Advances in Pure Mathematics, 2012, 2, 133138 http://dx.doi.org/10.4236/apm.2012.22020 Published Online March 2012 (http://www.SciRP.org/journal/apm) Making Holes in the Hyperspace of Subcontinua of Some Continua José G. Anaya, Enrique CastañedaAlvarado, Fernando OrozcoZitli Facultad de Ciencias, Universidad Autónoma del Estado de México, Toluca, México Email: {jgao, eca}@uaemex.mx, forozcozitli@gmail.com Received November 25, 2011; revised December 23, 2011; accepted December 30, 2011 ABSTRACT CX, Let be a metric continuum. Let is said to make a hole in CX, if is not unico herent. In this paper, we characterize elements ACX CX such that makes a hole in , where CX is either a smooth fan or an Elsa continuum. Keywords: Continuum; Elsa Continuum; Fan; Hyperspace; Property b); Unicoherence; Whitney Map 1. Introduction A connected topological space Z is unicoherent if when ever = AB , where and are connected and closed subsets of B , the set B is connected. Let be a unicoherent topological space and let be an element of z . We say that makes a hole in z if z is not unicoherent. A compactum is a nonde generate compact metric space. A continuum is a con nected compactum with metric . Given a continuum d , the hyperspace of all nonempty subcontinua of is denoted by and it is considered with the Haus dorff metric. It is known that the hyperspace CX CX is unicoherent (see [1, Theorem 19.8, p. 159]). In the papers [2] and [3] the author present some par tial solution to the following problem. Problem. Let be a hyperspace of such that is unicoherent. For which elements, X, does make a hole in ? In the current paper we present the solution to that pro blem when is either a smooth fan or an Elsa con tinuum and = CX . 2. Preliminary We use and to denote the set of positive integers and the set of real numbers, respectively. Let be a topological space and let be a subset of . We denote intA the interior of in . An arc is any homeomorphic space to the closed unit interval . Let in a topological space [0,1] ,pq , ,pq will denote an arc, where and are the end points of p q ,pq . A free arc in a continuum is an arc ,pq such that pq p ,q, is open in . A point in a connected topological space is a cut point of (noncut point of) Z provided that z 1 : is disconnected (is connected). A map is a continuous function. A map ZS, where is a connected topological space and is the unit circle in the Euclidean plane , has a lifting if there exists a map such that 1 S 2 :hZ=exp h exp 1 S , where is the map from onto defined by exp=cos 2π,sin2πttt. A connected topological space 1 : has property b) if each map ZS has a lifting. By an end point of , we mean an end point in the classical sense, which means a point of p that is a noncut point of any arc in that contains . A sub space of a topological space p Y is a deformation retract of if there exists a map :ZI Z such that, for each z Z , ,0 =xx , 1= ZY and, for each ,1 = Y , yy zZ . We say that a topolo gical space Z is contractible if there exists , such that z is a deformation retract of . It is known that each contractible normal topological space has property b), and so it is unicoherent (see [4, Theorems 2 and 3, pp. 69 and 70]). 3. Smooth Fans A point of a continuum p is a ramification point provided that is a point which is a common end point of three or more arcs in p that are otherwise disjoint. A fan is an arcwise connected, hereditarily unicoherent continuum with exactly one ramification point (here ditarily unicoherent means each subcontinuum is uni coherent). The ramification point of a fan will be called the vertex of the fan. If is a fan and yX , then , C opyright © 2012 SciRes. APM
