Rudin数学分析原理答案

MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1Generally,a\solution"issomethingthatwouldbeacceptableifturnedinintheformpresentedhere,althoughthesolutionsgivenareoftenclosetominimalinthisrespect.A\solution(sketch)"istoosketchytobeconsideredacompletesolutionifturnedin;varyingamountsofdetailwouldneedtobefllledin.Problem1.1:Ifr2Qnf0gandx2RnQ,provethatr+x;rx62Q.Solution:Weprovethisbycontradiction.Letr2Qnf0g,andsupposethatr+x2Q.Then,usingthefleldpropertiesofbothRandQ,wehavex=(r+x)¡r2Q.Thusx62Qimpliesr+x62Q.Similarly,ifrx2Q,thenx=(rx)=r2Q.(Here,inadditiontothefleldpropertiesofRandQ,weuser6=0.)Thusx62Qimpliesrx62Q.Problem1.2:Provethatthereisnox2Qsuchthatx2=12.Solution:Weprovethisbycontradiction.Supposethereisx2Qsuchthatx2=12.Writex=mninlowestterms.Thenx2=12impliesthatm2=12n2.Since3divides12n2,itfollowsthat3dividesm2.Since3isprime(andbyuniquefactorizationinZ),itfollowsthat3dividesm.Therefore32dividesm2=12n2.Since32doesnotdivide12,usingagainuniquefactorizationinZandthefactthat3isprime,itfollowsthat3dividesn.Wehaveprovedthat3dividesbothmandn,contradictingtheassumptionthatthefractionmnisinlowestterms.Alternatesolution(Sketch):Ifx2Qsatisflesx2=12,thenx2isinQandsatisfles¡x2¢2=3.Nowprovethatthereisnoy2Qsuchthaty2=3byrepeatingtheproofthatp262Q.Problem1.5:LetA‰Rbenonemptyandboundedbelow.Set¡A=f¡a:a2Ag.Provethatinf(A)=¡sup(¡A).Solution:Firstnotethat¡Aisnonemptyandboundedabove.Indeed,Acontainssomeelementx,andthen¡x2A;moreover,Ahasalowerboundm,and¡misanupperboundfor¡A.Wenowknowthatb=sup(¡A)exists.Weshowthat¡b=inf(A).That¡bisalowerboundforAisimmediatefromthefactthatbisanupperboundfor¡A.Toshowthat¡bisthegreatestlowerbound,weletc>¡bandprovethatcisnotalowerboundforA.Now¡c<b,so¡cisnotanupperboundfor¡A.Sothereexistsx2¡Asuchthatx>¡c.Then¡x2Aand¡x<c.SocisnotalowerboundforA.Problem1.6:Letb2Rwithb>1,flxedthroughouttheproblem.Comment:Wewillassumeknownthatthefunctionn7!bn,fromZtoR,isstrictlyincreasing,thatis,thatform;n2Z,wehavebm<bnifandonlyifm<n.Similarly,wetakeasknownthatx7!xnisstrictlyincreasingwhennisDate:1October2001.12MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1anintegerwithn>0.Wewillalsoassumethattheusuallawsofexponentsareknowntoholdwhentheexponentsareintegers.Wecan’tassumeanythingaboutfractionalexponents,exceptforTheorem1.21ofthebookanditscorollary,becausethecontextmakesitclearthatwearetoassumefractionalpowershavenotyetbeendeflned.