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On the Classification of Inverse Limit Spaces of Tent Maps

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PAGE 1

ONTHECLASSIFICATIONOFINVERSELIMITSPACESOFTENTMAPSBySLAGJANAJAKIMOVIKADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOLOFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENTOFTHEREQUIREMENTSFORTHEDEGREEOFDOCTOROFPHILOSOPHYUNIVERSITYOFFLORIDA2005

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Copyright2005bySlagjanaJakimovik

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Idedicatethisworktomyfather,mymotherandmysister.

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ACKNOWLEDGMENTSIwishtoexpressmydeepestgratitudetomyadviser,Dr.JedKeesling,forhisguidance,hisgreatenthusiasmforourworktogetherthroughoutmyentirestudiesattheUniversityofFloridaandforbeingmymentorbothacademicallyandspiritually.MythanksalsogotoDr.LouisBlock,whocontributedgreatlytothiswork,forbeingverysupportiveandawonderfulpersontoworkwith.Thisdissertationisaresultofourjointwork.IalsowishtoexpressmygratitudetoLoisKailhofer.Inthedissertationwebuildsomeofourresultsonthebasisofcertainresultsfromoneofherpapers.Moreover,Iwishtoacknowledgehercontributiontotheproofofoneoftheresultsinthedissertation. iv

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TABLEOFCONTENTS page ACKNOWLEDGMENTS .............................. iv LISTOFFIGURES ................................. vi ABSTRACT ..................................... vii 1INTRODUCTION ............................... 1 2PRELIMINARIES ............................... 6 2.1GeneralDenitionsandResults .................... 6 2.1.1InverseLimitSpacesofMaps .................. 7 2.1.2Itineraries ............................ 8 2.1.3TentMaps ............................ 10 2.2DenitionsandResultsfromKailhofer'sPaper ............ 12 2.2.1ChainingsoftheCore ...................... 14 2.2.2Composantofc ......................... 16 2.2.3Fudged"Maphq;p ....................... 21 3ONTHECLASSIFICATIONOFINVERSELIMITSPACESOFTENTMAPSWITHPERIODICTURNINGPOINT ............... 23 3.1UnravelledComposantofaPointandaNewMetricd ........ 23 3.2HomeomorphismsofXsandtheShiftHomeomorphisms ..... 24 3.3NumberofFixedPointsoffks ..................... 28 3.4ProofoftheMainTheorem ...................... 29 4ISOTOPYTHEOREM ............................. 31 4.1ConstructionofNeighborhoods .................... 31 4.2ProofoftheIsotopyTheorem ..................... 37 5ONINGRAM'SCONJECTURE ....................... 39 5.1Three-StepProgram .......................... 39 5.2ProofofStep2 ............................. 40 5.3ProofofStep3 ............................. 43 5.4Ingram'sConjectureFollowsfromthePseudo-isotopyConjecture .. 49 5.5SucientConditionforthePseudo-isotopyConjecture ....... 49 REFERENCES .................................... 52 BIOGRAPHICALSKETCH ............................. 55 v

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LISTOFFIGURES Figure page 2{1Projectionsofp-wrappingpointsofCcandprojectionsofp+1-wrappingpointsofCc ........................ 17 2{2Projectionsandp-levelsofp-wrappingpointsofCc ........... 18 vi

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AbstractofDissertationPresentedtotheGraduateSchooloftheUniversityofFloridainPartialFulllmentoftheRequirementsfortheDegreeofDoctorofPhilosophyONTHECLASSIFICATIONOFINVERSELIMITSPACESOFTENTMAPSBySlagjanaJakimovikAugust2005Chair:JamesE.KeeslingMajorDepartment:MathematicsLetfsandftbetentmapsontheunitintervalwiths;t2;2].Inthisdissertationwegiveanewproofofthefactthatiftheturningpointsoffsandftareperiodic,thentheinverselimitspacesI;fsandI;ftarehomeomorphicifandonlyifs=t.ThistheoremwasrstprovedbyKailhofer.OurproofsimpliestheproofofKailhofer.Wealsoprovethatforatentmapfswithslopes2p 2;2]andaperiodicturningpoint,foranyhomeomorphismhoftheinverselimitspaceI;fsintoitself,thereexistsanintegerNandthereexistsanintegerksuchthathNandksareisotopic.Withnoassumptionontheorbitoftheturningpointoffs,wemakeaconjecturethatanyhomeomorphismoftheinverselimitspaceI;fsintoitselfispseudo-isotopictoaniterateoftheshifthomeomorphism.Wegiveasucientconditionforourconjecturetohold.Finally,weprovethatIngram'sconjecturefollowsfromourconjecture. vii

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CHAPTER1INTRODUCTIONGivenacontinuousmapfofaone-dimensionalspacetoitself,onemayformaninverselimitspacebyusingfrepeatedlyasthebondingmap.Spacesformedinthiswaycommonlyappearasattractorsindynamicalsystems[ 1 3 7 15 20 35 ].Thismotivatesthestudyofsuchinversesystems.Inthecaseofsolenoids,thereisawell-knowncharacterizationseeAartsandFokkink[ 1 ],andKeesling[ 25 ].Considertheinverselimitspacefortheinversesystemwheretheinversesystemspacesareeachtheinterval[0;1]andthebondingmapsareeachsometentmapfsx=minfsx;s)]TJ/F22 11.955 Tf 11.992 0 Td[(xgforx2[0;1]ands2[1;2].Thisinverselimitspacehasbeenstudiedextensively.Anyunimodalmapwithoutwanderingintervals,restrictiveintervals,orperiodicattractorsisconjugatetoatentmapseedeMeloandvanStrien[ 27 ].Asconjugatemapshavehomeomorphicinverselimitspaces,thefamilyoftentmapsismoreinclusivethanitseemsatrstglance.Thus,itisnaturaltotrytodeterminewhentwosuchinverselimitsarehomeomorphic.Ingram[ 22 ]posedthefollowingquestion.If1
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2 Inthistheorem,Is;fsandIt;ftarethecoresoftheinverselimitspaceslim)]TJ 16.259 5.83 Td[(fI;fsg1i=0andlim)]TJ 16.258 5.83 Td[(fI;ftg1i=0,respectively.Thetheoremimpliesthatiftheinverselimitspaceslim)]TJ 16.258 5.83 Td[(fI;fsg1i=0andlim)]TJ 16.258 5.83 Td[(fI;ftg1i=0arehomeomorphic,thens=tunderthegivenassumptions. Remark1.0.1 Onecanextendthesameresultforthewholeintervals2;2]inthefollowingway.Fors2;p 2],therearetwointervalsJ1andJ2inthecoreIsoffswithpairwisedisjointinteriorssuchthatf2sjJ1andf2sjJ2aretopologicallyconjugatetofs2jIs2.Itfollowsthatfors2;p 2],thecoreoftheinverselimitspacelim)]TJ 16.259 5.83 Td[(fI;fsg1i=0isdeterminedbythecoreoftheinverselimitspacelim)]TJ 16.258 5.83 Td[(fI;fs2g1i=0.Therefore,itisenoughtoconsidertentmapswithslopesinp 2;2].SonjaStimac[ 32 ]generalizesresultsontheclassicationofinverselimitspacesoftentmapswithnitecriticalturningorbit,thatis,tentmapswithperiodicturningpointorstrictlypre-periodicturningpoint.Inthispaper,shedoesnotprovetheperiodiccase,forwhichshereferstoKailhofer[ 23 24 ].Stimacsuggeststhattheapproachshedevelops,withsomechangesindicatedlater,yieldstheproofoftheperiodiccaseaswell.Sheusedmethodsofsymbolicdynamicsandcodingtoprovethefollowingresult.Fortentmapsfsandftwithslopess;t2p 2;2]andsuchthateachofthetentmapshaspre-periodicturningpoint,iflogs logtisnotarationalnumber,thantheinverselimitspaceslim)]TJ 16.258 5.83 Td[(fI;fsg1i=0andlim)]TJ 16.259 5.83 Td[(fI;ftg1i=0arenothomeomorphic.InthedissertationwegiveasimpliedproofofKailhofer'stheorem.OurproofusessomeofKailhofer'sresults[ 23 ]togetherwithsomenewresults.InthesecondsectionofChapter2wegivedenitionsintroducedbyKailhoferandresultswhichweusetoproveTheorem 3.2.5 .Forsomeofherresultswegivenewproofsandforsomeofherresultswegivemodiedproofs,inwhichcaseweindicateso.Fortheproofstotherestofherresultswhichweincludeinthedissertation,wereferthereadertoKailhofer[ 23 ].ThecontentsofthisentiresectionservethepurposeofunderstandingonlytheproofofTheorem 3.2.5 ,whichisakeystepinprovingour

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3 mainresult.Nowhereelseinthedissertationdowerefertodenitionsorresultsfromthissection.Oneoftheresultsprovedinthedissertationisofparticularinterestinitself.Isotopytheorem.Lets2p 2;2]suchthattheturningpointofthetentmapfsisperiodic.LetIs=[f2sc;fsc]bethecoreoffs.LetXs=Is;fsbetheinverselimitofthecore.LethbeanyhomeomorphismofXs.Thenthereexistapositiveintegernandanintegerksuchthathnisisotopictoks,wheresistheshifthomeomorphismonXs.Thisresultwasthemotivationtostartworkingonthisproblem.Withthisinmind,KeeslingsuggestedreadingKailhofer'spaper[ 23 ]andconsideringthefollowingproblem.Fors2p 2;2]suchthatthetentmapfshasperiodiccriticalpoint,givenahomeomorphismhofthecoreXsintoitselfsuchthathleavestheendpointsinvariant,doesthereexistanintegerNsuchthathandtheshifthomeomorphismNsstayataboundeddistanceoneachcomposantofthecore?Thedistancewerefertohereisanewmetricddenedoneachcomposantofthecore,whichispossibletodeneonlyafterestablishingthatanypropersub-continuumofthecoreisanarc.AfterhervisittoUniversityofFloridainthefall2003,KailhofersuggestedtheoutlineoftheproofofTheorem 3.2.5 .ThisresultallowsustoproveaweakenedversionoftheIsotopytheoreminChapter3.Pseudo-isotopytheorem.Lets2p 2;2]suchthattheturningpointofthetentmapfsisperiodic.LetIs=[f2sc;fsc]bethecoreoffs.LetXs=Is;fsbetheinverselimitofthecore.LethbeanyhomeomorphismofXssuchthathleavestheendpointsinvariant.ThenthereexistsanintegerNsuchthathispseudo-isotopictoNs.Inotherwords,foratentmapfswithslopes2p 2;2]withperiodicturningpoint,foranyhomeomorphismhofthecoreintoitselfwhichleavestheendpointsinvariant,thereexistsanintegerNsuchthathandNspermutethecomposantsofXsinthesameway.IfhandNswereisotopic,thenitwouldeasilyfollowthatthe

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4 composantsofXsarepermutedbytheminthesameway.InthesimpliedproofofKailhofer'stheorem,weonlyneedthePseudo-isotopytheorem.Therestfollowsfromthefollowingtworesults.Foratentmapfswithslopes2p 2;2]withperiodicturningpoint,foranyintegerm,thenumberofcomposantsofthecoremappedtothemselvesbymsisequaltothenumberofpointsinIsmappedtothemselvesbyfms.Fortentmapsfsandftwithslopesp 2
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5 composantofthecoreofatentmapwithperiodicturningpoint;inthiscasethedmetricgivesthetopologyofthetail.Wealsogivetheprooftothesecondandthethirdstep,aswellastheproofofIngram'sconjectureundertheassumptionthatthepseudo-isotopyconjectureholds.Inthesecondstepweprovethateachcomposantofthecoremappedtoitselfbyksforsomeintegerk,hasauniquepointmappedtoitselfbyks.Inthethirdstepweprovethatiffsandftaretentmapswithslopes1
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CHAPTER2PRELIMINARIES2.1GeneralDenitionsandResultsWedenoteN=f0;1;2;g,N+=f1;2;g,R=;1andR+=[0;1.LetXbeatopologicalspace.XisanarcifthereexistsahomeomorphismfromXonto[0;1].AmaximalconnectedsubspaceofXiscalledacomponentofX.Acompactconnectedmetricspaceiscalledacontinuum.LetXbeacontinuum.Thecomposantofx2Xistheunionofallpropersub-continuaofXthatcontainx.Composantsaredense-thatis,everypointofacontinuumXisalimitpointofanycomposantofX[ 21 ,Theorem3-44].EverycomposantofacontinuumXistheunionofacountablenumberofpropersubcontinuaofX[ 21 ,Theorem3-45].LetXbeacontinuum.Asub-continuumTofXisanendcontinuuminXifforanytwosub-continuaAandBofXsuchthatTAandTB,itistruethateitherABorBA.Apointx2XisanendpointofXiffxgisanend-continuuminX.Notethatanyhomeomorphismfromonecontinuumtoanothermustsendendpointstoendpoints.Thusthecardinalityofthesetofendpointsofacontinuumisatopologicalinvariant.AcontinuumXisindecomposableifitisnottheunionoftwopropersub-continua.AnindecomposablecontinuumXhasuncountablymanycomposants[ 21 ,Theorem3-46],andthecomposantsofanindecomposablecontinuumXaredisjoint[ 21 ,Theorem3-47].Indecomposablecontinuarstappearedaspathologicalexamples.Buttheyhavebecomejusttheoppositeofwhatsomemathematiciansinthebeginningconsideredasbeingmonstrousthingscreatedbysettheoretictopologistsforsomeevilbutpurelymathematicalpurpose"[ 26 ,page107]. 6

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7 2.1.1InverseLimitSpacesofMaps Denition2.1.1 LetfXi;dig1i=0beacollectionofcompactmetricspaceseachwithametricdiboundedby1,andsuchthatforeachi,fi:Xi+1!Xiisacontinuousmap.TheinverselimitspaceoftheinverselimitsystemfXi;fig1i=0istheset lim)]TJ 16.259 5.83 Td[(fXi;fig1i=0=fx=x0;x1;jx21i=0Xi,fixi+1=xi,i2Ng,withametricdgivenby dx;y=P1i=0dixi;yi 2i.Foreachi2N,idenotestheprojectionmapfrom1i=0XiintoXi.IfXi=Xandfi=fforalli,theinverselimitspaceisdenotedX;f.Usually,thesubscript/superscriptnotationfjiusedintheliteratureoninverselimitsdenotesthemappingofthej-thfactorspacetothei-thfactorspace.But,inthedissertation,asalreadyadoptedinpapersconsideringinverselimitspacesusingsinglebondingmapsfromparameterizedfamilies,weusefkstodenotethek-foldcompositionofthemapfswithitself,wheresistheparameter. Remark2.1.2 Aninverselimitspacelim)]TJ 16.258 5.829 Td[(fXi;fig1i=0isacontinuumifXiisacontinuumforeveryi2N[ 30 ,Theorem2.4]. Denition2.1.3 Themap:X;f!X;fdenedby x0;x1;=fx0;x0;x1;iscalledtheshifthomeomorphism,ortheinducedhomeomorphism. Denition2.1.4 Acontinuousmapf:[a;b]![a;b]iscalledunimodalifthereexistsauniquepointcsuchthatfj[a;c]isincreasingandfj[c;b]isdecreasing.Thepointciscalledaturningpoint,oracriticalpoint. Denition2.1.5 Letf:I!Ibeamapoftheintervalintoitself.Apointxissaidtobeaperiodicpointoffwithperiodk>0iffkx=xandfnx6=xfor0
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8 Inotherwords,!x=k0cl[nkfnx. Denition2.1.7 Letf:I!Ibeamapoftheintervalintoitselfandletxbeaperiodicpointoffwithperiodk.Thesetofpointsx;fx;f2x;;fk)]TJ/F21 7.97 Tf 6.587 0 Td[(1xiscalledtheorbitofxunderf.Thebasinofxisthesetofpointswhose!-limitsetcontainsx.Wesaythatxisanattractingperiodicpointandthatitsorbitisanattractingperiodicorbitifitsbasincontainsanopenset.2.1.2ItinerariesLetf:[a;b]![a;b]beaunimodalmapwithturningpointc. Denition2.1.8 Foreachx2[a;b],theforwarditineraryofx,denotedI x=b0b1b2,isasequenceofR'sandL'ssuchthat 1. bi=Riffix>c, 2. bi=Liffix
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9 1. Ifk=0,thenI fxforallx2[a;b]nfcg,hencebyLemma 2.1.13 itfollowsthatthekneadingsequenceismaximal. Denition2.1.16 Letf:[a;b]![a;b]beaunimodalmap.LetAbeanadmissiblesequence.WesaythatAisdominatedbythekneadingsequenceKf,A<
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10 3. kA0,thenb0b1bi)]TJ/F21 7.97 Tf 6.587 0 Td[(1=ai)]TJ/F21 7.97 Tf 6.587 0 Td[(1ai)]TJ/F21 7.97 Tf 6.587 0 Td[(2a1a0,whereI 0fc=a0a1.DeneBf=fBxjx2I;fg. Remark2.1.20 Supposethatcisperiodicwithperiodn0.Letx2I;f.Ifxi6=cforalli2N,orifxi=cforinnitelymanyi2N,thenxhasexactlyonebackwarditinerary.Ifxi=cfornitelymanyi2N,thenxhastwobackwarditinerariesthatdieratonlyonecoordinate,maxfi2Njxi=cg.2.1.3TentMaps Denition2.1.21 Theone-parameterfamilyofmapsfs:I!Idenedby fsx=minfsx;s)]TJ/F22 11.955 Tf 11.955 0 Td[(xg,forx2Iands2[p 2;2]isknownasthefamilyoftentmaps.Thetentmapfsisunimodalforalls2[p 2;2].

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11 Remark2.1.22 Fors>1,ifx1,themapx!I sxisaninjectivemap.Denoteci=fiscfori=0;1;2;;n0)]TJ/F15 11.955 Tf 12.552 0 Td[(1.DenoteIL=[c2;c];IR=[c;c1]andIs=[c2;c1].Fors2;2]theintervalIs=[c2;c1]isinvariantunderfsandfsislocallyeventuallyontoonIs,thatis,foreverynon-degenerateintervalJIsthereexistsann>0suchthatfnsJ=Is.TheintervalIsisknownasthecoreoffs.Todenotethen-thcoordinateintheinverselimitspace,weuseIninsteadofIsn.ThesetXs=Is;fsiscalledthecoreoftheinverselimitspaceI;fs.Fors2p 2;2],theinverselimitspaceI;fsisequaltotheunionofXsandanopenrayhavingXsasitslimitset.WeknowthatXsisindecomposablefors2p 2;2].InthethirdchapterLemma 3.1.1 ,weprovethatundertheassumptionthatcisperiodic,everypropersub-continuumofXsisanarc-hence,everycomposantisaunionofarcs.ThecomposantCxofx2XsisthesetofallpointsinXswithbackwarditinerarieseventuallyidenticaltoBx-thatis,y2Cxifandonlyifthebackwarditinerariesofxandydierinatmostnitelymanycoordinates.ThisfollowsfromLemma 3.1.1 ,aswell.Fors2p 2;2]suchthattheturningpointcoffshasperiodn0,denoteci=ci;ci)]TJ/F21 7.97 Tf 6.586 0 Td[(1;;c1;c;cn0)]TJ/F21 7.97 Tf 6.587 0 Td[(1;;ci+11fori=0;1;2;;n0)]TJ/F15 11.955 Tf 11.955 0 Td[(1.Thefollowinglemmaisawell-knownresultandappearsinseveralpublications.BargeandMartin[ 8 ]describethebasicconstructionofendpointsintheinverselimitspaceX;fofamapf:X!X. Lemma2.1.23 Fors2p 2;2]suchthattheturningpointcoffshasperiodn0,theendpointsofXsarec;c1;;cn0)]TJ/F21 7.97 Tf 6.587 0 Td[(1.Itiseasytoobservethatthereareatleastcountablymanydistinctinverselimitspaces,sinceforanypositiveintegern0,thecoreoftheinverselimitspaceof

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12 atentmapwithperiodicturningpointofperiodn0hasexactlyn0endpoints,andendpointsaretopologicalinvariants.Infact,BargeandDiamond[ 6 ]provedthatuncountablymanydistinctinverselimitspacescanbeformedwithunimodalmapsasbondingmaps.Intheremainderofthischapter,considers2p 2;2]xedsuchthattheturningpointcoffshasperiodn0.2.2DenitionsandResultsfromKailhofer'sPaperInthissubsectionwegivesomegeneraldenitionsandthedenitionsandresultsfromKailhofer'spaper[ 23 ]necessarytoderivetheproofofTheorem 3.2.5 .Wegiveproofstosomeoftheresults.TheseproofsareeitherdierentfromtheonesgivenbyKailhofer[ 23 ],ortheyaremodiedversionsofherproofs,inwhichcaseweindicateso.Lets2p 2;2]besuchthattheturningpointcisperiodicofperiodn0underthetentmapfs. Denition2.2.1 Letw=w0w1w22Bfs.Dene Aw=fx2Xsjix2Iwiforalli2Ng, Anw=nsAw.Inotherwords,AwisthesetofpointsinXswithbackwarditineraryw.ThefollowinglemmaplaysanimportantroleinunderstandingthenatureofthecomposantsofthecoreXs,soweincludetheproofasgivenbyKailhofer[ 23 ]. Lemma2.2.2 [ 23 ,Lemma4]Letw2Bfs.Thereexist0i6=j
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13 NotethatthesetAwofpointsinXswithbackwarditineraryexactlywistheinverselimitspacefInw;fwn+1g1n=0.SinceeachofthespacesInwisacontinuum,by[ 30 ,Theorem20],AwisacontinuumandijAwisahomeomorphismforalli2N.Toprovethislemma,wehavetoshowthefollowing. 1. Inwisanon-degenerateintervalforalln2N, 2. fsIn+1w=Inwforeverynn0,and 3. fnsInw=[ci;cj]forsome0i;j
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14 2.2.1ChainingsoftheCoreAnopencoverU=fUigni=1ofatopologicalspaceXissaidtobeachainingofthespaceXifUiUj6=;ifandonlyifji)]TJ/F22 11.955 Tf 12.023 0 Td[(jj1.IfU=fUigni=1andV=fVjgmj=1arechainingsofatopologicalspaceX,wesaythatUrenesV,UV,ifforevery1in,thereis1jmsuchthatUiVj. Denition2.2.8 Letpbeapositiveintegerand0jwg.Ifx6=c1,setz=minfw2Pn;mjx
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15 Denition2.2.13 Letp;m;kbepositiveintegerswith0k
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16 2.2.2ComposantofcInthissubsectionwefocusourattentiononthecomposantofc,Cc.WeincludedenitionsofthefollowingconceptsintroducedbyKailhofer[ 23 ]:ap-wrappingpoint,thep-levelofap-wrappingpoint,ap-symmetricarc,ap-pseudo-symmetricarc,ap-gap,gapsofthesametypeandarcswiththesamep-shape. Denition2.2.15 Foreachpositiveintegerp,dene Wp=fx2Ccj9v6=w2Bfssuchthatx2Apn0vApn0wg[fcg.Ifx2Wp,thenxiscalledap-wrappingpoint.Thereisanaturalorderonthesetofallp-wrappingpointswithx
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17 5pCc5p+1Cc c c4 c1 c3 c2 c c4 c3 c2 c1 ccppppbbbbbbc""""""cQQQQQQQQcQQQQcrEEEErrEEEEEEEEErrrrEEEEEEEErrrrEEEEEEEEErrrrEEEEEEEErrrrEEEErrEEEEEEEEErrrrEEEEEEEErrrrEEEEEEEEErrrrEEEEEEEErrrrEEEErrEEEEErrEEEErrEEEEEEEEErrrrrpppp cFigure2{1.Projectionsofp-wrappingpointsofCcandprojectionsofp+1-wrappingpointsofCc Example2.2.17 LetTbethetentmapwithkneadingsequenceRLRRC.Figure 2{1 showsthe5p-projectionsofthep-wrappingpointsofCc,markedby,and5p+1-projectionsofthep+1-wrappingpointsofCc,markedby. Denition2.2.18 Letpbeapositiveinteger.Letxbeap-wrappingpointotherthanc.Letkbetheuniquenon-negativeintegersuchthat pn0+k=maxfnjnx=cg.Denethep-levelofxbyLpx=k.DeneLpc=1.ThesetfLpxjx2Wpnfcggisunbounded. Example2.2.19 LetTbethetentmapwithkneadingsequenceRLRRC.Figure 2{2 showsthe5p-projectionsofthep-wrappingpointsofCc,markedby,andthep-levelsofthecorrespondingp-wrappingpointsofCc.Notethatx2Wp+1ifandonlyifLpxn0. Proposition2.2.20 [ 23 ,Proposition29]Letpbeapositiveinteger.LetwandvbetwodistinctpointsinCcsuchthatwminfLpw;Lpvg.