J. G. ANAYA ET AL. 134 Copyright © 2012 SciRes. APM , y denotes the unique arc joining and . A fan y with vertex v is said to be smooth provided that if is a sequence in nn x =1 such that it converges to a point X =1 ,nn vx , then the sequence converges to ,vx in CX. To establish some notation, let be a smooth fan with vertex and let i be its end points set, where is an infinity indexing set. It fol lows from definition of smoothness that the set: v EX =:ie x X=,Xv:xNC is a natural homeomorphic copy of in CX. By the smoothness of , we have that the set: =,C v i e i X CX TC is a closed subspace of . Furthermore, each hy perspace ,Cvi e is a 2cell and ,=e j v,v i C v Cve for each which ,ij CX ,CvX are different. The set of all elements of such that it contains will be denoted by . Let CX. We say that is a simple arc if is an arc such that and, there exists a sequence of satisfying the following properties: X =AE CX n A im =1n 1) =l n A n n A n A AA and 2) for each , v int a) , b) and c) . n Since is embeded in the Cantor fan (see [5]), we can regard as embedded in the Euclidean plane such that v and each i is a convex arc, where . Note that for , i for each i. Throughout this section will de note the map from 2 ] i=re v eEXh =0 eE ,0 X [,ve =0r 0,1X onto TC defined by X ,=tx,hx t x. We assume in this section that if ,, i eabv , then the distance between and v a bis less than the distance between and . v Lemma 3.1. Let be a smooth fan with vertex . If v ,ab i eE is an arc contained in , where , then: 0 i X,ve 01) If ,b =1 nn x =int a, there exists a sequence of such that ,=mvb vli , n x and, for each n, 0 , ni ve . 2) If 0 i and ,ae b ,binta, then 0 ,i be is a free arc in . Proof. The proof of (1) is easy. In order to prove (2), we suppose that is not a free arc in 0 ,i be . Then there exists 00 ,, ii e be 0 yb such that 00 ,i yintbe . Hence, 00 ,i yintve =1 nn y 0 ,i . Then, there exists a sequence of ve 0=lim n yy such that . Since a smooth fan, 0 ,=lim, n vy vy . Notice that 0 ,,,ab vbvy . Let ,zintab =1 nn z 0. There exists a sequence of X such that 0 and, for each n, =lim n zz , n zvy n . Clearly ,vintab0 zv. Hence, . Let >0 be such that vBz 0 and 0,Bz ab 0 n. Let 0 00 ,, ni zBz abve be large enough such that 00 0 ,, ni n zve vy . Thus, . Since is a fan and 00 , ni yXve 0= n zv Q 1,1, 1,Q qQ , , this is a contradiction. □ Since the Hilbert cube, , is homogeneous (see [1, Theorem 11.9.1, p. 93]) and is con tractible, we have the following result. Lemma 3.2. Let . Then has property b). Qq Theorem 3.3. Let be a smooth fan with vertex . If v is a subcontinuum of such that vA and, for each eEX i, ,i ve, then does not make a hole in CX . Proof. We are going to prove that ,CvX A CX A is a deformation retract of . Notice that, for each BTCX , there exists ,0,1 BB xt X such that ,= BB hx tB ,if ,, =, 1,,if . BBB BBCvX HBt ttxx BTCX . We define Clearly is a map. Then, ,X ACv is a deformation retract of CX A. Since Q is ho meomorphic to ,CvX (see [6, Theorem 3.1, p. 282]), {},CvX A has property b) (see Lemma 3.2). Therefore CX A has property b) (see [2, Propo sition 9, p. 2001]). □ Lemma 3.4. Let be a smooth fan with vertex v and let 0 ,, i abCve be a simple arc contained in , for some 0 i eEX. Then 00 ,=, ii intb eb eb . Proof. Since ,ab A is a simple arc, there exists a sequence =1 nn of CX n that satisfies the required properties of the definition. Notice that, for each , n TCX 0 , ni and ve . Given n n a0 , ni bve , let , such that =, nnn ab. We need to prove the following claim.
