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Embedding cones over trees into their symmetric products
Topology and its Applications 231 (2017) 77–91
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Topology and its Applications www.elsevier.com/locate/topol
Embedding cones over trees into their symmetric products F. Corona-Vázquez, R.A. Quiñones-Estrella, H. Villanueva ∗ Facultad de Ciencias en Física y Matemáticas, Universidad Autónoma de Chiapas, Carretera Emiliano Zapata km 8.5, Rancho San Francisco, Ciudad Universitaria, Terán, C.P.29050, Tuxtla Gutiérrez, Chiapas, Mexico
a r t i c l e
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Article history: Received 9 November 2015 Received in revised form 10 March 2017 Accepted 8 September 2017 Available online 18 September 2017
a b s t r a c t Given a metric continuum X and a positive integer n, let Fn (X) be the hyperspace of nonempty sets of X with at most n points and let Cone(X) be the topological cone of X. We say that a continuum X is cone-embeddable in Fn (X) if there is an embedding h from Cone(X) into Fn (X) such that h(x, 0) = {x} for each x in X. In this paper, we characterize trees X that are cone-embeddable in Fn (X). © 2017 Elsevier B.V. All rights reserved.
Keywords: Cone Continuum Embedding Hyperspace Symmetric product Tree
1. Introduction A continuum is a nonempty, compact, connected metric space. If X is a continuum and n is a positive integer, then C(X) denotes the hyperspace of subcontinua of X and Fn (X) the hyperspace of nonempty subsets of X with at most n points, both endowed with the Hausdorff metric. Given a topological space X, the topological cone of X, denoted by Cone(X), is the quotient space obtained from X × [0, 1] by shrinking X × {1} to a point, called the vertex of Cone(X), and denoted by νX . An element of Cone(X), different from νX , will be denoted as an ordered pair (x, t), where x ∈ X and t ∈ [0, 1). For convenience, we will consider that the vertex νX of Cone(X), is equal to any ordered pair of the form (x, 1), where x ∈ X. The problem of determining conditions under which Cone(X) is homeomorphic (or topologically equivalent) to C(X) has been studied by many authors. Moreover, some authors have studied the problem of characterizing those continua X for which there exists a finite-dimensional continuum Z such that C(X) is homeomorphic to Cone(Z). * Corresponding author. E-mail addresses: fl[email protected] (F. Corona-Vázquez), [email protected] (R.A. Quiñones-Estrella), [email protected] (H. Villanueva). http://dx.doi.org/10.1016/j.topol.2017.09.004 0166-8641/© 2017 Elsevier B.V. All rights reserved.
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Recently, A. Illanes and V. Martínez-de-la-Vega have studied the problem of determining those continua X for which there exists a continuum Z such that Fn (X) is homeomorphic to Cone(Z) ([5]). In particular, they consider the problem when X is a finite graph or a fan (a dendroid with exactly one ramification point) and Z is finite dimensional. In the rest of the paper, for a continuum X, the symbol H(X) denotes either C(X) or Fn (X) for some n ∈ N. Definition 1. Let X be a continuum. We say that X is cone-embeddable in H(X) if there exists an embedding h : Cone(X) → H(X) such that h(x, 0) = {x} for each x ∈ X. Such embedding h will be called a good embedding. The third author have studied the case H(X) = C(X), obtaining general properties of cone-embeddable continua in C(X) ([8]), results on compactifications ([9]) and on arc-smooth continua ([10]). In this paper we work the case H(X) = Fn (X), when X is a tree and n is a positive integer. 2. Basics definitions and facts In this section, we present basic concepts and the notation used through the paper. A finite graph is a continuum which can be written as the union of finitely many arcs, called edges, any two of which are either disjoint or intersect only in one or both of their end points. The vertices of a finite graph X, are the end points of the edges of X, the set of all vertices of X will be denoted by V (X). A tree is a finite graph containing no simple closed curves. Throughout the paper, G will denote a finite graph and T will denote a tree. Given a positive integer k, a simple k-od is a finite graph which is the union of k arcs emanating from a single point, v, and otherwise disjoint from one another. The point v is called the core of the simple k-od. Given a finite graph G, p ∈ G and a positive integer k, we say that p is of order k, denoted by OrdG (p) = k, if p has a closed neighborhood which is homeomorphic to a simple k-od having p as the core. If OrdG (p) = 1, then p has a neighborhood which is an arc having p as one of its end points and we will call it an end point of G. If OrdG (p) = 2, then p has a neighborhood which is an arc, p is not an end point of it, and we will call it an ordinary point of G. A point p ∈ G is a ramification point of G if OrdG (p) ≥ 3. In order to distinguish the vertices from the rest of the points of a finite graph which is not a simple closed curve, we assume that each vertex of G is either an end point of G or a ramification point of G. With this restriction, the two end points of an edge of G may coincide, in whose case such an edge is a simple closed curve. This kind of edges will be called loops. We will denote by Ram(G) the set of all ramification points of G and by E(G) the set of all end points of G. Given two different vertices p, q ∈ G, we say that p and q are adjacent provided that there exists an edge J of G such that p and q are the end points of J. Given a vertex p ∈ G, let NG (p) = {q ∈ V (G) : q is adjacent to p}. Note that for a point p ∈ T , OrdT (p) = |NT (p)|, where |A| denotes the cardinality of the set A. For each pair of distinct vertices p, q ∈ T we will denote by pq = [p, q] the only arc in T joining p and q, [p, q) will be the set {x ∈ pq : x = q}. We have in a similar way the natural description of (p, q). Let G be a finite graph and p ∈ Ram(G). We denote by LG (p) the set of end points adjacent to p in G. We define the ramification order of p in G as roG (p) = |NG (p) − LG (p)| = OrdG (p) − |LG (p)|, the number of ramification points adjacent to p in G. For a tree T such that | Ram(T )| ≥ 2 we say that a ramification point p of T is exterior if p is an end point of the finite graph Ep = T − {(p, e] : e ∈ LT (p)} obtained by removing all end points adjacent to p together with their corresponding edges; i.e., if roT (p) = 1. We will denote by Ext(T ) the set of exterior points. Observe that the existence of a non exterior ramification point is equivalent to the fact that | Ram(T )| ≥ 3.
