NotesonIntroductoryPoint-SetTopologyAllenHatcherChapter1. BasicPoint-SetTopology . . . . . . . . . . . . . . . 1Topological Spaces 1, Interior, Closure, and Boundary 5, Basis for a Topology 7,MetricSpaces9, Subspaces10, ContinuityandHomeomorphisms12, ProductSpaces13, Exercises 16Chapter2. Connectedness . . . . . . . . . . . . . . . . . . . . 18Path-connected Spaces 19, Cut Points 20, Connected Components and Path Com-ponents21, The Cantor Set25, Exercises 28Chapter3. Compactness . . . . . . . . . . . . . . . . . . . . . 30CompactSetsinEuclideanSpace31, HausdorSpaces34, NormalSpaces36,Lebesgue Numbers 38, Innite Products 30, Exercises 41Chapter4. QuotientSpaces . . . . . . . . . . . . . . . . . . . 44Exercises 52Basic Point-Set Topology 1Chapter1. BasicPoint-SetTopologyOne wayto describe thesubject ofTopology isto say thatit isqualitativegeom-etry. Theideaisthatifonegeometricobjectcanbecontinuouslytransformedintoanother, thenthetwoobjectsaretobeviewedasbeingtopologicallythesame. Forexample, a circle anda square are topologically equivalent. Physically, a rubber bandcanbestretchedintotheformofeitheracircleorasquare, aswell asmanyothershapes which are also viewedas being topologically equivalent. On the otherhand, agure eight curve formed by two circles touching at a point is to be regarded as topo-logically distinct from a circle or square. A qualitative property that distinguishes thecirclefromthegureeightisthenumberofconnectedpiecesthatremainwhenasinglepointisremoved: Whenapointisremovedfromacirclewhatremainsisstillconnected, a single arc, whereas for a gure eight if one removes the point of contactof itstwo circles, whatremains is two separate arcs, two separate pieces.The term used to describe two geometric objects that are topologically equivalentishomeomorphic. Thusacircleandasquarearehomeomorphic. Concretely, ifweplaceacircleCinsideasquareS withthesamecenterpoint, thenprojectingthecircleradiallyoutwardtothesquaredenesafunctionf: CS , andthisfunctioniscontinuous: small changes inxproduce small changes inf(x). The functionfhas an inverse f1: SCobtainedby projecting the square radially inward to the circle, andthisiscontinuousaswell. Onesaysthat f isahomeo-morphism betweenCandS .One of the basic problems of Topology is to determine when two given geometricobjects are homeomorphic. This canbe quite dicult in general.OurrstgoalwillbetodeneexactlywhatthegeometricobjectsarethatonestudiesinTopology. Thesearecalledtopologicalspaces. Thedenitionturnsouttobeextremelygeneral, sothatmanyobjectsthataretopologicalspacesarenotverygeometric at all, in fact.Topological SpacesRatherthanjumpdirectlyintothedenitionofatopologicalspacewewillrstspend a little time motivating the denition by discussing the notion of continuity of afunction. One could say that topological spaces are the objects for which continuous2 Chapter 1functionscan be dened.For the sake of simplicity and concreteness let us talk about functionsf: RR.There are two denitions of continuity for such a function that the reader may alreadybefamiliarwith, the- denitionandthedenitionintermsoflimits. Butitisathird denition, equivalent to these two, that is the one we want here. This denitionis expressed in terms of the notion of an open set inR, generalizing the familiar ideaof an open interval (a, b).Denition. AsubsetOof Risopenifforeachpointx Othereexistsaninterval(a, b)that containsxand iscontainedinO.Withthisdenitionanopeninterval certainlyqualiesasanopenset. Otherexamples are:Ritself is an open set, as are semi-innite intervals(a, )and(, a).The complement of a nite set inRisopen.If Aistheunionoftheinnitesequencexn=1/n, n=1, 2, ,togetherwithitslimit0thenRAis open.Anyunionofopenintervalsisanopenset. Theprecedingexamplesarespecialcasesofthis. Theconversestatementisalsotrue: everyopenset OisaunionofopenintervalssinceforeachxOthereisanopeninterval (ax, bx) withx (ax, bx) O, andOis the union of all these intervals(ax, bx).The empty setis open, since the condition for openness is satised vacuouslyasthere are no pointsxwhere the condition could failto hold.Here are some examples of sets which are not open:Aclosedinterval [a, b]isnotanopensetsincethereisnoopenintervalabouteitheraorbthat is contained in[a, b]. Similarly, half-open intervals[a, b)and(a, b]are not open setswhena < b.A nonempty nite set is not open.Now forthe nice denition of a continuous function in terms of open sets:Denition. A functionf: RRis continuousif for each open setOinRthe inverseimagef1(O) =_x R f(x) O_is also an open set.Toseethatthiscorrespondstotheintuitivenotionofcontinuity,considerwhatwouldhappenifthisconditionfailedtoholdforafunctionf . Therewouldthenbeanopenset Oforwhichf1(O)wasnotopen. Thismeanstherewouldbeapointx0f1(O)forwhichtherewasnointerval (a, b)containingx0andcontainedinf1(O). Thisisequivalenttosayingtherewouldbepointsxarbitrarilyclosetox0thatareinthecomplementof f1(O). For xtobeinthecomplementof f1(O)Basic Point-Set Topology 3meansthat f(x)isnotinO. Ontheotherhand, x0wasinf1(O)sof(x0)isinO. SinceOwasassumedtobeopen,thereisaninterval (c, d)about f(x0)thatiscontained inO. The pointsf(x)that are not inOare therefore not in(c, d)so theyremain at least a xed positive distance fromf(x0). To summarize: there are pointsxarbitrarilyclosetox0forwhichf(x) remainsatleastaxedpositivedistanceawayfromf(x0). Thiscertainly says thatf isdiscontinuous atx0.Thisreasoningcanbereversed. Areasonableinterpretationofdiscontinuityoffatx0would be that there are pointsxarbitrarily close tox0for whichf(x)staysat least a xed positive distance away fromf(x0). Call this xed positive distance.LetObetheopenset (f(x0) , f(x0) + ). Thenf1(O)containsx0butitdoesnot contain any pointsxforwhichf(x)isnot inO, andwe are assuming there aresuch pointsxarbitrarily close tox0, sof1(O)is not open since it does not containall pointsin some interval (a, b)aboutx0.Thedenitionwehavegivenforcontinuityoffunctions RRcanbeappliedmoregenerallytofunctions RnRnandevenRmRnonceonehasanotionofwhat open sets inRnare. The natural denition generalizing the casen = 1is to saythatasetOinRnisopenifforeachxOthereexistsanopenballcontainingxandcontainedinO, whereanopenballofradiusr andcenter x0isdenedtobethe set of pointsxof distance less thanr fromx0. Here the distance fromxtox0is measured as in linear algebra, as the length of the vectorxx0, the square root ofthe dot product of this vectorwith itself.ThisdenitionofopensetsinRndoesnotdependasheavilyonthenotionofdistanceinRnasmightappear. ForexampleinR2whereopenballsbecomeopendisks, wecoulduseopensquaresinsteadofopendiskssinceifapoint xOiscontainedin an open disk containedinOthen it is also containedin an open squarecontainedinthediskandhenceinO, andconversely, if xiscontainedinanopensquarecontainedinOthenitiscontainedinanopendiskcontainedintheopensquareandhenceinO. Inasimilarwaywecouldusemanyothershapesbesidesdisks andsquares, such asellipses orpolygons withany number of sides.After these preliminary remarks we now give the denition of a topological space.Denition. Atopologicalspaceisaset XtogetherwithacollectionOofsubsetsofX, calledopen sets, such that:(1) The union of anycollection of setsinOis inO.(2) The intersection of any nite collection of setsinOisinO.