Section 18. Continuous Functions

1 18. Continuous Functions1 Section isthefundamental concept in topology! When you hear that a coffee cup and a doughnut are topologically equivalent, this is really a claimabout the existence of a certain Continuous function (this idea is explored in depthin Chapter 12, Classification of Surfaces ). We start by reviewing some continuityideas from Analysis 1 (MATH 4217/5217). standard definition of continuity of a real valued function of a realvariable at a pointx0in the domain of the functionf,D(f), is as follows ( , page 2) a function andx0 D(f). Thenfiscontin-uous at pointx0iffor all >0 there exits ( )>0 such thatfor all|x x0|< ( ) andx D(f) we have|f(x) f(x0)|< .We then say thatfis Continuous a setA Riffis Continuous at each point following is a consequence of the previous definition (see Theorem 4-5in the Analysis 1 notes mentioned above):Theorem :X Y.

2 Thenfis Continuous onD(f) if andonly if for all open setsV Y, we havef 1(V) is open relative toD(f).Inspired by this result, we have the following as our (standard) definition of contin-uous Functions . We take Functions defined on all of topological spaceX, so thereis no need for the open relative to part of the above Continuous topological spaces. A functionf:X Yiscontinuousif for each open subsetVofY, the setf 1(V) is open Calculus 1, continuity is defined based on limits (which are defined using s and s), however we have not yet defined the limit of a function (though wehave defined limit points of sets ). Our definition is based entirely on open sets andwe should in fact state thatfis continuousrelativeto the topologies :X Y, letBbe a basis for the topology onY, and letSbe a subbasis for the topology onY.(1)fis Continuous iff 1(B) is open inXfor eachB B.

3 (2)fis Continuous iff 1(S) is open inXfor eachX :R R(withRhaving the standard topology) by Continuous (under our definition). Letx0 Rand >0. Then the intervalV= (f(x0) , f(x0)+ ) is open in the range spaceR. Therefore,f 1(V) is open in the domainspaceR. Nowx0 f 1(V) so there is some basis element (a, b) containingx0with(a, b) f 1(V) (recall that the standard topology onRisdefinedas the topologywith basis{(a, b)|a, b R, a < b}; see page 81). Choose = min{f(x0) a, b f(x0)}. Then forx (f(x0) , f(x0) + ) (that is, for|x f(x0)|< ) we havef(x) V(that is,|f(x) f(x0)|< ). So our definition of continuity implies the / definition of continuity. In fact, in the settingf:R R, the / definition ofcontinuity implies the open set definition (so there are equivalent); you are askedto show this in Exercise Continuous Functions3 Example the standard topology andR`have the lower limit topol-ogy.

4 Letf:R R`be the identity functionf(x) =x(which is of coursecontinuous when mappingR R). Thenfisnotcontinuous here since fora < b,[a, b) is open inR`forf 1([a, b)) = [a, b) is not open the following theorem it is shown (in the (1) (4)) thatf:X Yiscontinuous if and only if for eachx Xand each neighborhoodVoff(x) thereis a neighborhoodUofxsuch thatf(U) V. So the for all >0 has beenthe there exists >0 has been replaced with there is a neighborhoodU. Theimplication |x x0|< |f(x) f(x0)|< is replaced withx U f(x) topological spaces. letf:X Y. Then thefollowing are equivalent:(1)fis Continuous .(2)For every subsetZofX, one hasf(A) f(A).(3)For every closed subsetBofY, the setf 1(B) is closed inX.(4)For eachx Xand each neighborhoodVoff(x), there is a neighborhoodUofxsuch thatf(U) Continuous that, in a general sense, anisomorphism between two mathematicalobjects is a one to one and onto mapping which preserves structure.]]]

5 For example,in a group the structure is the binary operation, so we require (a b) = (a) (b).In a graph, the structure is connectivity so that if verticesvandware adjacent inG, then we require that (v) and (w) are adjacent in (G) (and conversely). In avector space the structure is linear combination, so we requirea ~v1+~v2=~w, thena (~v1) +b (~v2) =~w. In a topological space, the structure is the collection of opensets. So we wantUopen inXto imply (U) open inY(and conversely). Theterm isomorphism is not used in the topological setting, but the concept is thesame. We have the topological spaces. Letf:X Ybe a bijection(one to one and onto). If bothfandf 1:Y Xare Continuous , thenfis :X Yis a homeomorphism thenUis open inXif and only iff(U)is open inY. Any property inXthat is expressed entirely in terms of the topologyonXyields through the homeomorphism the corresponding property inY.