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Given a continuum X and n ∈ N, we consider the following hyperspaces of X: 2X = {A ⊂ X : A = ∅, A is closed}, C(X) = {A ∈ 2X : A is connected}, Fn (X) = {A ∈ 2X : |A| ≤ n}. All the hyperspaces are endowed with the Hausdorff metric ([6, Theorem 2.2]). The hyperspace Fn(X) is called the n-th symmetric product of X. The hyperspace F1 (X) is called the hyperspace of singletons of X and is homeomorphic to X. For a finite family of subsets U1 , . . . , Um of X let U1 , . . . , Um n = {A ∈ Fn (X) : A ⊂ U1 ∪ · · · ∪ Um and A ∩ Ui = ∅ for each i ∈ {1, . . . , m}}. It is known that the sets of the form U1 , . . . , Um n , where the sets U1 , . . . , Um are open in X, form a basis of the topology of Fn (X) (see [7, Theorem 0.13]), called the Vietoris topology. It is not difficult to prove that A1 , . . . , Am n is closed in Fn (X) whenever A1 , . . . , Am are closed in X. A booklet is a space homeomorphic to Y × I where Y is an simple k-od for some k ≥ 3 and I = (0, 1). By [1, Lemma 3.1], we have that booklets are nonplanar subsets. Lemma 2. If {a, b} ∈ F2 (G) has a neighborhood basis consisting of booklets, then {a, b} ∩ Ram(G) = ∅. Proof. It is not difficult to see that singletons of non ramification points have neighborhoods in F2 (G) which are 2-cells. For the case that a = b and {a, b} ∩ Ram(G) = ∅, we also have that {a, b} has a neighborhood which is a 2-cell by [2, Lemma 4.2]. 2 3. A finite graph AT induced by a tree T Given a tree T such that Ram(T ) = ∅, a point p ∈ Ram(T ) and an edge J of T , the set {p}, J 2 is an arc in F2 (T ), since it is homeomorphic to {p} × J. Using this, we will construct a finite graph AT induced by T in the following way: 1) The vertices of AT are the elements {p, q} ∈ F2 (T ) such that: 1.1) p, q ∈ Ram(T ) (the case p = q is allowed), or 1.2) p ∈ Ram(T ) and q ∈ E(T ). 2) The edges of AT are the arcs of the form {p}, J 2 , where p ∈ Ram(T ) and J is an edge of T . 3) The vertices V1 = {p1 , q1 } and V2 = {p2 , q2 } of AT are adjacent if the following two conditions are satisfied: 3.1) the set V1 ∩ V2 consists of exactly one ramification point of T , and 3.2) if V1 ∩ V2 = {p1 } = {p2 }, then the points q1 and q2 are the end points of an edge J of T . Since T has finitely many edges, finitely many vertices and each set of the form {p}, J 2 as described in 2) is an arc in F2 (T ), we have that AT is the union of finitely many arcs. Hence, AT is a finite graph. Moreover, by 3.2), if the vertices {p, q1 } and {p, q2 } are adjacent in AT and p ∈ Ram(T ), then the points q1 and q2 are adjacent in T . In particular, if p ∈ Ram(T ), e ∈ E(T ) and the points p and e are adjacent in T , then the vertices {p, e} and {p} of AT are adjacent too. The end points of AT are determined as follows E(AT ) = {{p, e} ∈ F2 (T ) : p ∈ Ram(T ), e ∈ E(T ) and e ∈ NT (p)}. Hence, by 1), the ramification points of AT are the following ones Ram(AT ) = {{p, q} ∈ F2 (T ) : p, q ∈ Ram(T )}.
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The ordinary points of AT are the elements of F2 (T ) of the form {p, q}, where p ∈ Ram(T ) and q is an ordinary point of T . Note that, if p ∈ Ram(T ) and q, r ∈ NT (p), then {p}, pq 2 ∪ {q}, pr 2 ∪ {r}, pq 2 ∪ {p}, pr 2 is a simple closed curve. Thus, AT is a tree if and only if T is an arc. Let {p, q} ∈ F2 (T ) such that {p, q} ∩ Ram(T ) = ∅. Suppose that p ∈ Ram(T ). If q ∈ Ram(T ) ∪ E(T ), then {p, q} is a vertex of AT . If q is an ordinary point of T , then {p, q} is an ordinary point of AT . In any case, we have that {p, q} ∈ AT if {p, q} is a point of F2 (T ) such that {p, q} ∩ Ram(T ) = ∅. On the other hand, it is clear by the definition of AT that if {p, q} ∈ AT , then {p, q} ∩ Ram(T ) = ∅. We conclude that the finite graph AT can be identified with the closed subset of F2 (T ), defined as BT = {{p, q} ∈ F2 (T ) : {p, q} ∩ Ram(T ) = ∅}. Hence, for any {p, q} ∈ F2 (T ) such that {p, q} ∩ Ram(T ) = ∅, it follows that {p, q} ∈ AT . As a consequence of the preceding lemma we get the following. Lemma 3. Let T be a tree which is cone-embeddable in F2 (T ) and let h : Cone(T ) → F2 (T ) be a good embedding. If p ∈ Ram(T ), then h({p} × [0, 1]) ⊂ AT . In particular, h(νT ) ∈ AT . Proof. Since p ∈ Ram(T ), for each t ∈ (0, 1), the point (p, t) ∈ Cone(T ) has a neighborhood basis consisting of booklets. By Lemma 2, h({p} × (0, 1)) ∩ Ram(T ) = ∅. By the identification between AT and BT , we infer that h({p} × (0, 1)) ⊂ AT . Using the continuity of h and the fact that AT is closed in F2 (T ), it follows that h({p} × [0, 1]) ⊂ AT . In particular, h(νT ) = h((p, 1)) ∈ AT . 2 For each ramification point p of T , let γp : [0, 1] → Cone(T ) be the embedding given by γp (t) = (p, t) ∈ Cone(T ). Note that γp ([0, 1]) is an arc from the base of Cone(T ) to its vertex. It is not difficult to see that the arcs γp satisfy: • Each point γp (t), with t ∈ (0, 1), has a neighborhood homeomorphic to a booklet. • For distinct p, q ∈ Ram(T ), γp ([0, 1]) ∩ γq ([0, 1]) = {νT }.
If there exists a good embedding h : Cone(T ) −→ F2 (T ) then by composing γp with h we will get arcs in AT , αp = h({p} × [0, 1]) = h(γp ([0, 1])), with similar properties; each point h(p, t), for t ∈ (0, 1) has a neighborhood, in h(Cone(T )), which is homeomorphic to a booklet and for distinct p, q ∈ Ram(T ), αp ∩ αq = h(νT ). This trivial observation will play a key role in the proof of Theorem 16, we introduce before some notation. Given v = {p, q} a vertex of AT , an edge-arc from v is an arc in AT joining v and one point of F1 (T ) and contains only one point of F1 (T ). A set of edge-arcs from v is called admissible if the intersection of any two of them is {v}. By the previous discussion, if h(νT ) = v, then v ∈ F2 (T ) − F1 (T ) and there exists an admissible set of k edge-arcs from v, where k = | Ram(T )|, hence OrdAT (v) ≥ | Ram(T )|. Let EA(v) = max{|K| : K is an admissible set of edge-arcs from v}. Note that | Ram(T )| ≤ EA(v).
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We proof the following lemma. Lemma 4. Let p, q ∈ Ram(T ) with p = q, and v = {p, q}. Then v is a ramification point of AT with the following properties: (1) roAT (v) = roT (p) + roT (q); (2) if p ∈ / NT (q), then | Ram(T )| ≥ roT (p) + roT (q) + 1. Proof. To show (1), let w = {a, b} ∈ NAT (v). Then {a, b} ∩ {p, q} consists of exactly one ramification point of T . Assume that p = a ∈ Ram(T ). Thus, b ∈ NT (q) and {p, b} is a ramification point of AT if b ∈ Ram(T ). Hence, {p, b} is a ramification point adjacent to v for each ramification point b adjacent to q. Similarly, assuming that q = a, {q, b} is a ramification point adjacent to v for each ramification point b adjacent to p. Since p = q, we get that roAT (v) = roT (p) + roT (q). In order to prove (2), we consider the arc α in T joining p and q. Since p and q are not adjacent, there exist p1 ∈ NT (p) and q1 ∈ NT (q) such that p1 , q1 ∈ α. In case that p1 = q1 , then there are at least roT (p) +roT (q) distinct ramification points which are different than p and q. Thus, | Ram(T )| ≥ roT (p) + roT (q) + 2. If p1 = q1 , there are at least roT (p) + roT (q) − 1 distinct ramification points which are different than p and q. Hence, | Ram(T )| ≥ roT (p) + roT (q) + 1. 2 Remark 5. Given v = {p, q} ∈ Ram(AT ), then EA(v) ≤ roAT (v) since no end point of AT different to v belongs to and edge-arc from v. 4. Three special trees In this section we introduce three special trees that will be important in order to characterize trees which are cone-embeddable in its second symmetric product. Definition 6. Let T be a tree. We say that T is a g-star, if there exists p ∈ Ram(T ) such that the closure in T of each connected component of T − {p} is a tree with at most one ramification point. Any such point p will be called a center of the g-star. The letter “g” in the term g-star comes from “generalized”. Each simple n-od is a g-star and the following figures show examples of g-stars with center p.
