedinChapter3toshowthattheSPMcomplexityofasubsetof2NisequaltoitseffectiveHausdorffdimension. Proposition1.4. LetP2N.Then,dim(P)=supfdim(X):X2Pg. Proof. Lets2fs2Q:PS[d]forsomec.e.s-galedg.Then,foranyX2P,s2fs2Q:fXgS[d]forsomec.e.s-galedg.Thus,dim(X)dim(P).Therefore,dim(P)supfdim(X):X2Pg.Toprovetheotherdirectionassumethatdim(P)>rwherer=supfdim(X):X2Pg.Then,thereexistst,r02Qsuchthatdim(P)>t>r0>r.Now,foreachX2PthereexistssX2QandansX-galedXwheresX
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CHAPTER2COUNTABLESUBSHIFTS 2.1PreliminariesInthissectionweintroducethenotionofperiodicityandsomebasicresultsconcerningperiodicelementsofCantorspace. Denition2.1. 1.AnelementXof2NissaidtobeperiodicifX=v!forsomenitestringv;theperiodofXistheminimallengthofvsuchthatX=v!.2.Xissaidtobeeventuallyperiodicifforsomestringsuandv,X=u_v!.3.AnelementZof2ZisperiodicifZ=v1forsomenitev;theperiodofZistheminimaljvjsuchthatZ=v1.4.Ziseventuallyperiodicifforsomeniteu,vandw,Z=w)]TJ /F14 7.97 Tf 6.59 0 Td[(!.u_v!.Thefollowingfactsaboutperiodicsequenceswillbeuseful. Lemma3. LetuandvbenitewordsandletX=v!wherejvjisminimal.IfX=u_v!,thenu=vmforsomem. Proof. SupposethatX=u_v!=v!.Ifu=,thenu=v0.Otherwise,u!=v!andthereforejujjvjbytheminimalityofjvjanditfollowsthatu=vmforsomem. Wewillneedthefollowingsimpleconnectionbetweenperiodicityandtheshift. Lemma4. 1.X22Nisperiodicifandonlyifn(X)=Xforsomen>0.2.X22Niseventuallyperiodicifandonlyifm+n(X)=m(X)forsomemandn>0.3.Z22Zisperiodicifandonlyifn(Z)=Zforsomen>0. Proof. 1.IfXisperiodicthenX=u!forsomeu2f0,1g.Then,juj(X)=u!=X.FortheotherdirectionassumethatX=n(X)forsomen>0.LetubetheprexofXoflengthn,thatisjuj=nandX=u_YforsomeY22N.Then, 24

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X=n(X)=n(uY)=Y.So,X=uX.AneasyinductionargumentshowsthatX=uk_Xforallk.Hence,X=u!.2.IfXiseventuallyperiodicthenX=v_u!forsomev,u2f0,1g.Then,jvj(X)=u!.So,bytherstpartjvj+juj(X)=juj(jvj(X))=jvj(X).Fortheotherdirectionassumethatm+n(X)=m(X)forsomemandn.LetX=vf_Ywherejvj=m.Then,bytherstpartweseethatYisperiodic.So,Y=u!forsomeu2f0,1g.Thus,X=v_u!.3.SupposerstthatZisperiodicandletZ=u1forsomenitestringuoflengthn.ThenwehaveZ[kn,(k+1)n)=uforallk2Z.So,n(Z[kn,(k+1)n)])=Z[(k+1)n,(k+2)n)]=uforallk2Z.Thus,n(Z)=u1=Z 2.2SubshiftsofRankTwoThissectioncontainsresultsonthecomputabilityanddecidabilityofsubshiftsofranktwo.WeexaminetheconnectionbetweentheshiftoperatorandtheCantor-Bendixsonderivative.ThisleadstothesurprisingresultthattherearenosubshiftsofCantor-Bendixsonrankexactly!+1.Thefollowinglemmaswillbeneeded. Lemma5. (a) IfQ2Nisanitesubshift,thenQcontainsaperiodicelementandeveryelementofQiseventuallyperiodic. (b) IfQ2Zisanitesubshift,theneveryelementofQisperiodic. Proof. (a)LetY2Q,whereQisanitesubshift.Thenforeachi,i(Y)2Q.SinceQisnite,theremustexistmandnsuchthatm+n(Y)=n(Y),sothatYiseventuallyperiodic.ThenX=m(Y)isaperiodicmemberofQ.For(b),letZ2Q.SinceQisnite,thereexistm
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Lemma6. LetP2N(orP2Z)beanyclosedset.ThenD(P)=(D(P))foranyordinal.Hence,ifP2Nisasubshift,then,forallX2P,rkP(X)rkP((X)).Furthermore,ifP2Zisasubshift,then,forallX2P,rkP(X)=rkP((X)). Proof. Thekeytotheproofisthecasewhen=1.LetP2Nbeasubshift.SupposerstthatX2D((P)).ThenthereisasequencefYngn2NPofdistinctmembersofPsuchthatlimn(Yn)=X.Foreachn,eitherYn(0)=0orYn(0)=1.Thusthereexistsi2f0,1gandaninnitesubsequencen0,n1,...suchthatYnk(0)=iforallk.ItfollowsthatlimkYnk=i_XandbelongstoD(P).Butthenwehave(i_X)=XandhenceX2(D(P)).Conversely,supposethatX2(D(P))andchooseY2D(P)suchthat(Y)=X.ThenthereisasequencefYngn2NPofdistinctmembersofPsuchthatlimnYn=Y.Itfollowsfromthecontinuityoftheshiftoperatorthatlimn(Yn)=(Y)=XandhenceX2D((P)).Forthespace2Z,theconverseargumentisthesame,buttherstdirectionissimpler,sincelimn(Yn)=XimpliesthatlimnYn=)]TJ /F5 7.97 Tf 6.58 0 Td[(1(X),sothat)]TJ /F5 7.97 Tf 6.59 0 Td[(1(X)2D(P)andhenceX2(D(P)).Theproofproceedsbyinductionon.Forthesuccessorcase,wehaveD+1((P))=D(D((P)))=D((D(P)))=(D(D(P)))=(D+1(P)).Forthelimitcase,wehaveD((P))=\
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Forthenalconclusion,supposethatrkP(X)=.ThenX2D(P)andhence(X)2(D(P))=D((P))D(P).ForP2Z,wenotethattheproofabovecanbemodiedtoshowthatD()]TJ /F5 7.97 Tf 6.59 0 Td[(1(P))=)]TJ /F5 7.97 Tf 6.59 0 Td[(1(D(P))andrkP(X)rkP()]TJ /F5 7.97 Tf 6.58 0 Td[(1(X)). Proposition2.1. ForanysubshiftQ2N(orQ2Z)andanyordinal,D(Q)isasubshift. Proof. LetQ2N(2Z).ForanyX2D(Q),itfollowsfromLemma 6 that(X)2D((Q))andhence(X)2D(Q). Wewillneedthefollowinglemmasrelatingsubshiftsof2Ztothecorrespondingsubshiftsof2N.Let0and1bethetwoprojectionmapsfrom2Zonto2NsothatifZ=X)]TJ /F3 11.955 Tf 7.09 -4.34 Td[(.Y,then1(Z)=Yand0(Z)=X;thatis,foralln,Y(n)=Z(n)andX(n)=Z()]TJ /F7 11.955 Tf 9.3 0 Td[(n)]TJ /F3 11.955 Tf 11.95 0 Td[(1). Lemma7. LetQ2Zbeasubshift. (a) LetSbethesetofnitestringsusuchthatnoelementofQhasuasfactorandletS)]TJ /F3 11.955 Tf 10.4 -4.34 Td[(=fu)]TJ /F3 11.955 Tf 10.41 -4.34 Td[(:u2Sg.ThenQ=QZS,1[Q]=QNSand0[Q]=QNS)]TJ /F9 11.955 Tf 6.75 1.39 Td[(. (b) 0[Q]and1[Q]aresubshiftsof2NandareeffectivelyclosedifQiseffectivelyclosed. (c) Q0[Q]1[Q]. Proof. LetSbeasdened.ItisclearthatQQZS.Ontheotherhand,weknowbyProposition 1.3 thatQ=QZRforsomeRandwemusthaveRS,sothatQZSQZR=Q.Observenextthat1[Q]isaclosedsetasthecontinuousimageoftheclosedsetQandiseffectivelyclosedifQiseffectivelyclosed,since1isacomputablemapping.Furthermore,1[Q]isasubshift.Thatis,ifY=1(Z),then(Y)=1((Z)).ThusbyProposition 1.3 ,wehave1[Q]=QNRforsomeRandwemayassumethatSR,sincecertainly1[Q]QNS.Ontheotherhand,ifu=2SthenforsomeZ2Q,uisafactorof 27

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ZandbyshiftingZifnecessarywecanobtainuasafactorof1(Z),sothatu=2R.Itfollowsthat1[Q]=QNS.Asimilarargumentholdsfor0[Q].Forpart(c),letZ=X)]TJ /F3 11.955 Tf 7.08 -4.34 Td[(.Y2Q.ThenZavoidsS,sothatXavoidsS)]TJ /F1 11.955 Tf 10.4 -4.34 Td[(andYavoidsS. Proposition2.2. (a) ForanysubshiftQ2Nofrank+1,D(Q)hasaperiodicelementandanyelementofQhavingrankiseventuallyperiodic. (b) IfQ2Zhasrank+1,thenallelementsofrankareperiodic. Proof. (a) LetQhaverank+1,sothatD+1(Q)=;andD(Q)isnite.ThenD(Q)isasubshiftbyProposition 2.1 andtheresultfollowsbyLemma 5 (a). (b) Thesameargumentworksasin(a). Nextweconsidersomeresultsonthedecidabilityofsubshifts. Lemma8. ForanyZ22N(orin2Z)whichiseventuallyperiodic,thesetoffactorsofZisdecidable. Proof. Wewillgivetheprooffor2N.Theprooffor2Zcaseissimilar.SupposethatX=vw!andletWbethesetoffactorsofX.Thenx2Wifandonlyifithasoneofthefollowingforms: (i) v[s,t)where0stjvj (ii) v[s,jvj)wnw[0,t)wheres
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Example2.2. LetENbeanysetwhichisc.e.butnotcomputableandletE=fn0,n1,...gbeacomputableenumerationwithoutrepetition.LetX=10n010n11....ThenXiscomputablebutthesetoffactorsofXisnotsince10n1isafactorofXifandonlyifn2E. Proposition2.3. Givenanynaturalnumberm,thereisanatmostcountabledecidablesubshiftP2N(2Z)suchthatitsrankisequaltomandallofitselementsareeventuallyperiodic. Proof. LetP0=;andforeachn,letPn+1bethesetofelementsof2Ncontainingatmostnones.TheneachPn+1isclearlyasubshiftandisdecidablesincev2TPn+1ifandonlyifvcontainsatmostnones.CertainlyP0=;hasrank0andP1=f0!ghasrankone.Foreachn,weclaimthatPnPn+1andD(Pn+1)=Pn.SupposerstthatX2PnandletX=v_0!wherevhasatmostn)]TJ /F3 11.955 Tf 12.81 0 Td[(1ones.ThenX=limiXiwhereXi=v_0i1_0!2Pn+1.HenceX2D(Pn+1).ThisshowsthatD(Pn+1)=PnanditfollowsbyinductionthatPn+1hasrankn.NextsupposethatX=2Pn.IfX=2Pn+1,thencertainlyX=2D(Pn+1),sowemayassumethatX2Pn+1.ThenXmusthaveexactlynones,sothatX=v_0!wherevhasexactlynones.ThenclearlyPn+1\[v]=fXg,sothatX=2D(Pn+1).Thesameconstructionalsoworksfor2Z. NotethatforthesequenceofsetsPndenedabove,SnPnisnotclosedandinfactisdensein2N.Thefollowinglemmaiswell-knownintheareaofcombinatoricsonwords.Forexample,itfollowseasilyfromTheorem1.3.13of[ 28 ,p.22].Wewillpresentanindependentproof. Lemma9. SupposeX22NorX22Zisnoteventuallyperiodic.Thenforanyk2N,thereareatleastk+1distinctfactorsoflengthkthatoccurinnitelyofteninX. 29