(3) BothandXare inO.4 Chapter 1The collectionOof open setsis calleda topologyonX.All three of these conditions hold for open sets inRas dened earlier. To checkthat(1) holds, supposethatwehaveacollectionofopensets Owheretheindexrangesoversomeindexset I , eitherniteorinnite. Apoint x_OliesinsomeO, whichisopensothereisaninterval (a, b) withx(a, b)O, hencex(a, b)_Oso_Oisopen. Tocheck(2)itsucesbyinductiontocheckthat the intersection of two open setsO1andO2is open. Ifx O1O2thenxliesin open intervals inO1andO2, and there is a smaller open interval in the intersectionofthesetwoopenintervalsthatcontainsx. ThisopenintervalliesinO1 O2, soO1O2is open. Finally, condition (3) obviously holds foropen setsinR.In a similar fashion one can check that open sets inR2or more generallyRnalsosatisfy (1)(3).NoticethattheintersectionofaninnitecollectionofopensetsinRneednotbeopen. Forexample, theintersectionof alltheopenintervals(1/n, 1/n)forn =1, 2, isthesinglepoint {0}whichisnotopen. Thisexplainswhycondition(2)isonly fornite intersections.It is always possible to construct at least two topologies on every setXby choos-ing the collectionOof open setsto be aslarge aspossible orassmall aspossible:ThecollectionOofallsubsetsof XdenesatopologyonXcalledthediscretetopology.Ifwelet Oconsistofjust Xitselfand, thisdenesatopology, thetrivialtopology.Thus we have three dierent topologies onR, the usual topology, the discrete topol-ogy,andthetrivialtopology. Herearetwomore,therstwithfeweropensetsthanthe usual topology, the second with more open sets:LetOconsist of the empty set together with all subsets ofRwhose complementis nite. The axioms (1)(3) are easily veried, and we leave this for the reader tocheck. Every set inOis open in the usualtopology, but not vice versa.Let O consist of all sets Osuch that for eachx Othere is an interval [a, b)withx[a, b)O. Properties(1)(3)canbecheckedbyalmostthesameargumentasfortheusual topologyonR, andagainweleavethisforthereadertodo.Intervals [a, b) arecertainlyinOsothistopologyisdierentfromtheusualtopology onR. Every interval (a, b)isinOsince it can be expressed asa unionof an increasing sequence of intervals[an, b)inO. It follows thatOcontains allBasic Point-Set Topology 5setsthatareopenintheusualtopologysincethesecanbeexpressedasunionsof intervals(a, b).Theseexamplesillustratehowonecanhavetwotopologies OandOonaset X,witheveryset that isopeninthe Otopologyisalsoopeninthe Otopology, soOO. InthissituationwesaythatthetopologyOisnerthanOandthat Oiscoarser thanO. ThusthediscretetopologyonXisnerthananyothertopologyandthetrivial topologyiscoarserthananyothertopology. InthecaseX=Rwehave interpolated three other topologies between these two extremes, with the nite-complement topology being coarser than the usual topology and the half-open-intervaltopology being ner than the usual topology. Of course, given two topologies on a setX, it neednot be true that eitherone isner orcoarser than the other.Here is anotherpiece of basic terminology:Denition. A subsetAof a topologicalspaceXisclosedif itscomplementX Aisopen.Forexample, inRwiththeusual topologyaclosedinterval [a, b] isaclosedsubset. Similarly, inR2withitsusualtopologyacloseddisk, theunionofanopendisk with itsboundary circle, is a closedsubset.Insteadofdeningatopologyonaset Xasacollectionofopensetssatisfyingthethreeaxioms, onecouldequallywell considerthecollectionofcomplementaryclosedsets, anddeneatopologyonXtobeacollectionofsubsetscalledclosedsets, such that the intersection of any collection of closed sets is closed, the union ofany nite collection of closed sets is closed, and both the empty set and the whole setXareclosed. Noticethattheroleofintersectionsandunionsisswitchedcomparedwiththeoriginal denition. Thisisbecauseofthegeneral settheoryfactthatthecomplement ofaunionistheintersectionofthecomplements,andthecomplementof an intersection is the union of the complements.Interior, Closure, and BoundaryConsideranopendiskDintheplane R2, consistingofall thepointsinsideacircleC. We would like to assign precise meanings to certain intuitive statements likethe following:Cis the boundary of the open diskD, andalso of the closeddiskD C.DistheinteriorofthecloseddiskD C,andD Cistheclosure oftheopendiskD.6 Chapter 1The key distinction between points in the boundary of the disk and points in its interioris that for points in the boundary, every open set containing such a point also containspointsinside thediskandpointsoutside thedisk, whileeachpoint intheinteriorofthe disk lies in some open set entirely containedinside the disk.Withthisobservationinmindletusconsiderwhathappensingeneral. Givenasubset AofatopologicalspaceX, thenforeachpoint xXexactlyoneofthefollowing three possibilities holds:(1) There exists an open setOinXwithx O A.(2) There exists an open setOinXwithx O X A.(3) Every open setOwithx OmeetsbothAandX A.Points xsuchthat (1) holdsformasubset of Acalledtheinterior of A, writtenint(A). Thepointswhere(2)holdsthenformint(X A). Pointsxwhere(3)holdsforma set called the boundary or frontierofA, writtenA. The pointsxwhere either(1)or(3)holdarethepointsxsuchthateveryopenset OcontainingxmeetsA.Such points are called limitpointsofA, andthe set of these limit points iscalled theclosureofA,writtenA. NotethatA A, sowe haveint(A) A A = int(A) A,thislastunionbeingadisjointunion. Wewillusethesymbol todenoteunionofdisjoint subsets when we want to emphasize the disjointness, soA = int(A)AandX= int(A) Aint(X A).Asanexample,inRwiththeusualtopologytheintervals(a, b), [a, b], [a, b),and(a, b]allhaveinterior(a, b), closure[a, b]andboundary{a, b}. Similarly,inR2withtheusualtopology, if AistheunionofanopendiskDwithanysubsetofitsboundarycircleCthenint(A)=D, A=D C, andA=C. Forasomewhatdierenttypeofexample, let A=QinX=RwiththeusualtopologyonR. Thenint(A) = andA = A = R.Proposition 1.1. For everyA Xthe following statements hold:(a) int(A)isopen.(b) Aisclosed.(c) Aisopen if and only ifA = int(A).(d) Aisclosed if and only ifA = A.Proof. (a)If xisapointinint(A) thenthereisanopenset OxwithxOxA.We haveOx int(A)sinceforeachy Ox, Oxisanopen setwithy Ox Asoy int(A). Itfollowsthat int(A) =_xOx,theunionasxrangesoverallpointsofint(A). This is aunion of open sets andhence open.Basic Point-Set Topology 7(b) SinceX= int(A) Aint(X A), we haveAas the complement ofint(X A),soAis closed, being the complement of an open set by part (a).(c)IfA = int(A)thenAisopenby(a). Conversely, ifAisopentheneveryx Aisinint(A)sincewecantakeO=Aincondition(1). ThusAint(A). Theoppositeinclusion alwaysholds, soA = int(A).(d) IfA = AthenAis closed by (b). Conversely, ifAis closed thenX Ais open, soeach point ofX Ais contained in an open setX Adisjoint fromA, which meansthatnopointof X Aisalimitpointof A, orinotherwordswehaveAA. WealwayshaveA A, soA = A. Asmall caution: Someauthorsusethetermlimitpoint inamorerestrictedsensethanweareusingithere,requiringthateveryopensetcontainingxcontainspoints ofAother thanxitself. Other names for this more restricted concept that onesometimes nds are point of accumulation and clusterpoint.One might have expectedthedenition of alimit point tobe expressed in termsof convergent sequences of points. In an arbitrary topological spaceXit is natural todene limnxn=xtomeanthatforeveryopenset Ocontainingxthepointsxnlie inOfor all but nitely many values ofn, or in other words there exists anN> 0suchthat xnOforall n>N. Itisobviousthat limnxn=ximpliesthat xalimitpointofthesubsetof Xformedbythesequenceofpointsxn. Howeverthereexisttopological spacesinwhichalimitpointofasubsetneednotbethelimitofany convergent sequence of points in the subset. For subspacesXofRnthis strangebehaviordoesnotoccursinceif xisalimitpointofasubset AXthenforeachn >0thereisapointxn Aintheopenballofradius1/ncenteredat x,andforthissequence of pointsxnwe have limnxn= x.Basisfor aTopologyMany arguments with open sets inRreduce to looking at what happens with openintervalssinceopensetsaredenedintermsofopenintervals. Asimilarstatementholdsfor R2andRnwithopendisksandballsinplaceofopenintervals. Ineachcasearbitraryopensetsareunionsofthespecialopensetsgivenbyopenintervals,disks, or balls. This idea is expressed by the following terminology:Denition. AcollectionBofopensetsinatopologicalspaceXiscalledabasisforthe topology if every open set inXisa union of setsinB.8 Chapter 1Atopologicalspacecanhavemanydierentbases. Forexample,inR2anotherbasis besidesthe basisof open disks isthe basis of open squareswith edgesparallelto the coordinate axes. Or we could take open squares with edges at45degree anglestothecoordinateaxes, orall opensquareswithoutrestriction. Manyothershapesbesides squares could also be used. Another variation would be to x a numberc> 0andtakejusttheopendisks ofradiuslessthanc . These disksalsoformabasisforthe topology onR2.IfBis a basis for a topology onX, thenBsatises the following two properties:(1) Every pointx Xlies in some setB B.