It is easy to see that if | Ram(T )| ≥ 3 then the center of the g-star is uniquely determined. Definition 7. Let T be a tree. We say that T is a bi-star (or a double centered star), if there are two different adjacent elements p, q ∈ Ram(T ) such that the closure in T of each connected component of T − pq is a tree with at most one ramification point. The ramification points p and q are called the centers of the bi-star.
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Remark 8. Each g-star with at least 2 ramification points is a bi-star. Not every bi-star is a g-star. Let T be a bi-star with centers p and q. If x is a vertex of T , we define Nr (x) = NT (x) ∩ (Ram(T ) − {p, q}). We will also consider the sets 1. Tp = T − {[v, w) : w ∈ Nr (q) and v ∈ NT (w) − q}, 2. Tq = T − {[v, w) : w ∈ Nr (p) and v ∈ NT (w) − p} and 3. Tp,q = (Tp )q = (Tq )p = Tq,p . Note that Tp , Tq and Tp,q are g-stars. We also observed that in Tp we have Nr (q) = ∅, in Tq we have Nr (p) = ∅ and in Tp,q we have Nr (p) = Nr (q) = ∅ and, indeed, Ram(Tp,q ) = {p, q}. For each k ∈ N, it will be very useful to introduce the following tree, denoted by RLk and called the ramified Lk , where k is its number of ramification points each one having order 3.
Remark 9. 1. If T is a bi-star and RLk is a subgraph of T then k ≤ 4. 2. If T is a g-star and RLk is a subgraph of T then k ≤ 3. Lemma 10. Suppose that the tree T contains a homeomorphic copy of RL4 , let p ∈ Ram(T ) − Ext(T ) and q ∈ Ext(T ). Then there exist a copy of RL4 with p and q as vertices. Proof. Suppose first that p and q are adjacent. In this case, note that p is the only ramification point adjacent to q. Since p is not an exterior point, roT (p) ≥ 2. If roT (x) = 1 for each x ∈ NT (p) − {q}, then T is a g-star with p as the center. Thus, there is no copy of RL4 in T , which is a contradiction. We conclude that there exists x ∈ NT (p) − {q} such that roT (x) ≥ 2. Then, there exists a ramification point y ∈ NT (x) − {p}. We obtain a copy of RL4 by using the ramification points q, p, x and y. In case that p and q are not adjacent, there is a ramification point x in the arc pq such that x = p and x = q. Since p is not an exterior point, roT (p) ≥ 2 and there exists a ramification point y ∈ NT (p) such that y ∈ / pq. Thus, there is a copy of RL4 with vertices q, x, p and y. 2 Note that if T contains a homeomorphic copy of RL4 and p, q ∈ Ram(T ) − Ext(T ) are such that p = q, then there exists a homeomorphic copy L of RL4 with p and q as vertices of L. If p, q ∈ Ext(T ) and p = q,
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then it is not possible to guarantee the existence of a homeomorphic copy L of RL4 with p and q as vertices of L. 5. The case F2 (T ) To characterize cone-embeddable trees T in F2 (T ), we will introduce some notation. For the particular case when the tree is the arc pq, fixing a homeomorphism f = fpq : [0, 1] → pq such that f (0) = p and f (1) = q, a good embedding h = hpq : Cone(pq) → F2 (pq) can be defined as follows: given x ∈ pq, let tx = f −1 (x); define h(x, t) = {f ((1 − t)tx ), f ((1 − t)tx + t)}. It is not difficult to see that h is a good embedding such that αp = {{p, x} : x ∈ pq} = {p}, pq 2 , αq = {{q, x} : x ∈ pq} = {q}, pq 2 and h(νpq ) = {p, q}. Recall that νpq denotes the vertex of Cone(pq). It is now easy to describe inductively how to extend this map if the tree T is a simple n-od with q as the ramification point:
We only need to write down the extension to pz with z ∈ E(T ) and q ∈ pz, described by hzpq : Cone(pz) −→ F2 (pz) such that hzpq (x, 0) = {x}, for each x ∈ pz; hzpq |Cone(pq) = hpq ; hzpq ({z} × [0, 1]) = {{x, z} : x ∈ pz} ∪ {{p, y} : y ∈ qz} = {z}, pz 2 ∪ {p}, qz 2 . For each x ∈ (q, z), hzpq ({x} × [0, 1]) is an arc in F2 (pz) from {x} to {p, q} and contained in {{a, b} ∈ F2 (pz) : {a, b} ∩ qz = ∅}, as shown in the figure below. 5. hzpq (qz × [0, 1]) = {{a, b} ∈ F2 (pz) : {a, b} ∩ qz = ∅}.
1. 2. 3. 4.
Doing this for each vertex which is adjacent to q and distinct from p, we get an good embedding hq : Cone(T ) → F2 (T ) defined as hq (x, t) = hzpq (x, t) if x ∈ pz and z is a vertex which is adjacent to q and distinct to p. Since Cone(pz1 ) ∩ Cone(pz2 ) = Cone(pq) and hzpq1 |Cone(pq) = hpq = hzpq2 |Cone(pq) for distinct z1 , z2 ∈ N (q) − {p}, by [4, Theorem 9.4, p. 83], hq is a continuous function. Moreover, it is not difficult to see that hq is a good embedding with the following properties: 1. hq (x, 0) = {x} for each x ∈ T ; 2. hq |Cone(pq) = hpq ; 3. hq |Cone(pz) = hzpq for each z ∈ NT (q) − {p}. Lemma 11. If T is a g-star with at most two ramification points, then T is cone-embeddable in F2 (T ).
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Proof. For the previous discussion, we only need to prove the case in which T has exactly two ramification points p and q. Such a g-star has the following form:
In this case, T can be written, in a natural way, as the union of two simple n-ods Rp and Rq having p and q as ramification points, respectively, and such that Rp ∩ Rq = pq. The preceding discussion give us two good embeddings hp : Cone(Rp ) → F2 (Tp ) and hq : Cone(Rq ) → F2 (Tq ) satisfying the properties above. Define h : Cone(T ) → F2 (T ) as h(x, t) =
hp (x, t), if (x, t) ∈ Cone(Rp ); hq (x, t), if (x, t) ∈ Cone(Rq ).
Note that Cone(T ) = Cone(Rp ) ∪ Cone(Rq ). Since Rp ∩ Rq = pq, then Cone(Rp ) ∩ Cone(Rq ) = Cone(pq). Moreover, hp |Cone(pq) = hpq = hq |Cone(pq) . Hence, h is well defined and, by [4, Theorem 9.4, p. 83], it is continuous. Since, hp and hq are one-to-one functions, we conclude that h is a good embedding. 2 Corollary 12. If T is a bi-star, then Tp,q is cone-embeddable in F2 (Tp,q ). Proposition 13. Let T be a bi-star with centers p and q. Assume that Nr (p) = ∅, Nr (q) = ∅ and, for each w ∈ Nr (p), we have OrdT (w) ≤ OrdT (q). Then T is cone-embeddable in F2 (T ). Proof. Since Tp,q is a g-star and Ram(Tp.q ) = {p, q}, by Lemma 11, we have that Tp,q is cone-embeddable in F2 (Tp,q ).