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Proof. Theproofproceedsbyinductiononk.Clearlythelemmaistruefork=1,sinceboth0and1willoccurinnitelyofteninanoteventuallyperiodicsequence.Nowsupposethatthelemmaholdsfork,sothatXhasatleastk+1differentfactorsoflengthkwhichoccurinnitelyofteninX.Iftherearemorethank+1suchfactorsu1,...,u`where`k+2,thenclearlyXisgoingtohaveatleastk+2factorsoflengthk+1,sinceeitherui_0orui_1wouldoccurinnitelyoftenasafactorofXforalli`.Thuswemaysupposethereareexactlyk+1distinctfactorsoflengthkthatoccurinnitelyofteninX;denotethembyu0,...,uk.Itsufcestoshowthatforatleastonei,bothui_0andui_1occurinnitelyofteninX.Suppose,bywayofcontradiction,thatthisisnotthecase.Thenforeachithereexistsei2f0,1gandthereexistsnsuchthat1.Foreverym>n,thefactorX(m),...,X(m+k)]TJ /F3 11.955 Tf 11.95 0 Td[(1)isinthesetfu0,...,ukg.2.ForeveryoccurrenceofuiwhichoccurspastXnuiisfollowedinXbyei.NowwecanshowthatXmustbeeventuallyperiodic,whichwillprovidethenecessarycontradiction.Nowconsiderthesequenceoffactorsxi=(X(n+i),x(n+i+1),...,X(n+i+k)]TJ /F3 11.955 Tf 12.44 0 Td[(1))andletjkbetheleastsuchthatxj=x0.Withoutlossofgeneralitythesequencehastheformu0,u1,...,uj,u0forsomejk,sothatxi=uiforijandxj+1=u0.Sinceeachuimustbefollowedbyei,itfollowsthatfori>j,ui+i=ui_e)iandthatu0=uj_ej.Thusxj+2=u1,xj+3=u2andsoon,sothatX=X(n+k)(e0e1...ej)!.Thiscontradictionshowsthatatleastoneofthefactorsuioflengthkmusthavetwopossibleextensionsoflengthk+1whichoccurinnitelyofteninX.Theresultfor2Zfromthecorrespondingresultfor2N.LetZ22ZandletZ=X)]TJ /F3 11.955 Tf 7.08 -4.34 Td[(.YwhereXandYarein2N.IfZisnoteventuallyperiodic,thenatleastoneofthetwoelementsXandYof2Narenoteventuallyperiodic.Theresultnowfollowsfromthe2Ncaseabove. 30

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Theorem2.3. ForanysubshiftQ2N(2Z)ofranktwo,everymemberofQiseventuallyperiodic. Proof. SupposeQ2Nisasubshift,andsuppose,bywayofcontradiction,thatX2QandthatXisnoteventuallyperiodic.Letkbearbitraryandletw0,...,wkbedistinctfactorsofXoflengthkwhichoccurinnitelyofteninX.Thenforeachik,thereareinnitelymanynsuchthatn(X)2[wi]\Q.SinceXisnoteventuallyperiodic,m6=nimpliesthatm(X)6=n(X).Thus[wi]\Qhasalimitpointforeachik.ItfollowsthatQhasatleastk+1limitpoints.Sincekwasarbitrary,Qhasinnitelymanylimitpointsandthusrk(Q)>2,acontradiction.Forthe2Zcase,theargumentissimilar.Again,foragivenk,wehavek+1-manydistinctfactorsofZoflengthkwhichoccurinnitelyofteninZ,sayw0,w1,...,wk.Foreachik,thisimpliesthatthereareaninnitenumberofdistinctmembersofQwhichhavewiasacentralblock.ThenbycompactnessQmusthavealimitpointhavingwiasacentralblock.Sinceeachwifor0ikisdistinct,weobtaink+1distinctlimitpoints.Sincekwasarbitrary,Qhasinnitelymanylimitpointsandthusrk(Q)>2. Wenoteherethatthereare01classesofranktwowithnon-computableelements[ 4 ].Hencewehavethefollowing. Corollary1. Thereisa01classofranktwowhichisnotdegree-isomorphictoanysubshiftofranktwo.Nextwewilldiscussthedecidabilityofranktwosubshifts. Theorem2.4. (a) ForanyTuringdegreed,thereisasubshiftQ2NofranktwosuchthatTQhasdegreed. (b) Foranyc.e.degreed,thereisa01subshiftQ2NofranktwosuchthatTQhasdegreed. Proof. LetAbeanysetofnaturalnumbersofdegreedandletQcontainlimitpoints0!and1_0!,alongwithisolatedpoints0n1_0!,forn>0and1_0n1_0!forn2A.ThenQ 31

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isaranktwosubshiftandwehave1_0n12TQifandonlyifn2A.ThusATTQ.Fortheotherdirection,justobservethatnostringwithmorethantwo1'sbelongstoTQandeverystringwithoneorno1'sbelongstoTQ.For(b),justtakeac.e.setBofdegreed,letA=N)]TJ /F7 11.955 Tf 12.68 0 Td[(BandconstructQasin(a). OurnexttheoremwillshowthatwecannotachievetheresultoftheTheorem 2.4 forsubshiftsof2Z.However,inthenextsectionwewillpresentasimilarresultforrankthreesubshiftsof2Z. Theorem2.5. LetQ2Zbeasubshiftofranktwo.ThenQcanbedecomposedasanitenumberofperiodicelementstogetherwithanitenumberofelementsoftypeu)]TJ /F14 7.97 Tf 6.59 0 Td[(!vw!andtheirorbitsundertheshiftmap. Proof. SinceQisofranktwo,D(Q)hasnitelymanypointsandtheyareperiodicbyLemma 5 .Next,considertheisolatedpoints.ByTheorem 2.3 theyareeventuallyperiodic.ThuseveryelementofQiseventuallyperiodicandhenceiscomputable.Itshouldbenotedthatalthougheverylimitpointisperiodic,theremayalsobeperiodicpointswhichareisolated.NeverthelesswewillarguethatQhasonlynitelymanyperiodicpoints.AssumeforacontradictionthatthesetAofperiodicpointsofQisinnite.Then,bycompactness,thereexistalimitpointofA,sayX.Moreover,Xmustbeperiodic,sinceitbelongstotheniteshiftD(Q);letX=v1.OurgoalistoconstructasequencefZk:k2!gofelementsofA,whichconvergestoanonperiodiclimit.WebeginwithaX=v1=limku1k,forasequencefuk:k2!gofdistinctuk,alldifferentfromv.Wemayassumewithoutlossofgeneralitythatjukjkjvj.Sincelimku1k=v1,wemayassumefurthermorethatvkuk. 32

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Nowforeachk,letnkkbethelargestsuchthatvnkukandletuk=vnkwkwherevwk.NowletZk=nkjvj(u1k)2AsothatZk=(wkvnk)1=(wkvnk))]TJ /F14 7.97 Tf 6.59 0 Td[(!.wk_(vnkwk)!.Itisimportanttorecallherethatwhenwewrite,ingeneralZ=U)]TJ /F3 11.955 Tf 7.08 -4.34 Td[(.v_W,thismeansthatZ(i)=v(i)fori
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WeclaimthatSisnite.Otherwise,bytheabove,therearexeduandwsuchthatfv2S:u)]TJ /F14 7.97 Tf 6.59 0 Td[(!.v_w!2Qgisinnite.Wemayassumethatjujandjwjisminimalhere,thatis,thereisnoproperprexu0ofusuchthatu!0=u!andsimilarlyforw.Thentherewillbeaninnitesequencef(vk):k2Ngsuchthat,foreachk,uisnotaprexofvk,wisnotasufxofvk,jvkj0.Butthisimpliesthatwisasufxofvk,whichisacontradiction.ItnowfollowsthatthisinnitesetfYi:i2NghasalimitpointZ2Qoftheformu)]TJ /F14 7.97 Tf 6.58 0 Td[(!.XwithuX.ButthismeansthatZisanonperiodiclimitpointofQ,acontradiction.HenceSmustbenite.ItfollowsthatQconsistsofnitelymanyperiodicpointstogetherwiththeshiftsofanitesetC=fXi=u)]TJ /F14 7.97 Tf 6.58 0 Td[(!i.vi_w!i:i
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Lemma10. LetQ2Nbeasubshift,letX=v!beaperiodicelementofQwithperiodkand,foreachi
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isnotaninitialsegmentofanyofthelimitpoints.Choosevtsothatuhasthelongestagreementwithv!tofthelimitpointsandchooseiandnsothatvnt_(vti)(1)]TJ /F7 11.955 Tf 12.1 0 Td[(vt(i))u.Thenvnt(vti)(1)]TJ /F7 11.955 Tf 12.68 0 Td[(vt(i))disagreeswitheverylimitpointsothatQi,nisniteandhenceu2A(vt). Corollary3. ForanysubshiftQ2Nofranktwo,thereissomenitensuchthatn(Q)isdecidable. Proof. ByTheorem 2.3 ,D(Q)isanitesetofeventuallyperiodicpoints.ForeachX2D(Q),m(X)isperiodicforsomem;justletnbethemaximummoverX2D(Q).ThenbyLemma 6 ,D(n(Q))=n(D(Q))andthuscontainsonlyperiodicpoints,sothatTheorem 2.6 applies. Inthenextsection,wewillshowthatarankthreesubshiftof2Ncanhavememberswhicharenoteventuallyperiodicandindeednotevencomputable.ThereisanotherinterestingconsequenceofLemma 10 .Recallthatbycompactnessnosubshiftin2Ncanhaverank!. Theorem2.7. Thereisnosubshiftofrank!+1in2N(orin2Z). Proof. LetQ2NbeasubshiftandsupposebywayofcontradictionthatQhasrank!+1.ThenD!+1(Q)=;andD!(Q)isnite.ThenthereisaperiodicelementXofrank!byProposition 2.2 .LetXhaveperiodkandletthesetsQi,nbedenedasinLemma 10 .SinceXhasrank!,thereissomensuchthatforalliandallmn,Qi,mhasrankn.Butthisimpliesthatrk(X)r+1,whichisthedesiredcontradiction. 36