(2) For each pair of setsB1, B2inBand each pointx B1B2there exists a setB3inBwithx B3 B1B2.TherststatementholdssinceXisopenandisthereforeaunionofsetsinB. Thesecond statement holds sinceB1B2isopen andhence isa union of sets inB.Proposition1.2. If Bisacollectionof subsets of aset Xsatisfying (1) and (2) thenBisa basis fora topology onX.The open sets in this topology have to be exactly the unions of sets inBsinceBisa basis forthis topology.Proof. LetObe the collection of subsets ofXthat are unions of sets inB. Obviouslythe union of any collection of sets inOis inO. To show the corresponding result fornite intersections it suces by induction to showthatO1O2 OifO1, O2 O. Foreachx O1O2we can choose setsB1, B2 Bwithx B1 O1andx B2 O2.By(2)thereexistsaset B3BwithxB3B1 B2O1 O2. Theunionofallsuch setsB3asxranges overO1O2isO1O2, soO1O2 O.Finally, XisinOby(1),andOsincewecanregardastheunionoftheempty collection of subsets of B. Terminology: NeighborhoodsWe have frequently had to deal with open setsOcontaining a given pointx. Suchan open set is called a neighborhood ofx. Actually, it is useful to use the followingbroader denition:Denition. Aneighborhoodofapoint xinatopologicalspaceXisanyset AXthat containsan open setOcontainingx.Basic Point-Set Topology 9The more restricted kind of neighborhood can then be described as an open neigh-borhood.Asan example ofthe usefulness ofthisterminology, we canrephrase the condi-tion forapointxtobe alimit pointof asetAto saythatevery open neighborhoodofxmeetsA. The word open here can in fact be omitted, for if every neighborhoodofxmeetsAtheninparticulareveryopenneighborhoodmeetsA,andconversely,ifeveryopenneighborhoodmeetsAthensodoeseveryotherneighborhoodsinceevery neighborhood containsan open neighborhood.Similarly, a boundary point ofAis a pointxsuch that every neighborhood ofxmeets bothAandX A.Metric SpacesThe topology onRnis dened in terms of open balls, which in turn are dened intermsofdistancebetweenpoints. Therearemanyotherspaceswhosetopologycanbedenedinasimilarwayintermsofasuitablenotionofdistancebetweenpointsin the space. Here is the ingredient neededto do this:Denition. A metricon a setXisa functiond: X XRsuch that(1) d(x, y) 0forall x, y X, withd(x, x) = 0andd(x, y) > 0ifx= y.(2) d(x, y) = d(y, x)for all x, y X.(3) d(x, y) d(x, z) +d(z, y)forall x, y, z X.Thislastconditioniscalledthetriangleinequality becausefortheusual distancefunction in the plane it says that length of one side of a triangle is always less than orequalto the sum of the lengths of the othertwo sides.GivenametriconXonedenestheopenballofradiusr centeredat xtobethe setBr(x) = { y X| d(x, y) < r }.Proposition. The collection of all ballsBr(x)forr> 0andx Xforms a basis for atopology onX.A space whose topology can be obtained in this way via a basis of open balls withrespect to a metric is calleda metricspace.Proof. Firstapreliminaryobservation: Forapoint yBr(x)theball Bs(y)iscon-tained inBr(x)ifs r d(x, y), since forz Bs(y)we haved(z, y) < sand henced(z, x) d(z, y) +d(y, x) < s +d(x, y) r .10 Chapter 1Nowto showthe condition to have a basis is satised, suppose we are given a pointy Br1(x1) Br2(x2). Then the observation in the preceding paragraph implies thatBs(y) Br1(x1) Br2(x2)foranys min{r1d(x1, y), r2d(x2, y)}. Dierent metrics on the same setXcan give rise to dierent bases forthe sametopology. Forexample, inR2theusualmetricdenedintermsoflengthsofvectorsbyd(x, y) = |x y|has balls which are disks, but anothermetric whose balls aresquares isd(x, y) = max{|x1y1|, |x2y2|}forx= (x1, y1)andy= (y1, y2). Wewill leave it to the reader to verify that this satises the properties of a metric. Anothermetricisd(x, y)=|x1 y1| + |x2 y2|,whichhasballsthatarealsosquares,butrotated45degrees from the squaresin the previous metric.Many important spaces in the subject of Topology are metric spaces, but the pointofviewofTopology istoignoreanyparticularchoiceofmetricasmuchaspossible,andjust focuson the open sets, on the topology itself.SubspacesWeturnnowtoatopicwhichoughttobesimple, andseemssimpleenoughatrst glance, but turns out to be a source of many headaches until one nally becomescomfortable withit.Givenatopology Oonaspace Xandasubset AX, wewouldliketousethetopologyonXtodeneatopologyOAonA. Thereisaneasywaytodothis:Justdeneaset OAtobeinOAifthereexistsanopenset OinOsuchthatO=A O. Axiom(1)holdssinceif O=A OforsetsOOthen_O=_(A O)=A __O_,so_OisinOA. Axiom(2)issimilarsince_O=_(AO) = A__O_,whichisinOAiftherearejustnitelymanyindices.Axiom (3) is obvious.The topologyOAonAis called the subspace topology, andAwith this topologyis called a subspace ofX. For example, if we takeXto beR2with its usual topology,then every subset ofR2becomes a topological space. In particular, geometric guressuchascirclesandpolygonscannowbeviewedastopological spaces. Likewise,geometricguresinR3suchasspheresandpolyhedrabecometopological spaces,with the subspace topology from the usual topology onR3.IfBis a basis for the topology onXandAis a subspace ofX, then we can obtaina basis for the subspace topology onAby taking the collectionBAof all intersectionsABasBranges over all the sets inB. This gives a basis forAbecause an arbitraryBasic Point-Set Topology 11open set in the subspace topology onAhasthe formA__B_=_(AB)forsome collection of basis setsB B. In particularthis saysthat forany subspaceXofRn, a basis for the topology onXis the collection of open setsX BasBrangesoverallopenballsinRn. Forexample,foracircleinR2theopenarcsinthecircleform a basisfor itstopology.IfXis a metric space, any subset A Xbecomes a metric space by restricting themetricX XRtoAA, since the three dening properties of a metric obviouslystillholdfortherestricteddistancefunction. ThefollowingPropositiongivessomestrong evidence that the subspace topology is anatural topology to use on subsets.Proposition. The metric topology on a subsetAof a metric spaceXis the same as thesubspace topology.Proof. Observerstthatforaball Br(x) inX, theintersectionA Br(x) consistsof allpointsinAofdistance less thanr fromx,so thisisa ballinAregarded asametric space in itself. Fora collection of such ballsBr(x)wehaveA__Br(x)_=__ABr(x)_TheleftsideofthisequationisatypicalopensetinAwiththesubspacetopology,andtherightsideisatypicalopensetinthemetrictopology,sothetwotopologiescoincide. AsubspaceAXwhosesubspacetopologyisthediscretetopologyiscalledadiscretesubspaceof X. ThisisequivalenttosayingthatforeachpointxAthereis an open set inXwhose intersection withAisjustx. For example, Zis a discretesubspace ofR, butQis not discrete. The sequence1/2, 1/3, 1/4without its limit0isa discrete subspace ofR, but with0it is not discrete.ForasubspaceAX, asubsetof AwhichisopenorclosedinAneednotbeopen or closed inX. However, there are times when thisis true:Lemma. Foranopenset AX, asubset BAisopeninthesubspacetopologyonAifandonlyif BisopeninX. Thisisalsotruewhenopen isreplacedbyclosedthroughout the statement.Proof. IfB AisopeninA,ithastheformAOforsomeopensetOinX. ThisintersectionisopeninXif AisopeninX. Conversely,if BAisopeninXthenAB= Bisopen inA. (Notethatthe converse doesnotuse theassumption thatAisopen inX.) The argument for closed setsis just the same. Closures behave nicely with respect to subspaces:12 Chapter 1Lemma. Given a spaceX, a subspaceY , and a subsetA Y , then the closure ofAinthe spaceYis the intersection of the closure ofAinXwithY .ThisamountstosayingthatapointyY isalimitpointof AinY(i.e. usingthe subspace topology onY ) if andonly ifyis a limit point ofAinX.Proof. Forapointy YtobealimitpointofAinXmeansthateveryopensetOinXthatcontainsymeetsA. SinceA Y ,thisisequivalenttoO YmeetingA,or in otherwords, that every open set inYcontainingymeetsA. Theanalogousstatement forinteriorsisnot true. Forexample, if Aisalinesegment in thex-axisinR2, then theinteriorofAin thex-axisisan open interval,but the interior ofAinR2is empty.Continuity and HomeomorphismsRecallthedenition: Afunctionf: XYbetweentopologicalspacesiscontin-uousif f1(O) isopeninXfor eachopenset OinY . Forbrevity, continuousfunctionsaresometimescalledmapsormappings. (Amapintheeverydaysenseofthewordisinfactafunctionfromthepointsonthemaptothepointsinwhateverregion is being represented by the map.)Lemma. Afunctionf : XY iscontinuousifandonlyif f1(C) isclosedinXforeach closedsetCinY .Proof. An evident set-theory fact isthatf1(YA) = X f1(A)for each subsetAof Y . Supposenowthat f iscontinuous. Thenforanyclosedset CY , wehaveY Copen,hencetheinverseimagef1(Y C)=X f1(C)isopeninX,soitscomplement f1(C) isclosed. Conversely, iftheinverseimageofeveryclosedsetisclosed, thenfor OopeninY thecomplement Y Oisclosedsof1(Y O)=X f1(O)isclosed andthusf1(O)isopen, sof iscontinuous. Here is anotherusefulfact:Lemma. Givenafunctionf: XYandabasisBforY ,thenf iscontinuousifandonly iff1(B)isopen inXforeachB B.Proof. OnedirectionisobvioussincethesetsinBareopen. Intheotherdirection,supposef1(B)is open for eachB B. Then any open setOinYis a union_BBasic Point-Set Topology 13ofbasissetsB, hencef1(O)=f1__B_=_f1(B) isopeninX, beingaunion of the open setsf1(B). Lemma. Iff: XYandg : YZarecontinuous, thentheircompositiongf: XZisalso continuous.Proof. Thisusesthe easyset-theory factthat(gf)1(A) = f1(g1(A))foranyA Z. Thus iff andgare continuous andAis open inZtheng1(A)isopen inYsof1(g1(A))isopen inX. This meansgf iscontinuous. Lemma. Iff : XYiscontinuousandAisasubspace ofX,thentherestrictionf|Aoff toAiscontinuous as afunctionAY .Proof. Foranopenset OY wehave(f|A)1(O)=f1(O) A, whichisanopenset inAsincef1(O)is open inX. Denition. Acontinuousmapf: XY isahomeomorphismifitisone-to-oneandonto, andits inverse functionf1: YXisalso continuous.