J. G. ANAYA ET AL. 135 Claim. is a free arc in 0 i ,be . Let . First, we suppose that there exists n0 n such that 0 n b 00 , nn b ,vb int a . Since 0 n b 00 , ni be 0 ,i be and 00 , in ea, is a free arc (see (2) of Lem ma 3.1). Hence, is a free arc in . Now, we assume that, for each , n , ni bb 00 ,, ii e be ,>0y0 n 0 e b yb. Let . Notice that =lim n bb and db. Then there exists 0 ,< n b d0 ,i ve ,e 0, n int A , such that . Since is a con ,bydb 2 vex arc of , we have that 00 , ni be 00 ni yb 00 , ni be . Since is a free arc in (see (2) of Lemma 3.1). Thus, is an open subset of 00 , ni be 00 , ni be such that 0 0 ,, i i bebe 00 ,, ni beb 00 , i be 00 ni ey , i . Hence is an open subset of be . This proves the claim. By Claim, is a free arc in 0 ,i be . Since ,=, i ve va i =1 mm y i 0 ,i intb e 00 , ,ab be b 00 ,i ntbe , . Sup pose that ei . Then there exists a se quence in such that e and, for each , . Since 0=lim im y i ,>0 i dbe 0=lim im ey m m0 , mi ybe ,=, i ve va 00 , ,ab be , and 0 , we may assume that, for each e , . Since 0 , mi yv is a smooth fan, lim, im vy 0,,zbe 0 ve ,= 00 0 , ii i be ve. Let m z0=limm zz . There exists a sequence such that, =1m and, for each , m ,zvy 00 ,, ii be be mm is an open set in . Since and 00 ,, ii e be 0 m 0 i be ,,= m vyv , ii b eb 0 zb zb , there exists such that . Then 00 ,, mi e 0 , i ve 00 mi zve int , this is a contradiction. Therefore . □ ,=be 00 Theorem 3.5. Let be a smooth fan with vertex v. If CX is a simple arc, then makes a hole in . CX 0 =, , i AabCve Proof. We may assume that , where 0 i eEX and 00,1t0 =atb such that . Let: 0 =(,, i CvXTCXintbeA and 0 =, 0,1 i hbeA 0 ,i Xintbe . By Lemma 3.4, is a smooth fan. Then 0 ,i TCX int beA is a connected and closed subset of CX A CX A . So, is a con nected and closed subset of . Notice that is homeomorphic to 0 0 ,0,1, i be bt . Since 0 0 ,0,1, i be bt is a connected subset of 0 0,1 , bt CX A , we have that is a connected subset of . Clearly is a closed subset of CX A. ,=CX abNotice that and: 00 =0,1, ,0 i hbbt hbe . Let: 100 0 =0,, ,0 i hb tbthbe and 200 =,1,hb tbt 12 = . Clearly, is a separation of . Then CX Av is not unicoherent. □ Theorem 3.6. Let X be a smooth fan with vertex , let 0 i eEX 00 ,ii avee 0 ,i ae and let . Then CX does not make a hole in . Proof. Let 00 :,[0,1], ii GCX aeCX ae be defined by: ,= :GAttaa A G G . It is easy to prove that is well defined. In order to show that is continuous, we define :0,1GCX CX by ,= :GAttaa A G =1 , nn n At . We prove that is continuous. Let be a se quence in 0,1CX and 00 ,0,1At CX such 00 ,=lim , that nn tAt. We suppose that there exists BCX =lim ,BGAt such that nn. We will show =,BGAt bB 00 . Let . Consider two sequences =1 nn b and =1 nn a of such that Copyright © 2012 SciRes. APM