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NextassumebywayofcontradictionthatQ2Zisasubshiftofrank!+1.LetSbethesetofnitestringsusuchthatnoelementofQhasuasfactor,andletQ0=QNS)]TJ /F1 11.955 Tf -456.92 -22.51 Td[(andQ1=QNS.ThenbyLemma 7 ,Q=QZS,0[Q]=Q0,and1[Q]=Q1.Itfollowsfrom[thenoteafter]Lemma 1 thatQ0andQ1haverank!.Thenbythepreviousparagraph,Q0andQ1musthaveniterank,sothat,byLemma 2 ,Q0Q1hasniterank.ButQQ0Q1,andhenceQmustalsohaveniterank. Itfollowsfromtheproofthatthereisnosubshiftofrank+1in2Nor2Z,foranylimitordinal. Example2.8. Thereisa01subshiftQ2Zofrank!+2.ForanyZ22Z,letm(Z)betheminimumsuchthat10m1isafactorofZandletn(Z)bethenumberof1'sinZ.NowdeneQsothatZ2Qiffn(Z)m(Z)orifn(Z)1.ThuseverymemberofQhasonlynitelymany1's.Itiseasytoseethat,foranyk,Dk(Q)=fZ:n(Z)m(Z))]TJ /F7 11.955 Tf 12.11 0 Td[(kg.ThismeansthatZ2Qisisolatedifn(Z)=m(Z)andingeneralhasrankkifn(Z)=m(Z))]TJ /F7 11.955 Tf 12.32 0 Td[(k.ItfollowsthatD!(Q)=fZ:n(Z)1gandthatD!+1(Q)=f01g.Thissameexamplealsoworksin2N. 2.3SubshiftsofRankThreeandFourInthissection,weexaminethecomplexityofsubshiftsofrankthreeandfourandthecomplexityoftheirelements.WealsogiveageneralresultshowingthatforanyclosedsetP2Nofarbitraryrank+1,thereisasubshiftQofrank+3andacomputableinjectionfromPintoQ.WebeginwiththecounterpartoftheTheorem 2.4 forsubshiftsof2Z. Theorem2.9. (a) ForanyTuringdegreed,thereisasubshiftQ2ZofrankthreesuchthatTQhasdegreed (b) Foranyc.e.degreed,thereisa01subshiftQ2ZofrankthreesuchthatTQhasdegreed. 37

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Proof. (a)LetANbeaninnitesetofdegreed.Q2Zisdenedasfollows.Foranyw2f0,1g,w2TQifandonlyif (i) whasatmosttwo1's. (ii) Fori>j,ifw(i)=w(j)=1,theni)]TJ /F7 11.955 Tf 11.95 0 Td[(j2A.ThenQhasthefollowingelements. (0) Foranyi2Zandn2A,QhastheisolatedelementXi,nwithXi,n(j)=1()j=i_j=i+n. (1) Foranyi,QhastheelementXiofrankone,whereXi(j)=1()j=i. (2) Qhasauniqueelementofranktwo,namely01.(b)Ifdisac.e.degree,letAbeaco-c.e.setofdegreedintheargumentabove.ThenTQisa01setandhenceQisa01class. WeobservethatforthesubshiftQconstructedinTheorem 2.9 correspondingtothesetA,QwillbedecidableifandonlyifthesetAiscomputable,sinceTQhasthesameTuringdegreeasA. Proposition2.4. Foranyincreasingsequencen0
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(ii) allwords0n1_0ni1suchthatnni; (iii) allwords10ni1_0nj1suchthatj6=i+1; (iv) allwords10ni1_0ni+1+1.Items(iii)and(iv)ensurethateachblock10ni1mustbefollowedbytheblock0ni+11andcanonlybeprecededbytheblock10ni)]TJ /F19 5.978 Tf 5.75 0 Td[(1(ifi>1).(b)ThesubshiftQ2Zwillhavethefollowingelements: (0) z(X)foreachz2Zisanisolatedelement (1) fYi:Yi(k)=1()k=i,i2Zgisthesetofrankoneelements. (2) 01istheuniquerank2elementofQ.HereweseethatQavoidsthesetSasaboveexceptfor1_0n01. Theorem2.10. ForanyTuringdegreed,thereisarankthreesubshiftQ2N(2Z)whichcontainsamemberofTuringdegreedandsuchthatTQhasTuringdegreed. Proof. LetA=fa0,a1,...gbeanyinnitesetofdegreedandletni=a0+a1++aiforeachi.NowapplyProposition 2.4 Thisresultcanbeimprovedasfollows. Theorem2.11. ForanycountablesetD=fdi:i2NgofTuringdegrees(including0),thereisarankthreesubshiftQ2N(or2Z)suchthatD(Q)=D. Proof. Letp0=2,p1=3,...enumeratetheprimenumbersinincreasingorder.Foreachi,chooseasetAi=fni,00andeachnmi,k,theisolatedelement0n1_0mi,k+11_0mi,k+21....TheargumentforQ2ZfollowsasintheproofofProposition 2.4 Foreffectivelyclosedsubshifts,theresultisquitedifferent. 39

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Theorem2.12. IfQ2Nisa01subshiftofrankthree,thenallofitsmembersarecomputable. Proof. D(Q)isasubshiftofranktwoandhenceallofitsmembersareeventuallyperiodicandthereforecomputable.TheremainingmembersofQareisolatedandthereforecomputablebyTheorem3.12of[ 4 ]. Resultsfor01subshiftsofrankfourorhigherwithnoncomputableelementscanbeobtainedfromthefollowinggeneralresult. Theorem2.13. LetbeacountableordinalandletP2Nbeanyclosedset(01class)ofCantor-Bendixsonrank+1.Thenthereexistsaclosedsubshift(01class)Q2N(andalsoQ12Z)ofrank+3,acomputableinjectionfromPintoQ,andfurthermore,acountable-to-onedegree-preservingmappingfromQnD+1(Q)ontoP.Furthermore,D+1(Q)isthesetofeventuallyperiodicpointsofQ. Proof. WewilluniformizethemethodofProposition 2.4 .Fortherststep,transformPintoaclosedsetinwhicheveryelementhastheform0n01_0n11...,bymappinganarbitraryX22Nto(X)=(0X(0)1_0X(0)+X(1)+11...).Notethat,foreachi,ni+1ni+1ni+2andhenceini2i+1.Thismapisa(truth-table)computableinjectionandthereforepreservescomputabilityandrank.NowdeneQ2Ntocontainthefollowingelements. (0) ForeachX2Pandeachi2N,Qcontainsalloftheshiftsi(X).ForanysuchY=i(X),therankofYinQwillberkQ(Y)=maxfrkP(X):Y=j(X)forsomej2NandsomeX2Pg (1) Foreachn,Qcontainstheelement0n1_0!,whichwillhaverank+1inQ. (2) Qcontainstheelement0!ofrank+2.AcrucialobservationisthatforanyY=0n01_0n112Q,fw:w_Y2Pgisnite.Thisisbecauseifw=0m,thenm
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ItisclearthatQisclosedundertheshiftoperator,butweneedtocheckthatQistopologicallyclosed.SupposethereforethatYisalimitofpointofQ.IfYcontainsatmostone1,thenY2Q,sowithoutlossofgeneralityYhasaprexoftype0n01_0n11.NowletY=limiYiwhereeachYi2Q.ItfollowsfromtheinitialassumptionsaboutPthatYhasinnitelymany1's,sayY=0n01_0n11_0n21...forsomeinnitesequencefni:i2Ng;thisisbecauseanyelementofQwhichextends0n01_0n11canhaveatmostn1+2zeroesbeforehavinga1.Nowwemayassumethat0n01...0ni1YiforeachiandthatYi=ei(Xi)forsomeXi2Pandei2N.Thatis,foreachi,Xibeginswithaninitialsegmentoftheform0m01_0m11...0mk1_0n01...0ni1.Sincem0
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P,thenmk=niforsomeimkandthereforej0n01...0ni1_0mj2+4++2mk+2+m+1=m2k+3mk+m+3Notethatthefunctionf,denedonnitestringsbyf(w)=m2k+3mk+m+3wheneverwisoftheform0n01...0ni1_0mandf(w)=jwjotherwise,iscomputable.Now,w2Rifandonlyifforallvoflengthf(w),ifwisafactorofv,thenv=2TP.HenceRisac.e.set,asdesired.ItremainstoexaminetherankoftheelementsofQ.FirstnotethatsincePQ,rkP(X)rkQ(X)forallX2PandfurthermorerkQ(e(X))rkQ(X)byLemma 6 ,sothatrkQ(e(X))rkP(X)forallX2P,e2N.ItfollowsthatrkQ(0n1_0!)+1andhencerkQ(0!)+2.Fortheotherdirection,werstobservethatforanyY=0n01_0n12Q,fX2P:e(X)=Yforsomeegisnite.ThisisbecauseeachsuchXhastheformw_Yforsomewandbythecrucialobservationabovethereareonlynitelymanypossiblesuchw.ThusfrkP(X):Y=e(X)forsomeegisniteandhasamaximum.Wenowshowbyinductionon=maxfrkP(X):X2P&e(X)=YforsomeegthatrkQ(Y)=.Supposerstthat=0,thatiswheneverY=e(X),thenXisisolatedinP.SupposebywayofcontradictionthatYisnotisolatedinQ.ThenonceagainwehaveY=limiei(Xi),forsomeei2N,sothatasabove(withoutlossofgenerality),therewillbeaxedeandalimitX=limiXi2PsuchthatY=e(X).ButXisisolatedinP,whichgivesthedesiredcontradiction.Nowsupposetheclaimholdsforallordinals
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above(withoutlossofgenerality)thatX=limiXiwhichwouldimplythatrkP(X)+1.Thisisthedesiredcontradiction.WemaynowconcludethateveryelementoftheformY=e(X)forX2P,e2NhasrankrkQ(Y).Itfollowsthateach0n1_0!hasrank+1andthatrkQ(0!)+2.ThecomputableembeddingofPintoQwasgivenbythemappingwhere(X)=(0X(0)1_0X(0)+X(1)+11...)whichthenledustoconstructPQ.CertainlyD+1(Q)=f0!,0n1_0!:n2NgisthesetofeventuallyperiodicpointsofQ.Finally,foranyY=e(X)inQwheree2N,X2P,Yhasthesametruth-tabledegreeasX2P.ToobtainQ12Z,let1(X)=0)]TJ /F14 7.97 Tf 6.59 0 Td[(!.1_(X).Theninpart(0),wehaveagaintheshiftsk(1(X)),nowwithk2Z.Inpart(1),wehaveallZwithexactlyoneoccurenceof1,andpart(2),wehave01. NotethatifPisperfect(thatis,ifD(P)=P),thenQisalsoperfectbutwecanstillsaythatmapsthesetofnon-eventually-periodicpointsofQontoP.WeobtainthefollowingcorollarytotheproofofTheorem 2.13 Corollary4. Forany01classP2N,thereisa01subshiftQ2NsuchthatD(Q)=D(P)[f0g.NotethatinfactthemappingintheproofofTheorem 2.13 istruth-tablecomputable,sothatCorollary 4 alsoappliestotruth-tabledegrees.Thenextcorollarymaynowbeobtainedfromstandardresultson01classes. Corollary5. ForanydegreebsuchthateitherbT00or00TbT0",thereisa01subshiftQ2Nofrankfour(andalsoa01subshiftQ12Z)suchthat1.EveryelementofQ(Q1)ofrank2or3iseventuallyperiodic.2.EveryelementofQ(Q1)ofrank1hasTuringdegreeb. Proof. ThiscorollaryfollowsfromTheorem 2.13 togetherwithCorollary3.2of[ 7 ]andTheorem2.1of[ 12 ].Givendegreebaspostulated,thereexists(bythecitedresults)a 43