[Tobeadded: someexamplesofhomeomorphisms, e.g.,anopeninterval (a, b)ishomeomorphic toR, an open ball inRnishomeomorphic toRn.]ProductSpacesGiven two setsXandY , their product is the setX Y= { (x, y)|x Xand yY }. Forexample R2=R R, andmoregenerally Rm Rn=Rm+n. If XandY aretopologicalspaces, wecandeneatopologyonX Y bysayingthatabasisconsistsofthesubsetsU V asUrangesoveropensetsinXandV rangesoveropensetsinY . Thecriterionforacollectionofsubsetstobeabasisforatopologyissatisedsince(U1 V1) (U2 V2)=(U1 U2) (V1 V2). ThisiscalledtheproducttopologyonX Y . ThesametopologycouldalsobeproducedbytakingthesmallerbasisconsistingofproductsU VwhereUrangesoverabasisforthetopologyonXandV rangesoverabasisforthetopologyonY . Thisisbecause(_U) (_V) =_,(UV).Forexample, abasisfortheproduct topologyonR Rconsistsoftheopenrectangles(a1, b1)(a2, b2). This is also a basis for the usual topology onR2, so theproduct topology coincideswith the usualtopology.More generally one can dene the productX1 Xnto consist of all orderedn-tuples(x1, , xn)withxiXiforeachi. Abasisfortheproducttopologyon14 Chapter 1X1Xnconsists of all productsU1Unas eachUiranges over open sets inXi, or just over a basis for the topology onXi. ThusRnwith its usual topology is alsodescribableastheproductof ncopiesof R, withbasistheopenboxes (a1, b1) (an, bn).Example. IfweviewpointsintheunitcircleS1inR2asangles, thenpolarco-ordinatesgiveahomeomorphismf: S1(0, )R2{0} denedby f(, r) =(r cos , r sin). Thisisone-to-oneandontosinceeachpointinR2otherthantheoriginhasuniquepolarcoordinates(, r). Toseethatf isahomeomorphism, justobserve thatit takesabasissetUV ,whereUisanopeninterval (0, 1) of valuesandV isanopen interval(r0, r1)ofr values, to an open polarrectangleandsuchrectanglesformabasisforthetopologyonR2 {0}asasubspaceof R2. Byre-strictingf toaproduct S1 [a, b]for0 0.18 Chapter 2Chapter2. ConnectednessSomespacesareinasensedisconnected,beingtheunionoftwoormorecom-pletelyseparatesubspaces. ForexamplethespaceXRconsistingofthetwoin-tervalsA=[0, 1] andB=[2, 3] shouldcertainlybedisconnected, andsoshouldasubspaceXof R2whichistheunionoftwodisjointcircles AandB. Astheseexamplesshow, itisreasonabletointerprettheideaof AandBbeingcompletelyseparate as saying not only that they are disjoint, but no point ofAis a limit point ofBand no point ofBis a limit point ofA. Since we are assuming thatXis the union ofAandB, this is equivalent to saying thatAandBeach contain all their limit points.In otherwords, AandBareboth closedsubsets ofX. SinceeachofAandBisthecomplementoftheother,itwouldbeequivalenttosaythatbothAandBareopensets. Thus we have arrived at the following basic denition:Denition. AspaceXisconnectedifitcannotbedecomposedastheunionoftwodisjointnonemptyopensets. ThisisequivalenttosayingXcannotbedecomposedasthe union of two disjoint nonempty closed sets.AthirdequivalentconditionisthattheonlysetsinXthatarebothopenandclosed areandXitself. For ifAwere any other set that was both open and closed,thenXwould be decomposed as the union of the disjoint nonempty open setsAandX A. Conversely, ifXwerethedisjoint union ofthenonempty opensetsAandBthenAwouldbeclosedaswellasopen,being equaltothecomplement ofB,andAwould be neithernorX.Example. The subspace ofRconsisting of the rational numbersQis not connected,sincewecandecomposeit astheunionof thetwoopensets Q(,2) andQ(2, ). Moregenerally, anysubspace XRthat isnot aninterval isnotconnected. Forif Xisnotaninterval, thenthereexistnumbers a 1,sincethelatterspaceshavenocutpointswhereasinReverypointisacutpoint. Itisatruemore generallythat Rmisnothomeomorphic toRnifm= n,butthisisamuch harder theorem. To distinguishR2fromR3forexample, one might try to usecut curves instead of cut points, motivated by the fact that the complement of a lineinR2is disconnected whereas the complement of a line inR3is connected. However,a hypothetical homeomorphismfromR2toR3might take a line to a very complicatedcurve which might conceivably wanderaround so badly that itscomplement wasnotconnected,unlike thecomplement ofastraight line. To prove thatthiscanthappentakes quite a bit of work, and the best way to do it is to develop some heavy machineryrst, the machinery in a branch of topology calledalgebraic topology.Connected Components andPath ComponentsWe will showthat if a space Xis not connected, then it decomposes as the union ofacollection ofdisjoint connectedsubspacesthataremaximalinthesense thatnoneofthemiscontainedinanylargerconnectedsubspace. Thesemaximal connectedsubspacesarecalledtheconnectedcomponentsof X. Similarly, if Xisnotpath-22 Chapter 2connected, it decomposes as the union of disjoint maximal path-connected subspacescalledpath components.Letusbeginbytalkingaboutpathcomponents. Forapoint xXlet P(x)bethe subspace ofXconsisting of pointsysuch that there is a path inXfromxtoy.NoticethatanytwopointsyandzinP(x)canbejoinedby apath, since iff isa path fromxtoyandgisapathfrom xtozthen we obtain a path from ytozby rst goingbackwardalongf fromytoxandthenforwardalonggfromxtoz.Here is the key factabout the subspacesP(x):Lemma. IfP(x) P(y) = thenP(x) = P(y).Proof. Supposethereisapoint zP(x) P(y). Thismeanstherearepathsfromxandytoz. If wisanypointinP(x), thenthereisapathfromxtow, andhence also a path fromytowby going rst fromytoz,thentox,thentow. ThusP(x) P(y).Thesameargumentshowsthat P(y)P(x), soP(x) = P(y). The lemma implies that the subspacesP(x)that are distinct provide a decompo-sition ofXinto a collection of disjoint subspaces. EachP(x)is path-connected, andif Aisanypath-connectedsubspaceof XthenAmustbecontainedinsomeP(x)sinceif xAthenallpointsinAcanbeconnectedtoxbyapath, soAP(x).ThisshowsthatthesubspacesP(x) arethemaximal path-connectedsubspacesofX. They are calledthe path componentsofX.Note that a continuous mapf: XYtakes each path component ofXinto somepathcomponent ofY . ThisisbecauseapathcomponentP(x)ispath-connectedsoits imagef(P(x))must also be path-connected, hencef(P(x))must be contained insomepathcomponentofY . InfactitwillbecontainedinP(f(x))sinceitcontainsthe pointf(x).Example. Let Xbethesubspaceof R2whichistheclosureof thegraphof thefunctionf(x)=sin(1/x) for x>0. ThusXconsistsofthisgraphtogetherwiththelinesegment Ainthey-axisfrom(0, 1)to(0, 1). Let Bdenotethegraphoff itself, soXisthedisjointunionof AandB. ObviouslyAandBarebothpath-connected. WewillshowthatXitselfisnotpath-connected,henceAandBarethetwo path components ofXConnectedness 23Letf: [0, 1]Xbe a path starting at a point inA. We knowthatf1(A)is closedsinceAis closed. We will showthatf1(A)is also open. This impliesf1(A) = [0, 1]since[0, 1]is connected andf1(A)is nonempty. The equationf1(A) = [0, 1]saysthatf([0, 1]) A, and thus there is no path inXjoining a point inAto a point inB.To show thatf1(A)is open lett0be any point inf1(A). Choose a small open diskDinR2centered atf(t0). ThenDXhas innitely many path-components, one ofwhich isDA. Sincefis continuous, f1(D)is an open set in[0, 1]containingt0,so there is an intervalI [0, 1]which is open in[0, 1], containst0, and is containedinf1(D). Thisintervalispath-connected,soitsimagef(I)isalsopath-connectedand therefore lies inside one of the path components ofDX. This path componentcontainsthepoint f(t0)Asoithastobethepathcomponent D A. Thissaysthatf(I) A, or in otherwordsI f1(A). Thusf1(A)containsaneighborhoodoft0in[0, 1]. Sincet0was an arbitrary point off1(A)we conclude thatf1(A)isopen in[0, 1], nishing the