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18 c1 c c3 c4 c2 -rEEEEEEEErrEEEEEEEEEEEEEEEErrrrEEEEEEEEEEEEErrrrEEEEEEEEEEEEEEEErrrrEEEEEEEEEEEEErrrrEEEEEEEErrEEEEEEEEEEEEEEEErrrrEEEEEEEEEEErr 1 1 3 1 0 2 0 4 0 2 0 6 0 2 0 4 0 2 0 1 8 1 0 2 0 4 0p-level:Figure2{2.Projectionsandp-levelsofp-wrappingpointsofCc Denition2.2.21 Fixapositiveintegerp.LetHbeanarcinthecomposantofcwithIntHWp=fh1;h2;;hn)]TJ/F21 7.97 Tf 6.586 0 Td[(1gand@H=fh0;hng. 1. ThearcHisp-symmetricifpn0h0=pn0hnandLphi=Lphn)]TJ/F23 7.97 Tf 6.587 0 Td[(iforallpositiveintegers0Lpwandsuchthatthereisno

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19 p-wrappingpointbetweenwandvwithp-levelhigherthanLpw.Furthermore,visanendpointofthearcH.Proof.ThefollowingproofisamodiedversionofKailhofer'sproof.LetL=Lpw.LetxLpw,Lpy>Lpwandnop-wrappingpointbetweenxandyhasp-levelhigherthanL.WeclaimthatneitherxnoryisaninteriorpointofH.Assumetheopposite,sayxisaninteriorpointofH.SinceHisp-symmetric,thereisap-wrappingpointx0inIntHsuchthatLpx=Lpx0.Then,pn0x=pn0x0.ByProposition 2.2.20 ,thereexistsap-wrappingpointzbetweenxandx0withLpz>minfLpx;Lpx0g=Lpx.ButLpx>L.Sincenop-wrappingpointbetweenxandyhasp-levelhigherthanL,wehavethateitherz=yorz>y.ThisputsyinIntHandyminfLpy;Lpy0g=Lpy.ButLpy>L.Thiscontradictsthefactthatnop-wrappingpointbetweenxandyhasp-levelhigherthanL.Thus,weprovedthatneitherxnoryisaninteriorpointofH.LetAbethearcwithendpointsxandy.Sincenop-wrappingpointbetweenxandyhasp-levelhigherthanLpw,pn0+LjAisahomeomorphism.Notethatpn0+Lw=candpn0+LAissymmetricaboutc.ConsiderthesmalleroftheintervalsJ1=[pn0+Lx;1)]TJ/F22 11.955 Tf 12.396 0 Td[(pn0+Lx]andJ2=[pn0+Ly;1)]TJ/F22 11.955 Tf 12.395 0 Td[(pn0+Ly],bothofwhicharesymmetricaboutpn0+Lw=c.WithoutlossofgeneralityassumeJ1J2.Sincefspn0+Lx=fs)]TJ/F22 11.955 Tf 12.527 0 Td[(pn0+Lx,wehavethatfLspn0+Lx=fLs)]TJ/F22 11.955 Tf 11.955 0 Td[(pn0+Lx.Ifweassumethatpn0+Ly=1)]TJ/F22 11.955 Tf 11.955 0 Td[(pn0+Lx,then pn0y=fLspn0+Ly=fLs)]TJ/F22 11.955 Tf 11.955 0 Td[(pn0+Lx=fLspn0+Lx=pn0x,hencebyProposition 2.2.20 itfollowsthatthereexistsap-wrappingpointbetweenxandywithp-levelhigherthanL,whichcontradictsourchoiceofxandy.Therefore,pn0+Ly6=1)]TJ/F22 11.955 Tf 12.061 0 Td[(pn0+Lx.Thenxisthepointwhoseexistenceweclaimed.ThearcA0=)]TJ/F21 7.97 Tf 6.586 0 Td[(1pn0+L[pn0+Lx;1)]TJ/F22 11.955 Tf 12.075 0 Td[(pn0+Lx]Aisp-symmetricwithcenterwandxasan

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20 endpoint.SinceHistheunionofallp-symmetricarcswithcenterw,A0H.Sincex=2IntH,xisanendpointofH. Remark2.2.26 Letpbeapositiveinteger.LetHbeap-symmetricarcinCcandletL=LpH.Proposition 2.2.25 impliesthatalltheinteriorpointsinHhavep-levelslessthanL,consequentlypn0+LjHisahomeomorphism. Denition2.2.27 Letpbeapositiveinteger.Thesetp;0partitionsthecomposantofcincountablymanyarcscalledp-gaps.Letpbeapositiveinteger.Foranyp-gapHofCc,c=2pn0IntHandpn0@H=fcg.Theintersectionofanytwop-gapsofCcisatmostonepoint. Lemma2.2.28 Foranypositiveintegerp,ap-gapofCcisp-symmetric.Proof.Fixp2N.LetHbeap-gapand@H=fy;zg.Letx2IntHbeap-wrappingpointwithlargestp-level,sayL.Thenpn0+LjHisahomeomorphismandpn0+Lx=c.SupposeHisnotp-symmetric.Thencisnotthecenterofsymmetryfortheintervalwithendpointspn0+Lyandpn0+Lz.Hencefspn0+Ly6=fspn0+Lz.Itfollowsthatthereisap-wrappingpointw2IntHsuchthatfspn0+Lwisequaltoeitherfspn0+Lyorfspn0+Lz.Aspn0y=pn0z=c,thisimpliesthatpn0w=cwhichcontradictsHbeingap-gap. Denition2.2.29 Letpandqbepositiveintegers.SupposeGisap-gapofthecomposantofcwithGWp=fg0;g1;;gngandHisaq-gapofthecomposantofcwithHWq=fh0;h1;;hmg.ThegapsGandHareofthesametypeifn=mandpn0gi=qn0hiforallpositiveintegers0in. Proposition2.2.30 [ 23 ,Proposition41]Fixp;q2N.LetGbeap-gapofCcandletHbeaq-gapofCc.IfLpH=LqG,thenGandHarethesametype.Proof.LetGWp=fg0;g1;;gngandHWq=fh0;h1;;hmg.SupposeLpH=LqG=N.Thenpn0+NjGandqn0+NjHarehomeomorphisms.Also,theintervalpn0+NGissymmetricaboutc=pn0+NG,andtheintervalqn0+NHissymmetricaboutc=qn0+NH.SinceGandHaregaps,theendpointsof

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21 pn0+NGandtheendpointsofpn0+NHareinf)]TJ/F23 7.97 Tf 6.586 0 Td[(Nsc,butnopointsintheinteriorofpn0+NGareinf)]TJ/F23 7.97 Tf 6.587 0 Td[(Nscandnopointsintheinteriorofqn0+NHareinf)]TJ/F23 7.97 Tf 6.586 0 Td[(Nsc.Hence,ifweassumedthatpn0+NG6=qn0+NH,saypn0+NGqn0+NH,thentheendpointsofpn0+NGareinteriorpointsofqn0+NH,whichcontradictsthefactthatnointeriorpointsofqn0+NHareinf)]TJ/F23 7.97 Tf 6.587 0 Td[(Nsc.Thereforepn0+NG=qn0+NH.Thus,n=mandfNspn0+Ngi=fNsqn0+Nhi,thatis,pn0gi=qn0hi. Denition2.2.31 Fixapositiveintegerp.LetGbeap-gapofCc.Thearcsbetweentwoconsecutivep-wrappingpointsinGarecalledp-legsofG.Therstp-gapinthecomposantofcisdenotedFp. Lemma2.2.32 Fixp2N.LetGbeap-gapofCcandLitsrstp-leg.Then 1. pn0G=pn0L[ 23 ,Lemma44] 2. Lcontainsap)]TJ/F15 11.955 Tf 11.955 0 Td[(1-gap[ 23 ,Lemma46] 3. Gcontainsatleastthreep)]TJ/F15 11.955 Tf 11.956 0 Td[(1-gaps[ 23 ,Corollary50] 4. Therstp)]TJ/F15 11.955 Tf 11.955 0 Td[(1-gapinGisthesametypeasFp)]TJ/F21 7.97 Tf 6.587 0 Td[(1[ 23 ,Prop.47]. Denition2.2.33 Fixapositiveintegerp.Dene'=LpFp. Remark2.2.34 SincethetypeofFpdoesnotdependonp,'doesnotdependonp.SinceFpiscontainedintherstlegofFp+1,thecenterofFpisnotap+1-wrappingpoint-hence'=LpFp
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22 elementofhq;j.Thus,hinducesaone-to-onemaphq;p:q!p,denedasfollows. Denition2.2.35 Fixm;n;p;q2NsuchthathLqn0;nLpn0;m.Ifw2q;jandhcj=cifor0i;j
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CHAPTER3ONTHECLASSIFICATIONOFINVERSELIMITSPACESOFTENTMAPSWITHPERIODICTURNINGPOINT3.1UnravelledComposantofaPointandaNewMetricdInthisentirechapterweworkwithtentmapswithslopesinp 2;2]suchthattheturningpointisperiodicofperiodn0.Thefollowinglemmaisawell-knownresult[ 13 15 ]. Lemma3.1.1 SupposethatAisapropersub-continuumofXs.Thereisanon-negativeintegerksuchthatkjAisahomeomorphism.Inparticular,Aisanarc.Moreover,foranytwopointsx,y2A,BxandByagreeaftertherstkentries.Proof.Ifthereisanonnegativeintegermsuchthatforeachj>m,c=2jA,thenfsjjAisahomeomorphismforeachj>m,andtheconclusionfollowseasily.So,wemayassumethatc2jAforarbitrarilylargeintegerj.Sincecisperiodicunderfs,itfollowsthatforeachnonnegativeintegerj,jAcontainsatleastoneofthepointsc0;c1;;cn0)]TJ/F21 7.97 Tf 6.586 0 Td[(1.Sincefsislocallyeventuallyonto,thereisanonnegativeintegerksuchthatforeachintegerj>k,jAcontainsexactlyoneofthepointsc0;c1;;cn0)]TJ/F21 7.97 Tf 6.587 0 Td[(1.Wecompletetheproofbyshowingthatforeachj>ktheelementoffc0;c1;;cn0)]TJ/F21 7.97 Tf 6.586 0 Td[(1gwhichisinjAisanendpointofjA.Inparticular,foreachj>k,cisnotintheinteriorofjA,sofsjjAisahomeomorphism.Supposej>k.Sincecisperiodic,thereisanintegerm>jsuchthatc12mA.Sincec1isanendpointofIs,itfollowsthatc1isanendpointofmA.SincecisnotintheinteriorofmA,fsjmAisahomeomorphism.Thusc2isanendpointofm)]TJ/F21 7.97 Tf 6.586 0 Td[(1A.Ifm)]TJ/F15 11.955 Tf 12.136 0 Td[(1>j,wemayrepeatthisargumentandconcludethatc3isanendpointofm)]TJ/F21 7.97 Tf 6.586 0 Td[(2A.Byrepeatingtheargumentinductively,itfollowsthattheelementoffc0;c1;;cn0)]TJ/F21 7.97 Tf 6.587 0 Td[(1gwhichisinjAisanendpointofjA. 23

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24 Letx2Xs.ByRemark 2.2.7 ,thereisanaturalorderontheelementsofthecomposantofx.Cxwiththeordertopologywillbecalledtheunravelledcomposantofx. Remark3.1.2 Notethatthemaphq;pisorder-preserving.Wewillputaspecicmetricontheunravelledcomposantwhichisderivedfromtheinverselimitsystem.Letx;ybeinthesamecomposantCXs.ThenthereisanarcACwithendpointsxandy.ByLemma 3.1.1 ,thereisanonnegativeintegerksuchthatkjAisahomeomorphism.Dene dx;y=skjkx)]TJ/F22 11.955 Tf 11.955 0 Td[(kyj.Notethatifmk,thendx;y=smjmx)]TJ/F22 11.955 Tf 11.54 0 Td[(myj.Thus,diswell-denedforeverypairofpointsinthesamecomposantC.WemayconsiderC;deitherasR+orRdependingonwhetherChasanendpointornot.Welet`denotethelengthofanarcunderthemetricd.3.2HomeomorphismsofXsandtheShiftHomeomorphisms Denition3.2.1 SupposeXisanindecomposablecontinuumandh;g:X!Xarehomeomorphisms.Wesaythathandgarepseudo-isotopicifforanyx2X,thereisapropersub-continuumAXsuchthathx2Aandgx2A.Inotherwords,twohomeomorphismsofanindecomposablecontinuumintoitselfarepseudo-isotopiciftheypermutethecomposantsinthesameway. Theorem3.2.2 Leth1;h2:Xs!XsbehomeomorphismswhichmapCctoitself.SupposethatthereexistsM>0suchthatdh1z;h2zMforallz2Cc.Thenh1Cx=h2Cxforallx2Xs.Furthermore,foreveryx2Xs,dh1x;h2xM.Proof.Letx2Xs.Ifx2Cc,thenh1Cc=h2Ccbyassumption.Supposethatx=2Cc.SinceCcisdenseinXs,thereisasequencefxng1n=1inCcwhichconvergestox.Thenhixnconvergestohixfori=1;2.ConsidertheuniquearcsAnCcwithendpointsh1xnandh2xn.Byassumption,

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25 `An=dh1xn;h2xnMforeachn.Letk>0beanintegersuchthatMskc1)]TJ/F22 11.955 Tf 12.677 0 Td[(c2.ThenkAnisapropersubsetof[c2;c1]sincethelengthofkAnislessthanM sk
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26 Remark3.2.4 Onemightbeledtoconjecturethatthenumberofdistincttypesofp-gapsinCcisn0)]TJ/F15 11.955 Tf 12.016 0 Td[(1foranyp2N.However,forssuchthatthekneadingsequenceisRLLRRRLC,thereareatleasteightp-gaps. Theorem3.2.5 Leth:Xs!Xsbeahomeomorphismwhichleavestheendpoints,cifor0in0)]TJ/F15 11.955 Tf 12.749 0 Td[(1,invariant.ThenthereexistsanintegerNandapositivenumberMsuchthatdhx;Nsx
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27 dhx;t)]TJ/F23 7.97 Tf 6.587 0 Td[(qn0sx=dhx;hq;px
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28 3.3NumberofFixedPointsoffksWeadoptthefollowingnotation.Ifkisapositiveinteger,weletFfksdenotethenumberofxedpointsoffksinIs. Lemma3.3.1 Supposep 20.OurrstclaimisthatthereisatmostoneperiodicpointineachcomposantofXs.Supposenot.ThereisacomposantCinXswithatleasttwodistinctperiodicpointsofs,sayx=x0;x1;x2;andy=y0;y1;y2;.Thenthereisapositive

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29 integerksuchthatksx=xandksy=y.Inparticular,fksxk)]TJ/F21 7.97 Tf 6.586 0 Td[(1=xk)]TJ/F21 7.97 Tf 6.586 0 Td[(1andfksyk)]TJ/F21 7.97 Tf 6.587 0 Td[(1=yk)]TJ/F21 7.97 Tf 6.587 0 Td[(1.Notethatxk)]TJ/F21 7.97 Tf 6.587 0 Td[(16=yk)]TJ/F21 7.97 Tf 6.587 0 Td[(1.Sincexandyareinthesamecomposant,theyhaveeventuallythesamebackwarditinerary.Thus,thereissomepositiveintegerN,suchthatforallnN,xnandynareonthesamesideofc.Bythiswemeaneitherbothxncandync,orbothxncandync.Sincexandyareperiodic,itfollowsthatforeachintegerj=0;1;;k)]TJ/F15 11.955 Tf 11.08 0 Td[(1,xjandyjareonthesamesideofc.Henceforeachintegerj0,fjsxk)]TJ/F21 7.97 Tf 6.587 0 Td[(1andfjsyk)]TJ/F21 7.97 Tf 6.587 0 Td[(1areonthesamesideofc.Thisisimpossiblesincexk)]TJ/F21 7.97 Tf 6.586 0 Td[(16=yk)]TJ/F21 7.97 Tf 6.587 0 Td[(1andfsisatentmapwithaslopes>1.Thisprovestherstclaim.Ournextclaimisthateachcomposantmappedtoitselfbymscontainsaxedpointofms.SupposeCisacomposantinXswithmsC=C.Ifx,y2C,thendmsx;msy=smdx;y.Hence)]TJ/F23 7.97 Tf 6.587 0 Td[(msisacontractionandhasaxedpoint.Thisprovesoursecondclaim.Itfollowsfromtheclaimsthatthenumberofcomposantsmappedtothemselvesbymsisequaltothenumberofpointsxedbyms.Bydenitionofs,thisnumberisequaltothenumberofxedpointsoffms,Ffms.3.4ProofoftheMainTheorem Theorem3.4.1 Lets;t2p 2;2suchthatfsandfthaveperiodicturningpoints.ThenXsandXtarehomeomorphicifandonlyifs=t.Proof.Itiswell-knownthatifXsandXtarehomeomorphicandtheturningpointoffsisperiodic,thentheturningpointofftisalsoperiodicwiththesameperiod.Thus,thereisnolossingeneralityinassumingthattheperiodoftheturningpointsforfsandftarebothperiodicofperiodn0.Supposes
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30 composantwithanendpointtoitself,thesameistrueforh.ThusNtalsomapseachendpointofXttoitself.ByLemma 3.3.2 ,thetotalnumberofcomposantsmappedtothemselvesbyn0sandhencebyhisFfn0s.Thus,thesameistrueforNt.ItfollowsthatjNjn0andn0dividesjNj.ButthenumberofcomposantsmappedtothemselvesbyNtisFfNt.ThusFfn0s=FfNt.Ontheotherhand,sinces
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CHAPTER4ISOTOPYTHEOREMInthischapterweprovetheIsotopytheoremasstatedintheIntroduction.Lets2p 2;2].Letfsbeatentmapwithperiodicturningpointwithperiodn0.Wehavealreadyshownthatforanyhomeomorphismg:Xs!Xssuchthatgleavesalltheendpointsfcign0)]TJ/F21 7.97 Tf 6.587 0 Td[(1i=0xed,thereisaksuchthatgandksarepseudo-isotopic-thatis,gandkspermutethecomposantsofXsinpreciselythesameway.Itisclearthatforanyhomeomorphismh:Xs!Xs,thereisann>0suchthathnleavescixedfori=0;1;2;:::;n0)]TJ/F15 11.955 Tf 12.032 0 Td[(1.Letkbetheintegersuchthathnandksarepseudo-isotopic.Wenowshowthathnandksareactuallyisotopic.4.1ConstructionofNeighborhoodsThefollowinglemmaisawell-knownfacttotheexpertsinthiseld. Lemma4.1.1 SupposeAisanarcinXsnotcontaininganyendpointofXs.ThenthereisaneighborhoodVofAhomeomorphictoCI,whereCisaCantorset.TheboundaryofVwillcorrespondtoCf0;1g.Moreover,thereisapositiveintegermsuchthatmmapseachcomponentofVhomeomorphicallyontoitsimageinIm.Proof.LetAbeanarcinXsnotcontaininganyendpointofXs.BytheproofofLemma 3.1.1 ,thereisapositiveintegermsuchthatforeachkm,noneofthepointsciareinkA.Inparticular,kjAisahomeomorphism.Letz2A,z=z0;z1;:::;zm;:::.LetC=fy2Xsjy0=z0;y1=z1;:::;ym=zmg.ThenCiscompact,totallydisconnected,andeverypointisalimitpoint.ThereforeCisaCantorset.LetJm=mA.Fixy2C.SinceJmfc0;c1;:::;cn0)]TJ/F21 7.97 Tf 6.587 0 Td[(1g=;,forthisy2C,thereisasequenceofintervalsfJig1i=msuchthatyi2JiforeachimandfsJi+1=Jiforeachim. 31

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32 WecanextendthesequencefJig1i=mtofJig1i=0byJ0=fmsJm;J1=fm)]TJ/F21 7.97 Tf 6.587 0 Td[(1sJm;:::;Jm)]TJ/F21 7.97 Tf 6.587 0 Td[(1=fsJm.Thenforalli2N,fsJi+1=Jiandyi2Ji.NowJmishomeomorphictoJy=lim)]TJ 16.258 6.052 Td[(fJi;fsgXsbytheprojectionm:Xs!Im.Letgy:Jm!Jybetheinverseofthishomeomorphism.Finally,let:CJm!Xsbedenedbyy;t=gyt.ThenV=CJmistherequiredneighborhood. Remark4.1.2 IntheaboveproofletxbeintheCantorsetC.Notethatthepointsz0andz1correspondingtox;0andx;1,respectively,areinthesamecomposant.Moreover,dz0;z1doesnotdependonx-thatis,thelengthsofthecomponentsofVareallthesameinthedmetric. Denition4.1.3 SupposefDig1i=1isasequenceofnonemptycompactsubsetsofametricspaceY.ThenlimsupfDig=fy2Yjy=limj!1yijforsomesubsequencefDijgandyij2Dijg. Lemma4.1.4 LetfAig1i=1beasequenceofarcsinXs.SupposethatAi!BintheHausdormetric.SupposealsothatthereisanM>0suchthat`AiMforalli.ThenBisanarcand`BM.Proof.LetNbesuchthatMs)]TJ/F23 7.97 Tf 6.586 0 Td[(N`IN 2=`Is 2=c1)]TJ/F23 7.97 Tf 6.586 0 Td[(c2 2.Thenforeveryk,NAkhaslengthatmost`IN 2.SinceAk!B,NAk!NB.Inparticular,NBisapropersubsetofIN.ItfollowsthatBisapropersub-continuumofXs.ByLemma 3.1.1 ,Bisanarc.Finally,choosejlargeenoughsothatjjBisahomeomorphism.Thenforeachk,sj`jAkM,andhence`B=sj`jBM. Lemma4.1.5 LetfAig1i=1beasequenceofarcsinXswithendpointsaiandbi,respectively.SupposethatthereisapositivenumberMsuchthatdai;biMeachi.Supposealsothatthesequencefaig1i=1convergestosomea2Xs.ThenB=limsupfAigisanarcinXsand`B2M.Proof.Letx2B=limsupfAig.ThenthereisasubsequencefAijg1j=1suchthatAij!DBintheHausdormetricwithx2D.ByLemma 4.1.4 ,`DM.So,