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01classPofranktwoandarealXofdegreebsuchthatD(P)=fXgandfurthermoreeveryotherelementofYofPiseventually0,thatis,Y=u_0!forsomeu.ApplyingTheorem 2.13 ,weobtaina01subshiftQofrankfoursuchthateveryelementofrank2iseventuallyperiodic.IfYhasrankoneinQ,thenY=e(X)forsomeeandhenceYhasthesameTuringdegreeasX. Notethatthosetheoremsdonotapplytotruth-tabledegreesandlikewisethiscorollarydoesnotholdfortruth-tabledegrees.Wehavenowseenthatmany03realscanbelongto01subshiftsofrankfour.Ontheotherhand,itiseasytoseethateverymemberofa01subshiftofrankfouris03. Proposition2.5. Forany01subshiftQofrankfour,everyelementofQis03. Proof. LetQbea01subshiftofrankfour.ThenD2(Q)hasranktwo,sothatitsmembersarealleventuallyperiodic.ThusanyelementofranktwoorthreeinQiscomputable.TheisolatedmembersofQarealsocomputable.Finally,supposethatXhasrankoneinQ.ThenXisisolatedinthe03classD(Q)andistherefore03. Ontheotherhand,anarbitrary01classofrankfourmaycontainmemberswhicharenot06andevena01classofrankthreemaycontainmemberswhicharenot04. 44

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CHAPTER3SPMCOMPLEXITYOFINFINITESTRINGS 3.1PreliminariesInthissection,weintroducethenotionofstrictprocessmachinedescriptionofinnitestrings.WedeneSPMcomplexityofinnitestringsandweprovethatitdoesnotdependonthechoiceofauniversalmachine.WealsoshowthatcomputablesequenceshavetheSPMcomplexity0. Denition3.1. LetMbeastrictprocessmachineandletY22N.AnM-descriptionofYisabinarysequenceX22NwiththepropertythatforallvYthereexistsuXsuchthatvM(u). Remark3.1. IfwehaveadescriptionX22NofY22NviaastrictprocessmachineMthenwehaveamonotoneincreasingsequenceofnaturalnumbersftngn2Nwheretn'saregivenbyM(Xn)=Ytn.Intuitively,thefasterftngdivergestoinnitythebetteristhedescription.Thenextdenitionsareaimedtocapturethisintuition. Denition3.2. LetXbeanM-descriptionofYforsomestrictprocessmachineM.Letc>0.Wesaythat1.XdescribesYwitharatec,if(91n)(cnjM(Xn)j).2.XdescribesYstronglywitharatec,ifcnjM(Xn)j)foralmostalln.3.ciscalledarateofdescription. Denition3.3. LetXbeanM-descriptionofYforsomestrictprocessmachineM.Letc1.Wesaythat1.Xc-compressesY,orYisc-compressiblebyX,ifXdescribesYwitharatec2.Xstronglyc-compressesY,orYisc-compressiblebyX,ifXdescribesYwitharatec.3.ciscalledarateofcompression. 45

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Remark3.2. NotethatifXdescribesYwitharatec,thenforall0
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isaMn-descriptionofYandSMn(Y)=liminfkk juj+nkjvj=1 njvj.Thus,bytakingnaslargeasdesiredwecandeneastrictprocessmachineMsothatSM(Y)isassmallaswelike. Proposition3.1. ForanycomputablesequenceY22Nandany">0,thereexistsastrictprocessmachineMsuchthatSM(Y)<". Proof. LetY22Nbecomputableand">0.Takec2Nsothat1 c<.WedeneMbyM(v)=Y(cjvj)foranyv2f0,1g.ThisdenesastrictprocessmachinesinceYiscomputable.Now,anyX22NisanM-descriptionofYsuchthatcnjM(Xn)j.Therefore,Yisstronglyc-compressiblebyM.Hence,SM(Y)<". WerecallthatasaresultofTheorem 1.8 wexedauniversalstrictprocessmachineanddenoteditbyS. Denition3.6. LetY22N.1.Theuniversalstrictmachine(SPM)complexityofYisdenedbysS(Y)anddenotedbys(Y).2.ThestronguniversalmachinecomplexityofYisdenedbySS(Y)anddenotedbyS(Y).Thenotionsofcomplexitydenedonnitestringsrequiresachoiceofauniversalmachine.However,manyresultswewanttoachievedonotdependonxingaparticularmachinesincethecomplexitiesgivenbyanytwouniversalmachineareequaluptoaconstant.Ontheotherhand,duetotheasymptoticnatureofSPMcomplexity,anytwouniversalmachinegiverisetothesamecomplexity.Next,weprovethis. Remark3.4. ForanyY22NthereexistsanSdescriptionXofY.Inparticular,uNYisanNdescriptionofYwhereNistheidentitymachineanduNisthecodingstring. Proposition3.2. LetMbeastrictprocessmachine.Thens(Y)sM(Y)foranyY22N.Inparticular,s(Y)1foranyY22N.Moreover,thesameresultholdsforstrongcompressibility.Inotherwords,S(Y)SM(Y)andS(Y)1foranyY22N. 47

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Proof. LetMandY22Nbegiven.IfthereisnoM-descriptionofYthensM(Y)=SM(Y)=1.Thus,thereisnothingtoprove.AssumeXisanM-descriptionofYandtherateofdescriptionisc.LetuMbethecodingstringforM,thatisforallv,S(uMv)=M(v).LetX0=uM_X.Weclaimthefollowing:Claim:Forallc0suchthat00.Then,bytheProposition 3.1 thereisastrictprocessmachineMsuchthatSM(Y)<".Thus,byProposition 3.2 ,S(Y)SM(X)<".Since">0wasarbitrary,theresultfollows. Weshowthattheconverseisnottrue. Proposition3.3. ForanyX22N,thereisaTuringequivalentY22N,thatisY=TXsuchthatS(Y)=0. 48

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Proof. WedeneastrictprocessmachineM:f0,1g!f0,1gasfollows.Foranyu2f0,1g,deneM(u)=vwherejvj=2jujandv(k)=8>><>>:u(n)ifk=2n)]TJ /F3 11.955 Tf 11.96 0 Td[(1forsomen0otherwiseNow,foranyX22N,XisadescriptionofsomeY22NsuchthatX=TY.Moreover,foranyc0,cnjM(Xn)j=2nforalmostalln.Thus,SM(Y)=inff1 c:c0g=0.Hence,S(Y)=0. 3.2EffectiveHausdorffDimensionandSPMComplexityInthissection,wewillshowthatSPMcomplexityofaninnitesequenceisequaltoitseffectiveHausdorffdimension.WerecallthatTheorem 1.11 statesthatforallinnitesequenceYdim(Y)=liminfnK(Yn) n=liminfnC(Yn) n Theorem3.5. ForanyY22N,dim(Y)s(Y). Proof. LetY22Nbec-compressiblebytheuniversalstrictprocessmachineS.We'llshowthatdim(Y)1 c+"forany">0.Onceweshowthis,theresultimmediatelyfollowsfromthedenitionofs(Y).Bythedenitionofc-compressibilitythereexistsadescriptionofY,sayX22NsothatcnjS(Xn)jforinnitelymanyn.Alsothereisacodingconstant,eSsothatC(un)n+eSforallnwhereun=S(Xn).Letnkbeasequencesatisfyingforallk1.cnkjS(Xnk)j=junkj2.eS jS(Xnk)j<".ThesecondconditionissatisedbythefactthatjS(Xn)jisanondecreasingsequencesinceSisaprocessmachineanditisunboundedsinceXdescribesY. 49

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Forallk2N,wehaveC(unk) junkj=C(unk) jS(Xnk)jnk+eS jS(Xnk)j1 c+"Hence,bytheTheorem 1.11 dim(Y)=liminfC(Yn) nliminfC(unk) junkj1 c+" Theorem3.6. LetY22N.Then,s(Y)dim(Y). Proof. TheplanistobuildastrictprocessmachinewhichhasafastdescriptionofagivenY22N.Hereafastdescriptionwouldmeanasgoodasprexfreecomplexity.Firstwewillconstructaprocessmachinethatworkswiththisideaoffastness.Then,wewillconvertthismachineintoastrictprocessmachine.BythisstepwewilllosesomeefciencyindescribingY.However,itturnsoutthatthislossisaffordable.Constructionoftheprocessmachine:RecallthatUistheuniversalprex-freemachinewexedforthedenitionofprex-freeKolmogorovcomplexity,thatisKU(v)=K(v).Oninputv,Mdoesthefollowing:1.MsimulatesUandtriestondadecompositionofv=v0v1...vkwhereU(vi)#=uiforeachi=0,...,k.2.Ifthereisadecompositionasdescribedintherststep,Moutputsukifu0u1...uk.Otherwise,Mleavesvundened.Weneedtomakefewcommentstoclarifytheconstruction:OntheinputvwerunUonallprexesofvandwaitforananswer.IfUhaltsonaninputv0v,thenitmustbeunique.Thatmeansifv=u_w0suchthatU(u)#thenu=v0.ThisistruebecauseUisaprex-freemachine.So,wecandecomposevasv0_u.Now,Mperformsthesametaskonu.Itcontinuesthiswaytillitgetsafull 50

