answers-to-topology-munkres

SOLUTIONSTOEXERCISESHerearesolutionstosomeoftheproblemsinMunkres.Theremaybeother,andperhapsbetter,ones.Theyaresometimesmoredetailedthanabsolutelynecessary,butnotexcessivelyso.Problem13.1.Foreachx∈A,chooseanopensetUxwithx∈Ux⊆A.ThenA=Sx∈AUx,whichisopenbecauseitisaunionofopensets.¤Problem13.3.WehaveX−X=∅,whichiscountable,soX∈Tc,andX−∅=X,so∅∈Tc.NextsupposethatUα∈TcforαinsomeindexsetJ.IfallUαareempty,soistheirunion,soSα∈JUα∈Tc.Otherwise,thereissomeβ∈JsuchthatX−Uβiscountable.SinceX−Sα∈JUα⊆X−Uβ,X−Sα∈JUαiscountable,andagainSα∈JUα∈Tc.Finally,supposeU,V∈Tc.IfeitherofUandVisempty,soisU∩V.Otherwise,X−UandX−Varecountable,andhencesoisX−(U∩V)=(X−U)∪(X−V),andU∩V∈Tc.ThiscompletestheproofthatTcisatopology.ThecollectionT∞isnotatopology,ingeneral.Forexample,takeX=R.Then(−∞,0)and(0,∞)areinT∞,but(−∞,0)∪(0,∞)=R−{0}isnot.(Infact,T∞isatopologyiffXisfinite,inwhichcaseitistheindiscretetopology,butasingleexampleisallthat’sneeded.)¤Problem13.8.(a)CertainlyBisacollectionofopensetsinthestandardtopology.SupposethatUisopeninthestandardtopologyandx∈U.Bydefinitionofthestandardtopology,therearerealnumbersabwithx∈(a,b)⊆U.Sinceax,thereisarationalnumberqwithaqx.(Thisisastandardfactabouttherealnumbers,knownasthedensityoftherationals;forcompleteness,Igiveaproofbelow.)Similarly,thereisarationalnumberrwithxrb.Then(q,r)∈Bandx∈(q,r)⊆U,andtheresultfollowsfromLemma13.2.(b)Foranyx∈R,thereexistrationalnumbersaandbwitha≤xb(forinstance,a=bxcandb=bxc+1),sothatx∈[a,b)∈C.Also,theintersectionofanytwoelementsofCiseitheremptyoranelementofC,andsoCisabasisforsometopologyTonR.Letxbeanirrationalnumber.Theinterval[x,x+1)isopeninthelowerlimittopology;weshowthatitdoesnotbelongtoT.Supppose,onthecontrary,thatitdoes.Sincex∈[x,x+1),thereexistsanelement[a,b)(aandbrational)ofCwithx∈[a,b)⊆[x,x+1).Nowx∈[a,b)impliesa≤x,while[a,b)⊆[x,x+1)impliesx≤a,sox=a,acontradiction.¤Densityoftherationals.Letaandbberealnumberswithab.Since0b−a,thereisapositiveintegernwith1nb−a.Letmbethesmallestintegersuchthatmna.Thenm−1n≤a,somn≤a+1na+(b−a)=b.Thustherationalnumbermnsatisfiesamnb.¤Whilewe’reatit,let’sprovethatbetweenanytworealsthereisanirrational.Densityoftheirrationals.Letaandbberealnumberswithab.Bywhatwasjustproved,thereisarationalnumberqwitha−√2qb−√2.Nowx=q+√2isirrational,andaxb.¤1Problem16.1.LetTXandTYbethetopologiesthatAinheritsasasubspaceofXandY,respectively.SupposefirstthatW∈TX.ThenthereisanopensetUofXwithW=U∩A.SetV=U∩Y.ThenVisopeninYandW=V∩A,soA∈TY.SupposeconverselythatW∈TY.ThenthereisasetVopeninYwithW=V∩A.SinceVisopeninY,thereisanopensetUofXwithV=U∩Y.ThenW=U∩A,soW∈TX.