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33 da;xM.FromthisitfollowsthatBmustbeapropersub-continuumandthusanarcwiththe`-lengthofBatmost2M. Lemma4.1.6 LetfAig1i=1beasequenceofarcsinXswithendpointsaiandbi,respectively.Supposethatai!aandbi!b.SupposealsothatthereisanM>0suchthatdai;biMforalli.ThenaandbareinthesamecomposantofXs.LetAdenotetheuniquearcwithendpointsaandb.SupposethatlimsupAidoesnotcontainanendpointofXs.ThenAi!AintheHausdormetric.Proof.SupposethatfAig1i=1isasequenceofarcsinXswithendpointsaiandbi,respectively.Supposethatai!aandbi!b.SupposealsothatthereisanM>0suchthatdai;biMforalli.BytheproofofTheorem 3.2.2 ,aandbareinthesamecomposantofXsandda;bM.LetAbetheuniquearcwithendpointsaandb.LetB=limsupfAig.ByLemma 4.1.5 ,Bisanarcwith`B2M.ByassumptionBdoesnotcontainanendpointofXs.So,letV=CIbetheneighborhoodofBgivenbyLemma 4.1.1 .ThenthereisanNsuchthatforallnN,AnVsinceBisthelimsupfAig.ThereforeforeachiN,AifyigIforsomeyi2C.Furthermore,AiisthearcinfyigIjoiningtheendpoints.Leta;b2fygI.Then limi!1Ai=A=BfygI. Denition4.1.7 ConsiderJCR2whereCisthestandardmiddle-thirdCantorsetandJ=[)]TJ/F15 11.955 Tf 9.298 0 Td[(1;1].DeneanequivalencerelationonJCbyt;1)]TJ/F22 11.955 Tf 9.298 0 Td[(t;1forallt2J.LetQ=JC=. Remark4.1.8 WewillthinkofQastheunionoftwosetsEandFdenedinthefollowingway.LetE=C[)]TJ/F22 11.955 Tf 9.298 0 Td[(C[1;2]R2.LetFbeaCantorsetofsemicircleswithcentersat0;1joiningeachpointofCf1gwiththecorrespondingpointof)]TJ/F22 11.955 Tf 9.299 0 Td[(Cf1g.NowintheCantorsetC,letC0bethesetofpointsinCintheintervalbetween0and1 3,inclusively.LetC1bethesetofpointsinCbetween2 3and7 9,inclusively.Forhigherk,CkbethesubsetofCcontainingthepointsbetween3k)]TJ/F21 7.97 Tf 6.587 0 Td[(1 3k

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34 and3k+1)]TJ/F21 7.97 Tf 6.586 0 Td[(2 3k+1,inclusively.ThenfCkg1i=0isadisjointcollectionofCantorsetswithCk!1inCandC=S1i=0Ci[f1g. Lemma4.1.9 SupposethatAisanarcinXswhichcontainsanendpointofXs.ThenthereisaneighborhoodVofAhomeomorphictoQ=CJ=.Proof.ThereisnolossofgeneralityinassumingthatcistheendpointofA.LetAwbethesetofpointsinCcwiththesamebackwarditinerary,w,andsuchthatc2Aw.WeknowthatAwisanon-degeneratearcwithcasoneendpointandsomezastheotherendpoint,andthat0jAwisahomeomorphismonto[c;ci]or[ci;c]forsome1in0g.ThesetD0iscompact,totallydisconnectedandeverypointisalimitpoint,soD0isaCantorset.Letx2D0.Thenn0x=candix6=cforalli>n0.TherearetwoarcsAxandBxinXscontainingxasanendpoint,suchthatn0Ax=n0Aw,andsuchthatn0Axandn0Bxaresymmetricaboutc.Similarly,foranyk2N,deneDk=fx2Xsjk+1n0x=candix6=cforalli>k+1n0g.Then,foranyk2N,thesetDkisaCantorset.Foranyx2Dk,therearetwoarcsAxandBxinXscontainingxasanendpoint,suchthatk+1n0Ax=k+1n0Aw,andsuchthatk+1n0Axandk+1n0Bxaresymmetricaboutc.LetV=SfAx[Bxjx2[1k=0Dkg[Aw.Leta;bbetheopenintervalcontainingcsuchthatfn0a;bis[c;ciorci;c].Theneachpointof)]TJ/F21 7.97 Tf 6.586 0 Td[(1n0a;bisinV.HenceeverypointofAwexceptzisaninteriorpointofV.Deneamaph:V!Qinthefollowingway.Foreveryk2N,hsendsDkhomeomorphicallyontoCk.Foreveryx2Dk,Axismappedlinearlyonto[)]TJ/F15 11.955 Tf 9.299 0 Td[(1;0]fhxg,andBxismappedlinearlyonto[0;1]fhxg,andAwismappedlinearlyto[)]TJ/F15 11.955 Tf 9.298 0 Td[(1;0]f1g.Thenhis1-1,continuousandonto-henceitisahomeomorphism.NowtheneighborhoodVthatwejustcreatedmaynotcontainthegivenarcA.However,fork>1,applyingtheshifthomeomorphismktimeswillcreatealonger

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35 andthinnerneighborhoodofthesameform,kn0sV,withS1k=0kn0sVdenseinXs.Thus,therewillbesomekforwhichkn0sVwillcontainA. Remark4.1.10 Intheaboveproof`Aw=`Ax=`Bx,foreveryx2Dk,foreveryk2N.Furthermore,therearearbitrarilysmallneighborhoodsofchomeomorphictoQforwhichthisistrue. Theorem4.1.11 Supposethath1andh2arehomeomorphismsofXssuchthath1c=h2c=c.SupposealsothatthereisanM>0suchthatforeachy2Cc,dh1y;h2yM.LetxibeasequenceofpointsinXsconvergingtoapointxinXs.LetAibetheuniquearcjoiningh1xiandh2xi,andletAbetheuniquearcjoiningh1xandh2x.ThenAi!AintheHausdormetric.Proof.Weassumethehypothesesandnotationofthetheorem.Case1.Supposethatthecomposantcontainingxdoesnotcontainanend-point.InthiscaseLemma 4.1.6 appliessincelimsupfAig1i=1mustbeinthecom-posantofxwhichdoesnotcontainanendpointofXs.Thus,wehaveAi!Ainthiscase.Case2.Nowsupposethatx2Cciforsomeiwithx6=ci.ByTheorem 3.2.2 h1xandh2xareonthesamecomposant,andthiscomposantmustbesomeCcjforsomej.LetJ=[e;cj]beanarcinCcjsuchthath1x2J,h2x2J,de;h1x>M+1,de;h2x>M+1.LetVbeaneighborhoodofJasinLemma 4.1.9 .Considerthearch)]TJ/F21 7.97 Tf 6.587 0 Td[(11J[h)]TJ/F21 7.97 Tf 6.587 0 Td[(12JinCci.LetWbeaneighborhoodofthisarcasinLemma 4.1.9 .ByshrinkingVinthe"vertical"directionifnecessary,wemayassumethath)]TJ/F21 7.97 Tf 6.586 0 Td[(11V[h)]TJ/F21 7.97 Tf 6.587 0 Td[(12VW.LetKbeacomponentofVwhichdoesnotcontaincj.BythecentralpointofKwemeantheuniquepointofKwhichcorrespondstoapointoftheform;yin[)]TJ/F15 11.955 Tf 9.299 0 Td[(1;1]CasinDenition 4.1.7 .Wemayassumethatxn2h)]TJ/F21 7.97 Tf 6.586 0 Td[(11Vh)]TJ/F21 7.97 Tf 6.587 0 Td[(12Vforeachn.Sinceh1xn!h1xandh1xn!h2x,itfollowsfromRemark 4.1.10 thatfornsucientlylarge,theddistancefromh1xntoeitherendpointofthecomponentofVcontainingh1xnis

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36 greaterthanM+1.Sincedh1xn;h2xnM,itfollowsthath1xnandh2xnlieonthesamecomponentofVfornsucientlylarge.Withoutlossofgenerality,weassumethatthisholdsforeachn.Foreachpositiveintegern,letKndenotethecomponentofVwhichcontainsh1xnandh2xn,andletwndenotethecenterpointofKn.Letyn=h)]TJ/F21 7.97 Tf 6.587 0 Td[(11wnandzn=h)]TJ/F21 7.97 Tf 6.586 0 Td[(12wn.ThenynandznlieonthesamecomponentofWasxn.Sincexn!x,yn!ci,andzn!ci,itfollowsthatfornsucientlylarge,xndoesnotliebetweenynandznonacomponentofW.Again,wemayassumethatthisholdsforeachn.Foreachpositiveintegern,letKn=[an;bn].Wemayassumethatdan;wn>M+1anddbn;wn>M+1foreachn.Weclaimthatforeachpositiven,h1xnandh2xnlieonthesamesideofwninKn.Weprovethisbycontradiction.Supposethath1xnandh2xnlieonoppositesidesofwnforsomen.RecallthatKn=[an;bn]andsupposewithoutlossofgeneralitythath1xnliesonthesamesideofwnasan.Thereisapointpn2Wwithh1pn=an.Moreover,pn,xn,yn,andznlieonthesamecomponentofW,andonthiscomponent,pnisononesideofxn,whileynandznareontheotherside.Now,usingthemonotonicityofh2onacomponent,weseethatthearcinXswithendpointsh1pnandh2pncontainsbothanandwn.Thisimpliesthatdh1pn;h2pn>M+1.Thisisacontradictionandtheclaimisestablished.SinceAiVforeachi,itfollowsfromthespecialformofVandtheclaimthatAi!A.Case3.Nowsupposethatx=ciforsomei.InthiscaseAisjustthepointfcjg.ThiscaseisroutineusingthestructureoftheneighborhoodofcgiveninLemma 4.1.9 .OneofCase1,Case2orCase3mustholdsotogethertheyprovethestatementofthistheorem.

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37 4.2ProofoftheIsotopyTheorem Theorem4.2.1 Supposethath1;h2:Xs!XsarehomeomorphismswhichleavetheendpointsofXsxed.SupposethatthereisanM>0suchthatdh1x;h2xMforallx2Cc.Thenh1andh2areisotopic.Proof.Letthath1;h2:Xs!XsbehomeomorphismswhichleavetheendpointsofXsxed.SupposethatthereisanM>0suchthatdh1x;h2xMforallx2Cc.LetH:XsI!Xsbedenedinthefollowingway.Letx2Xsandt2I.ByTheorem 3.2.2 ,thereisauniquearcAxconnectingh1xandh2x.Letm2NbesuchthatmjAxisahomeomorphismintoIm.Letgm:mAx!Axbetheinverseofthishomeomorphism.Dene Hx;t=gm)]TJ/F22 11.955 Tf 11.955 0 Td[(tmh1x+tmh2x.IfkjAxisahomeomorphism,thengk)]TJ/F22 11.955 Tf 12.732 0 Td[(tkh1x+tkh2x=gm)]TJ/F22 11.955 Tf 11.982 0 Td[(tmh1x+tmh2x.So,Hx;tiswell-dened.WenowshowthatHiscontinuous.Supposethatxi;ti!x;t.LetAibetheuniquearcwithendpointsh1xi;h2xi.Thenh1xi!h1xandh2xi!h2x.So,ifAistheuniquearcconnectingh1xandh2x,whoseexistenceisgivenbyTheorem 3.2.2 ,then,byTheorem 4.1.11 ,Ai!AintheHausdormetric.Case1.Thepointxisnotanendpoint.InthiscasethearcAconnectingh1xandh2xdoesnotcontainanendpoint.LetVbeaneighborhoodofAgivenbyLemma 4.1.1 .WeknowthatVisoftheformVCIwhereCisaCantorsetandIisaninterval.ThenthereisanNsuchthatfornN,thearcAnV.NowbyLemma 4.1.1 ,thereisanmsuchthatmisahomeomorphismoneachcomponentofVtoitsimageinIm.ThereforeforthismandforallnN, Hxn;tn=gm)]TJ/F22 11.955 Tf 11.955 0 Td[(tnmh1xn+tnmh2xnand Hx;t=gm)]TJ/F22 11.955 Tf 11.955 0 Td[(tmh1x+tmh2x.So,clearlyHxn;tn!Hx;t.

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38 Case2.Thepointxisanendpoint.Inthiscaseh1x=h2x=xsincetheendpointsareassumedtobexed.ThereforeA=fxgandthusAn!fxg.ThisimpliesthatHxn;tn!fxg=Hx;t.So,Hx;tisahomotopy.Wenowshowthatitisanisotopybyshowingthatforeacht,htx=Hx;tisone-to-oneandonto.Firstweshowthathtisone-to-one.NotethathtpermutesthecomposantsofXsthesamewaythath1andh2do.So,toshowthathtisone-to-oneitwillsucetoshowthathtrestrictedtoacomposantCxisone-to-one.NowCxisthearc-componentofx.Thisarc-componentwiththedmetricishomeomorphictoeitherRorR+.FixorderingsonCxandCh1x.Nowh1andh2arehomeomorphismsfromCxtoCh1xeitherpreservingorreversingtheordersofCxandCh1x.However,sincetheddistancebetweenh1andh2onCxisbounded,theseeitherbothpreservetheordersorbothreversetheordersonCxandCh1xinthesameway.Thus,htjCxisone-to-one.Toshowthathtisontoissimilar.WenowgivetheproofoftheIsotopytheoremasoutlinedatthebeginningofthischapter.ProofoftheIsotopytheorem.Leth:Xs!Xsbeahomeomorphism.LetnbesuchthathnleavestheendpointsofXsxed.ByTheorem 3.2.5 ,thereisanM>0andthereisak2Zsuchthatdhnx;ksxMforallx2Cc.ByTheorem 4.2.1 ,hnandksareisotopic.

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CHAPTER5ONINGRAM'SCONJECTURE5.1Three-StepProgramInthischapterweaddressthequestionofclassicationofinverselimitspacesoftentmapswithoutanyassumptionsontheorbitoftheturningpoint.Wepresentathree-stepprogramfromwhichIngrams'sconjecturewouldfollow.Ingram'sconjecture.If1
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40 5.2ProofofStep2 Theorem5.2.1 Letfsbeatentmapwithslopes2p 2;2].SupposeCXsisacomposantwiththepropertythatthereexistsanintegerksuchthatksC=C.Thenthereexistsauniquexedpointzoffksandthepointz=z;fk)]TJ/F21 7.97 Tf 6.586 0 Td[(1sz;;fsz12C.Proof.Let>0besuchthats2 2k s2 2k)]TJ/F21 7.97 Tf 6.586 0 Td[(1
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41 Sinceforeachpositiveintegernandallm,m)]TJ/F21 7.97 Tf 6.586 0 Td[(2nksA=m+2nkA,itfollowsthatforeachpositiveintegernandallm, P1n=0`m)]TJ/F21 7.97 Tf 6.587 0 Td[(2nksA=P1n=0`m+2nkA.AsforallmM,`m+2kA<1 s2 2k`mA,itfollowsthatforallmM, `m[1n=0)]TJ/F21 7.97 Tf 6.586 0 Td[(2nksAP1n=0`m+2nkA P1n=01 s2 2nk`mA=s2 2k s2 2k)]TJ/F21 7.97 Tf 6.587 0 Td[(10distanceapart.Thepointsz=z;f2k)]TJ/F21 7.97 Tf 6.587 0 Td[(1sz;;fsz1andz0=ksz=fksz;fk)]TJ/F21 7.97 Tf 6.586 0 Td[(1sz;;fsz;z;f2k)]TJ/F21 7.97 Tf 6.587 0 Td[(1sz;;fk+1sz1aretwodistinctpointsinthecomposantC.Thereexistsapropersub-continuumACwhichcontainsbothzandz0.Observethat)]TJ/F23 7.97 Tf 6.586 0 Td[(ksz=z0and)]TJ/F23 7.97 Tf 6.587 0 Td[(ksz0=z.Hence,foreachpositiveintegern,)]TJ/F23 7.97 Tf 6.587 0 Td[(nksAcontainsbothzandz0.Thus,foreachpositiveintegern,nkAcontainsbothzandfksz.But,sinceAisapropersub-continuumofXs,`nA!0asn!1,whichcontradictstheassumptionthatzandfkszareaxeddistance>0apart.Byanargumentsimilartotheoneintheparagraphabove,itiseasytoprovethatzisauniquexedpointofksinC. Theorem5.2.2 Letfsbeatentmapwithslopes2p 2;2].SupposeCXsisacomposantforwhichthereexistsanintegerksuchthatksC=C.Assumethateithertheturningpointcisnotperiodic,orthatifcisperiodicofperiodn0,thenci=2Cforeachi2f0;1;;n0)]TJ/F15 11.955 Tf 12.146 0 Td[(1g.ThenthereisacontinuousbijectionfromthereallineRtoC.

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42 Proof.Letz=z0;zk)]TJ/F21 7.97 Tf 6.586 0 Td[(1;;z11,wherezi=fisz0,i2f0;1;;k)]TJ/F15 11.955 Tf 12.266 0 Td[(1g,betheuniquexedpointofksinC.Let"=1 2minfjzi)]TJ/F22 11.955 Tf 12.354 0 Td[(cj;jzi)]TJ/F22 11.955 Tf 12.353 0 Td[(c1j;jzi)]TJ/F22 11.955 Tf 12.353 0 Td[(c2j;jzi)]TJ/F22 11.955 Tf 12.353 0 Td[(c3j:i=0;1;;k)]TJ/F15 11.955 Tf 12.188 0 Td[(1g.ThenthesymmetricintervalJ0=[z0)]TJ/F22 11.955 Tf 12.188 0 Td[(";z0+"][c2;c1]andc3=2J0.LetJ1bethecomponentoff)]TJ/F21 7.97 Tf 6.587 0 Td[(1sJ0inI1=1Xswhichcontainszk)]TJ/F21 7.97 Tf 6.587 0 Td[(1.ThenfsmapsJ1linearlyontoJ0.LetJ2bethecomponentoff)]TJ/F21 7.97 Tf 6.587 0 Td[(1sJ1inI2=2Xswhichcontainszk)]TJ/F21 7.97 Tf 6.586 0 Td[(2.Andsoon,foreachpositiveintegern,letJnbethecomponentoff)]TJ/F21 7.97 Tf 6.586 0 Td[(1sJn)]TJ/F21 7.97 Tf 6.587 0 Td[(1inIn=nXswhichcontainsz)]TJ/F23 7.97 Tf 6.587 0 Td[(nmodk.Ourchoiceof"suchthatforeachpositiveintegern,c3=2Jn,ensuresthatfsmapsJnlinearlyontoJn)]TJ/F21 7.97 Tf 6.587 0 Td[(1,thatis,nopieceofJngetscutoutbypullingback"Jntotheleftofc.LetJ=lim)]TJ 16.258 5.829 Td[(fJn;fsg1n=0.ThenJisapropersub-continuumofCcontainingz.Notethatforeachpositiveintegern,njJisahomeomorphism,andforeachpositiveintegern,fsjJnisahomeomorphism.HenceJisanarcinC.ConsiderthesequenceJ;ksJ;2ksJ;ofarcsinC.Weclaimthatitisanincreasingsequence.Letx=x0;x1;x2;2J.Inordertoprovethatx2ksJ,weneedtoshowthatthereexistsapointy2Jsuchthatksy=x.Lety=)]TJ/F23 7.97 Tf 6.587 0 Td[(ksx=xk;xk+1;.Thenksy=x.Itonlyremainstoshowthaty2J.Asxk2JkandJk=[z0)]TJ/F23 7.97 Tf 15.164 4.707 Td[(" sk;z0)]TJ/F23 7.97 Tf 15.163 4.707 Td[(" sk],wehavethatjxk)]TJ/F22 11.955 Tf 11.678 0 Td[(z0j" sk<".Hencexk2J0.Itfollowsthaty2J.WeprovedthatJksJ,henceforeachpositiveintegern, nksJn+1ksJ.LetACbeapropersub-continuumcontainingz.Then`nA!0asn!1.HencethereexistsapositiveintegermsuchthatmkA[z0)]TJ/F22 11.955 Tf 12.148 0 Td[(";z0+"].ThusAmksJ.ItfollowsthatCisanincreasingunionofarcs. Remark5.2.3 ForacomposantCXswiththepropertythatthereexistsanintegerksuchthatksC=C,andsuchthateithertheturningpointcisnotperiodic,orthatifcisperiodicofperiodn0,thenci=2Cforeachi2f0;1;;n0)]TJET1 0 0 1 107.865 35.613 cm0 g 0 G1 0 0 1 438.583 0 cm0 g 0 G

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43 1g,wecandeneametricdinthesamewaywediditinthecaseofatentmapfswithaperiodicturningpoint.Inthelattercaseweonlyusedthefactthatforanypropersub-continuumAXs,thereexistsapositiveintegermsuchthatforallnm,njAisahomeomorphism.TheproofofTheorem 5.2.2 showsthatthisholdsforanysub-continuumofC.5.3ProofofStep3 Theorem5.3.1 Let11. Corollary5.3.2 Let1
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44 Beforeweembarkonprovingournextresult,weneedtointroducetheconceptoftopologicalentropyandstateafewwell-known,butnon-trivialresultsabouttheentropyofmapsinthetentfamilyandtheentropyofmapsinthequadraticfamily. Denition5.3.3 Theone-parameterfamilyofmapsf:I!Idenedbyfx=x)]TJ/F22 11.955 Tf 12.476 0 Td[(x,forx2Iandfor04,isknownasthequadraticfamily.Sinceanyoftheseveralequivalentdenitionsofentropyofamapisratherlonganddoesnotrevealmuchinformationaboutitsnature,atleastnottotheinexperiencedreadertowhomitwouldbeoered,nordotheyshowwayshowtocomputeit,wewilltrytogiveashortdescriptionandstateseveralpropertieswiththeappropriatereferences.Foracomprehensivestudyofthisconceptwereferthereaderto[ 9 ,ChapterVIII]and[ 2 ,Chapter4].Thetopologicalentropyofamapisaquantitativemeasureofthecomplexityofthesystemmodeledbyiteratingthatmap.Itmeasureshowmanydistinctorbitsofagivenlengththereareforthismap,thatis,howthisnumbergrowswiththelength.Foraprecisedenitionofentropy,see[ 9 ,SectionVIII.1]or[ 2 ,Section4.1].Iffisapiecewisemonotonemapfromaclosedintervalintoitself,andcnde-notesthenumberofpiecesofmonotonicityoffn,socalledlaps,thenthetopologicalentropyoffis hf=limn!11 nlogcn,and1 nlogcnhfforanyn[ 2 ,Theorem4.2.4].Wesaythatapiecewisemonotonemapffromaclosedintervalintoitselfhasaconstantslopesifoneachofitslapsitisanewithaslopeofabsolutevalues.Iffisapiecewisemonotonemapfromaclosedintervalintoitselfwithaconstantslopes,thenhf=maxf0;logsg[ 2 ,Corollary4.3.13].Thus,thetopologicalentropyofatentmapfswiths>1is hfs=logs.