Let hp,q : Cone(Tp,q ) → F2 (Tp,q ) be the good embedding given in the proof of Lemma 11. Let w ∈ Nr (p). Since OrdT (w) ≤ OrdT (q), there exists an one-to-one function fw : (NT (w) − {p}) → (NT (q) − {p}). For each v ∈ NT (w) − {p}, there is a natural embedding hv : Cone(vfw (v)) → F2 (vfw (v)) such that hv |Cone(wfw (v)) = hp,q |Cone(wfw (v)) , {v}, vfw (v) 2 ⊂ hv ({v} × [0, 1]) and hv ({v} × [0, 1]) ∩ {fw (v)}, vfw (v) 2 = {v, fw (v)}. The diagram below describes the image of hv in F2 (vfw (v)), and the images of some arcs of the form {x} × [0, 1], for some points x ∈ vfw (v), in the hyperspace F2 (vfw (v)).
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Note that if v1 , v2 ∈ NT (w) − {p} and v1 = v2 , then hv1 (Cone(v1 fw (v1 ))) ∩ hv2 (Cone(v2 fw (v2 ))) = hp,q (Cone(wq)). Define h : Cone(T ) → F2 (T ) as h(x, t) =
hp,q (x, t), if x ∈ Tp,q ; if x ∈ vfw (v) for some w ∈ Nr (p) and v ∈ NT (w) − {p}. hv (x, t),
It is easy to see that h is the required good embedding. 2 Corollary 14. If T is a g-star with | Ram(T )| ≥ 3, then T is cone-embeddable in F2 (T ). Theorem 15. Let T be a bi-star with centers p and q. Suppose that the following two conditions are satisfied: 1) Nr (p) = ∅ and, for each w ∈ Nr (p), we have OrdT (w) ≤ OrdT (q); 2) Nr (q) = ∅ and, for each v ∈ Nr (p), we have OrdT (v) ≤ OrdT (p). Then T is cone-embeddable in F2 (T ). Proof. Let hp,q : Cone(Tp,q ) → F2 (Tp,q ) be a good embedding. Since both conditions 1) and 2) are satisfied, by Proposition 13 applied to Tp and Tq , there exist good embeddings hp : Cone(Tp ) → F2 (Tp ) and hq : Cone(Tq ) → F2 (Tq ) such that hp |Cone(Tp,q ) = hq |Cone(Tp,q ) = hp,q . Defining h : Cone(T ) → F2 (T ) as h(x, t) = hp (x, t) if x ∈ Tp and h(x, t) = hq (x, t) if x ∈ Tq , it follows, by [4, Theorem 9.4, p. 83], that h is a well defined continuous function and, then, a good embedding. Hence, T is cone-embeddable in F2 (T ). 2 The following results state necessary conditions for a good embedding for a tree containing RL4 . Theorem 16. Let T be a tree containing RL4 . If h : Cone(T ) → F2 (T ) is a good embedding, then h(νT ) = {p, q}, where p, q ∈ Ram(T ) − Ext(T ) and p ∈ NT (q). Proof. Let h(νT ) = v = {p, q}. Because of Lemma 3 we know that at least one of p and q is a ramification point of T . Suppose that p is a ramification point. Since there are at least four ramification points in T , we have that OrdA (v) ≥ 4. Thus, q is neither an end point of T nor an ordinary point of T . Hence q ∈ Ram(T ) (note that only three ramification points are needed for this conclusion). If p, q ∈ Ext(T ), then roT (p) = roT (q) = 1. Hence, by Lemma 4, | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + roT (q) = 2, which is a contradiction. Suppose that q ∈ Ext(T ) and p ∈ Ram(T ) − Ext(T ), then roT (q) = 1. Thus, by Lemma 4, | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + 1. Moreover, by Lemma 10 there exists a homeomorphic copy of RL4 having p and q as vertices.
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If p ∈ NT (q), there are at least roT (p) − 1 ramification points pairwise distinct and different from p and q. On the other hand, since p and q are vertices of a copy of RL4 and q is exterior, there exists a ramification point x ∈ / NT (p) such that x ∈ / {p, q}. Thus, there are at least roT (p) + 2 ramification points of T . Hence | Ram(T )| ≤ roT (p) + 1 < roT (p) + 2 ≤ | Ram(T )|, which is a contradiction. Assume that p and q are not adjacent, then, by Lemma 4, | Ram(T )| ≥ roT (p) + roT (q) + 1 = roT (p) + 2. We conclude again that | Ram(T )| ≤ roT (p) + 1 < roT (p) + 2 ≤ | Ram(T )|. This contradiction proves that q∈ / Ext(T ). We have proved that p, q ∈ Ram(T ) −Ext(T ). If p ∈ / NT (q), by Lemma 4, | Ram(T )| ≥ roT (p) +roT (q) +1. On the other hand, we have that | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + roT (q) < roT (p) + roT (q) + 1 ≤ | Ram(T )|. Which is absurd. 2 Lemma 17. Let T be a bi-star with centers p, q and suppose that T is not a g-star. If h : Cone(T ) → F2 (T ) is a good embedding, then 1. αw = w, wp 2 ∪ w, pq 2 ∪ q, wp 2 for each w ∈ Nr (p); and 2. αz = z, zq 2 ∪ z, pq 2 ∪ p, zq 2 for each z ∈ Nr (q). Proof. Note that, since T is no a g-star, p and q are the only two non exterior ramification points of T . Then, by Theorem 16, h(νT ) = v = {p, q}. The proof will be given in three steps. Claim 1. For each w ∈ Nr (p), there exists tw ∈ (0, 1) such that h(w, tw ) = {w, p}. Moreover, w, wp 2 = h({w} × [0, tw ]). Since w ∈ Ram(T ), by Lemma 3, h({w} × [0, 1]) ⊂ A. Observe that the vertices of A adjacent to {w} are of the form {w, a} for some a ∈ NT (w) ∩ E(T ) or {w, p}. Inasmuch as {w, a} is an end point of A, it does not belong to an edge-arc from {v}. This proves Claim 1. Note that similar claim can be proved for z ∈ Nr (q). Claim 2. For each w ∈ Nr (p) there exists sw ∈ (tw , 1) such that h(w, sw ) = {w, q}. Moreover, w, pq 2 = h({w} × [tw , sw ]). Note that the vertices of A adjacent to {w, p} are one of the following: 1. 2. 3. 4.
{p, a}, for some a ∈ NT (w) ∩ E(T ); {w, b}, for some b ∈ Nr (p); {w, e}, for some e ∈ NT (p) ∩ E(T ); or {w, q}.
Since vertices in 1 and 3 are end points of A, they do not belong to an edge-arc from v. Thus, there are two possibilities: w, pb 2 ⊂ h({w × [tw , 1]}) or w, pq 2 ⊂ h({w × [tw , 1]}). Suppose that w, pb 2 ⊂ h({w × [tw , 1]}) for some b ∈ Nr (p). That is, assume that there exists s ∈ (tw , 1) such that h(w, s ) = {w, b}; thus h({w} × [tw , s ]) = w, pb 2 . We already know that w, wp 2 = h({w × [0, tw ]}). Then, b = w. Note that the vertices adjacent to {w, b} are the following: 1. 2. 3. 4.
{w, c}, for some c ∈ NT (b) ∩ E(T ); {b, a}, for some a ∈ NT (w) ∩ E(T ); {p, b}; or, {p, w}.