argument.TodeterminewhetherthespaceXinthisexampleisconnectedwewillusethefollowing generalfact:Lemma. If a subspaceAof a spaceXisconnected, then so isA.Proof. SupposeAis the disjoint union of subsetsBandCwhich are closed inAandhence also closed inXsince Ais closed inX. ThenABandACare disjoint closedsetsinAwhoseunionisA,sooneofthesetwosetsmustequal A,sayA B=A.This saysA B, soA B= B. Since we originally hadA = B Cthis impliesB= AandC= . Fromthislemmawecanconcludethatthespace XintheprecedingexampleisconnectedsinceitistheclosureofthesubspaceBwhichispath-connectedandhenceconnected. ThusXisanexample ofaspacewhichisconnectedbut notpath-connected. Anotherexample, verysimilartothisone, wouldbetheclosureofthegraphofthefunctionr =/(+ 1) for 0, inpolarcoordinates. Thisspaceconsists of a circle togetherwith a curve that spirals out to it from inside.Now we turn to the question of decomposing a spaceXinto maximal connected24 Chapter 2subspaces.Lemma. IfAis a subset of a spaceXwhich is both open and closed, then any connectedsubspaceC XwhichmeetsAmust be contained inA.Proof. If AisopenandclosedinXthenC AisopenandclosedinC. If Cisconnectedthisimplies thatC A = C, which saysthatCiscontainedinA. Proposition. If{C}is a family of connected subspaces of a spaceX, any two of whichhave nonempty intersection, then_Cisconnected.Proof. LetY=_CandletAYbeopenandclosedinY . ThenA Cisopenand closed inCfor each, hence is eitherorC. IfA = choose a pointx A.SinceAY andY istheunionofalltheCswehavexCforsome. ThusAC= ,so bytheprecedinglemma wemust haveC A. AnyotherCmeetsCby assumption, hence meetsAand so is contained inAby the Lemma again. ThusY , theunionoftheCs, iscontainedinA. WewereassumingAY , sowehaveA= Y . SinceAwasanynonemptyclosedandopensetinY ,weconcludethatYisconnected. Wecanapplytheproposition tothecollectionofallconnectedsubsetsCofXthatcontainagivenpoint x, toconcludethattheunionofallthesesubsetsiscon-nected. Call this unionC(x). It is obviously the largest connected set inXcontainingxsinceitcontainseveryconnectedsetthatcontainsx. ObviouslyC(x)isalsothelargest connectedset containinganyotherpointy C(x),sincealargerconnectedset containingywould also be a larger connected set containingx. It follows thatXisthe disjoint union of allthe dierent setsC(x). These are the maximalconnectedsets inX, the connected componentsofX.NotethattheconnectedcomponentsofXareclosedsubsetsbytheearlierfactthat if asubspace AXisconnectedthensoisitsclosure. Connectedcompo-nentsdonothavetobeopen,however. Forexample,considertherationalnumbersQ, topologizedasasubspaceof R. Weknowfromearlierthattheonlyconnectedsubspaces ofRare intervals, including the degenerate intervals[a, a]. SinceQcon-tainsnonondegenerateintervals,itfollowsthattheonlyconnectedsubspacesof Qarepoints. Thustheconnectedcomponentsof Qarejusttheindividual pointsofQ, sinceconnectedcomponentsaremaximal connectedsubspaces. Theconnectedcomponents ofQare therefore not open subsets ofQ.Pathcomponentsdonothavetobeeitheropenorclosed,asonecanseeintheexampleoftheclosureofthegraphof sin(1/x). ThisgraphisapathcomponentConnectedness 25thatisopenbutnotclosed,whiletheremaininglinesegmentofthespaceisapathcomponent thatis closed but not open.Eachpathcomponentofaspaceiscontainedinaconnectedcomponentsincepath-connectedsubspaces are connected, henceare containedin maximal connectedsubspaces,theconnectedcomponents. Onecommonlyencounteredsituationwherepathcomponentsarethesameasconnectedcomponentsisgivenbythefollowingresult:Proposition. IfeachpointinaspaceXhasaneighborhoodwhichispath-connected,then the path components ofXare also the connected components.Proof. Thehypothesisguaranteesthatpathcomponentsareopensubsets. Thisim-pliesthattheyarealsoclosedsincethecomplement ofapathcomponentisaunionof path components andhence isopen. Any connected set must intersect some pathcomponentandhencebecontainedinitbythepreviouslemma, sotheconnectedcomponents are containedin the path components. This proposition applies for example to an open setXinRn, since each point inXthenhasaneighborhoodwhichisaball,whichispath-connected. Noticethattheproposition denitely does not apply to the closure of thesin(1/x)graph.TheCantor SetAspaceXwhoseconnectedcomponentsarepointsissaidtobetotallydiscon-nected. This is equivalent to saying that the only connected subspaces ofXare points,since connected components are maximal connected subspaces. As a trivial example,a space with the discrete topology is certainly totally disconnected since every subsetisopen andclosed, so no subspace with more than one point canbe connected.For a subspace ofRto be totally disconnected means that it contains no intervals,otherthanpoints, sincetheonlyconnectedsubspacesof Rareintervals. Wemen-tionedtheexampleof Qpreviously, buttherearemanymoretotallydisconnectedsubspaces ofR. For a start, one can have subspaces such as the integers where everypointisisolated, beingbothopenandclosed. Thisisjustsayingthatthesubspacetopology isthe discrete topology, so one hasa discrete subspace.Let usaskthefollowingquestion: IsthereatotallydisconnectedsubspaceofRwhichisclosedinRbutcontainsnoisolatedpoints. Suchasubspacewouldthuscombine features of both the integers (a closed subspace) and the rationals (a subspace26 Chapter 2withnoisolatedpoints). Theanswertothequestionisyes, andthereisafamousexample which we willnow describe, known asthe Cantor set.TheCantorset Cisasubspaceof [0, 1] constructedasanintersectionofaninnitesequenceofsetsC0 C1 C2 . Westart withC0= [0, 1],andweformC1beremovingtheopeninterval (1/3, 2/3), soC1=[0, 1/3] [2/3, 1]. NextweformC2byremovingtheopenmiddlethirds(1/9, 2/9) and(7/9, 8/9) ofthetwointervals ofC1. This leavesthe four closedintervals[0, 1/9], [2/9, 1/3], [2/3, 7/9],and[8/9, 1]. Nowwerepeatthesameprocessoverandoveragain, ateachstageremoving the open middle thirds of each interval createdat the previous stage. ThusCnconsistsof2nclosedintervalsoflength1/3n. Finally wesetC=_nCn. Thisisa closed subset ofR, being the intersection of the closed subsetsCn.Theset Cistotallydisconnectedsinceitwasconstructedsoastocontainnointervalsotherthanpoints. Namely, if Ccontainedanintervalofpositivelengththen this interval would be contained in eachCn, butCncontains no interval of lengthgreaterthan1/3nsoif nischosentobelargeenoughsothat 1/3nislessthan,then there is no interval of lengthinCn.Nowletuscheckthat Ccontainsnoisolatedpoints. Eachpoint xinCisintheintersectionofalltheCnssoitisintheintersectionofthesequenceofintervalcomponents Inof Cnthat containx. Ifwelet xnbeeitherendpointof Inthenx=limxnsincethelengthsoftheInsareapproaching0. Wecanassumex=xnfor allnsince we have two choices for eachxnso we can always choose one of themdierentfromxif xhappenstobeanendpointofsomeIn. AlltheendpointsoftheIns are contained inC, so this shows that eachx Cis a limit of a sequence ofpointsxn Cdierent fromx. This shows thatCcontainsno isolatedpoints.Thereisanicewayofcharacterizingwhichpointsof [0, 1]lieinCintermsofbase3decimals. Themiddle-thirdinterval (1/3, 2/3) consistsofbase3decimals.a1a2 witha1=1, excludingonlythedecimal .1=1/3. Thus C1consistsofthedecimals.a1a2 witha1=1, including1/3as .0222 and1as .222.InsimilarfashionC2consistsofdecimals.a1a2 witha1=1anda2=1. Moregenerally Cnconsists of decimals.a1a2with none ofa1, , anequal to1. HenceCconsistsexactlyofalldecimals.a1a2 withnoaiequalto1. Thepointsof Cthat areendpointsof aninterval component of some Cnarethedecimalsendingeitherin all 0s or all 2s.ItisinterestingtothinkoftheprocessofconstructingCinphysical termsastakingalengthofstringandrepeatedlycuttingitintoshorterpieces. Ifwethinkoftherstpieceastheinterval [0, 1]andcutitatthepoint 1/2,thenitbecomestwoConnectedness 27pieces of string each with two endpoints. In eect the single point 1/2 has become twopoints,thetwonewendpointsofthehalfpieces,theintervals[0, 1/2]and[1/2, 1].Atthenextstageonecutseachofthesetwopiecesinhalf, producingfourseparatepieces [0, 1/4], [1/4, 1/2], [1/2, 3/4], and[3/4, 1], andsoonforlaterstages. Inorder to make all these pieces disjoint subsets ofRone can image the string as beingstretchedsotightlythateachtimeitiscut, itpullsapartatthecutandshrinkstotwo-thirds of its length, so after the rst cut, [0, 1/2]shrinks to[0, 1/3]and[1/2, 1]shrinks to[2/3, 1]. Then at the next stage we cut[0, 1/3]at its midpoint