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45 AnecientalgorithmtocomputetheentropyofunimodalmapsoftheintervalintoitselfwasgivenbyBlock,Keesling,LiandPeterson[ 10 ],andsuchanalgorithmforbimodalmapsoftheintervalintoitselfwasgivenbyBlockandKeesling[ 11 ].OnthespaceofallC1piecewisestrictlymonotonemapswithagivennumberoflaps,withtheC1topology,thetopologicalentropyiscontinuous[ 2 ,Corollary4.5.15].ThisfactwasprovenbyMilnorandThurstonin[ 28 ].Fortentmapsfswithslopes>1,thetopologicalentropyisstrictlyincreasingasafunctionoftheslopes.Forthequadraticfamily,thetopologicalentropyisanincreasingfunctionoftheslopes,butnoreal"proofisknown;allknownproofsofthisfactusecomplexicationofthemaps.Wewillstatethoseresultsforeasierreferral. Theorem5.3.4 Sullivan,Milnor,DouadyandHubbard[ 27 ,ChapterII,Theorem10.1]Iftheturningpointsofthequadraticmapsfandf0areeventuallyperiodicandtheirkneadinginvariantsareequal,then=0. Corollary5.3.5 Letf:I!IbeaunimodalfamilyconsistingofC1mapsdependingcontinuouslyontheparameter.Iftheassumptionthattheturningpointsoffandf0areeventuallyperiodicandtheirkneadinginvariantsareequalimpliesthat=0,thenthefollowingstatementsaretrue. 1. Thefunction!Kfismonotone[ 27 ,ChapterII,Corollary1]. 2. Thefunction!hfismonotone[ 27 ,ChapterII,Corollary2]. Theorem5.3.6 Letf1beamapinthequadraticfamilywithkneadingsequenceDC.LetBbeanLoranRsuchthatDB0suchthatforeach21)]TJ/F22 11.955 Tf 10.861 0 Td[(;1+andeachpointxwithindistanceofanypointintheorbitofc,wehavethatjfn0xj<1.Thus,foreach21)]TJ/F22 11.955 Tf 12.141 0 Td[(;1+,f

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46 hasuniqueattractingperiodicorbitwithindistanceofanypointintheorbitofc,andask!1,fkcapproachesthisorbit.Let2=1)]TJ/F22 11.955 Tf 12.866 0 Td[(.Consider22;1.ByTheorem 5.3.4 andCorollary 5.3.5 ,KfKf1.Letxbethepointnearcintheattractingperiodicorbit.ThenIfx=DA1,whereAisanLoranRdependingonwhichsideofc,xison.Byconstruction,Kf=Ifc=Ifx=DA1.But,byTheorem 5.3.4 andCorollary 5.3.5 ,Kfismonotoneasafunctionof.Thus,DA1=DB1.Theentropydependsonlyonthekneadingsequence.Moreover,theentropyisacontinuousfunctionoftheparameterinthequadraticfamily[ 2 ,Corollary4.5.5].Hencetheentropyisconstanton2;1].Thus,hf=hf1forall22;1].NotethattheproofofTheorem 5.3.6 showsthatthereisanopeninterval1;3suchthatforall21;3,Kf=DB1andhf=hf1,whereBisanLoranRsuchthatDB>DC. Theorem5.3.7 Letp 2
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47 DsBs1
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48 Case4.Foreachofthetentmapsfsandft,theturningpointisperiodic.ThenKfs=DsCandKft=DtCforsomenitesequencesDsandDtofR'sandL's.Letybeaxedpointoffns.Ifcisnotintheorbitofyunderfs,thentheforwarditineraryofyunderfsisI sy=S1forsomesequenceSoflengthnofL'sandR's,andkS1
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49 5.4Ingram'sConjectureFollowsfromthePseudo-isotopyConjecture Theorem5.4.1 Ifthepseudo-isotopyconjectureholds,thenIngram'sconjectureholds.Proof.Supposethepseudo-isotopyconjectureistrue.ByRemark 5.1.1 ,itisenoughtoconsiderparametersp 2FfNs.Thus,weprovedthatifStep1holds,thenforp 2
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50 Observethatfsc1;1]=[0;c2.Sincefsreachesitsmaximumc1atc,nopointsfromI=[0;1]mapintoc1;1].Thus,ifx=x0;x1;2I;fsandxm2[0;c2forsomepositiveintegerm,thenxk2[0;c2foreachpositiveintegerkm. Lemma5.5.1 Tisanopenraywithendpoint0=;0;2I;fs.I;fsisacompacticationofT.Proof.FirstweprovethatTisanopenraywithendpoint0=;0;.Denotex0=c2.LetJ0=[0;x0].Thereexistsapointx12J0suchthatfsx1=x0.LetJ1=[0;x1].ThenfsmapsJ1linearlyontoJ0.Thereexistsapointx22J1suchthatfsx2=x1.LetJ2=[0;x2].ThenfsmapsJ2linearlyontoJ1.WeformJ3;J4;inthesamemanner.LetJ=lim)]TJ 16.259 5.829 Td[(fJn;fsg1n=0.ThenJisanarcinI;fswithendpoints0andx=x0;x1;.Notethat)]TJ/F21 7.97 Tf 6.587 0 Td[(10[0;c2=Jnfxg.ThereforeJnfxgisopeninI;fs.SinceT=[1k=0ksJnx=[1k=0ksJisaunionofincreasingrayswithacommonendpoint0,itfollowsthatTisanopenraywithendpoint0.NextweprovethatTisdenseinI;fs.Letx=x0;x1;2Xs.Foranypositiveintegerm,thereexistapositiveintegerkandy2[0;c2suchthatfksy=xm.Thenthepointsxandym=x0;x1;;xm;fk)]TJ/F21 7.97 Tf 6.587 0 Td[(1sy;;fsy;y;agreeontherstm+1coordinatesandym2T.Thusthesequenceofpointsfymg1m=0inTconvergestothepointx2Xs.WecandenethemetricdonTasfollows.LetxandybepointsinT.ByLemma 5.5.1 ,Tisaray,hencethereisanarcATwithendpointsxandy.Then,thereexistsanonnegativeintegerksuchthatkjAisahomeomorphism.Dene dx;y=skjkx)]TJ/F22 11.955 Tf 11.955 0 Td[(kyj.Thisissimilartohowdwasdenedwhentheturningpointofthetentmapwasperiodic.Thereisadierence;nowdactuallygivesthetopologyofT. Theorem5.5.2 Letfs:I!I,fsx=minfsx;s)]TJ/F22 11.955 Tf 12.314 0 Td[(xgbeatentmapwithslopes2p 2;2].Leth:I;fs!I;fsbeahomeomorphism.SupposethereexistanumberM>0andanintegerksuchthatdhx;ksxM,forallx2T.ThenhjXsandksjXsarepseudo-isotopic.

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51 Proof.Letx2Xs.ByLemma 5.5.1 ,thetailTisdenseintheinverselimitspaceI;fs,hencethereexistsasequencefxng1n=1inTconvergingtothepointx.Thenthesequencehxnconvergestothepointhxandksxnconvergestothepointksx.ConsidertheuniquearcsAnTwithendpointshxnandksxn.Byassumption,`An=dhxn;ksxnMforeachn.Letm>0beanintegersuchthatMsmfsc)]TJ/F22 11.955 Tf 12.416 0 Td[(f2sc.Thenforeachn,`mAnM sm
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REFERENCES [1] J.M.AartsandR.J.Fokkink,Theclassicationofsolenoids,Proc.Amer.Math.Soc.111991,1161-1163 [2] L.Alseda,J.LlibreandM.Misiurewicz,CombinatorialDynamicsandEntropyinDimensionOne,AdvancedSeriesinNonlinearDynamics5,WorldScientic,RiverEdge,NJ,1993. [3] M.Barge,Horseshoemapsandinverselimits,PacicJ.Math.121986,29-39. [4] M.Barge,K.BrucksandB.Diamond,Self-similarityininverselimitspacesofthetentfamily,Proc.Amer.Math.Soc.124996,3563-3570. [5] M.BargeandB.Diamond,Homeomorphismsofinverselimitspacesofone-dimensionalmaps,Fund.Math.146995,171-187. [6] M.BargeandB.Diamond,Inverselimitspacesofinnitelyrenormalizablemaps,TopologyAppl.83998,103-108. [7] M.BargeandS.Holte,Nearlyone-dimensionalHenonattractorsandinverselimits,Nonlinearity8995,29-42. [8] M.BargeandJ.Martin,EndpointsofinverselimitspacesanddynamicsinContinuawiththeHoustonProblemBook,165-182,Lect.NotesinPureandAppl.Math.170,Dekker,NewYork,1995. [9] L.BlockandW.Coppel,DynamicsinOneDimension,Lect.NotesinMath.1513,SpringerVerlag,NewYork,1992. [10] L.Block,J.Keesling,S.LiandK.Peterson,Animprovedalgorithmforcomputingthetopologicalentropy,J.ofStat.Physics,55989929-939. [11] L.BlockandJ.Keesling,Computingthetopologicalentropyofmapsoftheintervalwiththreemonotonepieces,J.ofStat.Physics,66992893-912. [12] M.Brown,Someapplicationsofanapproximationtheoremforinverselimits,Proc.Amer.Math.Soc.,11960,478-483. [13] K.BrucksandH.Bruin,Subcontinuaofinverselimitspacesofunimodalmaps,Fund.Math.160999,219-246. [14] K.Brucks,B.Diamond,M.Otero-EspinarandC.Tresser,DenseorbitsoftheturningpointoftentmapsinContinuumTheoryandDynamicalSystems57-61,Contemp.Math.117,AMS,Providence,1991. 52

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53 [15] K.BrucksandB.Diamond,AsymbolicrepresentationofinverselimitspacesforaclassofunimodalmapsinContinuawiththeHoustonProblemBook,Lect.NotesinPureandAppl.Math.170995,Dekker,NewYork,207-226. [16] K.BrucksandM.Misiurewicz,Thetrajectoryoftheturningpointisdenseforalmostalltentmaps,Ergod.Th.Dynam.Sys.16996,1173-1183. [17] H.Bruin,Inverselimitspacesofpost-criticallynitetentmaps,Fund.Math.165,00,125-138. [18] P.ColletandJ.-P.Eckman,IteratedMapsontheIntervalasDynamicalSystems,Birkhauser,Boston,1980. [19] W.Debski,Ontopologicaltypesofthesimplestindecompasablecontinua,Colloq.Math.49985,203-211. [20] S.Holte,Generalizedhorseshoemapsandinverselimits,PacicJ.Math.156992,297-305. [21] J.HockingandG.Young,Topology,DoverPublications,Inc.,NewYork,1961. [22] W.Ingram,Inverselimitson[0;1]usingtentmapsandcertainotherpiecewiselinearbondingmapsinContinuawiththeHoustonproblembook,Lect.NotesinPureandAppl.Math.170995,Dekker,NewYork,253-258. [23] L.Kailhofer,Apartialclassicationofinverselimitsoftentmapswithperiodiccriticalpoints,Fund.Math.123002,235-265. [24] L.Kailhofer,Aclassicationofinverselimitsoftentmapswithperiodiccriticalpoints,Fund.Math.177003,95-120. [25] J.Keesling,Thegroupofhomeomorphismsofasolenoid,Trans.Amer.Math.Soc.172972,119-131. [26] J.Kennedy,AbriefhistoryofindecomposablecontinuainContinuawiththeHoustonProblemBook,103-126,Lect.NotesinPureandAppl.Math.170,Dekker,NewYork,1995. [27] W.deMeloandS.vanStrien,One-dimensionalDynamics,Springer-Verlag,NewYork,1993. [28] J.MilnorandW.Thurston,Oniteratedmapsoftheinterval,Lect.NotesinMath.1342988,465-563. [29] J.Mioduszewski,Mappingsofinverselimits,Coll.Math.10963,39-44. [30] S.Nadler,ContinuumTheory,DekkerInc.NewYork,1992. [31] S.Stimac,Structureofinverselimitspacesoftentmapswithnitecriticalorbit,preprint004.

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54 [32] S.Stimac,Aclassicationofinverselimitspacesoftentmapswithnitecriticalorbit,preprint004. [33] R.SwansonandH.Volkmer,Invariantsofweakequivalenceinprimitivematrices,ErgodicTheoryDynam.Systems20000611-626. [34] W.Watkins,Homeomorphicclassicationofcertaininverselimitspaceswithopenbondingmaps,PacicJ.Math.103982,589-601 [35] R.Williams,One-dimensionalnonwanderingsets,Topology6967,473-487

PAGE 62

BIOGRAPHICALSKETCHIwasbornonJuly19,1971inSkopje,Macedonia.IobtainedmyundergraduatedegreeinmathematicsattheDepartmentofMathematicsattheFacultyofNaturalSciencesandMathematicsattheUniversityStCyrilandMethodiusinSkopje,Macedonia,inAugust1995.IobtainedmyMasterofSciencedegreeatthesamedepartment,inAugust2001.SinceSpring1996,IhavebeenanemployeeatthePedagogicalFacultyattheUniversityStCyrilandMethodiusinSkopje,Macedonia.Intheperiodof1999-2001,IwasaparticipantinajointEuropeanprojectaimedatrestructuringthecurriculumatthePedagogicalFacultyattheUniversityStCyrilandMethodiusinSkopje,Macedonia.Withintheproject,IgotacquaintedwithsomecharacteristicsoftheeducationalsystemsintheNetherlands,Denmark,SwedenandGreatBritain.Mymaininterestsaretopologyanddynamicalsystems,aswellasmatheducationforelementaryschoolteachersandmatheducationingeneral. 55


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ON THE CLASSIFICATION OF INVERSE LIMIT SPACES OF TENT MAPS


By

SLAGJANA JAKIMOVIK
















A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA


2005

































Copyright 2005

by

Slagjana Jakimovik















I dedicate this work to my father, my mother and my sister.















ACKNOWLEDGMENTS

I wish to express my deepest gratitude to my adviser, Dr. Jed Keesling, for his

guidance, his great enthusiasm for our work together throughout my entire studies at

the University of Florida and for being my mentor both academically and spiritually.

My thanks also go to Dr. Louis Block, who contributed greatly to this work, for

being very supportive and a wonderful person to work with. This dissertation is a

result of our joint work.

I also wish to express my gratitude to Lois Kailhofer. In the dissertation we

build some of our results on the basis of certain results from one of her papers.

Moreover, I wish to acknowledge her contribution to the proof of one of the results in

the dissertation.

















TABLE OF CONTENTS


ACKNOWLEDGMENTS . . . . . . .

LIST OF FIGURES . . . . . . .

A B ST R A CT . . . . . . . .

1 INTRODUCTION . . . . . . .

2 PRELIMINARIES . . . . . . .

2.1 General Definitions and Results . . . .
2.1.1 Inverse Limit Spaces of Maps . . . .
2.1.2 Itineraries . . . . . .
2.1.3 Tent M aps . . . . . .
2.2 Definitions and Results from Kailhofer's Paper . .
2.2.1 C i iiii 2.2.2 Composant of c . . . . .
2.2.3 "Fu.I ., 1I M ap hq,p . . . . .

3 ON THE CLASSIFICATION OF INVERSE LIMIT SPACES OF


.

.

.

.

.

.
.
.
....
.
.

.

TENT


MAPS WITH PERIODIC TURNING POINT. .


3.1 Unravelled Composant of a Point and a New Metric d .
3.2 Homeomorphisms of X8 and the Shift Homeomorphism .
3.3 Number of Fixed Points of fk . . . . .
3.4 Proof of the Main Theorem . . . . .....

4 ISOTOPY THEOREM . . . . . .....

4.1 Construction of Neighborhoods . . . .....
4.2 Proof of the Isotopy Theorem . . . .....

5 ON INGRAM'S CONJECTURE . . . . ......

5.1 Three-Step Program . . . . . .....
5.2 Proof of Step 2 . . . . . .....


5.3 Proof of Step 3 . . . . . . .....
5.4 Ingram's Conjecture Follows from the Pseudo-isotopy Conjecture .
5.5 Sufficient Condition for the Pseudo-isotopy Conjecture . .

REFERENCES . . . . . . . . .

BIOGRAPHICAL SKETCH . . . . . . .....

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LIST OF FIGURES
Figure page

2-1 Projections of p-wrapping points of C, (*) and projections of (p+1)-
wrapping points of C, (o) . . . . . . 17

2-2 Projections and p-levels of p-wrapping points of C . . ... 18














Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Phil. .. 1hi

ON THE CLASSIFICATION OF INVERSE LIMIT SPACES OF TENT MAPS

By
Slagjana Jakimovik

August 2005

C'! i': James E. Keesling
Major Department: Mathematics

Let fs and ft be tent maps on the unit interval with s,t c (1,2]. In this

dissertation we give a new proof of the fact that if the turning points of fs and ft

are periodic, then the inverse limit spaces (I, fs) and (I, ft) are homeomorphic if and

only if s = t. This theorem was first proved by Kailhofer. Our proof simplifies the

proof of Kailhofer. We also prove that for a tent map fs with slope s e (/2-, 2] and

a periodic turning point, for any homeomorphism h of the inverse limit space (I, fs)

into itself, there exists an integer N and there exists an integer k such that hN and

a. are isotopic. With no assumption on the orbit of the turning point of fs, we make

a conjecture that any homeomorphism of the inverse limit space (I, fs) into itself

is pseudo-isotopic to an iterate of the shift homeomorphism. We give a sufficient

condition for our conjecture to hold. Finally, we prove that Ingram's conjecture

follows from our conjecture.














CHAPTER 1
INTRODUCTION

Given a continuous map f of a one-dimensional space to itself, one may form

an inverse limit space by using f repeatedly as the bonding map. Spaces formed in

this way commonly appear as attractors in dynamical systems [1, 3, 7, 15, 20, 35].

This motivates the study of such inverse systems. In the case of solenoids, there is a

well-known characterization (see Aarts and Fokkink [1], and Keesling [25]).

Consider the inverse limit space for the inverse system where the inverse system

spaces are each the interval [0, 1] and the bonding maps are each some tent map

fs(x) = min{s x, s (1 x)} for x E [0,1] and s E [1, 2]. This inverse limit space has

been studied extensively. Any unimodal map without wandering intervals, restrictive

intervals, or periodic attractors is conjugate to a tent map (see de Melo and van

Strien [27]). As conjugate maps have homeomorphic inverse limit spaces, the family

of tent maps is more inclusive than it seems at first glance. Thus, it is natural to try

to determine when two such inverse limits are homeomorphic. Ingram [22] posed the

following question.

If 1 < s < t < 2, is the inverse limit space lim{I, fs}j0 o/i, ,,, ',q.. 'li different

from the inverse limit space lim{I, ft}o ?

The positive answer to this question became known as Ingram's conjecture.

Partial results exist [5, 12, 17, 19, 23, 24, 29, 31, 32, 34]. Using topological

methods Lois Kailhofer [23, 24] proved the following result.

Theorem (Kailhofer). Suppose that s, t E [V2, 2]. Let fs and ft be tent maps

with periodic turning points. If K(fs) / K(ft), then (Is, f) and (It, ft) are not

homeomorphic.









In this theorem, (Is, fs) and (It, ft) are the cores of the inverse limit spaces
lim{I, fs}o and lim{I, ft}ro, respectively. The theorem implies that if the inverse

limit spaces lim{I, fs}o0 and lim{I, ft}ro are homeomorphic, then s = t under the

given assumptions.
Remark 1.0.1 One can extend the same result for the whole interval s e (1, 2]

in the following way. For s c (1, V2], there are two intervals J1 and J2 in the core

Is of fs with pairwise disjoint interiors such that f f|j and f)2 | are topologically
conjugate to fs2 11.2 It follows that for s E (1, V2], the core of the inverse limit
space lim{I, fs}j is determined by the core of the inverse limit space lim{I, fs2} 0.
Therefore, it is enough to consider tent maps with slopes in (V2, 2].

Son i Stimac [32] generalizes results on the classification of inverse limit spaces

of tent maps with finite critical (turning) orbit, that is, tent maps with periodic
turning point or strictly pre-periodic turning point. In this paper, she does not prove

the periodic case, for which she refers to Kailhofer [23, 24]. Stimac -ir--.- -I that the
approach she develops, with some changes indicated later, yields the proof of the

periodic case as well. She used methods of symbolic dynamics and coding to prove

the following result. For tent maps fs and ft with slopes s, t c (V2, 2] and such that
each of the tent maps has pre-periodic turning point, if logs is not a rational number,

than the inverse limit spaces lim{I, fs}jo and lim{I, ft}ro are not homeomorphic.

In the dissertation we give a simplified proof of Kailhofer's theorem. Our proof
uses some of Kailhofer's results [23] together with some new results. In the second

section of C'!i pter 2 we give definitions introduced by Kailhofer and results which
we use to prove Theorem 3.2.5. For some of her results we give new proofs and for

some of her results we give modified proofs, in which case we indicate so. For the

proofs to the rest of her results which we include in the dissertation, we refer the
reader to Kailhofer [23]. The contents of this entire section serve the purpose of

understanding only the proof of Theorem 3.2.5, which is a key step in proving our









main result. Nowhere else in the dissertation do we refer to definitions or results

from this section.

One of the results proved in the dissertation is of particular interest in itself.

Isotopy theorem. Let s C (/2, 2] such that the turning point of the tent map

fs is periodic. Let Is = [fj(c), fs(c)] be the core of f,. Let X, = (Is, fs) be the

inverse limit of the core. Let h be 'i.i1 homeomorphism of X,. Then there exist a

positive integer n and an integer k such that h" is isotopic to a-, where U is the

shift homeomorphism on Xs.

This result was the motivation to start working on this problem. With this in

mind, Keesling -i .-.- -1-. 1 reading Kailhofer's paper [23] and considering the following

problem. For s c (V2, 2] such that the tent map fs has periodic critical point,

given a homeomorphism h of the core X, into itself such that h leaves the endpoints

invariant, does there exist an integer N such that h and the shift homeomorphism

aN stay at a bounded distance on each composant of the core? The distance we refer

to here is a new metric d defined on each composant of the core, which is possible

to define only after establishing that any proper sub-continuum of the core is an

arc. After her visit to University of Florida in the fall 2003, Kailhofer -ir-.-', -I .1 the

outline of the proof of Theorem 3.2.5. This result allows us to prove a weakened

version of the Isotopy theorem in C'!i Ipter 3.

Pseudo-isotopy theorem. Let s c ( 2] such that the turning point of the

tent map fs is periodic. Let Is = [f,2(c), fs(c)] be the core of fs. Let X, = (Is, fs)

be the inverse limit of the core. Let h be i.1 homeomorphism of Xs such that

h leaves the endpoints invariant. Then there exists an integer N such that h is

pseudo-isotopic to ua.