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Vertices in 1 and 2 are end points of A, thus they do not belong to an edge-arc from v. By applying Claim 1 to b ∈ Nr (p), there exists tb < 1 such that h(b, tb ) = {b, p}. Thus, {p, b} ∈ / h({w} × [0, 1]). On the other hand, if h(w, s ) = {p, w} = h(w, s ) for some s > s , this contradicts that h is one-to-one. This contradiction shows Claim 2. Similar claim can be proved for z ∈ Nr (q). Claim 3. For each w ∈ Nr (p), q, wp 2 = h({w} × [sw , 1]). Observe that the vertices of A adjacent to {w, q} are one of the following: 1. 2. 3. 4.
{q, a}, for some a ∈ NT (w) ∩ E(T ); {w, b}, for some b ∈ Nr (q); {w, e}, for some e ∈ NT (q) ∩ E(T ); or {p, q}.
Note that vertices in 1 and 3 are end points of A, then they do not belong to an edge-arc from v. By assuming that h(w, s ) = {w, b} for some s ∈ (sw , 1) and b ∈ Nr (q), we can proceed as above to prove that h(w, s ) = {p, b} for some s ∈ (s , 1). Thus, by applying Claims 1 and 2 to the vertex b, we have that h(b, sb ) = {p, b} = h(b, s ) for some sb ∈ (0, 1), which contradicts the injectivity of h. This proves Claim 3. The lemma follows by using similar arguments for z ∈ Nr (q). 2 In order to prove the converse of Theorem 15, and gives the complete characterization of trees T which are cone-embeddable in F2 (T ), we will need the following notion used in [2]. Let Z be a continuum and W an open subset of Z. For each open subset U of Z, let c(U, W) = |{C ⊂ Z : C is component of U ∩W}|, if this number is finite and c(U, W) = ∞, otherwise. For each p ∈ clZ (W), define σ(p, W) = min({m ∈ N : p has a basis of neighborhoods β in Z such that c(U, W) = m for each U ∈ β} ∪ {∞}). The following lemma is easy to prove. Lemma 18. Let h : Z → Y be a homeomorphism and W an open subset of Z. If p ∈ clZ (W), then σ(p, W) = σ(h(p), h(W)). Theorem 19. Let T be a bi-star which is not a g-star and let p and q the centers of T . If T is cone-embeddable in F2 (T ) then deg(v) ≤ deg(p)
for all v ∈ Nr (q),
deg(w) ≤ deg(q)
for all w ∈ Nr (p).
Proof. Suppose to the contrary that deg(w) > deg(q) for some w ∈ Nr (p). Let h : Cone(T ) → F2 (T ) be a good embedding. Let W = Cone(T ) − {{x} × [0, 1] : x ∈ Ram(T )} ⊂ Cone(T ). Note that W is an open subset of Cone(T ). Let m = deg(w). For each t ∈ (0, 1), (w, t) has a basis of neighborhoods β whose elements are m-booklets. For each U ∈ β, U ∩ W has exactly m components and (w, t) ∈ clCone(T ) (C) for each component C of U ∩ W. Thus, by [2, Lemma 5.2], σ((w, t), W) = m, for each t ∈ (0, 1). Since h is an embedding, we have that σ(h(w, t), h(W)) = m for each t ∈ (0, 1), in h(Cone(T )). Thus, it is not difficult to see that σ(h(w, t), F2 (T ) − A) ≥ m for each t ∈ (0, 1). On the other hand, take t ∈ (0, 1) such that h(w, t) = {x, q} for some x ∈ wp, which exists by Lemma 17. By [2, Lemma 5.4 (b)], σ(h(w, t), F2 (T ) − A) = deg(q) < m ≤ σ(h(w, t), F2 (T ) − A). This contradiction completes the proof. 2
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To complete the characterization for n = 2, by Theorem 15, Theorem 19 and Remark 9, it remains only to work the case where T has RLk as a subgraph, for k ≥ 5. Theorem 20. If T is a tree containing RLk , k ≥ 5, as a subgraph then it is not cone-embeddable in F2 (T ). Proof. Suppose to the contrary that there exists a good embedding h : Cone(T ) → F2 (T ). By Theorem 16, h(νT ) = v = {p, q} where p, q are adjacent non exterior ramification points. Then, by Lemma 4 and Remark 5, | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + roT (q). On the other hand, since p and q are adjacent, there are roT (p) + roT (q) ramification points, all pairwise distinct and adjacent to p or q. Since T contains a copy of RL5 , there is a ramification point r ∈ / NT (p) ∪ NT (q). Then, T has at least roT (p) + roT (q) + 1 ramification points. Thus, | Ram(T )| ≤ roT (p) + roT (q) < | Ram(T )|, which is a contradiction. This completes the proof. 2 6. The case n ≥ 3 In this section, we will prove that each tree T is cone-embeddable in Fn (T ) for n ≥ 3. Since F3 (T ) ⊂ Fn (T ) for each n ≥ 3, it is suffices to prove this claim for n = 3. Given two points x, y ∈ T , let length(xy) denote the length of the arc xy ⊂ T . We assume that length(J) = 1 for each edge J of T . We may suppose that the metric of T is given by d(x, y) = length(xy). It is clear that if z ∈ xy, then d(x, y) = d(x, z) + d(z, y). A continuum X is said to be unicoherent if whenever A and B are subcontinua of X such that X = A ∪B, then A ∩ B is connected. A continuum X is hereditarily unicoherent if each subcontinua of X is unicoherent. A dendroid is a hereditarily unicoherent and arcwise connected continuum. A dendroid X is said to be smooth at p ∈ X if whenever (xn ) is a sequence in X convergent to x ∈ X, then the sequence of arcs (pxn ) converges to the arc px. A dendrite is a locally connected dendroid. It is known that every dendrite, and so, any tree, is smooth at each of its points ([3, Corollary 4]). Let p ∈ E(T ) and q ∈ NT (p). We label the vertices of T as follows. Let v10 = q. Let {v11 , v21 , . . . , vn1 1 } = NT (q) − {p} be the adjacent vertices to q, distinct to p. Note that d(vi1 , q) = 1 for i ∈ {1, . . . , n1 }. Let {v12 , v22 , . . . , vn2 2 } be the set {v ∈ V (T ) : d(v, q) = 2}. Note that each vi2 is adjacent to vj1 , for some j ∈ {1, . . . , n1 }. Similarly, let {v13 , . . . , vn3 3 } = {v ∈ T : d(v, q) = 3} and note that each vi3 is adjacent to vj2 for some j ∈ {1, . . . , n2 }. Following this process, we label all vertices of T .
Thus, for each x ∈ T − [p, q), there exist a non negative integer m such that x ∈ vim vjm+1 , for some i ∈ {1, . . . , nm } and some j ∈ {1, . . . , nm+1 }. Theorem 21. If T is a tree, then T is cone-embeddable in Fn (T ) for n ≥ 3. Proof. We will show that there exists a good embedding h : Cone(T ) → F3 (T ). Since F3 (T ) ⊂ Fn (T ) for n ≥ 3, the theorem will follow.