and the twopieces[0, 1/6]and[1/6, 1/3]shrink to[0, 1/9]and[2/9, 1/3], and similarly for thepiece[2/3, 1], andso on.All thiscuttingandshrinkingcanbedescribedveryneatlyintermsofdecimalexpansions. Initially we cut[0, 1]at1/2producing two copies of the point1/2. If weuse base2decimals, then these two copies of1/2can be distinguished by regardingone of them as.0111and the other one as.1000. Normally these two decimalsareregardedasthesamenumber, butnowwewanttokeepthemdistinct, viewingthem as the two new endpoints of the cut string. Similarly at the next stage the point1/4duplicates itself as the two decimals.00111and.01000, and the point3/4duplicates itself as.10111and.11000, etc. The process of shrinking pieces totwo-thirds of their lengths can then be achieved by simply changing the decimal base2to3andreplacing all decimaldigits1by2.Hereisaquestionforthereadertoponder: Whathappensifwedonotreplacedecimaldigits1by2butsimplyconsiderallbase3decimalsconsistingof0sand1s, with no2s?How does the resulting set compare with the Cantorset?WecandeterminethetotallengthoftheCantorsetbyaddingupthelengthsofthe complementary deleted intervals. First we deleted an interval of length1/3, then2intervalsoflength1/9, then4intervalsoflength1/27, andso on. Thusthetotallengthdeletedisthesumoftheseries13+29+427+ , whichisageometricserieswithinitialterm1/3andratio2/3so thesumis1. ThusthetotallengthofCis0.Anotherwaytoseethisistoobservethatthelengthof Cnis2/3ofthelengthofCn1, so the length ofCnis(2/3)n, which approaches0asngoesto innity.One can imagine many similar constructions of sets that look much like the Can-tor set. For example, instead of removing middle thirds at each step one could removemiddle halves or middle quarters, or any other fractions. The fractions could even varyfrom one step to the next. Or insteadof removing a single middle interval from eachinterval,onecouldremoveseveralinterioropenintervals,providedthatthelengths28 Chapter 2oftheremainingintervalsapproachzero. Interestinglyenough, all theseconstruc-tionsproduceCantorsets thatarehomeomorphictotheoriginalCantorset C. Infact, itispossibletolistseveralpropertiesthattheCantorsethas, includingbeingtotallydisconnectedandhavingnoisolatedpoints,suchthateveryspacewiththesepropertiesishomeomorphic totheCantorset. Wewillstudytheotherpropertiesinthenextcouplechapters, andeventually(hopefully)provethetheoremthatthelistof properties characterizesthe Cantorset up to homeomorphism.Exercises1. Let D1andD2betwoopendisksinR2whoseclosuresD1andD2intersectinexactlyonepoint, sotheboundarycirclesofthetwodisksaretangent. Determinewhichofthefollowingsubspacesof R2areconnected: (a) D1 D2. (b) D1 D2.(c) D1D2.2. Showthat if a subspaceAof a spaceXisconnectedthen so isits closureA.3. Show the subspaceX R2consisting of points(x, y)such that at least one ofxandyisrational isconnected.4. Bycountingcutpointsandnon-cutpointsshowthatnotwoofthefollowing fourgraphsinR2arehomeomorphic. (Eachgraphisaclosedsubspaceof R2,theunionof a nite number of closed line segments.)5. ShowthatifaspaceXhasonlyanitenumberofconnectedcomponents, thenthese components are open subsets ofX. (We already know they are closed.)6. Fromthefactthataninterval [a, b]isconnected,deducetheIntermediateValueTheorem: If f: [a, b]Riscontinuousandf(a)0, sowecouldaskwhetherXisinfacttheunionofanitecollectionoftheseballsB(x)ofxedradius. Togeneralizethisideatoarbitraryspaceswhichneednothaveametric,wereplace balls by arbitrary open sets, and this leads to the following general denition:Denition. AspaceXiscompactifforeachcollectionofopensetsOinXwhoseunion isX, there exist a nite number of theseOswhose union isX.More concisely, one says that every open cover ofXhas a nite subcover, whereanopencoverof XisacollectionofopensetsinXwhoseunionisX, andanitesubcover is anite subcollection whose union isstill X.Forexample, Risnotcompactbecausethecoverbytheopenintervals(n, n)for n=1, 2, hasnonitesubcover, sinceinnitelymanyoftheseintervalsareneededtocoverall of R. Anotheropencoverwhichhasnonitesubcoveristhecollection of intervals(n1, n+1)forn Z.In a similar vein, the interval(0, 1)fails to be compact since the cover by the openintervals(1/n, 1) for n1hasnonitesubcover. Ofcourse, theredoexistopencoversof (0, 1) whichhavenitesubcovers, forexamplethecoverby(0, 1) itself,oralittlelesstrivially, thecoverbyall opensubintervalsofxedlength, say1/4,which has the nite subcover(0, 1/4), (1/8, 3/8), (1/4, 1/2), (3/8, 5/8), (1/2, 3/4),(5/8, 7/8), (3/4, 1). To be compact means that every possible open cover has a niteCompactness 31subcover. Thiscouldbedicult tocheckinindividual cases, sowewill developgeneraltheorems to test forcompactness.CompactSets in Euclidean SpaceSpaceswithonlynitelymanypointsareobviouslycompact, ormoregenerallyspaceswhosetopologyhasonlynitelymanyopensets. However, suchspacesarenotveryinteresting. Ourgoal inthissectionwill betocharacterizeexactlywhichsubspaces ofRnare compact. We start with an important specialcase:Theorem. A closedinterval [a, b]iscompact.Proof. This will be somewhat similar in avor to the proof we gave that closed intervalsareconnected. Thecasea=bistrivial, sowemayassumea asuch that the interval [a, c]is containedin thisO, andhence[a, c]is containedinthe union of nitely manyOs. LetLbe the least upper bound of the set of numbersc [a, b]suchthat[a, c]iscontainedintheunion ofnitelymanyOs. We knowthat L>abytheprecedingremarks, andbythedenitionof LwecertainlyhaveL b.There is someO, call itO, that containsL. ThisOis open in[a, b], so sinceL > athere is an interval[L, L]contained inOfor some > 0. By the denitionof Lthereexistnumbersc0. Thenabasis for the new topology onXconsists rst of all of the usual open disksDr(x)ofradiusr and centerxthat are contained inX, and then in addition for each pointxin thex-axis we take the setsDr(x) = {x}(XDr(x))obtained by addingxto anopen half-disk centeredatx. It is easy to check that the condition forhaving a basisissatised, andweleavethisforthereadertodo. TheHausdorconditionisalsoeasilyveriedsince anytwodistinct pointsofXarecontainedin disjoint basissets.Notice that any subset of thex-axis is closed in this topology since its complement isopen, being a union of basis sets. However, if we letAbe the rational numbers in thex-axis andBthe irrational numbers, thenAandBare disjoint closed sets inXthatdonothavedisjointopenneighborhoods, sinceanyopensetUcontainingAwouldcontainbasissetsDr(a)foraAthatintersectallthebasissetsDs(b)forbBwith|a b| 0there existsa Awithd(x, a) < d(x, A)+. Thend(y, a) < d(y, x)+d(x, a) < ++d(x, A).Since d(y, a)d(y, A) thisimplies d(y, A)0suchthateveryball B(x)inXiscontainedinatleastonesetAof the cover.Theorem. Every open cover of a compact metricspaceXhas a Lebesgue number.Forexample, whatwouldaLebesguenumberforthefollowingopencoverofasquare be?Proof. SinceXis compact, each open cover has a nite subcover, and there is no lossCompactness 39ofgeneralitytondaLebesguenumberforthissubcover. Thuswemayassumethecover isby open setsU1, , Un.For aball B(x) tobecontainedin Uiisequivalent totherequirement thatd(x, X Ui). Thusweareledtoconsiderthefunctionsdi(x)=d(x, X Ui),and to ask whether there exists an > 0such that for eachx Xat least one of thenumbersd1(x), , dn(x)isatleast . Phraseddierently, weareaskingwhetherthereisan >0suchthat thefunctionf(x) =max{d1(x), , dn(x)} satisesf(x) forall x. Eachfunctiondi(x)iscontinuousbythe precedingLemma, andweleaveitasanexerciseforthereadertoverifythatthemaximumofanitesetofcontinuousfunctionsisagaincontinuous. (Byinductionitsucestodothecaseoftwofunctions.) Sof iscontinuous. SinceUiisopen, itscomplement X Uiisclosedandhencedi(x)>0forall xUi. Thusf(x)>0forall xXsincetheUis coverX. SinceXis compact, the imagef(X)is compact in(0, ), so there is alower bound > 0forthe values off onX. By the remarks atthe beginning of thisparagraph, thisis a Lebesgue number for the cover. Corollary. Let f: XY beacontinuousmapfromacompactmetricspaceXtoametricspaceY . Thenf isuniformlycontinuous: Foreach>0thereexistsa > 0such thatd(f(x), f(x)) < for all x, x Xwithd(x, x) < .Proof. Given> 0,coverXbytheopensetsf1(B/2(y))asyrangesoverY . LetbeaLebesguenumberforthiscoverofX. Thenf takeseachball B(x)tosomeball B/2(y), soif d(x, x) 0.(b) d(x, y) = min{d(x, y), 1}.