In other words, for a tent map fs with slope s c (v/2, 2] with periodic turning

point, for any homeomorphism h of the core into itself which leaves the endpoints

invariant, there exists an integer N such that h and a, permute the composants of

X, in the same way. If h and a. were isotopic, then it would easily follow that the









composants of X8 are permuted by them in the same way. In the simplified proof of

Kailhofer's theorem, we only need the Pseudo-isotopy theorem. The rest follows from

the following two results. For a tent map fs with slope s c (V2, 2] with periodic

turning point, for any integer m, the number of composants of the core mapped to

themselves by a' is equal to the number of points in Is mapped to themselves by

fm. For tent maps fs and ft with slopes -2 < s < t < 2 such that the turning

point is periodic of period no under both maps, the number of points in Is mapped

to themselves by fJ0o is strictly less than the number of points in Is mapped to

themselves by fto.

It is only in C'!i pter 4 that we actually prove the Isotopy theorem. For that

purpose we give a construction of neighborhoods of arcs in the core with the

necessary distinction between arcs which do not contain an endpoint of the core and

ones that do contain an endpoint of the core. A neighborhood of an arc in the core

which does not contain an endpoint is homeomorphic to the Cantor set cross an

interval, and a neighborhood of an arc in the core that contains an endpoint of the

core is homeomorphic to the Cantor set cross an interval with the endpoints on one

side of these intervals glued appropriately.

In the last chapter of the dissertation, we propose a three-step program for

proving Ingram's conjecture with no assumptions on the orbit of the turning points

of the tent maps fs and ft with s, t c (1, 2]. The first step consists of the following

conjecture.

Pseudo-isotopy conjecture. Let s c (v/2,2]. For wi; homeomorphism

h : (I, fs) -> (I, fs), there exists an integer k such that hx., and a Ix, are pseudo-

isotopic, where Xs is the core of the inverse limit space (I, fs).

We give a sufficient condition for the pseudo-isotopy conjecture to hold: there

exists an integer N such that h and ao stay bounded distance on the open ray

entwined with the core. We refer to this ray as the 1 &I". The distance referred

to is the d metric defined on the tail in the same fashion as it was defined on each









composant of the core of a tent map with periodic turning point; in this case the

d metric gives the topology of the tail. We also give the proof to the second and

the third step, as well as the proof of Ingram's conjecture under the assumption

that the pseudo-isotopy conjecture holds. In the second step we prove that each

composant of the core mapped to itself by a, for some integer k, has a unique point

mapped to itself by a,. In the third step we prove that if fs and ft are tent maps

with slopes 1 < s < t < 2, then for each positive integer n, the number of points

mapped to themselves by f,' is less or equal to the number of points mapped to

themselves by Jft; moreover, there exists a positive integer N such that the number

of points mapped to themselves by f is strictly less than the number of points

mapped to themselves by ft". The proof of the third step requires use of well known

results about the entropy of maps in the tent family and the entropy of maps in the

quadratic family. We include only a general description of the notion of entropy and

state the results we use with the appropriate references.














CHAPTER 2
PRELIMINARIES

2.1 General Definitions and Results

We denote N {0,1, 2,.. }, N+ {1, 2,. *}, R (-o, 00) and R+ [0, oc).

Let X be a topological space. X is an arc if there exists a homeomorphism

from X onto [0, 1]. A maximal connected subspace of X is called a component of

X. A compact connected metric space is called a continuum.

Let X be a continuum. The composant of x E X is the union of all proper

sub-continua of X that contain x. Composants are dense that is, every point

of a continuum X is a limit point of any composant of X [21, Theorem 3-44].

Every composant of a continuum X is the union of a countable number of proper

subcontinua of X [21, Theorem 3-45].

Let X be a continuum. A sub-continuum T of X is an end continuum in X

if for any two sub-continua A and B of X such that T C A and T C B, it is true

that either A C B or B C A. A point x E X is an endpoint of X if {x} is an

end-continuum in X. Note that any homeomorphism from one continuum to another

must send endpoints to endpoints. Thus the cardinality of the set of endpoints of a

continuum is a topological invariant.

A continuum X is indecomposable if it is not the union of two proper sub-

continua. An indecomposable continuum X has uncountably many composants

[21, Theorem 3-46], and the composants of an indecomposable continuum X are

di-.-iil [21, Theorem 3-47]. Indecomposable continue first appeared as pathological

examples. But they have become just the opposite of what some mathematicians

in the beginning considered -' being monstrous things created by set theoretic

topologists for some evil (but purely mathematical) pui ..p- [26, page 107].








2.1.1 Inverse Limit Spaces of Maps
Definition 2.1.1 Let {Xi, di}io be a collection of compact metric spaces each with
a metric di bounded by 1, and such that for each i, fi : Xi+1 -> Xi is a continuous
map. The inverse limit space of the inverse limit system {Xi, fi}i'o is the set
lim{Xi, fi}i {x = (xo, x1, )|x e UoXL, fdxx) = x, e N},
with a metric d given by


For each i E N, 7 denotes the projection map from IioXi into Xi.
If Xi = X and f, f for all i, the inverse limit space is denoted (X, f).
Usually, the subscript/superscript notation jf used in the literature on inverse
limits denotes the mapping of the j-th factor space to the i-th factor space. But, in
the dissertation, as already adopted in papers considering inverse limit spaces using
single bonding maps from parameterized families, we use ff to denote the k-fold
composition of the map fs with itself, where s is the parameter.
Remark 2.1.2 An inverse limit space lim{Xi, fi}io is a continuum if Xi is a
continuum for every i e N [30, Theorem 2.4].
Definition 2.1.3 The map a (X, f) (X, f) 1/, fi7, by
o(xo, x ,---) = (f (xo), xo, x ,---)
is called the shift homeomorphism, or the induced homeomorphism.
Definition 2.1.4 A continuous map f : [a, b] [a, b] is called unimodal if there
exists a unique point c such that f [a,c] is increasing and f [c,b] is decreasing.
The point c is called a turning point, or a critical point.
Definition 2.1.5 Let f : I --- I be a map of the interval into itself. A point x is
said to be a periodic point of f with period k > 0 if fk(x) x and fZ(x) / x for
0 < n < k.
Definition 2.1.6 Let f : I -> I be a map of the interval into itself. Let x C I. The
set of all points y c I such that ff (x) -> y for some sequence of integers nk 00,
is called the uw-limit set of x, and is denoted by uw(x).









In other words, w(x) = nk>O cl(Un>kf (x)).
Definition 2.1.7 Let f : I -+ I be a map of the interval into itself and let x be

a periodic point of f with period k. The set of points x, f(x), f2(x),... fk-l(x) is

called the orbit of x under f. The basin of x is the set of points whose U-limit set

contains x. We -.,1; that x is an attracting periodic point and that its orbit is an

attracting periodic orbit if its basin contains an open set.

2.1.2 Itineraries

Let f : [a, b] -+ [a, b] be a unimodal map with turning point c.

Definition 2.1.8 For each x c [a, b], the forward itinerary of x, denoted
I(x) = bobib2... is a sequence of R's and L's such that

1. bi R if f'(x) > c,

2. bi L if f'(x) < c, and
3. bi C if f'(x) c,

with the convention that the itinerary stops after the first C.
Definition 2.1.9 A sequence A of -,,;l,l..1- L, R, C is said to be admissible if

either A is infinite sequence of R's and L's, or if A is a finite (or pil) sequence of

R's and L's followed by a C.
Every itinerary is admissible. For two admissible sequences A and B, AB

denotes the concatenation of A and B, A" = AA ... A with A repeated n times, and
A =- AA ... with A repeated indefinitely.

Definition 2.1.10 The itinerary of f(c) is called the kneading sequence of f and

it is denoted K(f).
Definition 2.1.11 A finite sequence A of R's and L's is said to have even parity

if the number of R's in A is even. Otherwise A is said to have odd parity.
Definition 2.1.12 The parity-lexicographical ordering on the set of itineraries
is 1. I;,' as follows. Set L < C < R. Let I -= 'i'i .. and I' = be two

distinct itineraries and let k be the first index where the itineraries differ.









1. If k 0, then I < I' if and only if wo < vo.

2. If k > 1, and wa'i 1'-1_ = ,,,,"1 '" ,-1 has even y.p'1,; then I < I' if and

only if ,,, < Vk.

3. If k > 1, and wow1 ... ,- 1 vovi -1 has odd y.1' .,; then I < I' if and

only if ,, < i' -

Lemma 2.1.13 Let x,x' E [a, b].

1. If I(x) < I(x'), then x < x' [18, Lemma II.1.2].

2. If x < x', then I(x) < I(x') [18, Lemma II.1.3].

Definition 2.1.14 We /. i,.': the shift operation on the set of itineraries in

the following way. Let I = aoala2 ... be an itinerary such that I / C. D. /7i,

al = ala2a3 If I = C, cI is ;,,./. 7, u.,1

We use the same symbol, a, to denote both the shift homeomorphism on an

inverse limit space of a map, and the shift operation on the set of itineraries of a

unimodal map. The context in which it is used should indicate whether it denotes

the shift homeomorphism on an inverse limit space of a map or the shift operation

on the set of itineraries of a unimodal map.

Observe that I(x) I (f(x)) if x c.

Definition 2.1.15 A sequence A is called maximal if A is admissible and if

a kA < A for k = 1, 2, n when A is finite and has length n, and for k = 1, 2, -

when A is infinite.

For a unimodal map f : [a, b] -> [a, b] with turning point c, f(c) > f(x) for

all x c [a, b]\{c}, hence by Lemma 2.1.13 it follows that the kneading sequence is

maximal.

Definition 2.1.16 Let f : [a, b] [a,b] be a unimodal map. Let A be an admissible

sequence. We -.,;/ that A is dominated by the kneading sequence K(f),

A << K(f), if for all non-negative integers k,

1. a kA < K(f) if K(f) is infinite,

2. okA < (DL)0 if K(f) = DC and D has even j,.;.








3. k4A < (DR)< if K(f) = DC and D has odd pji,./;i
The following theorem is of great importance to us since we use it to prove two
key steps in both C'! ipter 3 and 5.
Theorem 2.1.17 [18, Theorem 11.3.8] Let f : [a, b] [a, b] be a unimodal map.
Let A be an admissible sequence. If I(a) < A << K(f), then there exists a point
x E [a, b] such that I(x) = A.
Definition 2.1.18 The modified forward itinerary of f(c), denoted I'(f(c)), is
1, i.,, ,1 as follows.
1. If K(f) a0oaa2 is infinite, let I'(f(c)) K(f).
2. If K(f) = ao ... ano-C is jinite, let '(f(c)) = (a ... a where

ano-1 replaces the terminal C in K(f) such that aoal ... ano-1 < K (f).
Definition 2.1.19 Let x = (xo,xi, ) be a point in the inverse limit space (I, f).
The backward itinerary of x, denoted B(x) = bobb2... is a sequence of R's and
L's such that
1. bi = R if xi > c and bi = L if xi < c
2. if xi = c for some i > 0, then bob1 b i- = a i-ai-2 alao, where
I'(f(c)) aoal .
D.f i.- Bf {B(x)|x C (I, f)}.

Remark 2.1.20 Suppose that c is periodic with period no. Let x c (I, f). If Xi / c
for all i C N, or if xi = c for infinitely many i c N, then x has exactly one backward
itinerary. If xi = c for finitely many i e N, then x has two backward itineraries that
differ at only one coordinate, max{i E N|xi = c}.

2.1.3 Tent Maps
Definition 2.1.21 The one-parameter f',. l,; of maps f I I I 17 ,.1 by

fs(x) -min{s -x,s (1 x)},
for x C I and s C [/2, 2] is known as the fi,.;l,/ of tent maps.
The tent map fs is unimodal for all s c [/2, 2].









Remark 2.1.22 For s > 1, if x < x' are two points in I, there exists a positive
integer k such that f([x,x']) = I and fk(x) is on one side of c and ff(x') is
on the other side of c. Hence 1s(x') / 1s(x'). By Lemma 2.1.13, it follows that
I,(x) < I,(x'). Therefore, for s > 1, the map x -- I,(x) is an injective map.
Denote ci = fj(c) for i = 0, 1, 2, no 1. Denote IL [c2, C], IR [c, C1
and Is [c2, Ci]. For s E (1, 2] the interval Is = [C2, C1 is invariant under fs and fs
is locally eventually onto on Is, that is, for every non-degenerate interval J C I,
there exists an n > 0 such that f,(J) = Is. The interval I, is known as the core
of fs. To denote the n-th coordinate in the inverse limit space, we use In instead of
(IJs). The set X, = (Is, fs) is called the core of the inverse limit space (I, fs).

For s C (/2, 2], the inverse limit space (I, fs) is equal to the union of Xs
and an open ray having Xs as its limit set. We know that X, is indecomposable
for s C (-/2, 2]. In the third chapter (Lemma 3.1.1), we prove that under the
assumption that c is periodic, every proper sub-continuum of X, is an arc hence,
every composant is a union of arcs.
The composant Cx of x C X, is the set of all points in X, with backward
itineraries eventually identical to B(x) that is, y C Cx if and only if the backward
itineraries of x and y differ in at most finitely many coordinates. This follows from
Lemma 3.1.1, as well.

For s C (/2, 2] such that the turning point c of f, has period no, denote

Ci (ci,ci_i,--- ,c,c,c0_no l ,c Ci+l) for i= 0,1,2, *.. no 1.
The following lemma is a well-known result and appears in several publications.
Barge and Martin [8] describe the basic construction of endpoints in the inverse limit

space (X, f) of a map f : X -- X.
Lemma 2.1.23 For s c (v/2, 2] such that the turning point c of f, has period no, the
endpoints of X, are c, Cl, .- Cno- .
It is easy to observe that there are at least countably many distinct inverse
limit spaces, since for any positive integer no, the core of the inverse limit space of









a tent map with periodic turning point of period no has exactly no endpoints, and

endpoints are topological invariants. In fact, Barge and Diamond [6] proved that

uncountably many distinct inverse limit spaces can be formed with unimodal maps

as bonding maps.

In the remainder of this chapter, consider s c (V2, 2] fixed such that the turning

point c of fs has period no.

2.2 Definitions and Results from Kailhofer's Paper

In this subsection we give some general definitions and the definitions and

results from Kailhofer's paper [23] necessary to derive the proof of Theorem 3.2.5.

We give proofs to some of the results. These proofs are either different from the ones

given by Kailhofer [23], or they are modified versions of her proofs, in which case we

indicate so.

Let s C (-2, 2] be such that the turning point c is periodic of period no under

the tent map fs.

Definition 2.2.1 Let w = <'uii1w2...* B>f. D .' i

A, = {x c XsT1i(x) cE w1, for all i c N},

Al _an(AA).

In other words, Aw is the set of points in X, with backward itinerary w.

The following lemma p1l i, an important role in understanding the nature of the

composants of the core Xs, so we include the proof as given by Kailhofer [23].

Lemma 2.2.2 [23, Lemma 4] Let w C Bf,. There exist 0 < i / j < no such that

To A is a homeomorphism onto [ci, cj].
Proof. Let w = n,'1 E Bf,. Let fe, denote the injective map fs| i for n C N.

For all n c N, define


1- = f 1 -1(f (( .. f, ( ..)).
11. Wn Jn-1 ( (wIWO) )).









Note that the set Aw of points in X8 with backward itinerary exactly w is the

inverse limit space {1., fw+i }Io0. Since each of the spaces I, is a continuum, by [30,

Theorem 20], A, is a continuum and 7iA| is a homeomorphism for all i E N.

To prove this lemma, we have to show the following.

1. I, is a non-degenerate interval for all n c N,

2. f8(Is 1) = I for every n > no, and

3. f7(Ijj) [ [c,cjc] for some 0 < i,j < no.
Suppose there exists a positive integer n such that I, is a singleton. Let

m = min{n c N| I is a singleton}; then 1,, {c2}. However, for every n E N there

exists t, e [c2, Ci such that f+l is monotone and [c2, tn] is a non-degenerate

interval. Then, by monotonicity of the map x -- I,(x), it follows that [c2, tT C 17,

which contradicts I, = {c2}. Thus the first claim holds.

Suppose there exists an n > no such that f8(I1+) C In. Then C3 is in the

interior of the interval I^. This leads to ci in the interior of In (no 2), which is a

contradiction with cl is an endpoint of Is. Thus the second claim holds.

The third claim follows from the construction of I". -

Corollary 2.2.3 [23, Corollary 5] Let w c Bf,. Then for each positive integer n,
- | j is a homeomorphism and 7rT(A1) = o(Aw,).

Remark 2.2.4 Note that for an interval J C [c2, Ci], fs(fl1(J)) C J if and only if

C3 is in the interior of J, where fL denotes the injective map f. j, but for n > no,

f((fW1)(J)) = (fL1)"-1(J).
Remark 2.2.5 Lemma 2.2.2 shows that each A, is a non-degenerate arc contained

in a single composant.

Remark 2.2.6 Note that if x E Xs is an endpoint of A, and x / c for any

i = 0, 1, ... no 1, then x has two backward itineraries.
Remark 2.2.7 Each composant C in X, has the property that there is a continuous

bijection either from R+ to C or from R to C.









2.2.1 Chainings of the Core

An open cover U = {U}i 1 of a topological space X is said to be a chaining of

the space X if U n Uj / 0 if and only if |i j| < 1. If U= {U}i 1 and V = {Vj}j

are chainings of a topological space X, we w that U refines V, U -4 V, if for every

1 < i < n, there is 1 < j < m such that Ui C Vj.

Definition 2.2.8 Let p be a positive integer and 0 < j < no. D. iu..

p,j ={x G X, | -, ,, (x) = jcj},
Ino-1) .


The elements of 4p are called p-special points.

Definition 2.2.9 Let n, m be positive integers. D. 7i,.

E, = T(o),

Pn,m= {z e Inx, y E E, 3k c {0,1, 2,..., 2m}

such that (x, y) n E, = 0 and z = kx+(2k)y}

For all positive integers n and m, E,, partitions I, into finitely many intervals

and Pn,m refines that partition by dividing each interval into 2" subintervals.

Remark 2.2.10 The definition of Pn,m and the fact that fs|iL and fs1iR are linear

maps ensure that fs(Pn+,m) = Pn,m, for all positive integers n and m.

Definition 2.2.11 Let n, m be positive integers and let x E Pn,m.

If x / C2, set y = max{w e Pn,mx > w}. If x / c1, set z = min{w e Pn,m x <
w}. D. fin,, (y,z) if x c (c2,ci),> [x, Z) if x = and l (y,z] if

x c. D. fii ,, -_-1(,m). Let

Ln,m {\n,m x c Pn,m},
[nim = {Lm XeC Pn,m}.

Lemma 2.2.12 [23, Lemma 16] Let n, m, i,j be positive integers. Then

1. Ln,m is a chaining of In and Ln,m is a 1,i',:,':,j of X,

2. [n,m r. fn, ij if n > i, m > j

3. If x GE o, then there exists a unique link E Cn,m, such that x e

4. mesh(In,m) -+ 0 as n --+ o and m -+ oc.







15
Definition 2.2.13 Let p, m, k be positive integers with 0 < k < no. D. 1pi.k lk to be

the link in the chaining tpno,m of X, which contains ck.

Proposition 2.2.14 [23, Proposition 25] Let p, m, k be positive integers with

0 < k < no. If D is a component of Cc n ~m, then D contains '.'. /a/ one element of

4)p,k.

Proof. By definition, the link lpom of the chaining Lpno,m of Ipno contains only

one point from the orbit of c, namely ck. Hence the only p-special points possibly

contained in D C ~~n are points from )p,k.

For each backward itinerary w of points in Cc, by Corollary 2.2.3, we have that
T, ..(APo) = To(Aw) = [ci, cj] for some i,j E {0, 1, no 1}, and 7, ., .. is a

homeomorphism. Thus D contains at most one endpoint of AV0. It follows that D

is contained in at most two distinct arcs APo and AW0 hence D contains at most

one element of )p,k.

Assume D contains no points from I)p,k. Then D contains no p-special points as

the only p-special points possibly in D come from I)p,k. Note that the fact that for

any backward itinerary w, 7 ..(Apo) = [ci, cj] for some i,j E {0, 1, no- 1} means

that the endpoints of AP,0 are p-special points. Hence D C AP,0 for some backward

itinerary w. Since lk, contains only one point from the orbit of c, namely ck, one

of the endpoints of 7, ..(AP,0) has to be exactly Ck. Let x be the endpoint of D with
., ,,(X) = Ck. As pkm -.(lo,m), there is a component D' / D of Cc n which

contains x. Let y e D. Since D and D' are two distinct components of Cc n fm,

there exists a point z between x and y such that z ~,. Then 7, ,,z) g lkf. So,

we have three distinct points x, z, y E AP0n where z is between x and y, and such

that 7, ,,(X) 7e (APto) n l 7 .. G,, (Ano) n Il, and 7, ,,(z) e 7, ,,(APo)

but 7, ,,(z) lkm. This is a contradiction with the fact that the projection 7, ,,|

is a homeomorphism and the set An is connected. -









2.2.2 Composant of c

In this subsection we focus our attention on the composant of c, Cc. We include

definitions of the following concepts introduced by Kailhofer [23]: a p-wrapping

point, the p-level of a p-wrapping point, a p-symmetric arc, a p-pseudo-symmetric

arc, a p-gap, gaps of the same type and arcs with the same p-shape.

Definition 2.2.15 For each positive integer p, 1, ,'.'

Wp = {xe cCcv / w e Bf, such that xe A0o n Ao} U {c}.
If x E Wp, then x is called a p-wrapping point.

There is a natural order on the set of all p-wrapping points with x < y if

h-'(x) < h-1(y) for any continuous bijection h : R+ Cc.

The following lemma gives a very useful characterization of p-wrapping points.

Lemma 2.2.16 [23, Lemma 24] Let p be a positive integer. The following hold.

1. Wp {x E Cc E n > pno such that -r,(x) = c}

2. Wp+i C p+1 C Wp

s. 7' (Wp) =wp+,.

Proof. 1. Let x e Xs. Then x c Wp if and only if there exist two distinct backward

itineraries v and w such that x c A n A ,o = P-(A,) n o Po(A,,). This means

that v and w differ at exactly one component, namely the one in which the last c

appears in x, and that component comes not earlier than pno. Thus r,(x2) = c for

some n > pno.

2. Suppose x c Wp+i. There exists a positive integer n > (p + 1)no such

that 7,,(x) = c. Then 7T(p+l)no(x) c= c for some i {0,, 1 no 1} hence

x C Dp+l,i C Dp+1. Thus Wp+i C 4p+i.

Since c is periodic of period no, 7(p+1l)no (x) c- for some i {0,, 1, no 1}

implies that rT,(x) = c for some positive integer n > pno. Thus 4p+l C Wp.
3. Suppose x (xo,x, -) e Wp. Then u7(x) (f (xo), ,s(x),x0o,x, ).

