About the uniqueness of the hyperspaces C(p,X) in some classes of continua
Florencio Corona-Vázquez, Russell Aarón Quiñones-Estrella, Javier Sánchez-Martínez
aa r X i v : . [ m a t h . GN ] D ec About the uniqueness of the hyperspaces C ( p, X ) insome classes of continua Florencio Corona–V´azquez, Russell Aar´on Qui˜nones–Estrella ∗ ,Javier S´anchez–Mart´ınez Universidad Aut´onoma de Chiapas, Facultad de Ciencias en F´ısica y Matem´aticas,Carretera Emiliano Zapata Km. 8, Rancho San Francisco, Ter´an, C.P. 29050, TuxtlaGuti´errez, Chiapas, Mexico.
Abstract
Given a continuum X and p ∈ X , we will consider the hyperspace C ( p, X ) ofall subcontinua of X containing p . Given a family of continua C , a continuum X ∈ C and p ∈ X , we say that ( X, p ) has unique hyperspace C ( p, X ) relativeto C if for each Y ∈ C and q ∈ Y such that C ( p, X ) and C ( q, Y ) are homeo-morphic, then there is an homeomorphism between X and Y sending p to q .In this paper we show that ( X, p ) has unique hyperspace C ( p, X ) relative tothe classes of dendrites if and only if X is a tree, we present also some classesof continua without unique hyperspace C ( p, X ); this answer some questionsposed in [2]. Keywords:
Continua, dendrites, hyperspaces.
Primary, 54B05, 54B20, 54F65
1. Introduction A continuum is a nonempty compact connected metric space. Givena continuum X , by a hyperspace of X we mean a specified collection ofsubsets of X . In the literature, some of the most studied hyperspaces are the ∗ Corresponding author
Email addresses: [email protected] (Florencio Corona–V´azquez), [email protected] (Russell Aar´on Qui˜nones–Estrella ), [email protected] (Javier S´anchez–Mart´ınez)
Preprint submitted to Elsevier December 10, 2019 ollowing: 2 X = { A ⊂ X : A is nonempty and closed } ,C ( X ) = { A ⊂ X : A is nonempty, connected and closed } ,C ( P, X ) = { A ∈ C ( X ) : P ⊂ A } , where P ∈ X . The collection 2 X is called the hyperspace of closed subsets of X whereasthat C ( X ) is called the hyperspace of subcontinua of X . These hyperspacesare endowed with the Hausdorff metric, see [7, p. 1]. The main objectsof interest in this paper are the hyperspaces C ( p, X ) := C ( { p } , X ), where p ∈ X .In general, given a hyperspace H ( X ) ∈ { X , C ( X ) } , X is said to haveunique hyperspace H ( X ) if for each continuum Y such that H ( X ) is home-omorphic to H ( Y ), it holds that X is homeomorphic to Y . In a similarsetting, the following concept was defined in [2]: Given a class of continua C , a continuum X ∈ C and p ∈ X , ( X, p ) is said to have unique hyperspace C ( p, X ) in (or relative to) C if for each Y ∈ C and q ∈ Y such that C ( p, X )is homeomorphic to C ( q, Y ), there is an homeomorphism h : X → Y suchthat h ( p ) = q .By a finite graph we mean a continuum X which can be written as theunion of finitely many arcs, any two of which are either disjoint or intersectonly in one or both of their end points. A tree is a finite graph withoutsimple closed curves. A dendrite is a locally connected continuum withoutsimple closed curves.The main result in [2] is the following: Theorem 1.1. [2, Theorem 4.14] Let X be a tree and p ∈ X . Then ( X, p ) has unique hyperspace C ( p, X ) in the class of trees. In the same paper the authors posed the following questions:
Question 1.2. If X is a dendrite and p ∈ X , has ( X, p ) unique hyperspace C ( p, X ) in the class of dendrites? Question 1.3.