(c) d(x, y) = d(x, y)/(1 +d(x, y)).14. For a metric spaceXdened by a metricdshow that the functiond: X XRiscontinuous.15. Let Xbeametricspacewithmetricd. ForsubsetsA, BXdened(A, B)=inf{ d(a, b)|aAand bB }. ShowthatifAandBarecompactthenthereexistpoints aAandbB withd(a, b) =d(A, B), soinparticular d(A, B) >0ifA B=. ShowbyanexamplethatboththesestatementscanbefalseifAandBare only assumed to be closed insteadof compact.16. LetXbe the subset ofR2which is the union of the line segmentsLnfrom(0, 0)to(1, 1/n) for n=1, 2, , togetherwiththelimitingsegment Lfrom(0, 0) to(1, 0). Dene a topologyOonXby saying that a setO Xis inOifOLnis openinLnforeachn , whereLnisgiven thesubspace topology fromR2. ShowthatthistopologyonXisnormalbutisnotdenedbyanymetriconX. Hint: Givenametric onX, nd a sequence of pointsxn Ln, n = 1, 2, , converging to(0, 0)inthe topology denedby the metric but not in the topologyO.44 Chapter 4Chapter4. QuotientSpacesOurnextobjectiveistodescribeageneral procedureforbuildingcomplicatedspaces out of simpler spaces. This is the topological analog of how all sorts of objectsaremadeintherealworld. JustthinkofallthedierentwordsthereareintheEn-glish language for putting things together: gluing, pasting, taping, stapling, stitching,sewing, welding, riveting, soldering, brazing, bonding, nailing, bolting, clamping, ...To take two simple examples, we can construct either a cylinder or a Mobius bandby taking a rectangularstrip andgluing two opposite edgestogether:The gluing process can be described as a mapf: XYwhereXis the rectangle andY iseitherthecylinderortheMobiusband. Themapf isonto, butnotquiteone-to-onebecauseofthegluing: Twopointsonopposite edgesoftherectanglethatareglued togetherhave the same image underf , but otherthan this, f isone-to-one.IsitpossibletodescribethetopologyonY justintermsofthetopologyonXandthemapf ? Ifso,thenwecanconstructthespaceY justbysayingstartwiththespaceXandgluethistothat", sincethegluinginstructionsaretellingwhattheset Y andthemapf are. Inmanycasesthereisaneasywaytocharacterizeopensets inYin terms of open setsinX:Proposition. If f: XY iscontinuousandonto, withXcompactandY Hausdor,then a setU Yisopen inYif and only iff1(U)isopen inX.Proof. Continuity off saysthatf1(U)isopen ifUis open, so we needto see thatthe converse is true ifXis compact andYis Hausdor. Iff1(U)is open inXthenXf1(U)is closed and hence compact since Xis compact. The image f(Xf1(U))isthencompactinY ,andtherefore alsoclosedsinceYisHausdor. But sincef isonto we havef(X f1(U)) = YU. ThusYUis closed, soUisopen. Quotient Spaces 45ThisPropositionappliestothecylinderandMobiusbandexamples. OnesrstguessmighthavebeenthattheopensetsinY aretheimagesf(O)ofopensetsOinX. However, this isnt right because an open neighborhood of a pointxon the leftorrightedgeoftherectangleissenttoasetinY thatisnotanopenneighborhoodoff(x), but only half of an actualopen neighborhoodUoff(x). To get the otherhalfof Uonewouldhavetoenlargetheopenset OinXtoincludealsoanopenneighborhood of the other pointxwithf(x) = f(x). This amounts to replacingObyf1(U),asstatedin the Proposition.The previousProposition givesacharacterizationof thetopology onYin termsofthetopologyonX, assumingthatthetopologyonY isalreadygiven. ItisalsopossibletomakethischaracterizationintoadenitionofatopologyonY , asthefollowing shows:Proposition. Given a spaceX, a setY , and a functionf: XY , then the collection ofall setsU Ysuch thatf1(U)isopen inXisa topology onY .Proof. Theaxiomsfor unionsandintersections aresatisedsince f1(_U) =_f1(U) and f1(_U) =_f1(U). For theother twoaxioms wehavef1() = andf1(Y) = X. Noticethatwedidnotrequiref tobeontointhisProposition. Iff isnotontothen pointsy Y f(X)form open sets{y}since their inverse imagesf1(y)areemptyandhenceopen. SothesepointsarejustisolatedpointsofYhavingnothingtodowiththetopologyonf(X). Assumingf isontoisthusnotaveryseriousrestriction.Denition. If the mapf: XYisonto, the topology onYgiven by thisPropositionis called the quotient topologyonY , and the mapfis called a quotient map. One alsosaysthatYis aquotient space ofX.In these terms, the earlier Proposition then says that any continuous map from acompact space to a Hausdor space that isonto is a quotientmap.IntheexamplesofthetorusandMobiusbandasquotientspacesofarectanglewealreadyhadexistingmodelsofthequotientspace, butsometimesonewantsto46 Chapter 4constructaquotientspacewithouthavingamodel ofitavailablebeforehand. Onewould just like to start with a spaceXand say glue this to that" and know that theseinstructionswill automaticallyproduceaquotient spaceof X. Toseethat thisisalwayspossible,suppose rstthatwedohaveaquotientmapf: XY . Thepointsof Xthataregluedtogether toproduceapoint yY areexactlythepointsinf1(y). Since quotientmaps are onto, f1(y)isalwaysnonempty. Thus the pointsy Yare in one-to-one correspondence with the subsets f1(y)of X. These subsetsare disjoint for dierent choicesofy, andtheirunion is allofXsince each point ofXissentbyf tosome point ofY . Thusthesetsf1(y)formapartitionofXintodisjoint subsets. Forexample, forthe cylinderandMobiusbandthese subsets of therectangle each have either one or two points eithera pair of points on the left andright edges of the rectangle or asingle point not on eitherof these edges.Conversely, suppose we are given a partition ofXinto disjoint nonempty subsets.Forx Xwe denote the set in the partition containingxby[x]. If we simply deneY tobethesetofall thesesubsets[x], thenwehaveanatural functionf: XYsendingxto[x], andthisf isontosowecanuseittodeneatopologyonY byspecifying that a setU Yis open exactly whenf1(U)is open inX. This makesfintoaquotientmap. WewillusethenotationX/ forthisquotientspaceYwhosepointsarethesets[x] inthepartitionof X. HereistherelationonXgivenbyxxif[x]=[x],i.e., xandxlieinthesamesetinthepartitionof X. Thisisan equivalence relation, meaning that it satisesthe three properties(i) x xforall x X.(ii) Ifx ytheny x.(iii) Ifx yandy zthenx z.Conversely, everyequivalencerelationonXdeterminesapartitionof Xintotheequivalence classes[x] = { x X| x x }.In practice the partition ofXis often specied implicitly by saying which pointswewishtogluetogether, orinotherwords, whichpointsof Xwewishtobecomeidentied to single points inY . For example, to obtain the cylinder fromthe rectangle[0, 1] [0, 1]wemaketheidentications(0, t) (1, t)fort [0, 1],andtoobtaintheMobiusbandweinsteadidentify(0, t)(1, 1 t). IfidenticationsinaspaceXare specied by means of a relationthen this determines a quotient spaceX/ associated to the partition ofXinto the smallest subsets that contain all the speciedpairsx x. Inparticular,ifnoidenticationsarespeciedforapointx Xthenxbecomes a set[x] = {x}of the partition all by itself.Quotient Spaces 47Example. The torusS1S1can be realized as a quotient space of a rectangle[0, 1] [0, 1]by identifying both pairs of opposite edges via(0, t) (1, t)and(s, 0) (s, 1).Ifwerstidentifythetopandbottomedgesoftherectangleweobtainacylinder,andthenidentifyingthetwoendsofthecylinderproducesatorus. Noticethattheidentications of opposite edges of the rectangle force all four corners to be identiedtoasinglepoint. Inparticular, wearemakingtheidentication(0, 0)(1, 1)eventhoughthisisnot part of either of theoriginal identications (0, t) (1, t) and(s, 0) (s, 1),but follows since(0, 0) (0, 1) (1, 1).Givenaconcretequotientmapf: XYwecanalsoformtheabstractquotientspaceX/associatedtothepartitionof Xintothesubsetsf1(y), anditwouldbenicetoknowthatthisisreallythesameasY . Themapf inducesawell-denedfunctionf: X/ Ysendingf1(y)toy, andthisf isobviously one-to-one andonto.Proposition. This functionf: X/ Yisa homeomorphism.Proof. Letp: XX/ be the quotient mapx[x] = { x X| f(x) = f(x) }, sof= fp. Then a setU Yis open if and only iff1(U)is open (sincefis a quotientmap), but sincef1(U) = p1(f1(U))andpis a quotient map, this is equivalent tof1(U)being open. Example. Let us describe three ways to obtain the sphere S2as a quotient space. First,asshowninthegurebelow, wecanformS2fromtwocloseddisksbyidentifyingtheirboundarycirclestogethertoformasinglecircle, theequatorof S2. Thetwodisksthenbecometheupperandlowerhemispheresof S2. Moreformally, wecantakethetwodisksasthespaceX=D2 {1, 1}R2 R=R3, whereD2istheclosed unit disk inR2, and then the identicationsare(x, 1) (x, 1)forx D2,with the mapXS2given by vertical projection of