¤Problem16.9.LetT1bethedictionaryordertopologyonR×R,andT2theproducttopologyonRd×R.Aspointedoutonpage85ofMunkres,thesets(a×b,a×d)forbdformabasisforT1.(Thisisbecauseanyset(a×b,c×d)withacistheunionofallthesets(a×b,a×y)foryb,(x×y1,x×y2)foraxcandy1y2,and(c×y,c×d)foryd.)Nowthesingletonsets{a}formabasisforRd(Example3onpage79),andthesets(b,d)forbdformabasisforRbydefinitionoftheusualtopologyonR,sothesets{a}×(b,d)forbdformabasisforT2byTheorem15.1.But(a×b,a×d)={a}×(b,d),sothesamecollectionofsetsisabasisforbothT1andT2,sothesetopologiesareequal.Iobservedinclass(aftercorrectingmyself)thatthedictionaryordertopologyisstrictlyfinerthantheusualtopologyonR2,butitiseasiertoseethisusingthebasisofrectangles(a,b)×(c,d)(abandcd)fortheusualtopology,ratherthanthecirclebasis.Wehave(a,b)×(c,d)=Sx∈(a,b){x}×(c,d),whichisopeninT1.Ontheotherhand,{a}×(b,d)isnotopenintheusualtopology,becauseitisnon-emptyandcontainsnorectangle.¤Fortheproblemonthehandoutforthesecondassignment,seeTheoremN16.1inthesupplementarynotes.Problem17.21.WeshallwriteA0forX−A.(a)FirstweshowthatifUisopenthenU−=U−◦−.SinceUisopenandU⊆U−,U⊆U−◦andhenceU−⊆U−◦−.Conversely,U−◦⊆U−andU−isclosed,soU−◦−⊆U−.NowletA1beanysubsetofX,anddefineA2n=A−2n−1andA2n+1=A02nforn∈Z+.AlsoletB1=A01,andsimilarlydefineB2n=B−2n−1andB2n+1=B02nforn∈Z+.ClearlyanysetobtainedfromA1byrepeatedclosureandcomplementationiseitherAnorBnforsomen∈Z+.NowA3isopen,sinceitisthecomplementofA2=A−1,soA−3=A−◦−3.ButA−3=A4andA−◦−3=A◦−4=A0−0−4=A8.HenceA8=A4,andsoAn+4=Anforn≥4.SimilarlyBn+4=Bnforn≥4,andsoeverysetobtainablefromA1isoneofthefourteensetsA1,...,A7,B1,...B7.(b)LetusfirsttrysimplytofindA1sothatA1,...,A7aredistinct.WehaveA3open,asobservedabove,andA7=A−0−03=A−◦3,soweneedanopensetthatisdifferentfromtheinteriorofitsclosure.Asimpleexampleofsuchasetisanopenintervalorraywithapointremoved;weshallworkwithrayssincetheyhavetheconvenientpropertythatthecomplementofarayisaray.IfA3isobtainedbyremovingapointfromanopenraywithendpointa,thenA4,...,A7arethefourrayswithendpointa.(Forinstance,ifA3=(0,∞)−{1}thenA4=[0,∞),A5=(−∞,0),A6=(−∞,0],andA7=(0,∞).)NowA2,beingthecomplementofA3,willbeaclosedraywithendpointatogetherwithanisolatedpoint((−∞,0]∪{1}intheexample),andifA1isanynon-closedsetwithclosureofthisformthenA1,...,A7willbedistinct.NextwetrytogetB1,...,B7alsodistinct.IfA1isaray(eitheropenorclosed)withapointremoved,togetherwithanisolatedpoint,thenB1hasthesameform,2andeachofA1andB1isanon-closedsetwhoseclosureisaclosedraytogetherwithanisolate