J. G. ANAYA ET AL. 136 =limb ,bGAt aX nn 00 ,At BG 0 0 ,GA t A=1 nn a :0,1H by n and, for each , nnn , nn and nn . Taking subsequences if nece ssary, we may assume that there exists 0 such that . Then . Moreover, b = n im n a =ta b n bta 0 aA aA 0=la =limta 0 00 and, so bG . This proves that . Now, let 00 . Then 00 . Then there exists a sequence 00 ,At ta a in such that and, for each , 0nn aA=lim n aa n . So . Since, for each , , . Thus . 00 ta nn ta =l i G mta =,At nn , n A G n tta B n00 00 Hence, is a map. So BG v is a deformation retract of . 0 ,i ae 0 ,i ae 0 e CX Then is contractible. Therefore has property b) (see [2, Proposition 9, CX ,i CX a p. 2001]). □ Theorem 3.7. Let be a smooth fan with vertex , v 0 i eE 0 ,, i C ve let and let such that Xab 0 i,eab and is not a free arc of 0 ,i be . Then ,ab CX does not make a hole in . Proof. In light of Proposition 9 of [2, p. 2001], it suffices to prove that there exist two connected, closed subsets and of ,ab CX which have pro perty b) and the intersection of them is connected. We may assume that there exists 00,1t 0=tb a 0=0t such that . We consider two cases. Case 1. . Then ,=,vb ab . Let =,Xab TC and ,X ab=,Cv . Clearly has property b). By Theorem 3.1 of [6, p. 282], is a Hilbert cube. By Lemma 3.2, ,CvX has property b). Notice that ,X ab ,=abN C . Clearly ,Xab NC is homeomorphic to b . Since b is connected, ,ab is con nected. By Proposition 8 of [2], ) ,ab [,]=(CX ab 0>0t has property b). Case 2. . Consider the following sets: 00 1,bt =,hX t and 00 0, ,tbt =,CvX hX . Clearly and are connected, closed subsets of ,abCX and 00 ,t bt=hX . Notice that is homeomorphic to b . So, since b is connected, is connected. have property b). If we define Now, we are going to prove that and ,, =,1 hxt shxtts, we have 1hX 1hX is a deformation retract of . Since is con tractible, 1hX has property b) (see [2, Proposi tion 9, p. 2001]). Hence, has property b) (see [2, Proposition 9, p. 2001]). In order to prove that has property b), note that ,CvX is a deformation retract of . By Theorem 3.1 of [6, p. 282], ,CvX is homeomorphic to a Hilbert cube. Thus, ,CvX has property b). Hence, has property b) (see Proposition 9 of [2, p. 2001]). Therefore [,]=CX ab has property b). □ Classification Theorem 3.8. Let be a smooth fan with vertex and v CX. Then makes a hole in CX if and only if a simple arc. Proof. Let CX be such that makes a hole in CX. By Theorem 3 of [2, p. 2001] and by Theorem 3.3, is an arc ,pq . By Theorems 3.6 and 00 ,, , ii pq veve for some 3.7, 0 iX 0 ,i qe eE , and . In order to prove that is a free arc in ,pq =1 nn a n b is a simple arc, let , be sequen =1n ces in ,,pq pq 0 ,i qe q =limpa =limb and , respectively, such that and q. Then n n ,= , nn pqlima bn and, for each , ,vabnn , ,inta b nn ,, nn pqa b. Therefore and is a simple arc. The sufficiency follows from Theorem 3.5. 4. Elsa Continua A compactification of 0, 0, with an arc as the re mainder is called an Elsa continuum. The Elsa continua was defined by S. B. Nadler Jr., in [7]. A particular example of an Elsa continuum is the familiar sin(1/x) continuum. There are uncountably many topologically different Elsa continua, the different topological types being a consequence of different ways “patterns into” the remainder of the compactification [8, p. 184]. Let be a continuum. A Whitney map for CX is a continuous function :[0,1]CX that satisfies the following two conditions: BCX such that , B and 1) for any < B B , , =0x for each X and c) 2) =1X . A Whitney block in CX, respectively a Whitney level in 1, CX, is a set of the form t , res pectively 1t 01, where t. It is known that Whitney maps always exist (see [1, Theorem 13.4, p. 107]). Moreover, Whitney blocks and Whitney levels in Copyright © 2012 SciRes. APM