each disk onto the hemisphere itistangent to.48 Chapter 4A second way to realizeS2as a quotient space is to start with a single closed diskD2andgluethetophalfofitsboundarycircletothebottomhalf,eachpointintheupper half being identiedwith the point directly below it in the lower half.ParallelverticallinesinthediskthenbecomeparallelverticalcirclesinS2. ItwouldnotbediculttowritedownaformulaforanexplicitquotientmapD2S2thatdoes this, although there isreally no needfor explicit formulas.Thethirdconstruction alsorealizesS2asaquotientofD2,but thistime allthepointsofD2areidentiedtoasinglepointofS2,thesouthpole. Thisisachievedby a mapD2S2that sends each radial line segment to a meridian ofS2going fromthe north pole to the south pole. We can view this as taking a disk tangent tangent tothe sphere andwrapping it completely aroundthe sphere.This example diers from all the previous ones in that the mapf : XYis no longernite-to-one.Quotient Spaces 49Inthenextexamplewewillnothaveaready-mademodelofthequotientspace,so wewill be constructing a new space asa quotientof a known space.Example. Letusconstructasurface,commonlycalledtheKleinbottle,bymodifyingtheearlierconstructionofthetorusfromarectanglebyreversingtheorientationofone of the edges of the rectangle:By denition, the Klein bottle is the quotient space of the rectangle obtained by identi-fying opposite edges according to the orientations shown. We will use the notationK2forthisspace,thesuperscript indicatingthatitisa2-dimensional surface. Identify-ing the top and bottom edges of the rectangle gives a cylinder, but to identify the twoendsofthecylinderbydeformingthecylinderinR3requiresthatthecylinderpassthrough itself in order to make the orientations match, as in the third gure. Thus wehaveamapfromK2intoR3which isnot one-to-onebecausetwocirclesinK2havethesameimagecircleCinR3. Infact, itisatheoremthatthereisnoembeddingofK2intoR3, i.e., there isno subspace ofR3homeomorphic toK2. However, thereisanembedding intoR4. Therstthreecoordinatesofthisembedding arethemapK2R3indicatedinthegure. Forthefourthcoordinatewejustneedamapf: K2RthattakesdierentvaluesonthetwocirclesC1andC2of K2thatmaptoC. Intherectanglethese two circles are positioned as in the gure at the right. Wecan choosefto have a value > 0onC1and be0everywhereoutside asmall neighborhood ofC1.Example. LetP2be the set of all lines through the origin inR3. This is a metric spaceif we dene the distance between two lines through the ori-gin to be the smaller of the two angles between them, so thisangleliesintheinterval [0, /2]. Thersttwopropertiesrequired of a metric are obviously satised, and we will leaveitforthereadertocheckthetriangleinequality. Thereisanatural mapp: S2P2sendingapoint xS2tothelinethroughtheorigincon-tainingx. EachlinethroughtheoriginintersectsS2intwoantipodalpointsxand50 Chapter 4x, sopisontoandisexactlytwo-to-one. Wecanseethat piscontinuoussincep1ofan-ballinP2consistsoftwoantipodal -ballsinS2. Itfollowsthat pisaquotientmapsinceitisacontinuousmapfromacompactspaceontoaHausdorspace(metricspacesareHausdor). ThIsmeansthatwecanregardP2asthequo-tient space ofS2in which each pointxisidentied with the antipodal pointx, soP2= S2/(x x).Theclassicalnamefor P2istheprojectiveplane. Thiscomesfromthinkingofhow2-dimensional pictures(drawingsorpaintings, forexample) of 3-dimensionalobjectsareobtained. Eachpointofa3-dimensionalobjectprojectsontotheplaneofthepicturealongthelinefromthepointtotheviewerseye. Thepointsthatweseearethenreallyjustallthedierentlinesconvergingtooureye. Normallyonesangleofvisionissomethinglessthan180degrees,andapictureencompassescon-siderably less than180degrees. But imagine what might happen if we could see morethan180degrees at a time. Two points in space that were exactly180degrees apartwouldpresumablyberecognizedasdistinctpoints. Soperhapsifweweremakingamathematicalmodelofhowweseethingsweshouldnotidentifyantipodalpointsof S2butjustconsider S2itself. Thisleadstoakindofgeometrycalledsphericalgeometry. The other alternative of identifying antipodal points leads to a slightly dif-ferent geometry called projective geometry, which is in some ways closer to Euclideangeometry, and therefore preferable for some purposes.Each line through the origin inR3intersects the upper hemisphere ofS2in eitheroneortwopoints,thelatterpossibilityholdingonlyforlinesinthexy-plane. ThismeansthatwecanalsodescribeP2asthequotientspaceoftheupperhemisphereobtained by identifying antipodal points in its boundary, the equator. Since the upperhemisphereishomeomorphictoadiskviavertical projectionontotheunitdiskinthexy-plane, this means thatP2is describable as the quotient space of the diskD2with the identicationsx xforx D2.As with the Klein bottle, it is impossible to embedP2as a subspace ofR3, so onelooks instead for a mapP2R3that is almost an embedding. We will describe a mapthat is an embedding on the interior ofD2and folds the circleD2/(x x)onto aline segment. The image of this mapis shown in the gure on the right below:Quotient Spaces 51To describe the construction in words, we rst deform theD2in the gure on the leftso thatitbecomesa sphere withaslit. Theboundary circle ofthe slit isdividedintofour arcs and we would like to identify these four arcs in pairs, awithaandbwithbaccordingto theorientationsshown. Identifyingthe rst pairpresents noproblems,but to identify the second pair then forces all four arcs to coincide inR3, so that onehasaspherepinchedalongalinesegmentastheimageofthemapP2R3. Thepreimage of the interior of this line segment consists of two distinct open arcs inP2.Theupperhalfofthethirdgureisknownasacrosscap. ItistheimageofamapofaMobiusbandtoR3, theMobiusbandobtainedbyremovinganopendiskfromP2. Thinking ofP2asD2with antipodal boundary points identied, the Mobiusband is obtained by removing a smaller concentric disk fromD2, leaving an annulus,withantipodal pointsononeofitsboundarycirclesidentied. Thinkingof P2asS2/(x x), one rst removes symmetric disk neighborhoods of the two poles ofS2leavinganequatorialband, andthenoneidentiesantipodalpointsinthisbandtoproduce a Mobiusband.This example of the projective plane easily generalizes to higher dimensions. Then-dimensionalprojectivespacePnisthespaceoflinesthroughtheorigininRn+1,thequotientspaceSn/(xx), orequivalentlythequotientof Dnwithantipodalpointsof Dnidentied. Whenn=1thisisjust identifyingtheendpointsof asemicircle, soP1is homeomorphic toS1.It can easily happen that a quotient space of a Hausdor space is non-Hausdor.Here is a simple example:Example. LetX=R {0, 1},twodisjointcopiesof R,andmaketheidentications(x, 0)(x, 1)forall x>0. Thenthetwopointsx0= (0, 0)andx1= (0, 1)aredistinctinthequo-tient although any two neighborhoods U0of x0andU1ofx1in the quotient intersect since they contain points(x, 0) (x, 1)forx> 0.52 Chapter 4Quotientmapsarenotgenerallyopenmaps, takingopensetstoopensets, butthishappens to be the case in this example.If we make a small modication of this examplebysetting (x, 0) (x, 1) for x0ratherthanx>0thenthequotientbecomesHausdor,threelines inR2meeting at apoint.Hereisatechnical result that isusedquiteoftenwhenprovingthingsaboutquotient spaces:Lemma. Givenaquotientmapf: XY andaspaceZ, thenafunctiong : YZiscontinuous if and only ifthe compositiongf: XZis continuous.Proof. LetO Zbeopen. Thenbythedenitionofaquotientmap,thesetg1(O)isopeninY ifandonlyif f1(g1(O))=(gf)1(O) isopeninX. Thisissayingthatgis continuousif and only ifgf iscontinuous. ExercisesForthefollowingproblemsrecallthatamapf: XY isaquotientmapifitisontoand has the property that a setO Yisopen if and only iff1(O)is openinX.1. Showthat a continuous mapf: XYis a quotient map if there exists a continuousmapg : YXsuchthatthecompositionfg : YY istheidentitymap. Applythisto show thata projection mapX YX, (x, y)x, isa quotient map.2. Showthat the composition of two quotient maps isa quotient map.3. Show that there exist quotient mapsD2S1, S2D2, S2S1, S2S1S1, andS1S1S2. [Youdo notneedtogive formulasforthesemaps,aprecise geometricdescriptionwillbesucient. Carefulproofsofcontinuityarenotrequired, butthequotient mapsdo have to be continuous. Remember thatS1S1isa torus.]4. (a) Showthat there is a quotientmap(0, 1)[0, 1], but not[0, 1](0, 1).(b) IfCis the Cantor set, show there is a quotient mapC[0, 1]but not[0, 1]C.