(a)Letm;n;p;q2Z,withn>0andq>0.Provethatifmn=pq,then(bm)1=n=(bp)1=q.Solution:BytheuniquenesspartofTheorem1.21ofthebook,appliedtothepositiveintegernq,itsu–cestoshowthath(bm)1=ninq=h(bp)1=qinq:NowthedeflnitioninTheorem1.21impliesthath(bm)1=nin=bmandh(bp)1=qiq=bp:Therefore,usingthelawsofintegerexponentsandtheequationmq=np,wegeth(bm)1=ninq=hh(bm)1=niniq=(bm)q=bmq=bnp=(bp)n=hh(bp)1=qiqin=h(bp)1=qinq;asdesired.ByPart(a),itmakessensetodeflnebm=n=(bm)1=nform;n2Zwithn>0.Thisdeflnesbrforallr2Q.(b)Provethatbr+s=brbsforr;s2Q.Solution:Choosem;n;p;q2Z,withn>0andq>0,suchthatr=mnands=pq.Thenr+s=mq+npnq.BytheuniquenesspartofTheorem1.21ofthebook,appliedtothepositiveintegernq,itsu–cestoshowthathb(mq+np)=(nq)inq=h(bm)1=n(bp)1=qinq:Directlyfromthedeflnitions,wecanwritehb(mq+np)=(nq)inq=•hb(mq+np)i1=(nq)‚nq=b(mq+np):Usingthelawsofintegerexponentsandthedeflnitionsforrationalexponents,wecanrewritetherighthandsideash(bm)1=n(bp)1=qinq=hh(bm)1=niniqhh(bp)1=qiqin=(bm)q(bp)n=b(mq+np):Thisprovestherequiredequation,andhencetheresult.(c)Forx2R,deflneB(x)=fbr:r2Q\(¡1;x]g:Provethatifr2Q,thenbr=sup(B(r)).Solution:Themainpointistoshowthatifr;s2Qwithr<s,thenbr<bs.Choosem;n;p;q2Z,withn>0andq>0,suchthatr=mnands=pq.ThenMATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK13alsor=mqnqands=npnq,withnq>0,sobr=(bmq)1=(nq)andbs=(bnp)1=(nq):Nowmq<npbecauser<s.Therefore,usingthedeflnitionofc1=(nq),(br)nq=bmq<bnp=(bs)nq:Sincex7!xnqisstrictlyincreasing,thisimpliesthatbr<bs.Nowwecanprovethatifr2Qthenbr=sup(B(r)).Bytheabove,ifs2Qands•r,thenbs•br.ThisimpliesthatbrisanupperboundforB(r).Sincebr2B(r),obviouslynonumbersmallerthanbrcanbeanupperboundforB(r).Sobr=sup(B(r)).Wenowdeflnebx=sup(B(x))foreveryx2R.WeneedtoshowthatB(x)isnonemptyandboundedabove.Toshowitisnonempty,choose(usingtheArchimedeanproperty)somek2Zwithk<x;thenbk2B(x).Toshowitisboundedabove,similarlychoosesomek2Zwithk>x.Ifr2Q\(¡1;x],thenbr2B(k)sothatbr•bkbyPart(c).ThusbkisanupperboundforB(x).Thisshowsthatthedeflnitionmakessense,andPart(c)showsitisconsistentwithourearlierdeflnitionwhenr2Q.(d)Provethatbx+y=bxbyforallx;y2R.Solution:Inordertodothis,wearegoingtoneedtoreplacethesetB(x)abovebythesetB0(x)=fbr:r2Q\(¡1;x)g(thatis,werequirer<xratherthanr•x)inthedeflnitionofbx.(Ifyouareskeptical,readthemainpartofthesolutionflrsttoseehowthisisused.)Weshowthatthereplacementispossibleviasomelemmas.Lemma1.Ifx2[0;1)andn2Zsatisflesn‚0,then(1+x)n‚1+nx.Proof:Theproofisbyinductiononn.Thestatementisobviousforn=0.Soassumeitholdsforsomen.Then,sincex‚0,(1+x)n+1=(1+x)n(1+x)‚(1+nx)(1+x)=1+(n+1)x+nx2‚1+(n+1)x:Thisprovestheresultforn+1.Lemma2.inffb1=n:n2Ng=1.(Recallthatb>1andN=f1;2;3;:::g.)Proof:Clearly1isalowerbound.