J. G. ANAYA ET AL. 137 CX = are continua (see [1, Theorem 19.9, p. 160]). Throughout this section IR will denote a Elsa continuum, where is the remainder of and is homeomorphic to the halfray . R 0, 0,t 1 Lemma 4.1. Let be a Whitney map, :CX let and let I 0 tCI . Then 0,tA 1() = CI 1 0 Proof. We consider has p r operty b). . It is easy to prove that 1 10 =0,tA =0 10 has property b). Let . Since has property b) (see [9, 12.66, p. 269]) and 1 11 10 0=, has property b) (see [2, Proposition 8, p. 2001]). Let :0 11 ,0 tAS exp=hf 0:h be a map. Then there exists a map such that . 0 Given 10, tCRi R , it is an arc contained in and it is determined by its end point, A, lying near to the end point of . Let R A be an order arc in from A to CX i . Since has property b), there exists a map : A h xp=  A hf s that euch 0 =hi and AA hi A . We define A 1 0 , , .t CR h =1n B :0ht by 0 1 ,if , ,if 0 A hAA hA hAA T In order to prove that is continuous, let be a sequence of 0,t0 = n limB B A 1 0 such that for some 00 0,tA B 1 B . We consider two cases. Case 1. For each , . nn Since is a closed subset of 0 0,tA n B nn R BR p =0 n Bpq q ([,]) Cpq 1 0 =h B B , . Then . n Blimh Case 2. For each , is an arc contained in . We consider two subcases. Subcase 1. . 0 R n Let be such that , where de , notes the end point of . Then R is a Whitney map for ,Cpq . Since is an arc, it 1 ([,]) 0 Cpq 1 , ]) 0, qt 1[0,]t = has property b). By Lemma 4 of [10, p. 254], ([Cp has property b). Then there exists a map such that ([ , ]) : Cpq g ([,]) Cpq gf exp and 0 11 = BB h igi. Notice that ([ ,Cp 1 ] 0q 0 h and [, ]Cpq g 1 0 are liftings of 1 [,] 0Cpq f and 0 11 = BB h igi. Then 11 0[,] 0 [,]0=Cpq Cpq hg . 0n. Notice that Given n h y n g are lift liftings of n f and 0 == Bn BB BB nn nn hihigi . Then = n Bn hg . Hence, 000 0 ===lim =lim=lim . n Bn n n hBh BgBgB hB hB BI ,nm Subcase 2. . 0 We can consider that, for any , B nm ii , if nm . Since 0n and =limBB is a compact space, we may assume that there exists 0 0 B such that 0 iB =lim Bn. We can suppose, taking subsequence if it is necessary, that there exists a subcontinuum 0 ii of CX =lim such that 0 Bn . It is easy to show that 0 is either an order arc from b to or a one pointset. 0 0 B 0n, we have Given n is an homeo 0, n B morphism between and n . Let :0, nn gB B n be such that 1 =g Bn n . By Lemma 3.1 of [7, p. 330], we can assume that is a subset of . Let 2 =0 =0 0, Bn n n DXi B D3 . Notice that is a subset of the Euclidian space and 0XD D is a deformation retract of . Then has property b). 1 10 :0, We define Dt 1 if 0, ,,if ,0,. n nBn xt fxtgtxt iB by It is easy to prove that 1 is a map. Since has property b), there exists a map such that D 3:hD 3 exp =hff 301 11 1 ,0=,0 BB h fihi. Then and 301 {0} {0} = XX hhf 0 . Thus, given n, it can prove that 31 0, 0, =B Bnn BiB iB nn n hhf . Hence, =limn hB hB 0 This proves that . 10,tA = 0 Theorem 4.2. Let has property b). IR be an Elsa continuum and let CI. Then does not make a hole in CX . Proof. In light of Proposition 8 of [2], it suffices to prove that there exist two connected and closed subsets and of CX A, which have property b) and the intersection of them is connected. :0,1CX be a Whitney map. Let Let Copyright © 2012 SciRes. APM