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decomposition.Wenotethatv=2dom(M)ifv=v0v1...vkwwhereU(vi)#fori=0,...,kandU(u)"foranyprexuofw.Thesecondstepistheretomakesurethatwehaveaprocessmachine.IfM(v)#andM(w)#wherevwthenv=v0...vkandw=v0...vkvk+1...vtwhereU(vi)#=uifori=0,...,t.Furthermore,u0u1ukut.Hence,M(v)=ukut=M(w).StricticationofM:ThestrictversionNwillworkalmostlikeM.Onaninputvitwilltrytodecomposevasbefore.Butthistimeitwillalsochecktheextensionsofv.Thatis,NwillsimulateUfortheinputsvuforallu.ThisisnotaproblembecausetherecanbeonlyoneprexorextensionofvwhereUhalts.Ifitdoesn'thalt,thenweleaveN(v)undened.WenotethatinthiscasethereisnoXwithvXsuchthatXdescribessomeYviaM.IfUhaltsforanextension,weknowthatMisdenedonthisextension.Ifitndsadecompositionofvthenwerepeattheprocessasbefore.IfvhasaperfectdecompositionthenwegetN(v)=M(v).Ifv=v0...vnwhereU(vi)#fori=0,...,n)]TJ /F3 11.955 Tf 12.99 0 Td[(1andU(vn)",thenwehadM(v)"before.Now,wewillsearchforU(vnu)forapossibleanswer.Ifthereisnoanswerthenthereisnoproblem,N(v)willbeundenedwhichisconsistentwithM.Ifwegetananswer,thatmeansthereisapossibleextension,wewilldeneN(v)tobeM(v0...vn)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=U(vn)]TJ /F5 7.97 Tf 6.59 0 Td[(1).Now,N(v)=M(v)wheneverM(v)#.And,ifM(w)"butM(v)#wherewv,thenN(w)=M(u)whereuisthelongestprexofwsuchthatM(u)#.WealsoobservethatforallX,Y22N,XdescribesYviaMifandonlyifXdescribesYviaN.Thisendstheconstruction.WewillusethestrictprocessmachineNdenedaboveandndadescriptionXofYsothats(Y)dim(Y)+"forall">0.Let">0begiven.WewillutilizeTheorem 1.11 asfollows.Sincedim(Y)=liminfnK(Yn) nthereexistasequenceftkgsatisfying 51

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1.dim(Y)=limkK(Ytk) tk2.k)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xi=0ti0wasarbitrarys(Y)dim(Y)follows. 52

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ByTheorem 3.5 andTheorem 3.6 wehavethefollowing: Theorem3.7. LetY22N.Then,s(Y)=dim(Y). 3.3SPMComplexityofSubsetsofCantorSpaceInthissection,weextendthenotionofSPMcomplexitytothesubsetsofCantorspace.WerelatetheSPMcomplexitytoeffectiveHausdorffdimensionbyshowingthattheyareequivalent.WewanttoxastrictprocessmachineMandmakesomeobservations.ThedomainofMisatree,sayT.ThecasethatTisniteisnotinteresting.So,weassumethatitisinnite.Therefore,theremustbeaninnitepaththroughT,sayX.ItisnaturaltoaskwhetherXisaM-descriptionforsomeY22N.Theanswerisnegativeasthenextexampleillustrates. Example3.8. LetT=f0n:n2Ng[f0n1k:n,k2NgbethedomainofM.M(0n1k)=8>>>>>>>>>><>>>>>>>>>>:ifn=k=01ifk=0,n6=0(10)nifn6=0,k6=0(01)kifn=0k6=0Wehave[T]=f0!g[f0n1!:n2Ngand1!isadescriptionfor(01)!.Ontheotherhand,0!isnotadescriptionforanyY22N.Intheaboveexample,wenotethat(10)!doesn'thaveadescription.However,foralln2N,thereisau2dom(M),namely0n1,sothatM(u)=(10)n.Motivatedbythisobservationwemakethefollowingdenition. Denition3.7. LetMbeastrictprocessmachine.1.ThedescriptionspaceofM,denotedbyD(M),isthesetfY:forsomeX,XisanM-descriptionofYg 53

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2.IfthesetfuY:M(v)=uforsomev2dom(M)gisinnite,thenwesaythatYisrepresentedintherangeofM.3.ThesetofsequencesrepresentedintherangeofMiscalledtherepresentationspaceofManddenotedbyR(M).NotethatthedescriptionspaceofMisasubsetoftherepresentationspaceofM.Theconverseisnottrueingeneral.InExample 3.8 ,therepresentationspaceofMisf(01)!,(10)!gwhereasthedescriptionspaceofMisf(01)!g.ThisexamplealsoshowsthatthedomainofMhasnodeadendsisnotsufcienttohaveD(M)=R(M).LetMbeastrictprocessmachinesatisfyingthefollowing:(8v)(9w)(v2dom(M)!(vw^M(v)M(w)))ThisconditionsaysthateverystringinthedomainofMisaninitialsegmentofadescription.However,thisconditionisnotsufcienttoproveD(M)=R(M)asseeninthefollowingexample. Example3.9. LetMdenedasfollows:M(0n1m)=8>>>>>><>>>>>>:ifm=n=00ifm=0,n6=00n1m)]TJ /F5 7.97 Tf 6.58 0 Td[(1ifm6=0Informally,Mdeletesa`1'ifthereisone,outputs`0'otherwise.Now,weseethattheconditionstatedaboveissatised.However,0!isrepresentedbutnotdescribed. Denition3.8. LetMbeastrictprocessmachineandFbeapartialfunctiondenedon2NsuchthatX2dom(F)()fjM(Xn)j:n2Ngisunbounded,F(X)=YwhereXistheM-descriptionofYThen,wesaythatMisarepresentationofF. 54

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Proposition3.4. LetMbeastrictprocessmachinethatrepresentsFsuchthatdom(M)=Ext(T)forsometreeT.Then,D(M)=R(M)ifdom(F)=[T]. Proof. WenotethatFisacontinuousfunctionwhereitsdomainisaclosedsetandthereforeitsrange,D(M)isclosedaswell.Now,letY2R(M).Then,thereexistsanincreasingsequencefnkg,asequencesofnitestringsfukg,andsequencesofinnitestringsfXkg[T],fYkgD(M)sothatforallk,Ynk=M(uk)whereukXkandF(Xk)=Yk.WealsoseethatYisalimitpointoffYkg.Thus,Y2D(M). Denition3.9. LetM:f0,1g!f0,1gbeastrictprocessmachinerepresentingF:2N!2N.LetP,Q2NsuchthatQ=fF(X):X2Pgandletc>0.Then,wesaythat1.PisanM-descriptionofQ.2.PdescribesQwitharatecif(8X2P)(91n)(cnjM(Xn)j).3.PdescribesQstronglywitharatecifforallX2Pandforalmostalln,cnjM(Xn)j.Asbefore,weusethetermcompressibilitywhenc1.And,weintroduceanotionofcomplexityforQ2Naswedidforbinarysequences. Denition3.10. LetMbeastrictprocessmachineandQ2N.1.TheM-complexityofQwithrespecttoP,denotedbysM(P,Q),isinff1 c:PdescribesQwitharatecg.2.McomplexityofQisinffsM(P,Q):PisanM-descriptionofQg,wedenoteitbysM(Q).Similarly,wedeneastrongversionasfollows: Denition3.11. LetMbeastrictprocessmachineandQ2N.1.ThestrongM-complexityofQwithrespecttoP,denotedbySM(P,Q),isinff1 c:PdescribesQstronglywitharatecg 55

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2.ThestrongMcomplexityofQ,denotedbySM(Q),isinffSM(P,Q):PdescribesQstronglygLetusremindthatwexedauniversalprocessmachineandcalleditS.UsingSwedenestrictprocessmachinecomplexityofasubsetof2N. Denition3.12. LetP2N.1.ThestrictprocessmachinecomplexityofPissS(P),denotedbys(P).2.ThestrongstrictprocessmachinecomplexityofPisSS(P),denotedbyS(P). Example3.10. 1.LetMbetheidentitymachine.Then,forallP2N,sM(P)=SM(P)=1.2.Letuk=(10k2)]TJ /F5 7.97 Tf 6.58 0 Td[(1)!andvk=(10k)]TJ /F5 7.97 Tf 6.58 0 Td[(1)!.LetQ=f(vk)!:k1gandP=f(uk)!:k1g.WedeneastrictprocessmachineMwhereforeachk1,(uk)!describes(vk)!.Foreachk1,letM(u)=ifu(uk)2,andM(u)=(vk)nif(uk)nu(uk)n+1.Largestpossibledescriptionrateoccurswhentheinputuis(uk)nforsomen.Inthatcase,1 kjuj=jM(u)j.So,sM(Xk,Yk)=k.Alsonotethat,PistheuniqueM-descriptionofQ.Ontheotherhand,sM(P,Q)=1sincethereisnoc>0sothatforallk,(uk)!describes(vk)!witharatec.ItfollowsthatsM(Q)=1. Theorem3.11. LetQbeaclosedsubsetof2Nsuchthat(Q)>0.Then,s(Q)=S(Q)=0. Proof. Assumethat(Q)>0andletb>0bearbitrary.Furthermore,assumeforacontradictionthatthereexistsP2NsuchthatPc-compressesQviaastrictprocessmachineMwherec>1.TakeN2NsothatcN>b+N.SincePdescribesQ,eachY2QisdescribedbysomeX2P,andforeachX2PthereisNX2NsuchthatjM(XNX)jcNX.WithoutlossofgeneralitywemayassumethatNX>N.For,bythedenitionofc-compressibilityjM(Xn)j>cnforinnitelymanynorforalmostallninthecaseofstrongcompressibility.Ineithercase,wecanalwaysndlargeenoughNX.Now,weseethatf[M(XNX)]:X2PgisanopencoverofQ.SinceQisclosedthereexistsanitesubcoversayf[M(XiNXi)]: 56

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ikg.Fortheconvenienceofthenotationletui=XNXiandvi=M(ui)forik.Moreover,wemayassumethatfui:ikgisaprex-freeset.For,ifuiujthenM(ui)M(uj).Therefore,[M(uj)][M(ui)]andujcanberemovedfromthesubcover.Now,(Q)Xik2jvijXik2)]TJ /F11 7.97 Tf 6.59 0 Td[(cjuijXik2)]TJ /F5 7.97 Tf 6.59 0 Td[((b+juij)Thelastinequalityfollowsbecausejuij>Nforeachik.Sincefui:ikgisaprex-freeset,weget(Q)2)]TJ /F11 7.97 Tf 6.59 0 Td[(bXik2juij<2)]TJ /F11 7.97 Tf 6.58 0 Td[(bSinceb>0wasarbitraryweconcludethat(Q)=0.Hence,theresultfollows. Lemma11. LetQ2NandMbeastrictprocessmachinerepresentingF:2N!2NsuchthatQIm(F).Then,forallY2QsM(Y)sM(Q). Proof. Wewillunfoldthedenitions.IfsM(Q)=1,thenthereisnothingtoprove.AssumethatsM(Q)<1.LetY2Qand">0bearbitrary.ThenthereexistsP2N,anM-descriptionofQsuchthatsM(Q)+" 2sM(P,Q).Also,thereisarateofdescriptioncwheresM(P,Q)+" 21 c.Since,PisadescriptionofQ,thereisX2PdescribingYwiththeratec.Now,bythedenitionsofsM(X,Y)andsM(Y),wehave1 csM(X,Y)sM(Y).Thus,sM(Q)+"sM(P,Q)+" 21 csM(Y).SinceY2Qand"werearbitrary,theresultfollows. Lemma12. LetQ2NandMbeastrictprocessmachinerepresentingF:2N!2NsuchthatQIm(F).ThensM(Q)supfsM(Y):Y2Qg. Proof. Let">0bearbitrary.IfsupfsM(Y):Y2Qg=1thenthereisnothingtoprove.LetsupfsM(Y):Y2Qg=<1.ForeachY2QwecanndadescriptionXYofYsuchthatsM(XY,Y)sM(Y)+" 2+" 2.BythedenitionofsM(XY,Y)foreachY2QthereexistscY>0suchthat1 cYsM(XY,Y)+" 2.So,foreachY2Qwehave1 cY+". 57