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We will define three functions, g1 , g2 , g3 : pq × [0, 1] → T by g1 (x, t) = x1 and g2 (x, t) = g3 (x, t) = x2 , where x1 , x2 ∈ pq are such that d(p, x1 ) = (1 − t)d(p, x) and d(q, x2 ) = (1 − t)d(q, x). Note that gi (x, t) ∈ pq for i ∈ {1, 2, 3}. We will show that g1 is continuous. Take a sequence ((yn , tn ))∞ n=1 in pq × [0, 1] and 1 (y, t) ∈ pq × [0, 1] such that lim(yn , tn ) = (y, t). Let yn = g1 (yn , tn ) ∈ pyn , then d(p, yn1 ) = (1 − tn )d(p, yn ). By compactness of T , we may assume that there exists y 1 ∈ T such that lim yn1 = y 1 . By the smoothness of T at p, lim pyn = py, thus y 1 ∈ py. By continuity of the distance function, d(p, y 1 ) = lim d(p, yn1 ) = lim(1 −tn )d(p, yn ) = (1 −t)d(p, y). Hence, g1 (y, t) = y 1 and g1 is continuous. Similarly, g2 = g3 is continuous. Note that g1 (x, 0) = g2 (x, 0) = g3 (x, 0) = {x}, g1 (x, 1) = {p} and g2 (x, 1) = g3 (x, 1) = {q} for each x ∈ T. Moreover, g1 (q, t) = q1 ∈ pq such that d(p, q1 ) = (1 − t)d(p, q), and g2 (q, t) = g3 (q, t) = {q} for each t ∈ [0, 1]. Take s < t. Let g1 (x, s) = x1 and g1 (x, t) = y1 . Then, d(p, x1 ) = (1 − s)d(p, x) > (1 − t)d(p, x) = d(p, y1 ). Thus, y1 ∈ px1 − {x1 }. In a similar way, we have that g2 (x, t) ∈ qg2 (x, s) − {g2 (x, s)}. Also, if x2 = g2 (x, t) ∈ xq, then d(p, x1 ) ≤ d(p, x2 ) ≤ d(p, q). Thus, x2 ∈ x1 q and x1 ∈ px2 . Suppose that x ∈ T − [p, q), there exist a non negative integer m such that x ∈ vim vjm+1 , for some i ∈ {1, . . . , nm } and some j ∈ {1, . . . , nm+1 }. Define the function f : T −[p, q) → R by f (x) = m +d(x, vim )j , if x = q and f (q) = 0. Note that, if m = 0, then vim ∈ vim vjm+1 for each vjm+1 ∈ NT (vim ) and vim ∈ vim vkm−1 for each vkm−1 ∈ NT (vim ). Since d(vim , vkm−1 ) = 1, then f (x) = m +d(vim , vim ) = m = (m −1) +1 = (m −1) + d(vim , vkm−1 )i , thus f is well defined. Observe that, if x ∈ qvj1 = v10 vj1 , then f (x) = d(x, v10 )j = d(x, q)j . Hence f is continuous at q. Moreover, it is not difficult to show that f is continuous in each vim vjm+1 and so continuous in T − [p, q). Define h1 , h2 , h3 : (T −[p, q)) ×[0, 1] → T by h1 (x, t) = x1 , h2 (x, t) = x2 and h3 (x, t) = x3 , where x1 ∈ px, x2 ∈ qx and x3 ∈ x1 x2 are such that d(p, x1 ) =(1 − t)d(p, x), d(q, x2 ) =(1 − t)d(q, x), d(x1 , x3 ) =tf (x) d(x1 , x2 ) =tf (x) d(h1 (x, t), h2 (x, t)), for t = 0, x3 =x1 if t = 0 and x = q, and x3 =q if x = q. It can be proved, in a similar way that for the functions gi, that h1 and h2 are continuous. Note that h1 (x, 0) = h2 (x, 0) = {x} for each x ∈ T − [p, q). Thus, h3 (x, 0) = {x} for each x ∈ T − [p, q). Moreover, by the smoothness of T at p and since the distance function and f are continuous, then h3 is continuous. Observe that h1 (q, t) = q1 ∈ pq such that d(p, q1 ) = (1 −t)d(p, q), and h2 (q, t) = q. Thus, h1 (q, t) = g1 (q, t) and h2 (q, t) = g2 (q, t). Moreover, h3 (q, t) = g3 (q, t). For t = 1, h1 (x, 1) = p, h2 (x, 1) = q and h3 (x, 1) = x3 ∈ pq such that d(p, x3 ) = d(p, q), this implies that h3 (x, 1) = q. Let x1 = h1 (x, t) and x2 = h2 (x, t). We will prove that d(p, x1 ) ≤ d(p, x2 ). Since x1 ∈ px and x2 ∈ xq ⊂ xp, then d(p, x1 ) = (1 − t)d(p, x) = (1 − t)d(p, q) + (1 − t)d(q, x) = (1 − t)d(p, q) + d(q, x2 ) ≤ d(p, q) + d(q, x2 ) = d(p, x2 ). It follows from x3 = h3 (x, t) ∈ x1 x2 , that d(p, x1 ) ≤ d(p, x3 ) ≤ d(p, x2 ).
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Moreover, take s < t and let y2 = h2 (x, s). Then d(q, x2 ) = (1 − t)d(q, x) < (1 − s)d(q, x) = d(q, y2 ). Thus, x2 ∈ qx − {y2 }. Hence, h2 (x, t) ∈ / pq for t < 1. Define the function h : T × [0, 1] → F3 (T ) by h(x, t) =
if x ∈ pq; {g1 (x, t), g2 (x, t), g3 (x, t)}, {h1 (x, t), h2 (x, t), h3 (x, t)}, if x ∈ T − [p, q).
Since gi and hi are continuous for i ∈ {1, 2, 3} and gi (q, t) = hi (q, t) for i ∈ {1, 2, 3} and t ∈ [0, 1], then h is continuous. We will show that h is one-to-one in T × [0, 1). Take x, y ∈ T and s, t ∈ [0, 1) such that h(x, t) = h(y, s). Note that h(x, t) ⊂ pq if and only if x ∈ pq. Thus, x, y ∈ pq or x, y ∈ / pq. Suppose that x, y ∈ pq. Let xi = gi (x, t), yi = gi (y, s) for i ∈ {1, 2, 3}. If h(x, t) = h(y, s), since d(p, x1 ) ≤ d(p, x2 ) and d(p, y1 ) ≤ d(p, y2 ), then xi = yi for i ∈ {1, 2, 3}. Thus, (1 − t)d(p, x) = d(p, x1 ) = d(p, y1 ) = (1 − s)d(p, y) and (1 − t)d(q, x) = d(q, x2 ) = d(q, y2 ) = (1 − s)d(q, y). Note that d(q, x) = d(p, q) − d(p, x). Hence, (1 − t)d(q, x) = (1 − t)d(p, q) − (1 − t)d(p, x) = (1 − t)d(p, q) − d(p, x1 ) and (1 − s)d(q, y) = (1 − s)d(p, q) − d(p, y1 ). We conclude that (1 − t)d(p, q) = (1 − s)d(p, q) and, then, s = t. Since (1 − t)d(p, x) = (1 − s)d(p, y), it follows from t < 1, that d(p, x) = d(p, y). Thus, x = y. Assume now that x, y ∈ / pq. Let xi = hi (x, t) and yi = hi (y, s) for i ∈ {1, 2, 3}. If h(x, t) = h(y, s), since d(p, x1 ) ≤ d(p, x3 ) ≤ d(p, x2 ) and d(p, y1 ) ≤ d(p, y3 ) ≤ d(p, y2 ), then xi = yi for i ∈ {1, 2, 3}. Then (1 − t)d(p, x) = d(p, x1 ) = d(p, y1 ) = (1 − s)d(p, y), (1 − t)d(q, x) = d(q, x2 ) = d(q, y2 ) = (1 − s)d(q, y). Hence, d(p, x) − td(p, x) = d(p, y) − sd(p, y), d(q, x) − td(q, x) = d(q, y) − sd(q, y). Since d(p, x) = d(p, q) + d(q, x), we have that d(p, q) + d(q, x) − td(p, q) − td(q, x) = d(p, q) + d(q, y) − sd(p, q) − sd(q, y). Thus, td(p, q) = sd(p, q). Then, s = t. Since (1 − t)d(q, x) = (1 − s)d(q, y), it follows from t < 1, that d(q, x) = d(q, y). Hence, there exist a non negative integer m such that x ∈ vim vjm+1 and y ∈ vkm vlm+1 for some i, k ∈ {1, . . . , nm } and some j, l ∈ {1, . . . , nm+1 }. Moreover, note that d(x, vim ) = d(y, vkm ). Since xi = yi for i ∈ {1, 2, 3}, tf (x) d(x1 , x2 ) = d(x1 , x3 ) = d(y1 , y3 ) = tf (y) d(y1 , y2 ). Thus, tf (x) = tf (y) . Since h(x, t) ∈ F1 (T ) if and only if t = 0, we may assume that t = 0, since, otherwise we have that {x} = h(x, 0) = h(y, 0) = {y} and x = y. Hence, m + d(x, vim )j = f (x) = f (y) = m + d(y, vkm )l . Thus,
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d(x, vim )j = d(y, vkm )l . Since d(x, vim ) = d(y, vkm ), we conclude that j = l and then i = k. It follows that x = y. We have proved that h is one-to-one in T × [0, 1). Note that h(x, 1) = {p, q} for each x ∈ T . Consider the quotient map π : T × [0, 1] → Cone(T ). Note that h is constant on each fiber of π. By Transgression Theorem ([4, Theorem 3.2, p. 123]), h = h ◦ π −1 : Cone(T ) → F3 (T ) is continuous. It is not difficult to show that h is one-to-one. Then h is a one-to-one map from a compact space into a Hausdorff space. It follows that h is an embedding. Moreover, h (x, 0) = h(x, 0) = {x} for each x ∈ T . We conclude that h is a good embedding. This completes the proof. 2 The following questions remain open for dendroids. Question 22. For which dendrites X, X is cone-embeddable in F2 (X)? Question 23. If X is a dendrite, is it true that X is cone-embeddable in Fn (X) for n ≥ 3? Question 24. Let X be a dendroid, if X is cone-embeddable in Fn (X) for some positive integer n, is it true that X is smooth? Acknowledgements The authors wish to thank to the referee for the valuable comments and suggestions that improve the paper. The third author was partially supported by Apoyo a la Incorporación de Nuevos PTC, PRODEP–SEP, DSA/103.5/14/10906. The authors thank the support given by FCFM–PROFOCIES–SEP–2014. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