(Indeed,(b1=n)n=b>1=1n,sob1=n>1.)Weshowthat1+xisnotalowerboundwhenx>0.If1+xwerealowerbound,then1+x•b1=nwouldimply(1+x)n•(b1=n)n=bforalln2N.ByLemma1,wewouldget1+nx•bforalln2N,whichcontradictstheArchimedeanpropertywhenx>0.Lemma3.supfb¡1=n:n2Ng=1.Proof:Part(b)showsthatb¡1=nb1=n=b0=1,whenceb¡1=n=(b1=n)¡1.Sinceallnumbersb¡1=narestrictlypositive,itnowfollowsfromLemma2that1isanupperbound.Supposex<1isanupperbound.Thenx¡1isalowerboundfor4MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1fb1=n:n2Ng.Sincex¡1>1,thiscontradictsLemma2.Thussupfb¡1=n:n2Ng=1,asclaimed.Lemma4.bx=sup(B0(x))forx2R.Proof:Ifx62Q,thenB0(x)=B(x),sothereisnothingtoprove.Ifx2Q,thenatleastB0(x)‰B(x),sobx‚sup(B0(x)).Moreover,Part(b)showsthatbx¡1=n=bxb¡1=nforn2N.Thenumbersbx¡1=nareallinB0(x),andsupfbxb¡1=n:n2Ng=bxsupfb¡1=n:n2Ngbecausebx>0,sousingLemma3inthelaststepgivessup(B0(x))‚supfbx¡1=n:n2Ng=bxsupfb¡1=n:n2Ng=bx:Nowwecanprovetheformulabx+y=bxby.Westartbyshowingthatbx+y•bxby,whichwedobyshowingthatbxbyisanupperboundforB0(x+y).Thusletr2Qsatisfyr<x+y.Thenthereares0;t02Rsuchthatr=s0+t0ands0<x,t0<y.Chooses;t2Qsuchthats0<s<xandt0<t<y.Thenr<s+t,sobr<bs+t=bsbt•bxby.ThisshowsthatbxbyisanupperboundforB0(x+y).(NotethatthisdoesnotworkusingB(x+y).Ifx+y2Qbutx;y62Q,thenbx+y2B(x+y),butitisnotpossibletoflndsandtwithbs2B(x),bt2B(y),andbsbt=bx+y.)Wenowprovethereverseinequality.Supposeitfails,thatis,bx+y<bxby.Thenbx+yby<bx:ThelefthandsideisthusnotanupperboundforB0(x),sothereexistss2Qwiths<xandbx+yby<bs:Itfollowsthatbx+ybs<by:Repeatingtheargument,thereist2Qwitht<ysuchthatbx+ybs<bt:Thereforebx+y<bsbt=bs+t(usingPart(b)).Butbs+t2B0(x+y)becauses+t2Qands+t<x+y,sothisisacontradiction.Thereforebx+y•bxby.Problem1.9:DeflnearelationonCbyw<zifandonlyifeitherRe(w)<Re(z)orbothRe(w)=Re(z)andIm(w)<Im(z).(Forz2C,theexpressionsRe(z)andIm(z)denotetherealandimaginarypartsofz.)ProvethatthismakesCanorderedset.Doesthisorderhavetheleastupperboundproperty?Solution:Weverifythetwoconditionsinthedeflnitionofanorder.Fortheflrst,letw;z2C.Therearethreecases.Case1:Re(w)<Re(z).Thenw<z,butw=zandw>zarebothfalse.Case2:Re(w)>Re(z).Thenw>z,butw=zandw<zarebothfalse.MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK15Case3:Re(w)=Re(z).Thiscasehasthreesubcases.Case3.1:Im(w)<Im(z).Thenw<z,butw=zandw>zarebothfalse.Case3.2:Im(w)>Im(z).Thenw>z,butw=zandw<zarebothfalse.Case3.3:Im(w)=Im(z).Thenw=z,butw>zandw<zarebothfalse.Thesecasesexhaustallpossibilities,andineachofthemexactlyoneofw<z,w=z,andw>zistrue,asdesired.Nowweprovetransitivity.Lets<wandw<z.IfeitherRe(s)<Re(w)orRe(w)<Re(z),thenclearlyRe(s)<Re(z),sos<z.IfRe(s)=Re(w)andRe(w)=Re(z),thenthedeflnitionoftheorderrequiresIm(s)<Im(w)andIm(w)<Im(z).WethushaveRe(s)=Re(z)andIm(s)<Im(z),sos<zbydeflnition.Itremainstoanswerthelastquestion.Weshowthatthisorderdoesnothavetheleastupperboundproperty.LetS=fz2C:Re(z)<0g.ThenS6=?because