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Now,letc=inffcY:Y2Qg.Notethatc1 +">0sinceisnite.WealsoobservethatforeachY,XYdescribesYwiththeratec.Thus,P"=fXY:Y2QgdescribesQwitharatec.BythedenitionsofsM(Q)andsM(P",Q)wegetsM(Q)sM(P",Q)1 c+".Since">0wasarbitrary,theresultfollows. Now,thefollowingtheoremisimmediatefromthelemmas. Theorem3.12. LetQ2NandMbeastrictprocessmachinerepresentingF:2N!2NsuchthatQIm(F).ThensM(Q)=supfsM(Y):Y2Qg. Corollary8. LetP2N.Then,dim(P)=s(P). Proof. ThisfollowsimmediatelyfromTheorem 3.7 ,Theorem 3.12 andProposition 1.4 whichstatesdim(P)=supfdim(X):X2Pg. AlthoughthefollowingpropositioncanbeproveddirectlyfromthedenitionsweomittheirelementaryproofssincetheyimmediatelyfollowthemainresultsTheorem 3.12 andCorollary 8 Proposition3.5. ForanyP2N,thereexistsasubsetQof2Nwiths(Q)=0suchthatQiscomputablehomeomorphictoP. 58

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CHAPTER4COMPRESSIBILITYOFCOUNTABLESUBSETSOFCANTORSPACENextwewillturnourattentiontocompressibilityofcountableclosedsubsetsof2N.First,weneedtodevelopsomemachinery. 4.1PreliminariesInthispreliminarysection,weinvestigatethestructureofcountablesubsetsofCantorspace.Weintroducethenotionsofisolationnodeandn-decidability.Weprovideapartitionofacountableclosedsetusingmaximalsetofisolatedpoints. Denition4.1. LetP2NandX2P.Astringu2f0,1giscalledtheisolationnodeofX,anddenotedbyiso(X),ifuXistheshorteststringsatisfying[u]\Drk(X)(P)=fXg Notation4.1. LetA2N.Wedenotethesetfiso(X):X2Agbyiso(A). Proposition4.1. IftheCantor-BendixsonrankofP2Nisnite,thenforallX2P,iso(X)exists. Proof. LetP2Nsuchthatrk(P)isniteandletX2P.Assumeforacontradictionthatiso(X)doesnotexist.Thatis,forallkthereexistsYk2Drk(X)(P)nfXgsuchthatXkYk.ItfollowsthatXisalimitpointoffYk:k2Ng.ThisisacontradictionsinceXisisolatedinDrk(X)(P).Hence,iso(X)exists. Denition4.2. LetP2NsuchthatforallX2P,iso(X)exists.Theextendedlength-lexicographicalorderonP,denotedby<``,denedasfollows:X<``Y()iso(X)<`iso(Y) Notation4.2. LetP2Nsuchthatrk(P)=n.ThesetofrankkelementsofP,thatisDk(P)nDk+1(P)willbedenotedbyPkforeach0k
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Denition4.3. LetPbeasubsetof2NofniterankandletX2Pk.AsubsetAofPk)]TJ /F5 7.97 Tf 6.58 0 Td[(1iscalledasetofisolatedpointsofX,ifXistheuniquelimitpointofA.Moreover,if([iso(X)]\Pk)]TJ /F5 7.97 Tf 6.59 0 Td[(1)A,thenAiscalledamaximalsetofisolatedpointsofX. Proposition4.2. LetPbeasubsetof2NofniterankandletX2Pk.IfAisasetofisolatedpointsofX,thenAn[iso(X)]isaniteset.Moreover,anymaximalsetofisolatedpointsofXcanbewrittenasaunionofanitesetand[iso(X)]\Pk)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Proof. LetA=B[CwhereC=A\[iso(X)].IfBisaninniteset,thenitmusthavealimitpoint,sayX0sothatiso(X)X0.So,X06=XisalimitpointofA.Thatisacontradiction.Hence,Bisaniteset.Now,ifAismaximal,thenC=[iso(X)]\Pk)]TJ /F5 7.97 Tf 6.58 0 Td[(1. Proposition4.3. LetPbeasubsetof2Nofniterank.AndletMbeamaximalsetofisolatedpointsofXforsomenon-isolatedpointXofP.Thenforeachn>0,thesetMndenedbyfY2M:n=leastk(X(k+1)6=Y(k+1))gisaniteset. Proof. AssumethatMnisinniteforsomen2N.Thatmeans,thereisalimitpointX0ofMnsuchthatXn_(1)]TJ /F7 11.955 Tf 12.32 0 Td[(X(n))X0.Thus,MnMhasalimitpointotherthanX.ThisisacontradictionsinceMisasetofisolatedpointsofX.Hence,Mnisanitesetforeachn>0. Denition4.4. LetMbeamaximalsetofisolatedpointsofXwhereXisalimitpointofaniterankP2N.LetMn=fY2M:n=leastk(X(k+1)6=Y(k+1))gforeachn2N.Wedene`(P(X))andav(P(X))asfollows:1.`(P(X))=limcardMn2.av(P(X))=limPkncardMn n+1 Remark4.3. `(P(X))andav(P(X))arewell-dened.Toseethis,letMandNbetwomaximalsetofisolatedpointsofXwhereXisalimitpointofaniterankP2N.Let 60

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Mn=fY2M:n=leastk(X(k+1)6=Y(k+1))gandNn=fY2N:n=leastk(X(k+1)6=Y(k+1))gforeachn>0.Then,Proposition 4.3 togetherwithProposition 4.2 impliesthatlimcardMn=limcardNnandlimPkncardMn n+1=limPkncardNn n+1whenevertheindicatedlimitsexist.Hence,`(P(X))andav(P(X))areindependentofthechoiceofthemaximalsetofisolatedpoints.Notethatif`(P(X))existsthen`(P(X))=av(P(X)).Ontheotherhand,theexistenceofav(P(X))doesnotimplytheexistenceof`(P(X)). Example4.4. LetP=f0!g[f0n1!:n2NgandQ=f0!g[f02n1!:n2Ng.Then,0!istheuniquelimitpointofbothPandQ.Sinceiso(0!)=forbothPandQ,P0=f0n1!:n2NgandQ0=f02n1!:n2Ngaretheuniquemaximalsetofisolatedpointsof0!ofPandQ,respectively.So,`(P(0!))=limcardf0n1!g=1andav(P(0!))=1.Ontheotherhand,av(Q(0!))=1 2,whereas`(Q(0!))doesnotexist. Proposition4.4. LetP2Nsuchthatrk(P)=n+1.Then,thereisapartitionPn[[X2PnP0P(X)ofPwhereforeachX,P(X)isamaximalsetofisolatedpointsofX. Proof. Wewilldeneafunction`:PnPn!PnP0satisfying1.rk(`(X))=rk(X)+12.foranyY2PnP0,Ywillbetheuniquelimitpointof`)]TJ /F5 7.97 Tf 6.59 0 Td[(1(Y).Foreachk
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Next,letY2Pk,sayY=Yiandconsider`)]TJ /F5 7.97 Tf 6.59 0 Td[(1(Y)Pk)]TJ /F5 7.97 Tf 6.58 0 Td[(1.LetAbethesetfX2Pk)]TJ /F5 7.97 Tf 6.59 0 Td[(1:iso(Y)Xg.WenotethatYisalimitpointofAandAisasubsetof`)]TJ /F5 7.97 Tf 6.59 0 Td[(1(Y).So,Yisalimitpointof`)]TJ /F5 7.97 Tf 6.59 0 Td[(1(Y).Moreover,ifY02Pk,sayY0=Yj,isanotherlimitpointof`)]TJ /F5 7.97 Tf 6.58 0 Td[(1(Y),thenthesetC=[Y0jiso(Y)j]\`)]TJ /F5 7.97 Tf 6.58 0 Td[(1(Y)isinnite.Inparticular,itisnon-empty.Now,letX2C.Since`(X)=Yi,wehaveNiNjandsinceX2CwehaveNjjiso(Yi)j.Thus,Yijiso(Yi)j=Xjiso(Yi)j=Yjiso(Yi)j.Therefore,Y=Y0bythedenitionofiso(Y).Hence,Yistheuniquelimitpointof`)]TJ /F5 7.97 Tf 6.59 0 Td[(1(Y).Now,Pk)]TJ /F5 7.97 Tf 6.59 0 Td[(1=SX2PkP(X)isapartitionofPk)]TJ /F5 7.97 Tf 6.58 0 Td[(1whereP(X)=`)]TJ /F5 7.97 Tf 6.59 0 Td[(1(X)for0whereki=cardfX2P:wXandrk(X)=ig.Wesaythatfisthen-decidabilityfunctionofP. Example4.5. LetAbeacomputableset.Then,P=f0!g[f0n1!:n2Agis1-decidable.Its1-decidabilityfunctionisgivenasfollows:f(u)=8>>>>>><>>>>>>:1ifu=0nforsomen2N1ifu=0n1kforsomen2Aandk10otherwise Remark4.6. 1.IfPisn-decidable,thenPisk-decidableforallk
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4.2RankTwoSubsetsof2NInthissectionwewillconsidersubsetsof2Nofranktwo.Weinvestigatethenecessaryandsufcientconditionsforaranktwoclosedsettocompressanotherone.Wealsointroducethenotionofstronghomeomorphism. Remark4.8. IfaranktwoclosedsetP2Nhasauniquelimitpoint,sayX,thenPnfXgistheuniquemaximalsetofisolatedpointsofX. Proposition4.5. LetPandQbecountablesubsetsof2NofranktwowithXandYbeingtheiruniquelimitpointsrespectively.Assumethatav(P(X))andav(Q(Y))exist.FurthermoreassumethatthereexistsastrictprocessmachineMsothatPstronglyc-compressesQviaMforsomec1.Then,av(Q(Y))cav(P(X)). Proof. AssumethatPandQaregivenasinthehypothesis.LetFbethecontinuousfunctionwhereMisitsrepresentation.Claim:F(X)=YAssumeforacontradictionthatF(X)=Y0forsomeisolatedpointofQ.Then,thereexistsN2Nsuchthat[iso(Y0)]M(XN).ThisimpliesthatforallZ2PsuchthatXNZ,F(Z)=Y0sinceMisaprocessmachine.Ontheotherhand,thesetPn[XN]isnitebytheProposition 4.3 .ThisisacontradictionsincePdescribesQ.Wedeneafunctiongasg(n)=jM(Xn)j)]TJ /F3 11.955 Tf 21.23 0 Td[(1.Wenotethatg(n)isnon-decreasingandunbounded.SinceXstronglyc-compressesY,thereexistsN2NsothatcnjM(Xn)j=g(n)+1forallnN.LetPn(X)andQg(n)(Y)denotethesetsPn[Xn]andQn[Yg(n)]respectively.Informally,thesearethesetofisolatedpointsbranchingoffbeforethelevelnandg(n),respectively.Now,weobservethatcardQg(n)(Y)cardPn(X).Toseethis,letX02PnPn(X).Then,XnX0.So,M(Xn)Y0whereY0=F(X0).Then,bythedenitionofg(n)wegetYg(n)Y(g(n)+1)=M(Xn)Y0.So,Y0=2Qg(n)(Y).Now,card(Qg(n)(Y))card(Pn(X))implies 63