Since 7,F(x) c for some n > pno, we have that .,(o.("()) c where

n+no > (p+ 1)no. Thus uo (Wp) C Wp+. Similarly, Wp+l C o^o(Wp). w







17
Cl
C4 "






c
C
C4

5(p+1) ( C)C3

C2
Figure 2-1. Projections of p-wrapping points of C, (*) and projections of (p+1)-
wrapping points of C, (o)


Example 2.2.17 Let T be the tent map with kneading sequence RLRRC. Figure

2-1 shows the 5p-projections of the p-wrapping points of C,, marked by *, and

5(p + 1)-projections of the (p + 1)-wrapping points of Cc, marked by o.

Definition 2.2.18 Let p be a positive integer. Let x be a p-wr ,j,,: .'j point other

than c. Let k be the unique non-negative integer such that

pno + k = max{n7,(x) = c}.

DA fI,. the p-level of x by Lp(x) = k. DA f,.- Lp(c) = 00.

The set {Lp(x)\x c W \{c}} is unbounded.

Example 2.2.19 Let T be the tent map with kneading sequence RLRRC. Figure

2-2 shows the 5p-projections of the p-wrapping points of Cc, marked by *, and the

p-levels of the corresponding p-wrapping points of Cc.

Note that x CE Wp+ if and only if Lp(x) > no.

Proposition 2.2.20 [23, Proposition 29] Let p be a positive integer. Let w and

v be two distinct points in CC such that w < v and -, ,,(w) = 7, ,,(v). There

exists a p-wr' j,: .'j point z such that w < z < v. Moreover, if both w and v are

p-wrapping points, then there exists a p-wrapping point z such that w < z < v and

Lp(z) > min {Lp(w), Lp() }.







c- 18

C4









C2


p-level: 00 1 3 1 020 4 020 6 020 4 020 1 8 1 020 4 0


Figure 2-2. Projections and p-levels of p-wrapping points of C,


Definition 2.2.21 Fix a positive integer p. Let H be an arc in the composant of c

with Int(H) n Wp = {hi, h2, .. n-i} and OH = {ho, h,n}.

1. The arc H is p-symmetric if 7, ,,(ho) = 7 ,,(h,,) and Lp(hi) = Lp(h_-i) for

all positive integers 0 < i < n.

2. The arc H is p-pseudo-symmetric if 7, (hi) = 7 ,,(h,,-) for all positive

integers 0 < i < n.

Remark 2.2.22 Fix p e N. Let H be an arc in the composant of c. If H is

p-symmetric, then H is p-pseudo-symmetric.

Definition 2.2.23 Let p be a positive integer. Let H be a p-pseudo--',ini, I;, or

a p--ii, i,'. arc in the composant of c. Then n is even and the point h- is called
2
the center of H, denoted KH-

Remark 2.2.24 Let p c N and let H C Cc be an arc. If H is p-pseudo-symmetric,

then H is q-pseudo-symmetric for all q < p. If H is p-symmetric, then H is q-

symmetric for all positive integers q such that qno < pno + Lp(KH).

Proposition 2.2.25 [23, Proposition 34] Let p E N and let w C Wp\{c} such

that Lp(w) / 0. Let H be the union of all p--i.,in, I,. arcs with center w. There

exists a p-wr'.: ,',j point v E H such that Lp(v) > Lp(w) and such that there is no







19
p-wrapping point between w and v with p-level higher than Lp(w). Furthermore, v is

an end point of the arc H.

Proof. The following proof is a modified version of Kailhofer's proof. Let L =

Lp(w). Let x < y be p-wrapping points on each side of w such that Lp(x) > Lp(w),

Lp(y) > Lp(w) and no p-wrapping point between x and y has p-level higher than L.

We claim that neither x nor y is an interior point of H. Assume the opposite,
v x is an interior point of H. Since H is p-symmetric, there is a p-wrapping

point x' in Int(H) such that Lp(x) = Lp(x'). Then, (x) (- ,,(x'). By

Proposition 2.2.20, there exists a p-wrapping point z between x and x' with Lp(z) >

min{Lp(x), Lp(x')} = Lp(x). But Lp(x) > L. Since no p-wrapping point between

x and y has p-level higher than L, we have that either z y or z > y. This puts

y in Int(H) and y < x'. Since H is p-symmetric, there is a p-wrapping point y' in

Int(H) with x < y' such that Lp(y) = Lp(y'). By Proposition 2.2.20, there exists a

p-wrapping point z' between y and y' with Lp(z') > min{Lp(y), Lp(y')} = Lp(y). But

Lp(y) > L. This contradicts the fact that no p-wrapping point between x and y has

p-level higher than L. Thus, we proved that neither x nor y is an interior point of H.

Let A be the arc with endpoints x and y. Since no p-wrapping point between

x and y has p-level higher than Lp(w), -, 1,, 1iA is a homeomorphism. Note that
7, .. (w) = c and -, .., i,(A) is symmetric about c. Consider the smaller of the

intervals J. = ,., K(X), 1 -, (x)] and J2 = -, .., K(u), 1 -, .., (u)], both

of which are symmetric about 7 1, ,(w) c. Without loss of generality assume

J, C J2. Since f,(-, .1 (X)) fs(1 ., ,(x)), we have that ff(-., (X))
ffL(1 -, r,(x)). If we assume that -, ., (yu) 1 -, ., <(x), then

-., (y) fL(- ., i (y)) 1- (1 -, .. (x)) fL(-., (x)) ., .. (X),

hence by Proposition 2.2.20 it follows that there exists a p-wrapping point between x

and y with p-level higher than L, which contradicts our choice of x and y. Therefore,
7, .. c(y) 1 -, .. ,, c(x). Then x is the point whose existence we claimed. The arc
A' = -1, .,(r-, ,, c(x), 1 -, .. c(x)] n A is p-symmetric with center w and x as an







20
endpoint. Since H is the union of all p-symmetric arcs with center w, A' C H. Since

x Int(H), x is an endpoint of H. -

Remark 2.2.26 Let p be a positive integer. Let H be a p-symmetric arc in C, and

let L = Lp(KH). Proposition 2.2.25 implies that all the interior points in H have

p-levels less than L, consequently 7, .., r|H is a homeomorphism.

Definition 2.2.27 Let p be a positive integer. The set 4p,o partitions the composant

of c in coi"w..i'bl'; i,,.i; arcs called p-gaps.

Let p be a positive integer. For any p-gap H of Cc, c 7, ,,(Int(H)) and

-, ,,(OH) = {c}. The intersection of any two p-gaps of Cc is at most one point.

Lemma 2.2.28 For i;.,1 positive integer p, a p-g'j of Cc is p-z-i .:..

Proof. Fix p C N. Let H be a p-gap and OH = {y, z}. Let x C Int(H) be a p-

wrapping point with largest p-level, -w L. Then -, ,, |H is a homeomorphism and

i L,(X) c.

Suppose H is not p-symmetric. Then c is not the center of symmetry for

the interval with endpoints -, ., j(y) and ,(2). Hence f.(., ,, ()) /

fs( ..,i c(z)). It follows that there is a p-wrapping point w C Int(H) such that

fs(-, [,, i(w)) is equal to either fs(-, ., ,j(y)) or f(- .., (z)). As -, ,,(. ) =

-, ,,() c, this implies that 7, ,,(w) c which contradicts H being a p-gap. E

Definition 2.2.29 Let p and q be positive integers. Suppose G is a p-g'j' of the

composant of c with G n Wp = {go, gi, g}, and H is a q-'j'gr of the composant

of c with H n Wq = {ho, hi, hm}. The /./_- G and H are of the same type if

n = m and 7, ,,(g9) =- ,,(hi) for all positive integers 0 < i < n.

Proposition 2.2.30 [23, Proposition 41] Fix p, q C N. Let G be a p-'jl' of Cc and

let H be a q-,'j' of Cc. If Lp(KH) = Lq(ro), then G and H are the same type.

Proof. Let G n Wp {go, gi, g,} and H n Wq {ho, hi, ... hm}. Suppose

LP(KH) = Lq(rc) = N. Then -, ,,, 'IG and ,,, rIH are homeomorphisms. Also, the

interval -, ., (G) is symmetric about c -= ,(, (Gc), and the interval .., (H)

is symmetric about c = ,,, '(KH). Since G and H are gaps, the endpoints of








-, .,, r(G) and the endpoints of -, .., ;(H) are in f- N(c), but no points in the
interior of -, ,,, r(G) are in f.N(c) and no points in the interior of ,,, '(H) are
in f N(c). Hence, if we assumed that .., (G) / -, ., r(H), v ., (G) C
- ,,+N(H) then the endpoints of -, ., (G) are interior points of ,, i(H), which

contradicts the fact that no interior points of ,., r(H) are in f;N(c). Therefore
-, .., r(G) ., (H). Thus, n m and f(- ., (gi)) f(-. ,,+N(hi)), that
is, -, .,,( ) ,,(h ). -
Definition 2.2.31 Fix a positive integer p. Let G be a p-'j' of C,. The arcs
between two consecutive p-wr i'j,:,,j points in G are called p-legs of G.
The first p-gap in the composant of c is denoted Fp.
Lemma 2.2.32 Fix p E N. Let G be a p-gI' of Cc and L its first p-leg. Then
1. -, ,, (G) ., ,, (L) [23, Lemma 44]
2. L contains a (p 1)-,.,i. [23, Lemma 46]
3. G contains at least three (p l)-g,,,- [23, Corollary 50]
4. The first (p 1)-,t.) in G is the same type as Fp- 1 [23, Prop. 47].
Definition 2.2.33 Fix a positive integer p. D. iu. = Lp(RF).
Remark 2.2.34 Since the type of Fp does not depend on p, o does not depend on
p. Since Fp is contained in the first leg of Fp+,, the center of Fp is not a (p + 1)-
wrapping point hence =- Lp(RKF) < no. Note also that -, .. (K) = Cy.

2.2.3 "Fudged" Map hq,p

Now, let us consider a homeomorphism h : (I, fs) -- (I, fs) which leaves the
endpoint of Cc fixed, that is, h(c) = c. In general, if h(c) -= C, for some positive
integer 0 < i < no, consider the map h a o h.
Fix m, n, p, q E N such that h(Lqno,n) -~ Lpno,m. If h(cj) = ci for 0 < i, j < no,
then h(0no,n) C f0',m- This implies that h(4q,j) C 0,m. By Proposition 2.2.14,
every component of no0,n contains exactly one element of 4q,j. Since two consecutive
points of 4q lie in two different links, each component of o,m contains at most one









element of h(4)qj). Thus, h induces a one-to-one map hq,p 4 q -- 4)p, defined as

follows.

Definition 2.2.35 Fix m, n,p, q E N such that h(--qno,n) Cpno,m- If w E 4)q,j and

h(cj) = c, for 0 < i, j < no, then hq,p(W) is 1, 1;, as the unique element of 4p,i that

lies in the same component of fgo,,m as h(w).

If G is an arc in the composant of c with G = {x, y} C 4q, let hq,p(G) be the

arc with endpoints hq,p(x) and hq,p(y).

Definition 2.2.36 Fix a positive integer q. Suppose G and H are arcs in Cc with

G W {go,gi, ,g} and H Wq {ho, h, hm}. The arcs G and H

are said to have the same q-shape if n m and if -7 ,,(g) = T7 ,,(hi) for each

0 < i < n. The arcs G and H are said to have a mirrored q-shape if n m and if

- ..(gi) = 7 ..(h-_) for each 0 < i < n.

Proposition 2.2.37 [23, Proposition 65] Fix positive integers m, n,p, q such that

h(qno,n) -4 pno,m. Suppose G and H are arcs in Cc with endpoints in 4)q. If G and

H have the same q-shape (a mirrored q-shape), then hq,p(G) and hq,p(H) have the

same p-shape (a mirrored p-shape).

Corollary 2.2.38 [23, Corollary 67] Fix positive integers m, n,p,q such that

h(Iqno,n) -< pno,m. If H is a q-pseudo--imm, ,.5 arc in the composant of c with

OH C 4q, then hq,p(H) is p-pseudo- Imm, i,/ .

Lemma 2.2.39 [23, Lemma 68] Let p E N. Let G and H be distinct p-pseudo-
-,iii, 1 ,'. arcs in the composant of c such that c c G and c c H. Then G C H if

and only if Lp(Kc) < Lp(KH).

Theorem 2.2.40 [23, Theorem 70] Fix m, n,p, q, u, v e N such that h(qno,Tn) -

ILpno,m -< h(uno,,). There exists an integer t such that

4)t,o D hq,p(4)q+l,o) D 4 t+1,o D hq,p(4q+2,0) D )t+2,0 D ".
Corollary 2.2.41 [23, Corollary 71] Fix m, n,p, q, u, v N such that h(qno,n) <

pno,m -< h(uno,v). If hq,p(Fq) = for some t c N, then hq,p(4q+k,o) = 4t+k,o for all
k cN+.














CHAPTER 3
ON THE CLASSIFICATION OF INVERSE LIMIT SPACES OF TENT MAPS
WITH PERIODIC TURNING POINT
3.1 Unravelled Composant of a Point and a New Metric d

In this entire chapter we work with tent maps with slopes in (/2, 2] such that

the turning point is periodic of period no. The following lemma is a well-known
result [13, 15].

Lemma 3.1.1 Suppose that A is a proper sub-continuum of X,. There is a non-

negative integer k such that 7Tk A is a homeomorphism. In particular, A is an arc.
Moreover, for wn; two points x, y c A, B(x) and B(y) agree after the first k entries.

Proof. If there is a nonnegative integer m such that for each j > m, c 7ji(A),
then fls [(A) is a homeomorphism for each j > m, and the conclusion follows easily.

So, we may assume that c E 7rj (A) for arbitrarily large integer j. Since c is periodic

under fs, it follows that for each nonnegative integer j, 7r (A) contains at least one of
the points c0o, Cl, co--1. Since f, is locally eventually onto, there is a nonnegative

integer k such that for each integer j > k, 7rj(A) contains exactly one of the points

co, Ci, cno-1. We complete the proof by showing that for each j > k the element
of {co, Cl, 0no-1} which is in 7rj(A) is an endpoint of 7rj(A). In particular, for

each j > k, c is not in the interior of 7rj(A), so fIs, (A) is a homeomorphism.
Suppose j > k. Since c is periodic, there is an integer m > j such that

Ci E 7Tm(A). Since c, is an endpoint of Is, it follows that c, is an endpoint of m(A).
Since c is not in the interior of m(A), fs, (A) is a homeomorphism. Thus c2 is an
endpoint of 7r_m1(A). If m 1 > j, we may repeat this argument and conclude that

C3 is an endpoint of Tm,-2(A). By repeating the argument inductively, it follows that
the element of {co, Cl, c0o-1} which is in 7rj(A) is an endpoint of 7rj(A). ]









Let x e Xs. By Remark 2.2.7, there is a natural order on the elements of

the composant of x. Cx with the order topology will be called the unravelled

composant of x.

Remark 3.1.2 Note that the map hq,p is order-preserving.

We will put a specific metric on the unravelled composant which is derived from

the inverse limit system. Let x, y be in the same composant C C Xs. Then there

is an arc A C C with endpoints x and y. By Lemma 3.1.1, there is a nonnegative

integer k such that 7Tk A is a homeomorphism. Define

d(x,y) =k 1k(X) -T k(Y)|-
Note that if m > k, then d(x, y) = _7. (x) 7m(y) Thus, d is well-defined for

every pair of points in the same composant C. We may consider (C, d) either as R+

or R depending on whether C has an endpoint or not.

We let denote the length of an arc under the metric d.

3.2 Homeomorphisms of X, and the Shift Homeomorphism uo

Definition 3.2.1 Suppose X is an indecomposable continuum and h, g : X -+ X

are homeomorphisms. We -.,1; that h and g are pseudo-isotopic if for wi.;1 x E X,

there is a proper sub-continuum A C X such that h(x) E A and g(x) c A.

In other words, two homeomorphisms of an indecomposable continuum into

itself are pseudo-isotopic if they permute the composants in the same way.

Theorem 3.2.2 Let hi, h2 : X, -+ X, be homeomorphisms which map Cc to

itself. Suppose that there exists M > 0 such that d(hi(z), h2(2)) < M for all

z c Cc. Then hi(Cx) = h2(C) for all x Xs. Furthermore, for every x Xs,

d(hi(x), h2()) < M.

Proof. Let x c Xs. If x c Cc, then hi(Cc) h2(Cc) by assumption.

Suppose that x Cc. Since Cc is dense in Xs, there is a sequence {xX}C 1

in Cc which converges to x. Then hi(xT) converges to hi(x) for i = 1, 2. Consider

the unique arcs A C Cc with endpoints hl(xT) and h2(xn). By assumption,








f(A,) = d(hi(x,), h2(xn)) < M for each n. Let k > 0 be an integer such that
M < Sk (c C2). Then 7k(A,) is a proper subset of [c2, C1 since the length of

Fk(A,) is less than 8 < c1 c2.
Let C(Xs) denote the space of nonempty sub-continua of X8 with the Hausdorff
metric. Then the projection 7k : X, -- [c2, c1] induces a continuous map 7k "

C(Xs) C([c2, C1]). Since C(Xs) is a compact metric space, the sequence {An}
of arcs in C, has a convergent subsequence {As, } converging to a subcontinuum
A E C(Xs). Note that hi(xj) converges to hi(x) and h2(xnj) converges to h2(x).
Hence hi(x), h2(x) c A. Since the induced map 7k : C(Xs) -> C([c2, ci]) is continuous,
7fk(A) has length at most M < Ci C2. Thus, A must be a proper sub-continuum of
X,. Therefore, hi(x) and h2(x) are in the same composant of X,. This implies that
hi(C1) h2(C).
Since x is arbitrary, the above also proves the last statement of the theorem. E
Recall that the shift homeomorphism, o-, : X, Xs, is defined by o-s((xo, x, ))

(fs(xo), xo, Xi, -).
Lemma 3.2.3 There is a positive integer B such that for 11;:1 p cE N the number of
legs in a l'-'j, is at most B.
Proof. Since f, is locally eventually onto, there is K E N such that if J is an
interval which contains two points in the orbit of c, then fK(J) L.
Fix p C N. Let H be a p-gap and L = Lp(KH). Let Hp be the arc connecting the
center of H and the right endpoint of H. Then -, ,, (Hip) is an interval with one
endpoint c and -, .., c HR is a homeomorphism. Note also that ffL( .,, c(HR))
-7 ..(Hp) = -, ..(H) is an interval with one endpoint c.
If fL',, ..i (HR) is linear, then there are at most two legs in H. Suppose

ff'L ,,, c(HR) is not linear. There is a least n c N such that f(-., ,,, c(HP))
contains two points in the orbit of c. This implies that H has at most 2L-, legs.
Since -, ,,(HR) is a proper subset of [c2, ci], we have that L n < K hence the
number of legs in H is at most 2K. F







26
Remark 3.2.4 One might be led to conjecture that the number of distinct types of

p-gaps in C, is no 1 for any p E N. However, for s such that the kneading sequence

is RLLRRRLC, there are at least eight p-gaps.

Theorem 3.2.5 Let h : X, Xs be a homeomorphism which leaves the endpoints,

ci for 0 < i < no 1, invariant. Then there exists an integer N and a positive

number M such that d(h(x), a,7x)) < M for all x Cc.

Proof. For convenience of referral, two points of any subset of Cc are said to be

.,i.i :ent in that set if the arc connecting those two points contains no other points of

that set. Note that if x and y are .,.Ii ,:ent in t)p, then d(x, y) < sKP.

By Lemma 2.2.12(5), given u,v E N, there are p, m E N such that Lpno,m -4

h(uno,0,), and there are q, r E N such that h(qno,r) -< tp0no,m. Fix p, m, q, r, u, v E N

such that h(qno,r) -< Lpno,m h(uno,v)-

Since Fq is q-symmetric, by Corollary 2.2.38, hq,p(Fq) is p-pseudo-symmetric and

hq,p(tFq) Khq,p(Fq) Let L = Lp( q,p(Fq)) and t be the largest positive integer with

the property tno < pno + L. Obviously t > p. Since h(ci) = ci for all 0 < i < no, we

have that ,.(F q,(Fq)) ( ..(q), which, by Remark 2.2.34, is equal to cy. From

Definition 2.2.33, we have that Lt(K ,) = p. Thus

Lp(KF,) = Lt(RFc) + (t P)no = + (t P)no = L L q,p(Fq).

Hence, by Lemma 2.2.39,

F = hq,p (Fq).

By Theorem 2.2.41 it follows that for every k c N+,

hq,p(+)q+k,o) 1)t+k,O.

By Remark 3.1.2, hq,p is order-preserving on the set )t+k,o for any k E N+. From

the definition of as and since a t-q)n is order preserving as well, it is easy to see that
s t-q)no q+k,O) -= t+k,o for any k E N+. Therefore, for every x cE 4q+1,0, we have

that hq,p(x) = t-q) (x). By Definition 2.2.35, for any x cE 4q+1,0, h(x) lies between

two .,li ,:ent p-special points, one of which is hq,p(x). Since the distance between two

special points is less than s'no, we have that









d(h(x(x), h (X)) < s-pn
for any x cE )q+1,0-
The length of any leg of a (t + 1)-gap is bounded by s(t+1)no as 7(t+1)no
restricted to the leg is a homeomorphism. Since the number of legs in a (t + 1)-gap
is bounded by B by Lemma 3.2.3, it follows that the length of a (t + 1)-gap is
bounded. Namely, if x and y are the endpoints of a (t + 1)-gap, then d(x, y) < 1,
where 1 = B- s(t+1)n.
Let N (t q)no and M so + 1. Let x G Cc. If X e 4)q+1,0, then
d(h (x), ,7N(X)) < spno < M.
If x 4)q+1,o, then there exist y and z .,i.11 ent in 4)q+1,o such that the (q + 1)-
gap whose end points are y and z, contains x. As y, z cE 4q+1,o, we have that
hq,p(y) = N, (y) and hq,p(z) = r(z). Since h is a homeomorphism, the arc
connecting h(y) and h(z) contains h(x). Similarly, the arc connecting acr(y) and
as(() contains (x).Thus,

d(h(x), N(X)) < max{d(7N(X), h(y)),d(7N(X), h(z))}.
Since oa sends a (q + 1)-gap to a (t + 1)-gap,
d(<(_),,7( y)) < 1.
Since h(y) lies between two .,li ,'ent p-special points, one of which is acr(y),
d(h(y),,N y)) < spn.
Therefore
d(<,(x), h(y)) < d((x), (y)) d((y(), h(y))
< d(N(z_), N(y)) + d((N(y),h(y))
< I + sPn = M.
Similarly, d( Corollary 3.2.6 (Pseudo-isotopy theorem) Let h : X, -- X, be a homeomor-
phism which leaves the endpoints, ci for 0 < i < no 1, invariant. Then there is an
integer N such that h(Cx) = a(C, ) for all x c Xs.









3.3 Number of Fixed Points of fk

We adopt the following notation. If k is a positive integer, we let F(ff) denote

the number of fixed points of ft in I.

Lemma 3.3.1 Suppose V2 < s < t < 2 and for each of the tent maps f8 and ft, the

turning point is periodic with period no. Then

F(f7o) < F(f0).