Are there a continuum X and p ∈ X such that ( X, p ) hasunique hyperspace in the class of continua?In this paper we give a negative answer to the Question 1.2, neverthelesswe show that, in the class of dendrites, the trees are the only continua havingunique hyperspace C ( p, X ) (see Theorem 3.12 below). We present also partialsolutions to Question 1.3 by showing some classes of continua without uniquehyperspace C ( p, X ). 2 . Definitions and preliminaries In this paper, dimension means inductive dimension as defined in [7,(0.44), p. 21], following the author of that book, we will denote by dim( X )and dim p ( X ) the dimension of the space X and the dimension of X at p ,respectively. For a subset A of X we denote by | A | the cardinality of A andwe use, as customary, the symbols Int X ( A ), Cl X ( A ) and F r X ( A ) to denotethe interior, closure and boundary of A in X , respectively. If there is noconfusion, we will write simply Int( A ), Cl ( A ) and F r ( A ). We will use thesymbol N to denote the set of all positive integers.Given a continuum X , p ∈ X and β be a cardinal number, we say that p is of order less than or equal to β in X , written ord( p, X ) ≤ β , providedthat for each open subset U of X containing p , there exists an open subset V of X such that p ∈ V ⊂ U and | F r ( V ) | ≤ β . We say that p is the order β in X , written ord( p, X ) = β provided that ord( p, X ) ≤ β and ord( p, X ) (cid:2) α for any cardinal number α < β . Since in this paper we are interested onlyin the finite and countable cases, we will write ord( p, X ) < ∞ if ord( p, X )is an integer and ord( p, X ) = ∞ in any other case (no matter wich infinitecardinal number is about).It is well known that, if Z is a finite graph and z ∈ Z , then ord( z, Z ) < ∞ (see [8, Theorem 9.10, p. 144]) and if Y is a dendrite and y ∈ Y , thenord( y, Y ) ≤ | N | = ℵ (see [8, Corollary 10.20.1, p. 173]).As customary, for a topological space Z , π ( Z ) denote the set of connectedcomponents of Z , also, if Z is connected, it is defined the component numberof z in Z , written c ( z, Z ), as the cardinality of π ( Z − { z } ) (see [8, Definition10.11]). Whit this notation, by [8, Theorem 10.13, 170], if X is a dendriteand x ∈ X then c ( x, X ) = ord( x, X ). In the case that ord( x, X ) = 1 we saythat x is an end point of X , if ord( x, X ) ≥ x is a ramificationpoint of X . We will denote by E ( X ) and R ( X ), respectively, the set of allend points and the set of all ramification points in X .An arc is any space which is homeomorphic to the closed interval [0 , dendroid is an arcwise continuum X , such that for each A, B ∈ C ( X ) holds A ∩ B ∈ C ( X ). In this paper, a dendroid X is say to be smooth at p ∈ X ifsatisfies the definition given in [1, p. 298].We denote by T and D the classes of all trees and all dendrites, respec-tively. By [8, Proposition 9.4, p. 142], each finite graph is locally connectedand then each tree is a dendrite, thus T ⊂ D .We denote by Q to the Hilbert Cube, i.e. Q is the product of infinite3ountable many copies of the unit interval [0 ,
1] with the product topology.
3. Dendrites with unique hyperspace C ( p, X ) We begin this section with a characterization of trees in the class ofdendrites using the structure of the hyperspaces C ( p, X ). Theorem 3.1.