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card(Qg(n)(Y)) g(n)g(n) ncard(Pn(X)) nSo,forallnN)card(Qg(n)(Y)) g(n)cn)]TJ /F3 11.955 Tf 11.96 0 Td[(1 ncard(Pn(X)) nHencebytakinglimitasn!1wegetav(Q(Y))cav(P(X)). Corollary9. LetPandQbecountablesubsetsof2NofranktwowithXandYbeingtheiruniquelimitpointsrespectively.Assumethatav(P(X))andav(Q(Y))exist.IfPd-compressesQforall1d0bearbitrary.Bytakingd=maxf1,c)]TJ /F14 7.97 Tf 13.97 4.71 Td[(" qgandapplyingProposition 4.5 wegetq(c)]TJ /F14 7.97 Tf 13.96 4.71 Td[(" q)qdp.Thus,qcp+".Since"wasarbitrary,theresultfollows. NotethattheconverseoftheProposition 4.5 isnottrue.Thatis,wecanhaveav(Q(Y))cav(P(X))butthereisnostrictprocessmachineMsothatPc-compressesQ.Thefollowingexampleillustratesthisfact. Example4.9. LetP=f0!g[f0n10!:n2Ng[f0n110!:n2NgandQ=f0!g[f0n1!:n2Ng[f10!,110!g.Wehave2av(P(0!))av(Q(0!)).BythewayofcontradictionassumethatMisastrictprocessmachinesothatP2-compressesQ.Letg(n)=jM(0n)j)]TJ /F3 11.955 Tf 17.95 0 Td[(1.Then,asintheproofofProposition 4.5 ,thereexistsN2Nsuchthatforalln>Nwehave2n)]TJ /F3 11.955 Tf 12.32 0 Td[(1g(n)andcard(Qg(n)(0!))card(Pn(0!)).Ontheotherhand,2n+1card(Qg(n)(0!))andcardPn(0!)=2n.Thatisacontradiction. Lemma13. LetPbea1-decidable,ranktwosubsetof2Nwithfbeingits1-decidabilityfunction.Thenthefollowingfunctionsarecomputable:1.(branchingnode)h1:f0,1g!f0,1gdenedash1(u)(iff(u)=1orf(u)=0ukif1f(u)<1 64

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wherekistheleastsuchthatf(u(k+1))<12.(totalnumberofisolatedpoints)h2:f0,1g!N[fegdenedash2(u)=(Pv2Af(v)iff(u)=1eotherwisewherev2A()(v(jvj)]TJ /F3 11.955 Tf 17.93 0 Td[(1)<`uandf(v)<1andf(vjvj)]TJ /F3 11.955 Tf 17.93 0 Td[(1)=1)3.(extension)h3:(f0,1gN)!f0,1g[fegsuchthath3(u,n)=(eiff(u)6=1orjujnvotherwisewhereuvandjvj=nandf(v)=1.4.(isolationlevel)h4:N!f0,1g[fegdenedbyh4(k)=(uifk>0ek=0whereuislength-lexicographicallylastnitestringwithf(u)=1suchthath2(u)
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Proof. Forh1weneedtocheckf(u1),...,f(ujuj).Thisiscomputableandh1iswelldened.Forh2,wenotethatthesetAdenedaboveiscomputable.Computabilityofh3isgivenbythefollowingargument.Oninput(u,n)wecomputef(u)andjuj.Iff(u)6=1orjujnweoutpute.Otherwise,wendallextensionsofuoflengthnandcomputetheirfvalue.Bythedenitionof1-decidabilityofP,thereexistsauniqueextensionofuinPwithlengthn.Thatis,thereexistsauniquevoflengthnextendinguwithf(v)=1.Weleth3(u,n)=v.Forh4,itissufcienttonotethatfh2(u)0. 66

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5.Theorderonthebranchfunctionh5aimstoordertheisolatedpointsbranchingoffathesamelevel.Onaninputnodeuwithf(u)=1,h5backtracksuandcountsthenumberofisolatedpointstotheleftoftheisolatedpointthatextendsu.Forexample,h5(0n10k)=1andh5(0n110k)=2foranyn,k2N.6.Thefunctionh6aimstoenumeratetheisolatedpoints.Ifanodeuisaninitialsegmentofasingleisolatedpoint,thatisiff(u)=1,theindexnumberfunctionrstcomputeswhereuisbranchedoffusingbranchingnodefunctionandthencountsthenumberofisolatedpointsbelowthatlevelusingtheh2function.Finally,itaddsh5(u)togettheindexofu.Forexample,h6(0n10k)=2n+1andh6(0n110k)=2n+2foranyn,k2N.7.Thefunctionh7onagiveninputucomputesthelargestksuchthatuk2TP.Forexample,h7(0n110m11k)=n+m+3foranyn,m,k2N. Theorem4.11. LetPandQbe1-decidable,ranktwosubsetsof2NwithXandYbeingtheiruniquelimitpoints,respectively.Then,thereexistsabijectionF:P!QrepresentedbyacomputablestrictprocessmachineMsatisfying1.PisanM-descriptionofQ2.F(X)=Y3.forallc1,eachZ2PnfXgstronglyc-compressesF(Z).Hence,SM(F(Z))=0.Moreover,ifav(P(X))andav(Q(Y))exist,thenlimjM(Xn)j n=av(P(X)) av(Q(Y)). Proof. Letf:f0,1g!N[f1gandg:f0,1g!N[f1gbethe1-decidabilityfunctionsofPandQ,respectively.ConstructionofM:Lethiandh0ibethecomputablefunctionsprovidedbyLemma 13 forfandg,respectively.Oninputu2f0,1gwedeneMasfollows:Case1:f(u)=0Inthiscase,weleaveM(u)undened.Case2:f(u)=1LetM(u)=h04(h2(u)+1).Case3:1
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Inthiscase,Xnuiso(X0)forsomen2NandforsomeisolatedpointX0.So,werstndn,thatisn=jh1(u)janddeneM(u)=M(un).Sincef(un)=1,Case2applies.Case4:f(u)=1andf(ujuj)]TJ /F3 11.955 Tf 17.93 0 Td[(1)>1Inthiscase,u=iso(X0)forsomeisolatedpointX02P.WendtheindexofuinP,sayk=h6(u),andmatchifwithitscounterpartinQ.Thatis,wedeneM(u)=vwhereh06(v)=k.Thisiscomputable,sinceh06isone-to-oneonthesetofstringswithg(v)=1andthereforecanbefoundbyanitesearchinincreasinglexicographicalorder.Case5:f(u)=1andf(ujuj)]TJ /F3 11.955 Tf 17.93 0 Td[(1)=1Inthiscase,uisanextensionofanisolationnode,sayu0=iso(X0)uforsomeisolatedpointX02P.MwillmaputosomeextensionofM(u0).Letnbetheleastsuchthatf(un)=1.So,unisanisolationnodeandthereforeCase4applies.Thus,wecancomputeM(un)=v0.WedeneM(u)=vwherev=h03(v0,jv0j+2juj).ThiscompletestheconstructionofM.Claim:Misacomputablestrictprocessmachine.FromtheconstructionitisclearthatMiscomputable.Letu,u02TPsuchthatu0uandM(u)=v.Iff(u)=1,thenf(u0)=1aswell.Bythedenitionofh2,h2(u0)h2(u)sinceu0uandthereforebythedenitionofh04andbytheuniquenessoflimitpointofQ,M(u0)M(u)follows.Ifuissothat1
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So,weconcludethatMisastrictprocessmachine.Claim:Q=fF(Z):Z2Pg,thatisPisanM-descriptionofQ.LetY0beanisolatedelementofQ.Letv0=iso(Y0).Then,thereexistsu0=iso(X0)forsomex02Psuchthath6(X)=h6(Y).BythedenitionofM,M(u0)=v0.Letu=X(ju0j+n)forsomen2NthenM(u)=vwherev=h03(v0,jv0j+2juj).Now,bythedenitionofh03,v0vandjvj=jv0j+2jujandg(v)=1.Sincev0=iso(Y0),v0vandg(v)=1wegetv=Y0jv0j+2juj.Thus,foranyk2N,wemaytaken=jiso(X0)j+k.Then,M(X0n)Y0k.Thatis,X0describesY0.Moreover,X0stronglyc-compressesY0.Toseethis,letc1.Forn>jiso(X0)jwehavejM(X0n)j=jiso(Y0)j+2n.So,thereexistsN2Nsothatforalln>N,cnjiso(Y0)j+2n.Thus,fromthedenitionofstrongMcomplexityofY0itfollowsthatSM(X0,Y0)=0.Hence,SM(Y0)=0forallisolatedY02Q.Next,wewillshowthatXdescribesyviaM,andanalyzetherateofdescription.First,wenotethatPnfXgisamaximalsetofisolatedpointsofXandh2(Xn)iscountingthenumberofisolatedpointsbelowleveln.So,h2(Xn)isnon-decreasingasafunctionofnandincreasesunboundedlyasn!1.Wealsonotethatav(P(X))=limh2(Xn) nwheneverthislimitexists.AsimilarobservationholdsforQaswell.ConsidervYandletk=h02(v).Then,thereexistsuXsuchthath2(u)=k0wherek0>h02(Y(jvj+t))>kforsomet>0.BythedenitionofM,M(u)=h04(k0+1),say,M(u)=v0.Then,bythedenitionofh04,v0islength-lexicographicallylaststringwithg(v0)=1suchthath02(v0)
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h02(Y(m+1))h2(Xn)+1.Then,m nh02(Ym) m
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Case5:f(u)=0andf(u(h7(u)))=1Inthiscase,uisanextensionofanisolationnodeofanisolatedpointZsuchthatiso(Z)uZ.WerstcomputewhereubranchesoffofZusingthefunctionh7.So,weletu0=u(h7(u)).Wecanwriteu=u0_wforsomew2f0,1gsuchthatjwj=k+1.Bythedenitionofu0,f(u0w(0))=0andf(u0_(1)]TJ /F7 11.955 Tf 13.03 0 Td[(w(0))=1.So,M(u0)andM(u0_(1)]TJ /F7 11.955 Tf 13.15 0 Td[(w(0)))aredenedeitherbyCase3orCase4.LetM(u0)=v0andM(u0_(1)]TJ /F7 11.955 Tf 13.29 0 Td[(w(0)))=v0_w1forsomew12f0,1gsuchthatjw1j=t+1.Wenotethatt>0bythedenitionofMinCase4.Now,wedeneM(u)=v0w(0)...w(t)]TJ /F3 11.955 Tf 11.95 0 Td[(1)_(1)]TJ /F7 11.955 Tf 11.96 0 Td[(w(t))(w(1))2(w(2))4...(w(k))2k.Case6:f(u)=0andf(u(h7(u)))>1Letu0=u(h7(u)).ThenbyCase2,M(u0)=Ymforsomem.Wendlength-lexicographicallyrstvwithh06(v)=1suchthatYmvandY(m+1)v.ThisiscomputablesincebyCase1weknowthatthereexistsY0suchthatYmY0butY(m+1)Y0.Moreover,wenotethatv=iso(Y0)forsomeY0.So,forsomei2f0,1g,g(v_i)=0.WedeneM(u)=v_i_u(0)_u(1)2...u(k)2kwherejuj=k+1.Weneedtocheckthattheextendedversionisstillastrictprocessmachine.Letu0=w_u1andu=w_u1_u2whereju1j=k+1andju2j=s+1andw=uh7(u)=u0h7(u0).Firstconsiderthecasewhenf(w)=1.Now,M(w_(1)]TJ /F7 11.955 Tf 12.3 0 Td[(u1(0))=w0forsomew0andM(u0)=w0(jw0j)]TJ /F3 11.955 Tf 17.93 0 Td[(1)_(1)]TJ /F7 11.955 Tf 11.96 0 Td[(w0(jw0j)]TJ /F3 11.955 Tf 17.93 0 Td[(1)_(u1(1))2_(u1(2))4_..._(u1(k))2kM(u0)_(u2(0))2k+1_..._(u2(s))2k+s+1=M(u)Nextconsiderthecasewhenf(w)>1.Letju0j=t+1.Then,M(u0)=v_i_u0(0)_..._(u0(t))2tM(u0)_(u2(0))2k+1_..._(u2(s))2k+s+1=M(u) 71