J. G. ANAYA ET AL. Copyright © 2012 SciRes. APM 138 =tA , 1 =,1tA and 1 =0,t A. Clearly CX A A = , and are connected and closed subsets of . CX The sufficiency follows from Theorem 1 of [2, p. 2001]. REFERENCES By Lemma 13 of [2, p. 2004], has property b) and, by Lemma 4.1, has property b). In order to show that is connected, notice that and 1t A= 1 tCI 1 tCR tCR t . By Corollary 3 of [11, p. 386], [1] A. Illanes and S. B. Nadler Jr., “Hyperspaces: Fundamen tals and Recent Advances,” Marcel Dekker, Inc., New York, 1999. [2] J. G. Anaya, “Making Holes in Hyperspaces,” Topology and Its Applications, Vol. 154, No. 10, 2007, pp. 2000 2008. doi:10.1016/j.topol.2006.09.017 11 =ttCI 1 and [3] J. G. Anaya, “Making Holes in the Hyperspace of Sub continua of a Peano Continuum,” Topology Proceedings, Vol. 37, 2011, pp. 114. 1 approximates the whole continuum CI. Hence, = is connected. Theorem 4.3. Let IR be an Elsa continua. If [4] S. Eilenberg, “Transformations Continues en Circon férence et la Topologie du Plan,” Fundamenta Mathe maticae, Vol. 26, 1936, pp. 61112. CX such that is homeomorphic to , then does not make a hole in CX. Proof. In light of Proposition 2.4 of [3, p. 3], it suffices to prove that there exists a closed neighborhood of [5] C. Eberhart, “A Note on Smooth Fans,” Colloquium Ma thematicum, Vol. 20, 1969, pp. 8990. in such that CX bd CX has property b) and CX is connected (bd denotes the boundary of in ). CX [6] C. Eberhart and S. B. Nadler Jr., “Hyperspaces of Cones and Fans,” Proceedings of the American Mathematical Society, Vol. 77, No. 22, 1979, pp. 279288. doi:10.1090/S00029939197905420985 :CX0,1 Let ,1I be a Whitney map. Let 1 = [7] S. B. Nadler Jr., “Continua Whose Cone and Hyperspace Are Homeomorphic,” Transactions of the American Ma thematical Society, Vol. 230, 1977, pp. 321345. doi:10.1090/S00029947197704641910 . Clearly is a closed neigh borhood of . Since 1 =I CX bd , is connected. By [12, Theorem 4.3, p. 217], is a 2cell. Moreover, CX bd is an element of its manifold boundary (see [11, Lemma 2, p. 386]). Then is contractible. Therefore [8] S. B. Nadler Jr., “Arc Components of Certain Chainable Continua,” Canadian Mathematical Bulletin, Vol. 14, No. 2, 1971, pp. 183189. doi:10.4153/CMB19710338 = has pro perty b) (see [2, Proposition 9, p. 2001]). [9] S. B. Nadler Jr., “Continuum Theory: An Introduction,” Marcel Dekker, Inc., New York, 1992. Classification [10] A. Illanes, “Multicoherence of Whitney Levels,” Topology and Its Applications, Vol. 68, No. 3, 1996, pp. 251265. doi:10.1016/01668641(95)00064X Theorem 4.4. Let IR be an Elsa continuum and let CX. Then makes a hole in CX if and only if is a free arc such that . pq int pq,pq [11] W. J. Charatonik, “Some Counterexamples Concerning Whitney Levels,” Bulletin of the Polish Academy of Sci ences Mathematics, Vol. 31, 1983, pp. 385391. Proof. Let CX be such that makes a hole in . By Theorem 3 of [2, p. 2001] and Theorems 4.2 and 4.3, CX is an arc contained in R. So, pq [12] C. B. Hughes, “Some properties of Whitney continua in the hyperspace C(X),” Topology Proceedings, Vol. 1, 1976, pp. 209219. is a free arc in . By Theorem 4 of [2, p. 2001], . q,pqint p