Proof. Since each point in the orbit of the turning point is a fixed point of f,0 and

the same holds for f10, we need only consider fixed points of f'o and t10o which are

not in the orbit of the turning point. Suppose y is a such a fixed point of fjo. Then

the forward itinerary 1(y) = S' for some sequence S of length no of L's and R's. By

Theorem 2.1.17 there is a fixed point z of ft0 with 1(z) = S.

We complete the proof by showing that there is a sequence T of length no of

L's and R's such that there is a fixed point of ft' with itinerary T', but no fixed

point of f7,0 has this itinerary. The itinerary of fs(c) is of the form DC where D is a

sequence of length no 1 of L's and R's. We can modify fs to construct a unimodal

map g with the same kneading sequence as fs such that on the orbit of c, g = fs,

but for a small non-degenerate interval J with right endpoint g(c) each point of J

is periodic under g with period no. The itinerary of a point in J other than g(c) is

of the form T', where T is a sequence of L's and R's of length no. Moreover, T is

shift maximal and T is either DR or DL. It follows that no fixed point of fJ0O has

itinerary T', but by Theorem 2.1.17 there is a fixed point of f70o with itinerary T'.

0
Lemma 3.3.2 Let s e ( 2,2]. For 1t:' integer m, the number of composants

mapped to themselves by a' is F(fm).

Proof. Without loss of generality we may assume that m > 0.

Our first claim is that there is at most one periodic point in each composant of

Xs. Suppose not. There is a composant C in X, with at least two distinct periodic

points of a-, w- x = (xo, xi, X2, ) and y = (yo, Yi, Y2, ). Then there is a positive









integer k such that a(x) x= and f(y) = y. In particular, f (xk-1) = k-1 and

ftk(yk-1) U/k-1. Note that Xk-1 / Uk-1. Since x and y are in the same composant,
they have eventually the same backward itinerary. Thus, there is some positive

integer N, such that for all n > N, x, and y,, are on the same side of c. (By this we

mean either both x, > c and y,. > c, or both xn, < c and yn, < c.) Since x and y are

periodic, it follows that for each integer j = 0, 1, k 1, xj and yj are on the same

side of c. Hence for each integer j > 0, fl(xk-1) and f,(yk-1) are on the same side of

c. This is impossible since xk-1 / Uk-1 and fs is a tent map with a slope s > 1. This

proves the first claim.

Our next claim is that each composant mapped to itself by a" contains a fixed

point of 0a. Suppose C is a composant in X8 with 0-(C) = C. If x, y c C, then

d(am (x), m()) smd(x, y). Hence a-crm is a contraction and has a fixed point. This

proves our second claim.

It follows from the claims that the number of composants mapped to themselves

by 0-m is equal to the number of points fixed by e"-. By definition of a8, this number

is equal to the number of fixed points of f,, F(f,). 1

3.4 Proof of the Main Theorem

Theorem 3.4.1 Let s,t c (/02,2) such that fs and ft have periodic turning points.

Then Xs and Xt are homeomorphic if and only if s = t.

Proof. It is well-known that if X, and Xt are homeomorphic and the turning point

of fs is periodic, then the turning point of ft is also periodic with the same period.

Thus, there is no loss in generality in assuming that the period of the turning points

for fs and ft are both periodic of period no. Suppose s < t. Assume there is a

homeomorphism g : X, -- Xt.

Consider the map h : Xt -4 Xt, h g o ao -1. Then h is a homeomorphism,

and it maps each composant with an endpoint to itself. By Corollary ??, there

is an integer N such that h(Cx) = tC) for all x E Xt. Since aj0 maps each







30
composant with an endpoint to itself, the same is true for h. Thus oa also maps

each endpoint of Xt to itself. By Lemma 3.3.2, the total number of composants

mapped to themselves by a '0 and hence by h is F(fO). Thus, the same is true for

oa. It follows that N N > no and no divides INI. But the number of composants

mapped to themselves by oa is F(ftN). Thus F(f~,) F(fN). On the other hand,

since s < t, by Lemma 4.7, F(fJ) < F(ft' ). Hence F(ftN) < F(ft' ), which is a

contradiction. -














CHAPTER 4
ISOTOPY THEOREM

In this chapter we prove the Isotopy theorem as stated in the Introduction. Let

s E (/2, 2]. Let fs be a tent map with periodic turning point with period no. We

have already shown that for any homeomorphism g : X, -- X, such that g leaves

all the endpoints {c-}010 fixed, there is a k such that g and a. are pseudo-isotopic

- that is, g and a, permute the composants of X, in precisely the same way. It is

clear that for any homeomorphism h : X, -- Xs, there is an n > 0 such that h"

leaves ci fixed for i = 0, 1, 2,... no 1. Let k be the integer such that h" and ao are
k
pseudo-isotopic. We now show that h" and ar are actually isotopic.

4.1 Construction of Neighborhoods

The following lemma is a well-known fact to the experts in this field.

Lemma 4.1.1 Suppose A is an arc in Xs not containing iw..; endpoint of Xs. Then

there is a neighborhood V of A homeomorphic to C x I, where C is a Cantor set.

The boundary of V will correspond to C x {0, 1}. Moreover, there is a positive integer

m such that Tm maps each component of V homeomo',id.. all/ onto its 'Ia./.:' in Im.

Proof. Let A be an arc in X, not containing any endpoint of Xs. By the proof

of Lemma 3.1.1, there is a positive integer m such that for each k > m, none of

the points ci are in 7k(A). In particular, Tk IA is a homeomorphism. Let z c A,

z (zo,zi,... ,z,...). Let C {y e XAyo = zo,Y z,1... ,y, zm}. Then C

is compact, totally disconnected, and every point is a limit point. Therefore C is a

Cantor set.

Let Jm = 7Tm(A). Fix y E C. Since Jm n {co, cl,..., co-0} = 0, for this

y c C, there is a sequence of intervals { Ji}im such that yi E Ji for each i > m and

fs(Ji+i) = Ji for each i > m.









We can extend the sequence { Ji}~m to { Ji}'o by Jo = (m(Jm), J1 =

f 1(Jm), -1 = fs(Jm). Then for all i e N, f,(J+i) Ji and y eG Ji.
Now Jm is homeomorphic to J(y) l= im{Ji, f} C X, by the projection

7,m : Xs Im- Let gy : Jm J(y) be the inverse of this homeomorphism.

Finally, let : C x Jm -- X, be defined by (y, t) = gy(t). Then V -= (C x Jm)

is the required neighborhood. -

Remark 4.1.2 In the above proof let x be in the Cantor set C. Note that the

points z0 and z1 corresponding to (x, 0) and (x, 1), respectively, are in the same

composant. Moreover, d(zo, zi) does not depend on x that is, the lengths of the

components of V are all the same in the d metric.

Definition 4.1.3 Suppose {Di}i1 is a sequence of no,, m1,1*I compact subsets of a

metric space Y. Then limsup{Di} = {y Y y = limjo y i for some subsequence

{Dei } and yji e Di 1}.
Lemma 4.1.4 Let {A }-1 be a sequence of arcs in Xs. Suppose that Ai -+ B in the

Hausdorff metric. Suppose also that there is an M > 0 such that f(Ai) < M for all i.

Then B is an arc and f(B) < M.

Proof. Let N be such that M s-N < W =() -(I1) -= Then for every k, 7rN(Ak)
2 2 2
has length at most ('N). Since Ak -- B, rN(Ak) -- 7N(B). In particular, 7TN(B) is a

proper subset of IN. It follows that B is a proper sub-continuum of Xs. By Lemma

3.1.1, B is an arc.

Finally, choose j large enough so that 7jI B is a homeomorphism. Then for each

k, s (7j (Ak)) < M, and hence (B) = sJ( (B)) < M. w

Lemma 4.1.5 Let {A }-1 be a sequence of arcs in X, with endpoints ai and bi,
". -"'. /:';. /;/ Suppose that there is a positive number M such that d(ai, bi) < M

each i. Suppose also that the sequence {}ai}1 converges to some a G Xs. Then

B =limsup{Ai} is an arc in X, and f(B) < 2. M.

Proof. Let x c B = limsup{Ai}. Then there is a subsequence {Aj,} 1, such that

Ai, -- D c B in the Hausdorff metric with x e D. By Lemma 4.1.4, f[(D) < M. So,







33
d(a, x) < M. From this it follows that B must be a proper sub-continuum and thus

an arc with the f-length of B at most 2 M. -

Lemma 4.1.6 Let {A }Ia be a sequence of arcs in X, with endpoints ai and bi,
". /,:'. l/;. /;/ Suppose that ai -+ a and bi -+ b. Suppose also that there is an M > 0

such that d(ai,bi) < M for all i. Then a and b are in the same composant of Xs.

Let A denote the unique arc with endpoints a and b. Suppose that lim sup Ai does not

contain an endpoint of X,. Then Ai -+ A in the Hausdorff metric.

Proof. Suppose that {Ai}Ia is a sequence of arcs in X, with endpoints ai and bi,

respectively. Suppose that ai -+ a and bi -+ b. Suppose also that there is an M > 0

such that d(ai, bi) < M for all i. By the proof of Theorem 3.2.2, a and b are in the

same composant of X, and d(a, b) < M. Let A be the unique arc with endpoints

a and b. Let B = limsup{Ai}. By Lemma 4.1.5, B is an arc with f(B) < 2 M.

By assumption B does not contain an endpoint of X,. So, let V = ((C x I)

be the neighborhood of B given by Lemma 4.1.1. Then there is an N such that

for all n > N, A, C V since B is the limsup{Ai}. Therefore for each i > N,

Ai C ({yi} x I) for some yi c C. Furthermore, Ai is the arc in ({yi} x I) joining

the endpoints. Let a, b GE ({y} x I). Then

limit, Ai = A =B C (({y} x I). -

Definition 4.1.7 Consider J x C C R2 where C is the standard middle-third Cantor

set and J = [-1,1]. D. iu,.- an equivalence relation ~ on J x C by (t, 1) ~ (-t, 1) for

all te J. Let Q =J x C/ .

Remark 4.1.8 We will think of Q as the union of two sets E and F defined in the

following way. Let E (C U (- C)) x [1, 2] C R2. Let F be a Cantor set of semicircles

with centers at (0, 1) joining each point of C x {1} with the corresponding point of

-C x {1}.

Now in the Cantor set C, let Co be the set of points in C in the interval

between 0 and inclusively. Let C1 be the set of points in C between 2 and ,

inclusively. For higher k, Ck be the subset of C containing the points between 3-









and 3f+12, inclusively. Then {Ck}jo is a disjoint collection of Cantor sets with

Ck -1 in C and C Uto C, u {Y.

Lemma 4.1.9 Suppose that A is an arc in X, which contains an endpoint of Xs.

Then there is a neighborhood V of A homeomorphic to Q = C x J/ .

Proof. There is no loss of generality in assuming that c is the endpoint of A. Let

Aw be the set of points in Cc with the same backward itinerary, w, and such that

c e A,. We know that A, is a non-degenerate arc with c as one endpoint and some

z as the other endpoint, and that -,,| is a homeomorphism onto [c, cj] or [ci, c] for

some 1 < i < no.

Define Do {x E X. 1 ,.(x) = c and 7i(x) / c for all i > no}. The set Do is

compact, totally disconnected and every point is a limit point, so Do is a Cantor set.

Let x E Do. Then ,,(x) c and 7rm(x) / c for all i > no. There are two arcs Ax

and Bx in X, containing x as an endpoint, such that ,(Ax) -= ,,(A,), and such

that ,, (Ax) and ,, (Bx) are symmetric about c.

Similarly, for any k E N, define Dk ={x E X, l (k+l).no() = c and wr(x) / c

for all i > (k + 1) no}. Then, for any k e N, the set Dk is a Cantor set. For any

x E Dk, there are two arcs Ax and Bx in XA containing x as an endpoint, such

that 71(k+I).no(AP) = 7(k+I).no(A,), and such that 7(k+I).no(Ax) and 71(k+I).no(BP) are

symmetric about c.

Let V = U {Ax UBx Ix E UoDk}jUA,. Let (a, b) be the open interval containing

c such that f' ((a,b)) is [c, ci) or (ci, c]. Then each point of .((a,b)) is in V.

Hence every point of A, except z is an interior point of V.

Define a map h : V -- Q in the following way. For every k E N, h sends Dk

homeomorphically onto Ck. For every x E Dk, Ax is mapped linearly onto [-1, 0] x

{h(x)}, and By is mapped linearly onto [0,1] x {h(x)}, and A, is mapped linearly to

[-1, 0] x {1}. Then h is 1-1, continuous and onto hence it is a homeomorphism.

Now the neighborhood V that we just created may not contain the given arc A.

However, for k > 1, applying the shift homeomorphism k times will create a longer









and thinner neighborhood of the same form, -no (V), with U-0o o-no < (V) dense in
Xs. Thus, there will be some k for which oa." (V) will contain A. w

Remark 4.1.10 In the above proof (A,) (A,) = (B,), for every x E Dk,

for every k E N. Furthermore, there are arbitrarily small neighborhoods of c

homeomorphic to Q for which this is true.

Theorem 4.1.11 Suppose that hi and h2 are homeomorphisms of Xs such that

hi(c) = h2(c) c= Suppose also that there is an M > 0 such that for each y Cc,
d(hi(y), h2(y)) < M. Let xi be a sequence of points in X, converging to a point x in

Xs. Let Ai be the unique arc joining hi(xi) and h2(x), and let A be the unique arc

joining hi(x) and h2(x). Then Ai -- A in the Hausdorff metric.

Proof. We assume the hypotheses and notation of the theorem.

Case 1. Suppose that the composant containing x does not contain an end-

point. In this case Lemma 4.1.6 applies since limsup{Ai}g1 must be in the com-

posant of x which does not contain an endpoint of Xs. Thus, we have Ai -+ A in this

case.

Case 2. Now suppose that x e Cc for some i with x / ci. By Theorem 3.2.2

hi(x) and h2(x) are on the same composant, and this composant must be some

-Cc for some j. Let J = [e, cj] be an arc in Cc such that hi(x) c J, h2(x) E J,

d(e, hi(x)) > M+ 1, d(e, h2(x)) > M+ 1. Let V be a neighborhood of J as in Lemma

4.1.9.

Consider the arc h 1(J) U h2 (J) in C,. Let W be a neighborhood of this

arc as in Lemma 4.1.9. By shrinking V in the "vertical" direction if necessary, we

may assume that h1 1(V) U h2 -(V) C W. Let K be a component of V which does

not contain cj. By the central point of K we mean the unique point of K which

corresponds to a point of the form (0, y) in [-1, 1] x C as in Definition 4.1.7.

We may assume that x2, h1 (V) n h21(V) for each n. Since hi(xT,) -+ hi(x)

and hi(xT) -- h2(x), it follows from Remark 4.1.10 that for n sufficiently large, the d

distance from hi(xT,) to either endpoint of the component of V containing hi(xT,) is







36
greater than M + 1. Since d(hi(xn), h2(xn)) < M, it follows that hi(xn) and h2(Xn)

lie on the same component of V for n sufficiently large. Without loss of generality,

we assume that this holds for each n.

For each positive integer n, let K, denote the component of V which contains

hi(xn) and h2(xn), and let n',, denote the center point of K,n. Let y,, h -(r',,) and

z,n h2 1(a,,). Then y,, and z,, lie on the same component of W as x,. Since x, -- x,

y,, -a- ci, and z,, -- ci, it follows that for n sufficiently large, x, does not lie between

y, and z, on a component of W. Again, we may assume that this holds for each n.

For each positive integer n, let Kn, [a, bn,]. We may assume that d(an, n',,) >

M + 1 and d(bn, n',,) > M + 1 for each n.

We claim that for each positive n, hi(xn) and h2(xn) lie on the same side of

t',, in Kn. We prove this by contradiction. Suppose that hi(x,) and h2(xn) lie on

opposite sides of n',, for some n. Recall that K, = [an, bn] and suppose without loss

of generality that hi(xn) lies on the same side of n',, as an. There is a point pn, W

with hl(pn) = an. Moreover, p., xn, y., and z., lie on the same component of W, and

on this component, p. is on one side of x,n, while y., and z., are on the other side.

Now, using the monotonicity of h2 on a component, we see that the arc in

X, with endpoints h1(pn) and h2(pn) contains both a,, and n',,. This implies that

d(hi(pn), h2(pPn)) > M + 1. This is a contradiction and the claim is established. Since

Ai C V for each i, it follows from the special form of V and the claim that Ai -- A.

Case 3. Now suppose that x = Ci for some i. In this case A is just the point

{cj}. This case is routine using the structure of the neighborhood of c given in

Lemma 4.1.9.

One of Case 1, Case 2 or Case 3 must hold so together they prove the statement

of this theorem. -









4.2 Proof of the Isotopy Theorem

Theorem 4.2.1 Suppose that hi, h2 : X, -- X, are homeomorphisms which leave the

endpoints of X, fixed. Suppose that there is an M > 0 such that d(hi(x), h2(x)) < M

for all x c CC. Then hi and h2 are isotopic.
Proof. Let that hi, h2 : X, X, be homeomorphisms which leave the endpoints

of X, fixed. Suppose that there is an M > 0 such that d(hi(x), h2(2)) < M for all

x e Cc. Let H : X, x I -l X, be defined in the following way.
Let x E X, and t E I. By Theorem 3.2.2, there is a unique arc Ax connecting
hi(x) and h2(x). Let m E N be such that -. i is a homeomorphism into 1,. Let

g, m 7r(Ax) -+ Ax be the inverse of this homeomorphism. Define
H(x, t) g= ((1 t) 7Tm(h i()) + t 7Tm(h2(X))).

If 7k \A. is a homeomorphism, then gk ((1 t) 7k(hl(x)) + t 7 2k(h2(x)))

g9m((1 t) 7lm(hi(')) + t i- 7m(h2(X))). So, H(x, t) is well-defined. We now show that
H is continuous.

Suppose that (xi, ti) (x, t). Let Ai be the unique arc with endpoints
hl(xi), h2(xi). Then hl (xi) -+ hi(x) and h2(xi) -i h2(x). So, if A is the unique arc

connecting hi(x) and h2(x), whose existence is given by Theorem 3.2.2, then, by
Theorem 4.1.11, Ai -+ A in the Hausdorff metric.

Case 1. The point x is not an endpoint. In this case the arc A connecting

hi(x) and h2(x) does not contain an endpoint. Let V be a neighborhood of A given
by Lemma 4.1.1. We know that V is of the form V m C x I where C is a Cantor

set and I is an interval. Then there is an N such that for n > N, the arc A, C V.

Now by Lemma 4.1.1, there is an m such that x7m is a homeomorphism on each
component of V to its image in Im. Therefore for this m and for all n > N,

H(xn, tn) = gm((1 tn) ,m(hi(xn)) + tn m(h2 (Xn)))
and

H(x, t) g= ((1 t) 7rm(hi(x)) + t Tm(h2(X))).

So, clearly H(x,,tn) -+ H(x,t).









Case 2. The point x is an endpoint. In this case hi(x) = h2(x) x since the

endpoints are assumed to be fixed. Therefore A = {x} and thus An -- {x}. This

implies that H(x,tn) {x} = H(x, t).

So, H(x, t) is a homotopy. We now show that it is an isotopy by showing that

for each t, ht(x) = H(x, t) is one-to-one and onto.

First we show that ht is one-to-one. Note that ht permutes the composants of

X8 the same way that hi and h2 do. So, to show that ht is one-to-one it will suffice

to show that ht restricted to a composant C1 is one-to-one. Now C1 is the arc-

component of x. This arc-component with the d metric is homeomorphic to either R

or R+. Fix orderings on C1 and Ch,(). Now h, and h2 are homeomorphisms from C1

to Ch,(x) either preserving or reversing the orders of Cx and Ch,(x). However, since

the d distance between hi and h2 on Cx is bounded, these either both preserve the

orders or both reverse the orders on C1 and Ch,(x) in the same way. Thus, ht lc is

one-to-one.

To show that ht is onto is similar. -

We now give the proof of the Isotopy theorem as outlined at the beginning of

this chapter.

Proof of the Isotopy theorem. Let h : X8 -- X, be a homeomorphism.

Let n be such that h" leaves the endpoints of X, fixed. By Theorem 3.2.5, there is

an M > 0 and there is a k E Z such that d(h"(x), r (x)) < M for all x E C,. By

Theorem 4.2.1, h' and a. are isotopic. -














CHAPTER 5
ON INGRAM'S CONJECTURE
5.1 Three-Step Program

In this chapter we address the question of classification of inverse limit spaces of
tent maps without any assumptions on the orbit of the turning point. We present a
three-step program from which Ingrams's conjecture would follow.
Ingram's conjecture. If 1 < s < t < 2, then the inverse limit spaces (I, fs)
and (I, ft) are not homeomorphic.
Remark 5.1.1 We remind the reader that the core of the inverse limit space (I, fs)
is indecomposable for s c (V2, 2], thus giving us enough information to work with. If
the result is proved for V2 < s < t < 2, then it can be extended to 1 < s < t < 2 as
discussed in Remark 1.0.1.
Step 1. (Pseudo-isotopy conjecture) Let fs be a tent map with slope
s E (v2, 2]. Suppose h : (I, fs) -- (I, fs) is a homeomorphism. Then there exists an
integer k such that hix, and o- x, are pseudo-isotopic.
Step 2. Let fs be a tent map with slope s c (v2, 2]. Suppose C C X, is a com-
posant with the property that there exists an integer k such that a(C) C. Then
there exists a unique fixed point z of ff and the point z = (z, f- (z)," fs(z))
C.
Step 3. Let fs and ft be tent maps with slopes 1 < s < t < 2. Then for
any positive integer n, F(fJ) < F(ft), where F(fJ) denotes the number of fixed
points of the map fJ. Furthermore, there exists a positive integer N such that
F(fN) < F(ftN).
Step 1 is a conjecture at present time, but Step 2 and Step 3 are true. At the
end of this chapter we give a sufficient condition for the pseudo-isotopy conjecture.









5.2 Proof of Step 2

Theorem 5.2.1 Let fs be a tent map with slope s c (V 2,2]. Suppose C C
X8 is a composant with the 1 -i '/;/ that there exists an integer k such that

~(C) = C. Then there exists a unique fixed point z of fk and the point z =

(z, fsk-1 I (Z fs Z C.
2
Proof. Let 6 > 0 be such that (2) 6 < c2-c2
(2)-1 2
Let x = (xo,x, ) E C. Since o (C) C, there is a proper sub-continuum

A C C containing x and s 2k(x). Consequently, for any positive integer n, the points
a -2nk(x) and s 2(n+1)k(x) are contained in -2nk(A) C C, a proper sub-continuum of

Xs.

Let M be a large enough positive integer so that f(7,{(A)) < 6 for all m > M.

This is possible since A is a proper sub-continuum of Xs, and f(7T,(A)) -+ 0 as
n -- oo.