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wherev=iso(Y0)forsomeisolatedpointandi2f0,1gsuchthatv_(1)]TJ /F7 11.955 Tf 11.96 0 Td[(i)Y0.Thus,Misastrictprocessmachinedenedonf0,1g.TheanalysisconcerningtheelementsofPandQisthesame.Weneedtoanalyzethecomplexityoftheotherelements.Claim:ForallZ22NnP,SM(F(Z))=0.LetY0=F(Z).SinceFisone-to-one,wehaveSM(Y0)=SM(Z,Y0).Weneedtoshowthatforanyc>0,ZdescribesY0stronglywitharatec.SinceZ=2P,thereexistsanodeu2TPsuchthatuZandZ(juj+1)=2TP.Inotherwords,h7(Z(juj+1))=juj.Letc>0begiven.LetN>2juj+c+4.Considern>N.Iff(u)=1thenbythedenitionM(Zn)=w_(u0(1))2...(u0(k))2kwhereZn=u_u0suchthatju0j=k+1andM(u)w.So,wehavejM(Zn)j2k+1wherek=n)-222(juj)]TJ /F3 11.955 Tf 17.93 0 Td[(1.Ontheotherhand,iff(u)>1thenbythedenitionjM(Zn)j>2n+1.Ineithercase,jM(Zn)j>2k+1>2k2>2kc>(juj+k+1)c=ncSo,weconcludethatZdescribesY0stronglywitharatec.Sincec>0wasarbitrary,theclaimfollows. Corollary10. LetPandQbe1-decidablesubsetsof2NofranktwowithXandYbeingtheiruniquelimitpoints.If`(P(X))and`(Q(Y))exist,saypandqrespectively,andp>q,thenthereisastrictprocessmachineMsuchthatPstronglyc-compressesQviaMforallcsuchthat1c


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Since`(P(X))and`(Q(Y))exist,thereisN2Nsothatforalln>Nthereareexactlyp-manyisolatedpointsbranchingoffatXn,andthereareexactlyq-manyisolatedpointsofQbranchingofatM(Xn),thatiscardfX0:X06=X,XnX0,X(n+1)X0g=pandcardfY0:Y06=Y,M(Xn)Y0,Y(jM(Xn)j+1)Y0g=q.Letcbegivensothatqcqc(N+q))]TJ /F11 7.97 Tf 6.58 0 Td[(M p)]TJ /F11 7.97 Tf 6.59 0 Td[(qc.Now,k>c(N+q))]TJ /F11 7.97 Tf 6.58 0 Td[(M p)]TJ /F11 7.97 Tf 6.59 0 Td[(qcimpliesc(N+q))]TJ /F7 11.955 Tf 11.95 0 Td[(M
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Corollary12. LetPandQbe1-decidable,ranktwosubsetsof2NwithXandYbeingtheiruniquelimitpoint,respectively.Then,PandQarehomeomorphicwherethehomeomorphismcanberepresentedbyastrictprocessmachine.Moreover,ifav(P(X))andav(Q(Y))existandequalthenPisstronglyhomeomorphictoQ. Proof. Bytheorem 4.11 thereexistabijectionFrepresentedbyastrictprocessmachineMsuchthatPdescribesQviaM.Similarly,thereexistsanotherbijectionGrepresentedbyastrictprocessmachineM0suchthatQdescribesPviaM0.WeobservefromtheproofofTheorem 4.11 thatF)]TJ /F5 7.97 Tf 6.59 0 Td[(1=G.Thus,Fisahomeomorphism.Moreover,ifav(P)=av(Q),thenwehavelimjM0(Xn)j n=1whereXistheuniquelimitpointofP.FortheisolatedpointsweneedtomodifytheproofofTheorem 4.11 asfollows.WearegivenastrictprocessmachineMwhereeachisolatedpointc-compressesitsimageunderFforanyc1.WewanttoxapositiveintegerkandrequirethatM(X0n)=nkforallisolatedpointsX02Pandalmostalln.ThiscanbeachievedbyasmallmodicationontheconstructionofMwherewedenedMontheextensionofanisolationnode(Case4).WedenedM(iso(X0)_u)=M(iso(X0))_vY0wherejvj=2juj.Theexponentialextensionofiso(Y0)wasnotnecessary,butsufcienttoachievec-compressionforanyc.Now,wedeneM0(iso(X0)_u)byh03(M0(iso(X0),N)whereN=maxfjM0(iso(X0))j,jujkjg.Thus,foralln>jiso(X0)j kwehavejM0(X0n)j=nk.Now,letk=1togetlimjM0(X0n)j n=1forallisolatedpointsX0ofP.Now,bythesameargumentwegetlimjM0(Zn)j n=1forallZ2Q.Hence,ifav(P)=av(Q),thenPandQarestronglyhomeomorphic. Example4.13. Letfun:n2Ngbeacomputableenumerationoff0,1g.DeneP=f0!g[f0n_1_un_0!:n2Ng.Then,PisstronglyhomeomorphictoQ=f0!g[f0n1!:n2Ng.ThehomeomorphismFprovidedbyCorollary 12 ,itsrepresentationMandtherepresentationNofF)]TJ /F5 7.97 Tf 6.59 0 Td[(1aredenedasfollows: 74

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F(X)=8>><>>:0!ifX=0!0n1!ifX=0n1un0!M(u)=8>>>>>><>>>>>>:0nifu=0n0n1if0nu0n1un0n10kifu=0n1un0kN(u)=8>>>>>><>>>>>>:0nifu=0n0n1unifu=0n10n1un0kifu=0n10k Proposition4.6. LetPbeasubshiftwithauniquelimitpoint.ThenPis1-decidable. Proof. SincePhasauniquelimitpoint,Pisranktwoanditslimitpointiseither0!or1!.Withoutlossofgeneralityassumethat0!istheuniquelimitpoint.WealsonotethatPisdecidablebyTheorem 2.6 .LetMn=fZ22N:0n_1_Z2Pg.Then,byProposition 4.3 Mnisnite.So,thesetA=fiso(Y):Y2Pand0n1Yforsomen
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cardinalityoffv2A:uvorvug.Ontheotherhand,ifu=0n1wforsomenNandw2f0,1g,thensimilartothepreviouscaseweletf(u)=swheresisthecardinalityoffv2B:wvorvwg.SinceTPiscomputableandthesetsAandBarenitethefunctionfiscomputable.Hence,fisthe1-decidabilityfunctionofP. Corollary13. LetPbea1-decidable,ranktwosubsetof2NwiththeuniquelimitpointXsuchthatav(P(X))exists.Then,Pisstronglyhomeomorphictoasubshift. Proof. Letav(P(X))=panddeneQ=f0!g[f0n1k0!:k=1,...pg.ThenQisasubshiftwiththeuniquelimitpoint0!andtherefore1-decidablebyProposition 4.6 .Also,wenotethatav(Q(0!))=p.HencePisstronglyhomeomorphictoQbyCorollary 12 Corollary14. LetPbeasubshiftwithauniquelimitpoint.Then,Pisstronglyhomeo-morphictoQ=f0!g[f0n1k0!:k=1,...pgforsomep2N. Proof. SincePisasubshiftwithauniquelimitpointbyProposition 4.6 Pis1-decidable.Also,`(P(X))existswhereXisthelimitpointofP.Say`(P(X))=panddeneQasinthestatement.WenotethatQis1-decidableand`(Q(0!))=p.HencePisstronglyhomeomorphictoQbyCorollary 12 76

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[44] KlausWeihrauch.Computableanalysis.TextsinTheoreticalComputerScience.AnEATCSSeries.Springer-Verlag,Berlin,2000.Anintroduction. [45] A.K.ZvonkinandL.A.Levin.Thecomplexityofniteobjectsandthebasingoftheconceptsofinformationandrandomnessonthetheoryofalgorithms.UspehiMat.Nauk,25(6(156)):85,1970. 80

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BIOGRAPHICALSKETCH FeritToskawasbornandraisedinIstanbul,Turkey.HeattendedBogaziciUniversityinIstanbulforhisundergraduatestudies,majoringinMathematicsandPhilosophy.In2004,hegraduatedwithhisBAdegree.Thesameyear,hestartedhisgraduatestudiesattheUniversityofFloridaDepartmentofPhilosophy.HethentransferredtotheDepartmentofMathematics.HecompletedhisMasterofSciencedegreeinDecember2008andhisDoctorofPhilosophydegreeinMay2013.Duringhistenureasagraduatestudent,Ferithasalsobeenacommittedrunner.Recordsindicateheranmoremarathonsthanheprovedtheorems.FeritismarriedtoGizemToskaandthecouplehasason,namedDerinMete. 81