Let X be a dendrite. The following conditions are equivalent:1. There exists p ∈ X such that dim C ( p, X ) < ∞ .2. X satisfies the next two conditions: • ord ( x, X ) < ∞ for all x ∈ X ; • ord ( x, X ) ≤ for all but finitely many x ∈ X .3. X is a tree.4. dim C ( x, X ) < ∞ for each x ∈ X .Proof. To see 1 . implies 2 . , suppose by the contrary that there exists x ∈ X such that ord( x, X ) = ∞ . By [10, Corollary 3.12, p. 1006], x = p . Since X − { x } has countable many components { A i } i ∈ N ∪{ } (we can assume that p ∈ A ) and for each i ∈ N , A i ∪ { x } ∈ C ( X ), we can consider a countablemany order arcs α i : [0 , → C ( X ) such that α i (0) = { x } and α i (1) = A i ∪ { x } . Let f : Q → C ( p, X ) given by f (( t i ) i ∈ N ) = A ∪ ∞ [ i =1 α i (cid:18) t i i (cid:19) , for each ( t i ) i ∈ N ∈ Q . It is easy to see that f is an embedding and therefore dim C ( p, X ) = ∞ ,which is a contradiction.To see the second point, we will proceed again by contradiction. We sup-pose that there exists a sequence { x n } n ∈ N of different points in X such thatord( x i , X ) ≥ n ∈ N . By [8, Lemma 9.11, p. 145], there exists asubcontinuum K of X such that ord( K, X ) = ∞ . Using the same idea ofthe first part, replacing K by { x } we have that dim C ( p, X ) = ∞ , which isimpossible.By [8, Theorem 9.10, p. 144] we have that 2 . and 3 . are equivalent.4o see, 3 . implies 4 . suppose that X is a tree. In this case, if n = P q ∈ R ( X ) ord( q, X ), then using [10, Theorem 3.11 and Corollary 3.12] we havethat dim C ( x, X ) ≤ n for each x ∈ X .Finally 4 . implies trivially 1 . Now, we show an extension of Theorem 1.1.
Lemma 3.2.
Let X ∈ T and p ∈ X . Then ( X, p ) has unique hyperspace C ( p, X ) relative to D .Proof. Let Y ∈ D and let q ∈ Y such that C ( p, X ) is homeomorphic to C ( q, Y ). Since dim C ( q, Y ) = dim C ( p, X ) < ∞ , by Proposition 3.1, weobtain that Y ∈ T . From Theorem 1.1, we conclude that there exists anhomeomorphism between X and Y sending p to q .Similarly to Theorem 3.1, in the following result we present another char-acterization of the trees in the class of dendrites by using dimension of C ( X ). Theorem 3.3.
Let X be a dendrite. The following conditions are equivalent:1. dim C ( X ) < ∞ ,2. X is a tree,3. dim C ( p, X ) < ∞ for some p ∈ X ,4. dim C ( p, X ) < ∞ for each p ∈ X .Proof. By Theorem 3.1 we have that 2 . , 3 . and 4 . are equivalent. We willprove that 1 . and 2 . are equivalent. First, suppose that dim C ( X ) < ∞ ,so for each x ∈ X , dim C ( x, X ) ≤ dim C ( X ) < ∞ and then X is a treeby Theorem 3.1. Now, if X is a tree, then by [6, Theorem 2.4, p. 791],dim C ( X ) < ∞ .Given a dendrite X , we say that p ∈ X is a semi-hairy point in X if p satisfies one of the following two conditions: • ord( p, X ) = ∞ , or • p ∈ Cl ( R ( X ) − { p } ).We denote by sh ( X ) the set of all semi-hairy points in X .5 emark 3.4. It follows from [8, Theorem 9.10, p. 144] that if X is adendrite, then X is a tree if and only if sh ( X ) = ∅ .The next result is an easy application of [3, Theorem 4, p. 221]. Proposition 3.5.
Let X be a dendrite. For a point p ∈ X are equivalent:1. p ∈ sh ( X ) ,2. dim { p } ( C ( p, X )) = ∞ ,3. C ( p, X ) is homeomorphic to Q ,4. p is not in the interior (relative to X ) of a finite graph in X . As a consequence of the last result, we are able to give an answer toQuestion 1.2, the answer is negative as the following example shows.