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Embedding cones over trees into their symmetric products F. Corona-Vázquez, R.A. Quiñones-Estrella, H. Villanueva ∗ Facultad de Ciencias en Física y Matemáticas, Universidad Autónoma de Chiapas, Carretera Emiliano Zapata km 8.5, Rancho San Francisco, Ciudad Universitaria, Terán, C.P.29050, Tuxtla Gutiérrez, Chiapas, Mexico
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Article history: Received 9 November 2015 Received in revised form 10 March 2017 Accepted 8 September 2017 Available online 18 September 2017
a b s t r a c t Given a metric continuum X and a positive integer n, let Fn (X) be the hyperspace of nonempty sets of X with at most n points and let Cone(X) be the topological cone of X. We say that a continuum X is cone-embeddable in Fn (X) if there is an embedding h from Cone(X) into Fn (X) such that h(x, 0) = {x} for each x in X. In this paper, we characterize trees X that are cone-embeddable in Fn (X). © 2017 Elsevier B.V. All rights reserved.
Keywords: Cone Continuum Embedding Hyperspace Symmetric product Tree
1. Introduction A continuum is a nonempty, compact, connected metric space. If X is a continuum and n is a positive integer, then C(X) denotes the hyperspace of subcontinua of X and Fn (X) the hyperspace of nonempty subsets of X with at most n points, both endowed with the Hausdorff metric. Given a topological space X, the topological cone of X, denoted by Cone(X), is the quotient space obtained from X × [0, 1] by shrinking X × {1} to a point, called the vertex of Cone(X), and denoted by νX . An element of Cone(X), different from νX , will be denoted as an ordered pair (x, t), where x ∈ X and t ∈ [0, 1). For convenience, we will consider that the vertex νX of Cone(X), is equal to any ordered pair of the form (x, 1), where x ∈ X. The problem of determining conditions under which Cone(X) is homeomorphic (or topologically equivalent) to C(X) has been studied by many authors. Moreover, some authors have studied the problem of characterizing those continua X for which there exists a finite-dimensional continuum Z such that C(X) is homeomorphic to Cone(Z). * Corresponding author. E-mail addresses: fl[email protected] (F. Corona-Vázquez), [email protected] (R.A. Quiñones-Estrella), [email protected] (H. Villanueva). http://dx.doi.org/10.1016/j.topol.2017.09.004 0166-8641/© 2017 Elsevier B.V. All rights reserved.
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Recently, A. Illanes and V. Martínez-de-la-Vega have studied the problem of determining those continua X for which there exists a continuum Z such that Fn (X) is homeomorphic to Cone(Z) ([5]). In particular, they consider the problem when X is a finite graph or a fan (a dendroid with exactly one ramification point) and Z is finite dimensional. In the rest of the paper, for a continuum X, the symbol H(X) denotes either C(X) or Fn (X) for some n ∈ N. Definition 1. Let X be a continuum. We say that X is cone-embeddable in H(X) if there exists an embedding h : Cone(X) → H(X) such that h(x, 0) = {x} for each x ∈ X. Such embedding h will be called a good embedding. The third author have studied the case H(X) = C(X), obtaining general properties of cone-embeddable continua in C(X) ([8]), results on compactifications ([9]) and on arc-smooth continua ([10]). In this paper we work the case H(X) = Fn (X), when X is a tree and n is a positive integer. 2. Basics definitions and facts In this section, we present basic concepts and the notation used through the paper. A finite graph is a continuum which can be written as the union of finitely many arcs, called edges, any two of which are either disjoint or intersect only in one or both of their end points. The vertices of a finite graph X, are the end points of the edges of X, the set of all vertices of X will be denoted by V (X). A tree is a finite graph containing no simple closed curves. Throughout the paper, G will denote a finite graph and T will denote a tree. Given a positive integer k, a simple k-od is a finite graph which is the union of k arcs emanating from a single point, v, and otherwise disjoint from one another. The point v is called the core of the simple k-od. Given a finite graph G, p ∈ G and a positive integer k, we say that p is of order k, denoted by OrdG (p) = k, if p has a closed neighborhood which is homeomorphic to a simple k-od having p as the core. If OrdG (p) = 1, then p has a neighborhood which is an arc having p as one of its end points and we will call it an end point of G. If OrdG (p) = 2, then p has a neighborhood which is an arc, p is not an end point of it, and we will call it an ordinary point of G. A point p ∈ G is a ramification point of G if OrdG (p) ≥ 3. In order to distinguish the vertices from the rest of the points of a finite graph which is not a simple closed curve, we assume that each vertex of G is either an end point of G or a ramification point of G. With this restriction, the two end points of an edge of G may coincide, in whose case such an edge is a simple closed curve. This kind of edges will be called loops. We will denote by Ram(G) the set of all ramification points of G and by E(G) the set of all end points of G. Given two different vertices p, q ∈ G, we say that p and q are adjacent provided that there exists an edge J of G such that p and q are the end points of J. Given a vertex p ∈ G, let NG (p) = {q ∈ V (G) : q is adjacent to p}. Note that for a point p ∈ T , OrdT (p) = |NT (p)|, where |A| denotes the cardinality of the set A. For each pair of distinct vertices p, q ∈ T we will denote by pq = [p, q] the only arc in T joining p and q, [p, q) will be the set {x ∈ pq : x = q}. We have in a similar way the natural description of (p, q). Let G be a finite graph and p ∈ Ram(G). We denote by LG (p) the set of end points adjacent to p in G. We define the ramification order of p in G as roG (p) = |NG (p) − LG (p)| = OrdG (p) − |LG (p)|, the number of ramification points adjacent to p in G. For a tree T such that | Ram(T )| ≥ 2 we say that a ramification point p of T is exterior if p is an end point of the finite graph Ep = T − {(p, e] : e ∈ LT (p)} obtained by removing all end points adjacent to p together with their corresponding edges; i.e., if roT (p) = 1. We will denote by Ext(T ) the set of exterior points. Observe that the existence of a non exterior ramification point is equivalent to the fact that | Ram(T )| ≥ 3.