6 Sucha property is called atopological propertyofX. Examples of such properties areopen/ closed , limit points of a set, limits of a sequence, a basis or subbasis for thetopology, and (as we will see in Chapter 3) connectedness Continuous :X Ybe an injective (one to one) Continuous map. LetZ=f(X) (so thatfis ontoZ) be considered a subspace ofY. Letf :X Zbethe restriction offtoZ(sof is a bijection). Iff is a homeomorphism ofXwithZ, thenf:X Yis atopological imbedding(or simplyimbedding) : ( 1,1) Rdefined byF(x) =x/(1 x2). ThenFiscontinuous and one to one (sinceF (x) = (1+x2)/(1 x2)2 0) and is continuousonR. SoFis a homeomorphism (whereRhas the standard topology and ( 1,1)has the subspace topology).Example we give an example of a Continuous bijective function which hasan inverse which is not Continuous . Consider [0,1) with the subspace topology (asa subspace ofRwith the standard topology) andS1={(x, y)|x2+y2= 1}withthe subspace topology (as a subspace ofR2with the standard topology).]

7 Definef: [0,1) S1asf(t) = (cos 2 t,sin 2 t). Thenfis a Continuous bijection butf 1is not Continuous sincef([0,1/2)) ={(cos 2 t,sin 2 t)|t [0,1/2)}and so theinverse image (under mappingf; that is, (f 1) 1([0,1/2)) =f([0,1/2))) of openset [0,1/2) is not Continuous Functions6 Theorem for Constructing Continuous ,Y, andZbe topological spaces.(a)(Constant Function) Iff:X Ymaps all ofXinto a single pointy0 Y,thenfis Continuous .(b)(Inclusion) ifAis a subspace ofX, the inclusion functionj:A Xiscontinuous.(c)(Composites) Iff:X Yandg:Y Zare Continuous , then the mapg f:X Zis Continuous .(d)(Restricting the Domain) Iff:X Yis Continuous and ifAis a subspace ofX, then the restricted functionf|A:A Yis Continuous .(e)(Restricting or Expanding the Range) letf:X Ybe Continuous . IfXis asubspace ofYcontaining the image setf(X), then the functiong:X Zobtained by restricting the range offis Continuous .]]]]]]

8 IfZis a space havingYas a subspace, then the functionsh:X Zobtained by expanding the rangeoffis Continuous .(e)(Local Formulation of Continuity) The mapf:X Yis Continuous ifXcanbe written as the union of open setsU such thatf|U is Continuous for each . following result, which is fairly easy to prove, will be extremely usefulin the Continuous Functions7 Theorem Pasting Lemma for closed BwhereAandBare closed inX. Letf:A Yandg:B Ybecontinuous. Iff(x) =g(x) for allx A B, thenfandgcombine (or paste )to give a Continuous functionh:X Ydefined by settingh(x) =f(x) ifx Aandh(x) =g(x) ifx you can see in the proof of the previous theorem, we can replace closed with open to get the The Pasting Lemma for open BwhereAandBare open inX. Letf:A Yandg:B Ybecontinuous. Iff(x) =g(x) for allx A B, thenfandgcombine (or paste )to give a Continuous functionh:X Ydefined by settingh(x) =f(x) ifx Aandh(x) =g(x) ifx we define a limit point of a set without an appeal to closeness, butonly by using properties of the topology (which means we onlyuse open sets ).

9 Gives examples on page 109 of applications of The Pasting Lemmainvolving piecewise defined Functions Continuous Functions8 Theorem into :A X Ybe given by the equationf(a) = (f1(a), f2(a)) wheref1:A Xandf2:Y B. Thenfis Continuous if and only if the functionsf1:A Xandf2:A Y, wheref1= 1 fandf2= 2 fforf:A X Yabove, are calledcoordinate states (see page 110): There is no useful criterionfor the conti-nuity of a mapf:A B Xwhosedomainis a product space. For example,we might expect that iff:R R Ris Continuous iff(x, y0) andf(x0, y) arecontinuous for eachy0 Randx0 R(that is, iffis Continuous in each variableseparately ) thenf(x, y) is Continuous . This is not the case, though, as shown byf(x, y) = xy/(x2+y2) if (x, y)6= (0,0)0if (x, y) = (0,0).This claim is justified in Exercise : 6/11/2016