As we chose 6 such that 6 < 2 we have that for all m > M,

( m+2k (A)) < -f ((A)).
Note also that for all m > M and all n > 0, 7,m(9-2nk()) Tm+2nk(X). Hence, for

all m > M and all n > 0,

d(Wm(U 2nk (X) T --(a 2(n+l)k(1))) < .
Since the series Y 20 converges, the sequence {7 m(7- 2nk(X))}c0 is
n rr)n s n-
Cauchy hence it converges to some Zm. By continuity of fs, fs(zm+l) = m for all
m > M. Define z, = ff- 'ZM) for all 0 < i < M. Then z (zo, zI,z2, ) is a point

in the inverse limit space (I, fs). Moreover, the sequence {j- 2nk (XI)} converges to z,

and a2k(z) z a2k(z).
As a-2nk(x) u2nk(A) n a-2(n- )k(A), we have that U -o 2nk(A) is connected,

hence cl(Uo 2nk(A)) is connected.
We know that for all m, it is true that

(1mF(Uo- 2n8 k(A))) < E-Zo (m(( o-2n(A))).








Since for each positive integer n and all m, m(os 2nk(A)) = m.+2nk(A), it follows
that for each positive integer n and all m,

ZEno o (, 2nk (A))) 8Z- tQm+2nk (A)).
As for all m > M, f(7'W+2k (A)) < )- ('wm(A)), it follows that for all m > M,

S(U 2 (A))) < ( )k +2nk(A)) <


Therefore for all m > M, f(7 (cl(U1 o-2nk (A)))) < 2, which tells us that

cl(Un- 2nk (A)) is a proper sub-continuum of X,. Note that z e cl(Un o- 2nk(A)) C
C.
Since z is a fixed point for the map ,2k, there exists a point z e I such that
Z (z, fk-1(z ), ... f,(z))00. Then z, f2k-1(z), ... f(z) e I are fixed points
for the map f2k. In fact, we claim that z e I is a fixed point for the map fk.
Suppose not. Then z and fk(z) are distinct points in I, and they are some c > 0
distance apart. The points z = (z, fk- (z),... fs(z))00 and z' = (2) =

(f(z), f- 1(z), fs(z), z, f2k- (z), fk+l(z)) are two distinct points in the
composant C. There exists a proper sub-continuum A C C which contains both z
and z'. Observe that c. k(2) = z' and u-k ('z) = z. Hence, for each positive integer
n, a-nk (A) contains both z and z'. Thus, for each positive integer n, Trnk(A) contains
both z and fk(z). But, since A is a proper sub-continuum of Xs, f(TT,(A)) -+ 0 as
n oo, which contradicts the assumption that z and fk(z) are a fixed distance
e > 0 apart.
By an argument similar to the one in the paragraph above, it is easy to prove
that z is a unique fixed point of a, in C. -
Theorem 5.2.2 Let fs be a tent map with slope s c ( /02,2]. Suppose C C Xs is
a composant for which there exists an integer k such that oa(C) =C. Assume that
either the turning point c is not periodic, or that if c is periodic of period no, then
ci C for each i C {0, 1,... no 1}. Then there is a continuous bijection from the
real line R to C.







42
Proof. Let z = (zo, Zk- 1 z), where zi = fj(zo), i {0, 1,... k 1}, be the

unique fixed point of a, in C. Let = min{|z c|, |z C1|, C2, C3
i 0,1,... k 1}. Then the symmetric interval Jo = [zo zo + ] C [c2, Ci] and

C3 0 JO.
Let Jd be the component of f8 (Jo) in 11 7- I(Xs) which contains Zk-1. Then
fs maps J1 linearly onto Jo. Let J2 be the component of f8 (Ji) in 12 72(Xs)
which contains zk-2. And so on, for each positive integer n, let J, be the component
of f8 ,(J_-) in I, 7,(Xs) which contains z(-n mod k)- Our choice of F such that for
each positive integer n, c3 Jn, ensures that fs maps J, linearly onto J,-\1, that is,
no piece of J, gets cut out by "pulling ,. 1 J, to the left of c.
Let J = lim{J, fs}~ 0. Then J is a proper sub-continuum of C containing z.
Note that for each positive integer n, T- 1 is a homeomorphism, and for each positive
integer n, fs j is a homeomorphism. Hence J is an arc in C.
Consider the sequence J, af (J), k(J),... of arcs in C. We claim that it is an
increasing sequence.
Let x = (xo,xx2, X) e J. In order to prove that x e of(J), we need to
show that there exists a point ye J such that ao (/) x x. Let y/ -k(x)

(ak, k+x" ). Then af(y') x. It only remains to show that y e J. As Xk e Jk and
Jk [z0 [ z z 9], we have that lXk Z0 < _- < F. Hence Xk E Jo. It follows that
y G J.
We proved that J C oaf(J), hence for each positive integer n,
akJ) C(n+l)k J)

Let A c C be a proper sub-continuum containing z. Then (7,(A)) -- 0 as
n -- oc. Hence there exists a positive integer m such that Trmk(A) C [zo zo + F].
Thus A C a7k(J). It follows that C is an increasing union of arcs. ]
Remark 5.2.3 For a composant C C X, with the property that there exists an
integer k such that af (C) = C, and such that either the turning point c is not
periodic, or that if c is periodic of period no, then ci C for each i {0, 1, ... no -









1}, we can define a metric d in the same way we did it in the case of a tent map fs

with a periodic turning point. In the latter case we only used the fact that for any

proper sub-continuum A C Xs, there exists a positive integer m such that for all

n > m, 7Tn\A is a homeomorphism. The proof of Theorem 5.2.2 shows that this holds

for any sub-continuum of C.

5.3 Proof of Step 3

Theorem 5.3.1 Let 1 < t < 2 be such that the turning point c is periodic under the

tent map ft. Suppose K(ft) = DC where D is a finite sequence of R's and L's. Let

B be an R or an L such that DB < DC. Let y c [f (c), ft(c)] be a point such that c

is not in the orbit of y under ft. Then It(y) < (DB)R .

Proof. Suppose It(y) > (DB)R Since (DB)R < It(y) < DC, we know that

It(y) = DBY, where Y is some sequence of R's and L's. Let Y = DIX, where D1
and X are sequences of R's and L's and the length of D1 is the same as the length of

D. Since crkt (y) < DC for all k > 0, we have that D1 < D. But It(y) > (DB) and

DB has even parity by the way B was chosen, hence DI > D. Therefore D1 = D.

By induction, we get that It(y) = (DB)<. Then for each j > 0, the points ff (y) and

f/(ft(c)) are in the same closed interval [f/(c),c] or [c, ft(c)]. This contradicts the
fact that ft is a tent map with t > 1. -

Corollary 5.3.2 Let 1 < s < t < 2 be such that the turning point c is periodic under

the tent map ft. Suppose K(ft) = DC where D is a finite sequence of R's and L's.

Let B be an R or an L such that DB < DC. Let y c [ f(c), fs(c)] be a point such

that c is not in the orbit of y under fs. Then I,_(y) < (DB)".

Proof. Since K(f,) < K(ft), the assumption that I,(y) > (DB)R gives us that

(DB) <_ I_(y) < DC. The same argument as in the proof of Theorem 5.3.1 leads to

a contradiction. -







44
Before we embark on proving our next result, we need to introduce the concept

of topological entropy and state a few well-known, but non-trivial results about the

entropy of maps in the tent family and the entropy of maps in the quadratic family.

Definition 5.3.3 The one-paramneter f, iil:Ji of maps f, : I -> I /. p[,,/ by

f,(x) = .p x (1 x), for x E I and for 0 < p < 4, is known as the quadratic

family.

Since any of the several equivalent definitions of entropy of a map is rather

long and does not reveal much information about its nature, at least not to the

inexperienced reader to whom it would be offered, nor do they show v--iva how to

compute it, we will try to give a short description and state several properties with

the appropriate references. For a comprehensive study of this concept we refer the

reader to [9, C'!i Ipter VIII] and [2, C'!i Ipter 4].

The topological entropy of a map is a quantitative measure of the complexity

of the system modeled by iterating that map. It measures how many distinct orbits

of a given length there are for this map, that is, how this number grows with the

length. For a precise definition of entropy, see [9, Section VIII. 1] or [2, Section 4.1].

If f is a piecewise monotone map from a closed interval into itself, and c, de-

notes the number of pieces of monotonicity of f, so called laps, then the topological

entropy of f is

h (f) = lim A, log cn,

and 1log c > h(f) for any n [2, Theorem 4.2.4].

We -iv that a piecewise monotone map f from a closed interval into itself has

a constant slope s if on each of its laps it is affine with a slope of absolute value

s. If f is a piecewise monotone map from a closed interval into itself with a constant

slope s, then h(f) = max{0,logs} [2, Corollary 4.3.13]. Thus, the topological

entropy of a tent map fs with s > 1 is


h(f) log s.







45
An efficient algorithm to compute the entropy of unimodal maps of the interval

into itself was given by Block, Keesling, Li and Peterson [10], and such an algorithm

for bimodal maps of the interval into itself was given by Block and Keesling [11].

On the space of all C1 piecewise strictly monotone maps with a given number

of laps, with the C1 topology, the topological entropy is continuous [2, Corollary

4.5.15]. This fact was proven by Milnor and Thurston in [28].

For tent maps fs with slope s > 1, the topological entropy is strictly increasing

as a function of the slopes. For the quadratic family, the topological entropy is an

increasing function of the slopes, but no "real" proof is known; all known proofs

of this fact use complexification of the maps. We will state those results for easier

referral.

Theorem 5.3.4 (Sullivan, Milnor, Douady and Hubbard) [27, C'!i Ilter II,

Theorem 10.1] If the turning points of the quadratic maps f, and f,, are ev, ,l./.ll,.

periodic and their kneading invariants are equal, then p p= '.
Corollary 5.3.5 Let f, : I -+ I be a unimodal fin.';/l, consisting of C1 maps
depending continue.; -l;, on the parameter p. If the assumption that the turning points

of fl and fy, are ev, ,.l.iall periodic and their kneading invariants are equal implies

that p p= ', then the following statements are true.

1. The function p K(f,) is monotone [27, C'!i Ilter II, Corollary 1].

2. The function p -+ h(f,) is monotone [27, C'! Filter II, Corollary 2].
Theorem 5.3.6 Let fy, be a map in the quadratic fji .,1;/ with kneading sequence

DC. Let B be an L or an R such that DB < DC. Then there is an open interval

(P2, 1) such that for all p e (P2, 1), K(f,) (= DB)" and h(f,) h(f,,).
Proof. Since K(f,) = DC, the turning point is periodic under f,,. Let n denote

the period of c. Then (f )' 0 at each point in the orbit of c. There exists 6 > 0

such that for each p E (i p + ) and each point x within distance 6 of any point

in the orbit of c, we have that I(f,)'(x)| < 1. Thus, for each p c (pi 6, pl + S), f,









has unique attracting periodic orbit within distance 6 of any point in the orbit of c,
and as k -0 o, f (c) approaches this orbit.
Let P2 i 6. Consider p E (p2, ). By Theorem 5.3.4 and Corollary

5.3.5, K(f,) < K(ft1). Let x, be the point near c in the attracting periodic orbit.
Then It(ft(x )) = (DA)', where A is an L or an R depending on which side of

c, x is on. By construction, K(ft) =I(ft(c)) -I(ft(x )) (DA)'. But, by
Theorem 5.3.4 and Corollary 5.3.5, K(f,) is monotone as a function of p. Thus,
(DA)- (DB)-.
The entropy depends only on the kneading sequence. Moreover, the entropy is
a continuous function of the parameter in the quadratic family [2, Corollary 4.5.5].
Hence the entropy is constant on (p2, 1]. Thus, h(ft) = h(f1) for all p c (p2, /ll-
Note that the proof of Theorem 5.3.6 shows that there is an open interval

(Pl, P3) such that for all p c (l, P3), K(f,) = (DB)< and h(f,) = h(f1), where B
is an L or an R such that DB > DC.
Theorem 5.3.7 Let 0/2 < s < t < 2 be such that for each of the tent maps f,

and ft, the turning point is periodic. Suppose K(fs) = DC and K(ft) = DtC,
where Ds and Dt are some finite sequences of R's and L's. Let Bs be an R or an
L so that DB, < DC and let Bt be an R or an L so that DtBt < DtC. Then

ak(DsBst) < (DtBt)0 for each k > 0.
Proof. By [18, Theorem III.1.1], there exists a parameter ps such that the kneading
sequence of the quadratic map f, is K(fj) = DC. By Theorem 5.3.4 and
Corollary 5.3.5, for Pi1 slightly smaller than pi, K(fp) = (DsBs)<. By Theorem
5.3.6, the entropy of the quadratic map ft, is h(f,) = h(ft), and we know that
h(f) = logs, hence h(ft) = logs. Similarly, for p2 slightly smaller than pt,
K(ft,) (DtBt)0 and the entropy of the quadratic map ft2 is h(f2) = h(ft) =
logt. Since s < t, h(fj) = logs < logt = h(f,). Since in the quadratic
family, the entropy depends monotonically on the kneading sequence, it follows that







47
(DsBs)0 < (DtBt)t. As (DsBs)< is the kneading sequence of the map f,,, we know
that (DSB,) is shift maximal, hence ak(DsBs)O < (DtBt)0 for each k > 0. w
Theorem 5.3.8 Let fs and ft be tent maps with slopes 1 < s < t < 2. Then for i;,
positive integer n, F(f") < F(ft").
Proof. Let n be a fixed positive integer. There are four cases to distinguish. In each
of these cases we will show that there is an injective function from the set of fixed
points of f,' to the set of fixed points of ft.
Case 1. For each of the tent maps fs and ft, the turning point is not periodic.
Then both K(fs) and K(ft) are infinite.
Let y be a fixed point of f,. Then c is not in the orbit of y under fs. The
forward itinerary of y under fs is I8 (y) = S' for some sequence S of length n of
L's and R's. Since sakSo < K(fs) < K(ft) for all k, by Definition 2.1.16, we have
that S- << K(ft). By Theorem 2.1.17 there is a point z with It(z) = S'. Since z
and ft(z) have the same itinerary, by Remark 2.1.22, it follows that ft(z) = z. By
Remark 2.1.22, it also follows that z is the unique point with It(z) = S.
Case 2. The turning point c is periodic under fs, and c is not periodic under ft.
Then K(ft) is infinite and K(fs) = DC for some finite sequence D of R's and L's.
Let y be a fixed point of f,. Then kj8(y) < K(fs) < K(ft) for all k. Since
K(ft) is infinite, this means that I,(y) << K(ft). As in Case 1, there exists a unique
point z with I_(z) = I(y) and z is a fixed point of ft.
Case 3. The turning point c is not periodic under fs, and c is periodic under ft.
Then K(fs) is infinite and K(ft) = DC for some finite sequence D of R's and L's.
Let y be a fixed point of f,. Then c is not in the orbit of y under fs. The
forward itinerary of y under fs is I8(y) = S' for some sequence S of length n of L's
and R's, and akSo < K(f,) < K(ft) for all k. By Corollary 5.3.2, orkS < (DB)<
where B is chosen to be an R or an L so that DB < DC. Thus S << K(ft). It
follows that there exists a unique point z with I,(z) = S- and f,(z) = z.









Case 4. For each of the tent maps fs and ft, the turning point is periodic.
Then K(fs) = DsC and K(ft) = DtC for some finite sequences D8 and Dt of R's

and L's.
Let y be a fixed point of fJ. If c is not in the orbit of y under fs, then the
forward itinerary of y under fs is I,(y) = S for some sequence S of length n of L's
and R's, and OakSc < K(f,) < K(ft) for all k. By Corollary 5.3.2, okS, < (DsBs)<
where Bs is chosen to be an R or an L so that DsB < DsC. Thus S << K(ft). It
follows that there exists a unique point z with It(z) = S- and f"(z) = z.
If c is not fixed by f, we are done.
Suppose c is fixed by f,. Then the length of DsC is n. Choose Bs to be an R or
an L so that DB8, < DC and choose Bt to be an R or an L so that DtBt < DtC.
By Theorem 5.3.7, ark(DsBs)O < (DtBt)0 for each k > 0. Thus (DsBs)" << K(ft).
It follows that there exists a unique point z with It(z) (DBs,) and ftj(z) z.
Note that no point y has I (y) = (DsBs)0.
Thus we proved that in all of the above cases there is an injective function from
the set of fixed points of f, to the set of fixed points of ft. F-
Theorem 5.3.9 Let fs and ft be tent maps with slopes 1 < s < t < 2. Then, there
exists a positive integer N such that F(ff) < F(fLN).
Proof. Let s < tl < t be such that c is periodic under the tent map ft,. Let N be
the period of c under ft,. In the proof of Theorem 5.3.8, we establish an injective
function from the set of fixed points of f to the set of fixed points of ftN. But the
turning point is not in the orbit of fixed points of ft which we obtained. Thus, since
c is a fixed point of fJ, we have that F(f) < F(f4N). Finally, by Theorem 5.3.8, we
have F(ft) < F( f).

Note that the case when c has the same period no under fs and ft has already
been proven in Lemma 3.3.1. -







49
5.4 Ingram's Conjecture Follows from the Pseudo-isotopy Conjecture

Theorem 5.4.1 If the pseudo-isotejl.: conjecture holds, then I..gain 's conjecture

holds.

Proof. Suppose the pseudo-isotopy conjecture is true. By Remark 5.1.1, it is enough

to consider parameters V2 < s < t < 2. Assume there exists a homeomorphism

h : (I, fs) (I, ft). By Theorem 5.3.9, there exists a positive integer N such that

F(fN) < F(fJ).

Consider the following diagram.
N
as

(I, fs) ( I, fs)


(I, ft) (, ft)
h- oaN oh
By assumption, the pseudo-isotopy conjecture is true hence there exists a

positive integer k such that h-1 o a8 o h and ak are pseudo-isotopic. It follows that
h-1 o JN o h and JaN are pseudo-isotopic.

By Theorem 5.2.1, the number of composants mapped to themselves by ao is

equal to F(f/), and the number of composants mapped to themselves by oN is

equal to F(ftkN). Now, since N and h-' onf oh are conjugate, and since h-' oKN oh

and aN are pseudo-isotopic, it follows that the number of composants mapped to

themselves by ao is equal to the number of composants mapped to themselves by

a(t. Hence F( f) F( N). But, this contradicts the fact

F(ftkN) > F(fN) > F N(f).

Thus, we proved that if Step 1 holds, then for v2 < s < t < 2, the inverse limit

spaces (I, f,) and (I, ft) are homeomorphic if and only if s = t. ]

5.5 Sufficient Condition for the Pseudo-isotopy Conjecture

Let T = (I, fs)\X, denote the 1 ,!" of the inverse limit space (I, fs) of the tent

map fs with 2/2 < s < 2.







50
Observe that fs((ci, 1]) = [0, c2). Since fs reaches its maximum ci at c, no points
from I= [0,1] map into (cl, 1]. Thus, if x = (xo, xi, ,- ) C (I, fs) and Xm C [0, C2) for
some positive integer m, then Xk C [0, C2) for each positive integer k > m.
Lemma 5.5.1 T is an open ,.,, with endpoint 0 (0,0,---) E (I, fs). (I, fs) is a
, .'*I I. 1./7, ,I: n, of T .
Proof. First we prove that T is an open ray with endpoint 0 (0, 0, ). Denote
Xo = C2. Let Jo = [0, xo]. There exists a point x, C Jo such that fs(xi) = o. Let
J] = [0, x1]. Then fs maps J] linearly onto Jo0. There exists a point x2 E J1 such that

fs(x2) 1. Let J2 [0, x2]. Then fs maps J2 linearly onto J1. We form J3, 4,...
in the same manner. Let J lim{J,, fs}j 0. Then J is an arc in (I, fs) with
endpoints 0 and x = xo,xi, ). Note that ro ([0, c2)) J\{x}. Therefore J\{x}
is open in (I, fs). Since T Uo 0((J\x) Uo ,7(J) is a union of increasing rays
with a common endpoint 0, it follows that T is an open ray with endpoint 0.
Next we prove that T is dense in (I, fs). Let x = (xo, xi, -) e Xs. For
any positive integer m, there exist a positive integer k and y C [0, c2) such that

fk(y) X= x. Then the points x and ym (xo, x, ',ft I(y), f' (y}),,)
agree on the first m + 1 coordinates and ym T. Thus the sequence of points

{Ym} =o in T converges to the point x C Xs. -
We can define the metric d on T as follows. Let x and y be points in T. By
Lemma 5.5.1, T is a ray, hence there is an arc A C T with endpoints x and y. Then,
there exists a nonnegative integer k such that Tk A is a homeomorphism. Define
d(x,y) = Sk (X) Tk()Y.
This is similar to how d was defined when the turning point of the tent map was
periodic. There is a difference; now d actually gives the topology of T.
Theorem 5.5.2 Let fs I -- I, fs(x) = min{s x, s (1 x)} be a tent map with
slope s C (0/2, 2]. Let h: (I,f) -- (I, fs) be a homeomorphism. Suppose there exist a
number M > 0 and an integer k such that d(h(x), crf(x)) < M, for all x E T. Then
h x, and o- xs are pseudo-isotopic.









Proof. Let x e Xs. By Lemma 5.5.1, the tail T is dense in the inverse limit space
(I, fs), hence there exists a sequence {x,n} in T converging to the point x. Then

the sequence h(x,,) converges to the point h(x) and o,7(x,) converges to the point

~Jf(x). Consider the unique arcs A, C T with endpoints h(x,) and ~Jf(x,). By
assumption, (A,n) = d(h(x,),uf (x,)) < M for each n. Let m > 0 be an integer such

that M < sm (f/(c) f,(c)). Then for each n, f(,m (A,)) < M < f,(c) fi(c),

hence for each n, 7Tm(A,) is a proper subset of [f.(c), fs(c)].
Let C((I, fs)) denote the space of nonempty sub-continua of the inverse limit

space (I, fs) with the Hausdorff metric. The projection 7m : (I, fs) -+ [fi(c), fs(c)]
induces a continuous map Tm : C((I, f)) -- C([f,(c),fs(c)]). Since C((I, f)) is a

compact metric space, the sequence {A,} of arcs in T has a convergent subsequence
{Aj,} converging to a subcontinuum A c C((I, fs)). Note that h(x), 7f(x) c A and

A C Xs. Since 7m : C((I,fs)) C([f,(c), f,(c)]) is continuous, 7, (A) has length at

most M < f,(c) f,(c). Thus, A must be a proper sub-continuum of X,. Therefore,
h(x) and o(r(x) are in the same composant of X,. By Definition 3.2.1, hIx, and ao Ix,

are pseudo-isotopic. -

Note that the proof of Theorem 5.5.2 is analogous to the proof of Theorem
3.2.2.














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BIOGRAPHICAL SKETCH

I was born on July 19, 1971 in Skopje, Macedonia. I obtained my undergraduate

degree in mathematics at the Department of Mathematics at the Faculty of Natural

Sciences and Mathematics at the University St Cyril and Methodius in Skopje,

Macedonia, in August 1995. I obtained my Master of Science degree at the same

department, in August 2001. Since Spring 1996, I have been an employee at the

Pedagogical Faculty at the University St Cyril and Methodius in Skopje, Macedonia.

In the period of 1999-2001, I was a participant in a joint European project aimed

at restructuring the curriculum at the Pedagogical Faculty at the University St

Cyril and Methodius in Skopje, Macedonia. Within the project, I got acquainted

with some characteristics of the educational systems in the Netherlands, Denmark,

Sweden and Great Britain. My main interests are topology and dynamical systems,

as well as math education for elementary school teachers and math education in

general.