¡12S,andSisboundedabovebecause1isanupperboundforS.WeshowthatSdoesnothavealeastupperboundbyshowingthatifwisanupperboundforS,thenthereisasmallerupperbound.First,bythedeflnitionoftheorderitisclearthatRe(w)isanupperboundforfRe(z):z2Sg=(¡1;0):ThereforeRe(w)‚0.Moreover,everyu2CwithRe(u)‚0isinfactanupperboundforS.Inparticular,ifwisanupperboundforS,thenw¡i<wandhasthesamerealpart,soisasmallerupperbound.Note:ArelatedargumentshowsthatthesetT=fz2C:Re(z)•0galsohasnoleastupperbound.OneshowsthatwisanupperboundforTifandonlyifRe(w)>0.Problem1.13:Provethatifx;y2C,thenjjxj¡jyjj•jx¡yj.Solution:Thedesiredinequalityisequivalenttojxj¡jyj•jx¡yjandjyj¡jxj•jx¡yj:Weprovetheflrst;thesecondfollowsbyexchangingxandy.Setz=x¡y.Thenx=y+z.Thetriangleinequalitygivesjxj•jyj+jzj.Substitutingthedeflnitionofzandsubtractingjyjfrombothsidesgivestheresult.Problem1.17:Provethatifx;y2Rn,thenkx+yk2+kx¡yk2=2kxk2+2kyk2:Interpretthisresultgeometricallyintermsofparallelograms.Solution:Usingthedeflnitionofthenormintermsofscalarproducts,wehave:kx+yk2+kx¡yk2=hx+y;x+yi+hx¡y;x¡yi=hx;xi+hx;yi+hy;xi+hy;yi+hx;xi¡hx;yi¡hy;xi+hy;yi=2hx;xi+2hy;yi=2kxk2+2kyk2:Theinterpretationisthat0;x;y;x+yaretheverticesofaparallelogram,andthatkx+ykandkx¡ykarethelengthsofitsdiagonalswhilekxkandkykareeach6MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK1thelengthsoftwooppositesides.Thereforethesumofthesquaresofthelengthsofthediagonalsisequaltothesumofthesquaresofthelengthsofthesides.Note:Onecandotheproofdirectlyintermsoftheformulakxk2=Pnk=1jxkj2.Thestepsareallthesame,butitismorecomplicatedtowrite.Itisalsolessgeneral,sincetheargumentaboveappliestoanynormthatcomesfromascalarproduct.MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK2Generally,a\solution"issomethingthatwouldbeacceptableifturnedinintheformpresentedhere,althoughthesolutionsgivenareoftenclosetominimalinthisrespect.A\solution(sketch)"istoosketchytobeconsideredacompletesolutionifturnedin;varyingamountsofdetailwouldneedtobefllledin.Problem2.2:Provethatthesetofalgebraicnumbersiscountable.Solution(sketch):Foreachflxedintegern‚0,thesetPnofallpolynomialswithintegercoe–cientsanddegreeatmostniscountable,sinceithasthesamecardinalityasthesetf(a0;:::;an):ai2Ng=Nn+1.Thesetofallpolynomialswithintegercoe–cientsisS1n=0Pn,whichisacountableunionofcountablesetsandsocountable.Eachpolynomialhasonlyflnitelymanyroots(atmostnfordegreen),sothesetofallpossiblerootsofallpolynomialswithintegercoe–cientsisacountableunionofflnitesets,hencecountable.Problem2.3:Provethatthereexistrealnumberswhicharenotalgebraic.Solution(Sketch):ThisfollowsfromProblem2.2,sinceRisnotcountable.Problem2.4:IsRnQcountable?Solution(Sketch):No.QiscountableandRisnotcountable.Problem2.5:ConstructaboundedsubsetofRwithexactly3limitpoints.Solution(Sketch):Forexample,use'1n:n2N“['1+1n:n2N“['2+1n:n2N“:Problem2.6:LetE0denotethesetoflimitpointsofE.ProvethatE0isclosed.ProvethatE0=E0.Is(E0)0alwaysequaltoE0?Solution(Sketch):ProvingthatE0isclosedisequivalenttoprovingthat(E0)0‰E0.Soletx2(E0)0andlet">0.Choosey2E0\(N"(x)nfxg).Choose–=min(d(x;y);"¡d(x;y))>0.Choosez2E\(N–(y)nfyg).Thetriangleinequalityensuresz6=xandz2N"(x).ThisshowsxisalimitpointofE.Hereisadifierentwaytoprovethat(E0)0‰E0.Letx2(E0)0and">0.Choosey2E0\(N"=2(x)nfxg).ByTheorem2.20ofRudin,thereareinflnitelymanypointsinE\(N"=2(y)nfyg).Inparticularthereisz2E\(N"=2(y)nfyg)withz6=x.Nowz2E\(N"(x)nfxg).ToproveE0=E0,itsu–cestoproveE0‰E0.WeflrstclaimthatifAandBareanysubsetsofX,then(A[B)0‰A0[B0.Thefastestwaytodothisistoassumethatx2(A[B)0butx62A0,andtoshowthatx2B0.Accordingly,letx2(A[B)0nA0.Sincex62A0,thereis"0>0suchthatN"0(x)\AcontainsnoDate:8October2001.12MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK2pointsexceptpossiblyxitself.Nowlet">0;weshowthatN"(x)\Bcontainsatleastonepointdifierentfromx.Letr=min(";"0)>0.Becausex2(A[B)0,thereisy2Nr(x)\(A[B)withy6=x.Theny62Abecauser•"0.Sonecessarilyy2B,andthusyisapointdifierentfromxandinNr(x)\B.Thisshowsthatx2B0,andcompletestheproofthat(A[B)0‰A0[B0.ToproveE0‰E0,wenowobservethatE0=(E[E0)0‰E0[(E0)0‰E0[E0=E0:AnalternateproofthatE0‰E0canbeobtainedbyslightlymodifyingeitheroftheproofsabovethat(E0)0‰E0.Forthethirdpart,theanswerisno.TakeE=f0g['1n:n2N“:ThenE0=f0gand(E0)0=?.(Ofcourse,youmustprovethesefacts.)Problem2.8:IfE‰R2isopen,iseverypointofEalimitpointofE?WhatifEisclosedinsteadofopen?Solution(Sketch):EverypointofanopensetE‰R2isalimitpointofE.Indeed,ifx2E,thenthereis">0suchthatN"(x)‰E,anditiseasytoshowthatxisalimitpointofN"(x).(Warning:Thisisnottrueinageneralmetricspace.)Noteverypointofaclosedsetneedbealimitpoint.TakeE=f(0;0)g,whichhasnolimitpoints.Problem2.9:LetE–denotethesetofinteriorpointsofasetE,thatis,theinteriorofE.(a)ProvethatE–isopen.Solution(sketch):Ifx2E–,thenthereis">0suchthatN"(x)‰E.SinceN"(x)isopen,everypointinN"(x)isaninteriorpointofN"(x),henceofthebiggersetE.SoN"(x)2E–.(b)ProvethatEisopenifandonlyifE–=E.Solution:IfEisopen,thenE=E–bythedeflnitionofE–.IfE=E–,thenEisopenbyPart(a).(c)IfGisopenandG‰E,provethatG‰E–.Solution(sketch):Ifx2G‰EandGisopen,thenxisaninteriorpointofG.ThereforexisaninteriorpointofthebiggersetE.Sox2E–.