FollandRealAnalysis-UCLA实分析

Chapter4.1,Page117Problem10:AtopologicalspaceXiscalleddisconnectedifthereexistnonemptyopensetsU,VsuchthatU\V=;andU[V=X;otherwiseXisconnected.WhenwespeakofconnectedordisconnectedsubsetsofX,werefertotherelativetopologyonthem.(a)Xisconnectedi;andXaretheonlysubsetsofXthatarebothopenandclosed.(b)IffEg2AisacollectionofconnectedsubsetsofXsuchthat\2AE6=;,then[2AEisconnected.(c)IfAXisconnected,thenAisconnected.=(d)Everypointx2XiscontainedinauniquemaximalconnectedsubsetofX,andthissubsetisclosed.(Itiscalledtheconnectedcompo-nentofx.)Solution.(a)IfXisdisconnected,thereexistnonemptydisjointopensetsUandVwithU[V=X.ThenUandVarebothclosedaswellsincetheyarecomplementsofopensets,sothereexistclopensets(bothUandV)otherthan;andX.Conversely,supposeFisaclopensetwithF6=;andF6=X.LetU=FandV=XnF.ThenVisnonemptysinceF6=XandopenbecauseFisclosed,soXisdisconnected.(b)LetX=[EandsupposeXisdisconnected,soX=U[VwhereUandVaredisjoint,open,andnonempty.ForeachE,wemusthaveeitherEUorEVsinceotherwiseE=(E\U)[(E\V)isadisconnectionofE.BecauseUandVarebothnonempty,thereexistE1UandE2V.ButthenE1\E2=;,contradictingthefactthat\Eisnonempty.HenceXmustbeconnected.(c)SupposeAisdisconnected.Then9disjointopenU;VXwithAU[VandA\UandA\Vnonempty.ThenA\UmustbenonemptysinceotherwiseXnUwouldbeaclosedsetcontainingAsothatA\Uwouldbeempty.Similarly,A\Vmustbenonempty.ThenA=(A\U)[(A\V)isadisconnectionofA.(d)Givenx2X,letCx=[x2UUconnectedU:ThenCxisconnectedbypart(b),sincex2\U.SinceeveryconnectedsetcontainingxisasubsetofCx,itisthemaximalconnectedsubsetcontainingx.Nowsupposey2C0x.BythemaximalityofCx,Cx[fygisdisconnected,soitiscontainedindisjointopensetsUandV.ThesecannotbothintersectCxsincethiswouldgiveadisconnectionofCx,sothereexistUandVopenanddisjointwithCxUandy2V.Buttheny2VC0x,soC0xisopen,soCxisclosed.Probem13:IfXisatopologicalspace,UisopeninX,andAisdenseinX,thenU=U\A.12Solution.ObviouslyU\AUsinceU\AU.Conversely,sinceA=(A\U)[(A\U0)whereU0=XnU,wehaveX=A=A\U[A\U0:NowU0isaclosedsetcontainingA\U0,soA\U0U0.ThereforeUA\U.ButthisimpliesUA\U.Chapter4.2,Page123Problem26:LetXandYbetopologicalspaces.(a)IfXisconnectedandf2C(X;Y),thenf(X)isconnected.(b)Xiscalledarcwiseconnectedifforallx0;x12Xthereexistsf2C([0;1];X)withf(0)=x0andf(1)=x1.Everyarcwiseconnectedspaceisconnected.(c)LetX=f(s;t)2R2:t=sin(s1)g[f(0;0)g,withtherelativetopol-ogyinducedfromR2.ThenXisconnectedbutnotarcwiseconnected.Solution.(a)Supposef(X)=U[VwhereU;Varedisjoint,nonempty,andopen.ThenX=f1(f(X))=f1(U)[f1(V)whichareclearlydisjointandnonempty,andareopenbecausefiscontinuous.(b)LetXbearcwiseconnected,andsupposeXisdisconnected,soX=U[VwhereU;Varedisjoint,nonempty,andopen.Letx12Uandx22V.LetCbeacurvefromx1tox2.ThenC=(C\U)[(C\V)andbotharenonemptybecausex12C\Uandx22C\V.ThiscontradictsthefactthatCisconnectedbypart(a)becauseitistheimageoftheconnectedset[0;1]underacontinuousfunction.(c)First,weshowthatXisconnected:SupposeU;VisadisconnectionofX.SupposeWLOGthat(0;0)2U.SinceVisnonemptyitintersectseitherf(s;sin(s1)):s0gorf(s;sin(s1)):s0g,WLOGtheformer,whichwecallX0.ButUintersectsX0aswellsinceitcontainsaneighborhoodoftheorigin.Then(X0\U)[(X0\V)isadisconnectionofX0.ButX0istheimageoftheconnectedset(0;1)Runderthecontinuousfunctionx7!(x;sin(x1))andhenceconnected,acontradiction.ThereforeXmustbeconnected.ToshowthatXisnotpathconnected,suppose:[0;1]!Xisapathwith(0)=(0;0)and(1)=(12;0).ThenI=x((t))isaconnectedsubsetoftherealline,wherex((a;b))=a,sincexandarebothcontinuous.SinceIcontains0and12,itcontainseverythinginbetween.Thisimpliesthattheimageofincludesthepointspn=(12n+=2;1)forn=1;2;:::.Butpn!(0;1)whichisnotinXandhencenotintheimageof.Thus,theimageofisnotclosedinR2;butinfactitmustbecompactandthereforeclosedbecauseiscontinuousand[0;1]iscompact.Thiscontradictionprovesthatthereisnopathfrom(0;0)to(12;0),soXisnotpathconnected.Problem27:IfXisconnectedforeach2A,thenX=Q2AXisconnected.Solution.Werstsolvetheproblemforniteproducts:3Lemma1.Theproductofnitelymanyconnectedspacesisconnected.Proof.Byinduction,itissucienttoprovethisfortheproductoftwospaces,sinceQnj=1Xj=(Qn1j=1Xj)Xn.SupposeXandYareconnected.IfXYisdisconnected,letUandVbedisjointopennonemptysetswithU[V=XY.Let(x0;y0)2Uand(x1;y1)2V.If(x0;y1)2U,letZ=f(x;y0):x2Xg,whichishomeomorphictoXbyExercise18.ThenZ\UandZ\Varebothnonempty,since(x0;y1)2U\Zand(x1;y1)2V\Z.ButthenZ=(U\Z)[(V\Z)isadisconnectionofZ,acontradictionbecauseXisconnected.Similarly,if(x0;y1)2V,letZ=f(x0;y):y2YgwhichishomeomorphictoYbyExercise18,andweobtainadisconnectionofYbythesameargument.HenceXYmustbeconnected.NowletX=QXbeanarbitraryproductofconnectedspaces.Chooseanyx2X.Foranynitesetf1;:::;ngofindices,letU1;:::;n=fy2X:(y)=(x)forall6=1;:::;ng:ThenU1;:::;nishomeomorphictoX1


XnbyExercise18,andthereforeconnectedbythelemma.SinceeachU1;:::;ncontainsx,theirunion,whichwecallU,isconnectedbyExercise10b.ThereforeUisalsoconnectedbyExercise10c.Finally,wewillshowthatU=X.LetVXbeanyopenset.Usingthebasisfortheproducttopology,Vmustcontainasetoftheform~V=U1


UmY