Example 3.6. If F ω and W are the dendrites known as hairy point [4,bottom p. 46] and null comb [4, Exercise 6.4 and Remark p. 50], then C ( v, F ω ) and C ( w, W ) are homeomorphic because both are homeomorphicto Q [4, part of Exercise 61 p. 48 and Exercise 6.4 p. 50], but F ω and W arenot homeomorphic. • v . . . F ω w · · · • W As a nice consequence of the above proposition and example, we get thefollowing more general result.
Proposition 3.7. If X is a dendrite and p ∈ sh ( X ) , then there exists adendrite Y and p ′ ∈ Y such that C ( p, X ) is homeomorphic to C ( p ′ , Y ) butthere is not an homeomorphism between X and Y sending p to p ′ .Proof. With the notation of Example 3.6, we have that C ( p, X ) is homeo-morphic to C ( v, F ω ) and idem to C ( w, W ), but X cannot be simultaneouslyhomeomorphic both F ω and W , just choose Y ∈ { F ω , W } such that Y is nothomeomorphic to X . 6iven a dendrite X and x, y ∈ X two different points, it is easy to see,by using [8, Lemma 10.24, p. 175], that there exist a unique arc containedin X with end points x and y , we will denote this arc by xy .Now, let X ∈ D − T , if there exists p / ∈ sh ( X ), we denote by sh ( X, p )the following subset of sh ( X ), sh ( p, X ) = { q ∈ sh ( X ) : pq ∩ sh ( X ) = { q }} . Lemma 3.8.
Let X ∈ D − T and let p / ∈ sh ( X ) , then:1. sh ( p, X ) = ∅ ;2. if Y ∈ D , p ′ ∈ Y and there exists f : X → Y be an homeomorphismsending p to p ′ , then f ( sh ( p, X )) = sh ( p ′ , Y ) .Proof. For the first part, let r ∈ sh ( X ) and take α : [0 , → pr be anhomeomorphism such that α (0) = p and α (1) = r . Let t = inf { t ∈ [0 ,
1] : α ( t ) ∈ sh ( X ) } . By using 4. of Proposition 3.5, we have that 0 < t , now thecontinuity of α implies that α ( t ) ∈ sh ( X ). By definition of t we get that α ( t ) ∈ sh ( p, X ).To the second part, since the order of points are preserved by homeo-morphism it is clear that f ( sh ( X )) ⊂ sh ( Y ) and consequently f ( sh ( X )) = sh ( Y ). The uniqueness of arcs connecting different points in a dendrite im-plies that the arc connecting p with any other point q ∈ sh ( p, X ) is sending by f into the arc connecting f ( p ) with f ( q ), and this arc contains no other pointof sh ( Y ) different to f ( q ). This means that f ( q ) ∈ sh ( p ′ , Y ). As before, thisshows that f ( sh ( p, X )) ⊂ sh ( p ′ , Y ) and therefore f ( sh ( p, X )) = sh ( p ′ , Y ).Given a dendrite X , we say that A ⊂ X is an edge in X if there existan homeomorphism α : [0 , → A such that α (0) , α (1) ∈ E ( X ) ∪ R ( X ) andfor each t ∈ (0 , α ( t ) , X ) = 2. We say that A is a free arc of X ifthe homeomorphism α can be choose in such a way that for each t ∈ (0 , α ( t ) ∈ Int X ( A ). Since each homeomorphism between dendrites preserves theorder of its points, we get the following result. Lemma 3.9.
Let f : X → Y be an homeomorphism between dendrites X and Y . Then A ⊂ X is a free arc of X if and only if f ( A ) is a free arc in Y .Also, if p ∈ X is an end point of a free arc in X then f ( p ) is an end pointof a free arc in Y .
7n the same setting of Proposition 3.7, we present the next complementaryresult.
