Stein and Shakarchi Real Analysis Solution;Stein实分

Chapter1.6,Page37Problem2:(a)ProvethatxisintheCantorsetixhasaternaryexpansionthatusesonly0'sand2's.(b)TheCantor-LebesguefunctionisdenedontheCantorsetbywrit-ingx'sternaryexpansionin0'sand2's,switching2'sto1's,andre-interpretingasabinaryexpansion.Showthatthisiswell-denedandcontinuous,F(0)=0,andF(1)=1.(c)ProvethatFissurjectiveonto[0;1].(d)ShowhowtoextendFtoacontinuousfunctionon[0;1].Solution.(a)ThenthiterationoftheCantorsetremovestheopensegment(s)con-sistingofallnumberswitha1inthenthplaceoftheternaryexpansion.Thus,thenumbersremainingafterniterationswillhaveonly0'sand2'sintherstnplaces.Sothenumbersremainingattheendarepre-ciselythosewithonly0'sand2'sinallplaces.(Note:Somenumbershaveanon-uniqueternaryrepresentation,namelythosethathavearepresentationthatterminates.Forthese,wechoosetheinnitelyre-peatingrepresentationinstead;ifitconsistsofall0'sand2's,itisintheCantorset.Thisworksbecauseweremoveanopenintervaleachtime,andnumberswithterminatingrepresentationsaretheendpointsofoneoftheintervalsremoved.)(b)First,weshowthatthisiswell-dened.Theonlypossibleproblemisthatsomenumbershavemorethanoneternaryrepresentation.How-ever,suchnumberscanhaveonlyonerepresentationthatconsistsofall0'sand2's.Thisisbecausetheonlyproblemsarisewhenonerep-resentationterminatesandanotherdoesn't.Nowifarepresentationterminates,itmustendina2ifitcontainsall0'sand2's.Butthentheotherrepresentationendswith12222...andthereforecontainsa1.NextweshowFiscontinuousontheCantorset;given0,chooseksuchthat12k.Thenifwelet=13k,anynumberswithinwillagreeintheirrstkplaces,whichmeansthattherstkplacesoftheirimageswillalsoagree,sothattheirimagesarewithin12kofeachother.TheequalitiesF(0)=0andF(1)=1areobvious;forthelatter,1=0:2222:::soF(1)=0:1111


=1.(c)Letx2[0;1].Chooseanybinaryexpansionofx,replacethe0'swith2's,andre-interpretasaternaryexpansion.Bypart(a),thiswillproduceamemberoftheCantorsetwhoseimageisx.(Note:Theirmaybemorethanonepreimageofx,e.g.F(13)=F(23)=12.)(d)First,notethatFisincreasingontheCantorsetC.NowletG(x)=supfF(y):yx;y2Cg:NotethatG(x)=F(x)forx2CbecauseFisincreasing.GiscontinuousatpointsnotinC,becauseCisopen,soifz2C,thereisaneighborhoodofzonwhichGisconstant.ToshowthatGiscontinuousonC,letx2CandusethecontinuityofF(partb)to12choose0suchthatjG(x)G(z)forz2C;jxzj.Choosez12(x;x),z22(x;x+)andlet0min(xz1;z2x).Thenforjyxj0,ify2CweautomaticallyhavejF(y)F(x)j.Ify=2Cbutyx,G(x)G(y)G(z1)G(x);similarly,ify=2Cbutyx,G(x)G(y)G(z2)G(x)+.Problem3:Supposethatinsteadofremovingthemiddlethirdoftheseg-mentateachstep,weremovethemiddle,where01.(a)ProvethatthecomplementofCistheunionofopenintervalswithtotallength1.(b)Provedirectlythatm(C)=0.Solution.(a)Atthenthstep(startingatn=0),weremove2nsegments,eachoflength12n.Thetotallengthofthesesegmentsis1Xn=02n12n=1Xn=01(1)n=11(1)=1:(b)IfCnisthesetremainingafterniterations,thenCnisaunionof2nsegmentsoflength12n.Som(Cn)=(1)n:Notethatm(Cn)!0.SinceeachCnisacoveringofCbyalmostdisjointcubes,theinmumofthemeasuresofsuchcoveringsis0.Problem4:Constructaclosedset^Csothatatthekthstageofthecon-structiononeremoves2k1centrallysituatedopenintervalseachoflength`k,with`1+2`2+


+2k1`k1:(a)If`jarechosensmallenough,thenP1k=12k1`k1.Inthiscase,showthatm(^C)0,andinfact,m(^C)=11Xk=12k1`k:(b)Showthatifx2^C,thenthereexistsasequencexnsuchthatxn=2^C,yetxn!xandxn2In,whereInisasub-intervalinthecomplementof^CwithjInj!0.(c)Proveasaconsequencethat^Cisperfect,andcontainsnoopeninterval.(d)Showalsothat^Cisuncountable.Solution.(a)LetCkdenotethesetremainingafterkiterationsofthisprocess,withC0beingtheunitsegment.Thenm([0;1]nCk)=kXj=12j`j3since[0;1]nCkisaunionofdisjointsegmentswiththistotallength.Thenm(Ck)=1kXj=12j`j:NowCk&^C,sobyCorollary3.3,m(^C)=limn!1m(Ck)=11Xk=12k`k:(b)Fork=1;2;:::,letJkbetheintervalofCkwhichcontainsx.LetInbetheintervalin^CcwhichisconcentricwithJn1.(Thus,atthenthstepoftheiteration,theintervalInisusedtobisecttheintervalJn1.)LetxnbethecenterofIn.Thenxn2^Cc.Moreover,jxnxjjJn1jsinceJn1containsbothxnandx.SincethemaximumlengthoftheintervalsinCntendsto0,thisimpliesxn!x.Finally,xn2In^Cc,andInJn1)jInj!0.(c)Clearly^CisclosedsinceitistheintersectionoftheclosedsetsCn.Toproveitcontainsnoisolatedpoints,weusethesameconstructionfromthepreviouspart.Letx2^C.Thistime,letxnbeanendpointofIn,ratherthanthecenter.(Wecanactuallytakeeitherendpoint,butforspecicity,we'lltaketheonenearertox.)BecauseInisconstructedasanopeninterval,itsendpointslieinCn.Moreover,successiveiterationswillnotdeletetheseendpointsbecausethekthiterationonlydeletespointsfromtheinteriorofCk1.Soxn2^C.WealsohavejxnxjJnasbefore,sothatxn!x.Hencexisnotanisolatedpoint.Thisprovesthat^Cisperfect.(d)Wewillconstructaninjectionfromthesetofinnite0-1sequencesinto^C.Todothis,wenumberthesub-intervalsofCkinorderfromlefttoright.Forexample,C2containsfourintervals,whichwedenoteI00,I01,I10,andI11.Now,givenasequencea=a1;a2;:::of0'sand1's,letIandenotetheintervalinCnwhosesubscriptmatchestherstntermsofa.(Forinstance,ifa=0;1;0;0;:::thenIa4=I01