(d)ProvethatXnE–=XnE.Solution(sketch):FirstshowthatXnE–‰XnE.Ifx62E,thenclearlyx2XnE.Otherwise,considerx2EnE–.RearrangingthestatementthatxfailstobeaninteriorpointofE,andnotingthatxitselfisnotinXnE,onegetsexactlythestatementthatxisalimitpointofXnE.NowshowthatXnE‰XnE–.Ifx2XnE,thenclearlyx62E–.Ifx62XnEbutxisalimitpointofXnE,thenonesimplyrearrangesthedeflnitionofalimitpointtoshowthatxisnotaninteriorpointofE.(e)Proveordisprove:¡E¢–=E.MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK23Solution(sketch):Thisisfalse.Example:takeE=(0;1)[(1;2).WehaveE–=E,„E=[0;2],and¡E¢–=(0;2).AnotherexampleisQ.(f)Proveordisprove:E–=E.Solution(sketch):Thisisfalse.Example:takeE=(0;1)[f2g.ThenE=[0;1][f2g,E–=(0;1),andE–=[0;1].ThesetsQandf0garealsoexamples:inbothcases,E–=?.Problem2.11:WhichofthefollowingaremetricsonR?(a)d1(x;y)=(x¡y)2.Solution(Sketch):No.Thetriangleinequalityfailswithx=0,y=2,andz=4.(b)d2(x;y)=pjx¡yj.Solution(Sketch):Yes.Someworkisneededtocheckthetriangleinequality.(c)d3(x;y)=jx2¡y2j.Solution(Sketch):No.d3(1;¡1)=0.(d)d4(x;y)=jx¡2yj.Solution(Sketch):No.d4(1;1)6=0.Also,d4(1;6)6=d4(6;1).(e)d5(x;y)=jx¡yj1+jx¡yj.Solution(Sketch):Yes.Someworkisneededtocheckthetriangleinequality.Youneedtoknowthatt7!t1+tisnondecreasingon[0;1),andthata;b‚0impliesa+b1+a+b•a1+a+b1+b:Dotheflrstbyalgebraicmanipulation.Thesecondisa+b1+a+b=a1+a+b+b1+a+b•a1+a+b1+b:(ThisiseasierthanwhatmostpeopledidthelasttimeIassignedthisproblem.)MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK3Generally,a\solution"issomethingthatwouldbeacceptableifturnedinintheformpresentedhere,althoughthesolutionsgivenareoftenclosetominimalinthisrespect.A\solution(sketch)"istoosketchytobeconsideredacompletesolutionifturnedin;varyingamountsofdetailwouldneedtobefllledin.Problem2.14:Giveanexampleofanopencoveroftheinterval(0;1)‰Rwhichhasnoflnitesubcover.Solution(sketch):f(1=n;1):n2Ng.(Notethatyoumustshowthatthisworks.)Problem2.16:RegardQasametricspacewiththeusualmetric.LetE=fx2Q:2<x2<3g.ProvethatEisaclosedandboundedsubsetofQwhichisnotcompact.IsEanopensubsetofQ?Solution(sketch):ClearlyEisbounded.WeproveEisclosed.ThefastwaytodothisistonotethatQnE=Q\h‡¡1;¡p3·[‡¡p2;p2·[‡p3;1·i;andsoisopenbyTheorem2.30.Todoitdirectly,supposex2QisalimitpointofEwhichisnotinE.Sincewecan’thavex2=2orx2=3,wemusthavex2<2orx2>3.Assumex2>3.(Theothercaseishandledsimilarly.)Letr=jxj¡p3>0.Theneveryz2Nr(x)satisflesjzj‚jxj¡jx¡zj>jxj¡r>0;whichimpliesthatz2>(jxj¡r)2=3.Thisshowsthatz62E,whichcontradictstheassumptionthatxisalimitpointofE.ThefastwaytoseethatEisnotcompactistonotethatitisasubsetofR,butisnotclosedinR.(SeeTheorem2.23.)Toprovethisdirectly,showthat,forexample,thesets'y2Q:2+1n<y2<3¡1n“formanopencoverofEwhichhasnoflnitesubcover.ToseethatEisopeninQ,thefastwayistowriteE=Q\h‡¡p3;¡p2·[‡p2;¡p3·i;whichisopenbyTheorem2.30.Itcanalsobeproveddirectly.Problem2.19:LetXbeametricspace,flxedthroughoutthisproblem.