Proposition 3.10. If X ∈ D−T and p / ∈ sh ( X ) , then there exists a dendrite Y and p ′ ∈ Y such that C ( p, X ) is homeomorphic to C ( p ′ , Y ) but there isnot an homeomorphism between X and Y sending p to p ′ .Proof. By Lemma 3.8 we have that sh ( p, X ) = ∅ . By [8, 10.37, p. 181] wecan suppose that X it is contained in the plane z = 0 of the Euclidean space R .We will divide the construction of Y in two cases. Case 1.
For each q ∈ sh ( p, X ) there is not a free arc of X containedin Cl ( X − pq ) and with q as an end point. Fix q ∈ sh ( p, X ) and define Y = X ∪ ( { q } × [0 , Y is a dendrite containing a free arcin Cl ( X − pq ) with q as end point. Case 2.
There is q ∈ sh ( p, X ) and a free arc A of X contained in Cl ( X − pq ) such that q is an end point of A . Set A the family of all free arcs B , such that one of its end points, say q B , belongs to sh ( p, X ), and B iscontained in Cl ( X − pr ). To construct Y , we will attach to each B ∈ A a countable family of arcs, we do this in the following way. For B ∈ A ,choose α B : [0 , → B a homeomorphism such that α B (0) = q B . Now, foreach n ∈ N , let L Bn = { α B ( n ) } × [0 , n ]. It is clear that B ∪ S ∞ i =1 L Bn ishomeomorphic to the null comb. • q B B ••••• · · · · · · L B L B L B We define Y = X ∪ S B ∈A ( B ∪ S ∞ i =1 L Bn ). Since each element in A is afree arc, it is easy to see that Y is a dendrite.In both cases, we have construct a denrite Y such that X ⊂ Y . Wedenote by p ′ = p ∈ Y . It is easy to see that sh ( p, X ) = sh ( p ′ , Y ). Therefore,by Lemmas 3.8 and 3.9, we have that there is not a homeomorphism between X and Y , sending p to p ′ . 8o finish the prove, we need to show that C ( p, X ) is homeomorphic to C ( p ′ , Y ), but this is in fact the case because for each q ∈ sh ( p, X ) = sh ( p ′ , Y ), C ( pq, X ) and C ( p ′ q, Y ) are both Hilbert Cubes .The following result is a consequence of propositions 3.7 and 3.10. Corollary 3.11.
Let X ∈ D − T and p ∈ X . There exist Y ∈ D and q ∈ Y such that C ( p, X ) and C ( q, Y ) are homeomorphic but X is not homeomorphicto Y . The next is the main result of this paper and its proof follows from Lemma3.2, Corollary 3.11 and Theorem 1.1. It gives a characterization of dendriteshaving unique hyperspace C ( p, X ) in the class D . Theorem 3.12.
Let X ∈ D and p ∈ X . Then ( X, p ) has unique hyperspace C ( p, X ) relative to D if and only if X ∈ T .
4. Uniqueness of C ( p, X ) in the class of continua Concerning Question 1.3, in this section we show that for each dendrite orfinite graph X and any p ∈ X , ( X, p ) has not unique hyperspace C ( p, X ) inthe class of continua. Acording Theorem 3.12, we have answered the questionwhen X ∈ D − T , thus it is sufficient to consider the case of finite graphs. Theorem 4.1.
For each finite graph X and any p ∈ X , there exists a con-tinuum Y , which is not finite graph, and y ∈ Y such that C ( p, X ) is homeo-morphic to C ( y, Y ) .Proof. Before we go into the proof, we want to remark a couple of thingsabout to the Knaster continuum, Z , with two end points a and b (see [5, p.205]). By using [9, Observation 5.18 and Theorem 5.21], it is easy to see that Z satisfies that C ( a, Z ) and C ( b, Z ) are arcs and C ( q, Z ) is a 2-cell for each q ∈ Z − { a, b } . 9 a Knaster with two end points • b Given a finite graph X and p ∈ X , suppose that C = { A i : i ∈ { , · · · , n }} is a finite family of arcs such that X = ∪C and any two elements in C are either disjoint or intersect only in one or both of their end points. Toconstruct Y we pick one of the arcs in C , say A , such that p / ∈ A andwe exchange this arc in X for a topological copy of Z , say Z , such thatthe end points of Z are the same that the end points of A and for each i ∈ { , · · · , n } , Z ∩ A i = A ∩ A i .With this construction, for y = p ∈ Y , C ( p, X ) and C ( y, Y ) are homeo-morphic but X and Y are not.We present now other classes of continua X such that for each p ∈ X ,( X, p ) has not unique hyperspace C ( p, X ) in the class of continua.Recall that, a continuum X is say to be decomposable if there exist A, B ∈ C ( X ) − { X } such that X = A ∪ B . A continuum X is say to be inde-composable if it is not decomposable. Also, X is hereditarily indecomposable(decomposable) if for each C ∈ C ( X ) (and | C | > C is indecomposable(decomposable).Since the Knaster continuum with two end points is indecomposable (see[5, p. 205]) we have the following result, wich uses a similar construction tothe given in the previous theorem. Theorem 4.2.
Let X be a hereditarily decomposable continuum with at leastone free arc. For any p ∈ X , there exists a no hereditarily decomposablecontinuum Y and q ∈ Y such that C ( p, X ) and C ( q, Y ) are homeomorphic. The following two results are applications of [3] (see the remark about(4) on page 222 and Theorem 8). 10 roposition 4.3.
Let X is a finite dimensional continuum such that thereexists q ∈ X in such a way that X locally connected in a neighborhood of q ∈ X and dim { q } C ( q, X ) = ∞ . Then for each p ∈ X there exists a continuum Y and y ∈ Y such that C ( p, X ) and C ( y, Y ) are homeomorphic but X and Y are not. Proposition 4.4.
Let X is a smooth dendroid at q ∈ X such that q is notin the interior of a finite tree in X . Then for each p ∈ X there exists acontinuum Y and y ∈ Y such that C ( p, X ) and C ( y, Y ) are homeomorphicbut X and Y are not Y . In both propositions, the construction of Y is done by just attaching acopy of Q , say K , at q ∈ X , in such a way that X ∩ K = { q } . It follows that C ( p, X ) = C ( p, Y ). Note that, in the case of Proposition 4.4, it is enough toattach a 2-cell instead of a copy of Q .In [9], an end point of a continuum X means a point p such that C ( p, X )is an arc (note that this definition of end point is quite different to the onegiven here). Using this definition of end point, we can obtain other classes ofcontinua without unique hyperspace C ( p, X ). Thus classes of continua aresummarized in the next theorem, which is an easy application of [2, Example1.1, p. 2], [9, Theorem 5.22 and Corollary 5.24] and the fact that C (0 , [0 , C ((1 , , S ) is a two cell. Theorem 4.5. If X belongs to some of one of the following classes of con-tinua: • hereditarily indecomposable, • indecomposable without end points and with all proper and nondegen-erate subcontinua being arcs, • indecomposable with exactly two end points and with all proper andnondegenerate subcontinua being arcs,then, for each p ∈ X , ( X, p ) has not unique hyperspace C ( p, X ) . We finish this paper with some open questions.
Question 4.6. If X is a tree and p ∈ X , it is true that ( X, p ) has uniquehyperspace C ( p, X ) in the class of dendroids?11 uestion 4.7. If X is a dendroid such that ( X, p ) has unique hyperspacefor some p ∈ X , should X be a tree? Question 4.8.
Is the condition of smoothness necessary in Proposition 4.4?
Acknowledgments.
The authors wish to thank Eli Vanney Robleroand Rosemberg Toal´a for the fruitful discussions. We also want to thankProfessors Fernando Mac´ıas Romero and David Herrera Carrasco for askingthe questions wich motivated us to write down this paper.
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