(a)IfAandBaredisjointclosedsubsetsofX,provethattheyareseparated.Solution(Sketch):WehaveA\B=A\B=A\B=?becauseAandBareclosed.Date:15October2001.12MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK3(b)IfAandBaredisjointopensubsetsofX,provethattheyareseparated.Solution(Sketch):XnAisaclosedsubsetcontainingB,andhencecontainingB.ThusA\B=?.InterchangingAandB,itfollowsthatA\B=?.(c)Fixx02Xand–>0.SetA=fx2X:d(x;x0)<–gandB=fx2X:d(x;x0)>–g:ProvethatAandBareseparated.Solution(Sketch):BothAandBareopensets(proof!),andtheyaredisjoint.SothisfollowsfromPart(b).(d)ProvethatifXisconnectedandcontainsatleasttwopoints,thenXisuncountable.Solution:LetxandybedistinctpointsofX.LetR=d(x;y)>0.Foreachr2(0;R),considerthesetsAr=fz2X:d(z;x)<rgandBr=fz2X:d(z;x)>rg:TheyareseparatedbyPart(c).Theyarenotempty,sincex2Arandy2Br.SinceXisconnected,theremustbeapointzr2Xn(Ar[Br).Thend(x;zr)=r.Notethatifr6=s,thend(x;zr)6=d(x;zs),sozr6=zs.Thusr7!zrdeflnesaninjectivemapfrom(0;R)toX.Since(0;R)isnotcountable,Xcan’tbecountableeither.Problem2.20:LetXbeametricspace,andletE‰Xbeaconnectedsubset.IsEnecessarilyconnected?Isint(E)necessarilyconnected?Solutiontotheflrstquestion(sketch):Thesetint(E)neednotbeconnected.TheeasiestexampletowritedownistotakeX=R2andE=fx2R2:kx¡(1;0)k•1g[fx2R2:kx¡(¡1;0)k•1g:Thenint(E)=fx2R2:kx¡(1;0)k<1g[fx2R2:kx¡(¡1;0)k<1g:Thissetfailstobeconnectedbecausethepoint(0;0)ismissing.Amoredramaticexampleistwocloseddisksjoinedbyaline,sayE=fx2R2:kx¡(2;0)k•1g[fx2R2:kx¡(¡2;0)k•1g(1)[f(fi;0)2R2:¡3•fi•3g:(2)Thenint(E)=fx2R2:kx¡(2;0)k<1g[fx2R2:kx¡(¡2;0)k<1g:Solutiontothesecondquestion:IfEisconnected,thenEisnecessarilyconnected.ToprovethisusingRudin’sdeflnition,assumeE=A[BforseparatedsetsAandB;weprovethatoneofAandBisempty.ThesetsA0=A\EandB0=B\EareseparatedsetssuchthatE=A0[B0.(TheyareseparatedbecauseA0‰AandB0‰B.)BecauseEisconnected,oneofA0andB0mustbeempty;withoutlossofgenerality,A0=?.ThenA‰EnE.ThereforeE‰B.ButthenA‰E‰B.BecauseAandBareseparated,thiscanonlyhappenifA=?.MATH413[513](PHILLIPS)SOLUTIONSTOHOMEWORK33Alternatesolutiontothesecondquestion:IfEisconnected,weprovethatEisnecessarilyconnected,usingthetraditionaldeflnition.Thus,assumethatE=A[BfordisjointrelativelyopensetsAandB;weprovethatoneofAandBisempty.ThesetsA0=A\EandB0=B\EaredisjointrelativelyopensetsinEsuchthatE=A0[B0.BecauseEisconnected,oneofA0andB0mustbeempty;withoutlossofgenerality,A0=?.ThenA‰EnEandisrelativelyopeninE.Nowletx2A.Thenthereis">0suchthatN"(x)\E‰A.SoN"(x)\E‰EnE,whichimpliesthatN"(x)\E=?.Thiscontradictsthefactthatx2E.ThusA=?.Problem2.22:ProvethatRnisseparable.Solution(sketch):ThesubsetQniscountablebyTh