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Given a continuum X and n ∈ N, we consider the following hyperspaces of X: 2X = {A ⊂ X : A = ∅, A is closed}, C(X) = {A ∈ 2X : A is connected}, Fn (X) = {A ∈ 2X : |A| ≤ n}. All the hyperspaces are endowed with the Hausdorff metric ([6, Theorem 2.2]). The hyperspace Fn(X) is called the n-th symmetric product of X. The hyperspace F1 (X) is called the hyperspace of singletons of X and is homeomorphic to X. For a finite family of subsets U1 , . . . , Um of X let U1 , . . . , Um n = {A ∈ Fn (X) : A ⊂ U1 ∪ · · · ∪ Um and A ∩ Ui = ∅ for each i ∈ {1, . . . , m}}. It is known that the sets of the form U1 , . . . , Um n , where the sets U1 , . . . , Um are open in X, form a basis of the topology of Fn (X) (see [7, Theorem 0.13]), called the Vietoris topology. It is not difficult to prove that A1 , . . . , Am n is closed in Fn (X) whenever A1 , . . . , Am are closed in X. A booklet is a space homeomorphic to Y × I where Y is an simple k-od for some k ≥ 3 and I = (0, 1). By [1, Lemma 3.1], we have that booklets are nonplanar subsets. Lemma 2. If {a, b} ∈ F2 (G) has a neighborhood basis consisting of booklets, then {a, b} ∩ Ram(G) = ∅. Proof. It is not difficult to see that singletons of non ramification points have neighborhoods in F2 (G) which are 2-cells. For the case that a = b and {a, b} ∩ Ram(G) = ∅, we also have that {a, b} has a neighborhood which is a 2-cell by [2, Lemma 4.2]. 2 3. A finite graph AT induced by a tree T Given a tree T such that Ram(T ) = ∅, a point p ∈ Ram(T ) and an edge J of T , the set {p}, J 2 is an arc in F2 (T ), since it is homeomorphic to {p} × J. Using this, we will construct a finite graph AT induced by T in the following way: 1) The vertices of AT are the elements {p, q} ∈ F2 (T ) such that: 1.1) p, q ∈ Ram(T ) (the case p = q is allowed), or 1.2) p ∈ Ram(T ) and q ∈ E(T ). 2) The edges of AT are the arcs of the form {p}, J 2 , where p ∈ Ram(T ) and J is an edge of T . 3) The vertices V1 = {p1 , q1 } and V2 = {p2 , q2 } of AT are adjacent if the following two conditions are satisfied: 3.1) the set V1 ∩ V2 consists of exactly one ramification point of T , and 3.2) if V1 ∩ V2 = {p1 } = {p2 }, then the points q1 and q2 are the end points of an edge J of T . Since T has finitely many edges, finitely many vertices and each set of the form {p}, J 2 as described in 2) is an arc in F2 (T ), we have that AT is the union of finitely many arcs. Hence, AT is a finite graph. Moreover, by 3.2), if the vertices {p, q1 } and {p, q2 } are adjacent in AT and p ∈ Ram(T ), then the points q1 and q2 are adjacent in T . In particular, if p ∈ Ram(T ), e ∈ E(T ) and the points p and e are adjacent in T , then the vertices {p, e} and {p} of AT are adjacent too. The end points of AT are determined as follows E(AT ) = {{p, e} ∈ F2 (T ) : p ∈ Ram(T ), e ∈ E(T ) and e ∈ NT (p)}. Hence, by 1), the ramification points of AT are the following ones Ram(AT ) = {{p, q} ∈ F2 (T ) : p, q ∈ Ram(T )}.
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The ordinary points of AT are the elements of F2 (T ) of the form {p, q}, where p ∈ Ram(T ) and q is an ordinary point of T . Note that, if p ∈ Ram(T ) and q, r ∈ NT (p), then {p}, pq 2 ∪ {q}, pr 2 ∪ {r}, pq 2 ∪ {p}, pr 2 is a simple closed curve. Thus, AT is a tree if and only if T is an arc. Let {p, q} ∈ F2 (T ) such that {p, q} ∩ Ram(T ) = ∅. Suppose that p ∈ Ram(T ). If q ∈ Ram(T ) ∪ E(T ), then {p, q} is a vertex of AT . If q is an ordinary point of T , then {p, q} is an ordinary point of AT . In any case, we have that {p, q} ∈ AT if {p, q} is a point of F2 (T ) such that {p, q} ∩ Ram(T ) = ∅. On the other hand, it is clear by the definition of AT that if {p, q} ∈ AT , then {p, q} ∩ Ram(T ) = ∅. We conclude that the finite graph AT can be identified with the closed subset of F2 (T ), defined as BT = {{p, q} ∈ F2 (T ) : {p, q} ∩ Ram(T ) = ∅}. Hence, for any {p, q} ∈ F2 (T ) such that {p, q} ∩ Ram(T ) = ∅, it follows that {p, q} ∈ AT . As a consequence of the preceding lemma we get the following. Lemma 3. Let T be a tree which is cone-embeddable in F2 (T ) and let h : Cone(T ) → F2 (T ) be a good embedding. If p ∈ Ram(T ), then h({p} × [0, 1]) ⊂ AT . In particular, h(νT ) ∈ AT . Proof. Since p ∈ Ram(T ), for each t ∈ (0, 1), the point (p, t) ∈ Cone(T ) has a neighborhood basis consisting of booklets. By Lemma 2, h({p} × (0, 1)) ∩ Ram(T ) = ∅. By the identification between AT and BT , we infer that h({p} × (0, 1)) ⊂ AT . Using the continuity of h and the fact that AT is closed in F2 (T ), it follows that h({p} × [0, 1]) ⊂ AT . In particular, h(νT ) = h((p, 1)) ∈ AT . 2 For each ramification point p of T , let γp : [0, 1] → Cone(T ) be the embedding given by γp (t) = (p, t) ∈ Cone(T ). Note that γp ([0, 1]) is an arc from the base of Cone(T ) to its vertex. It is not difficult to see that the arcs γp satisfy: • Each point γp (t), with t ∈ (0, 1), has a neighborhood homeomorphic to a booklet. • For distinct p, q ∈ Ram(T ), γp ([0, 1]) ∩ γq ([0, 1]) = {νT }.
If there exists a good embedding h : Cone(T ) −→ F2 (T ) then by composing γp with h we will get arcs in AT , αp = h({p} × [0, 1]) = h(γp ([0, 1])), with similar properties; each point h(p, t), for t ∈ (0, 1) has a neighborhood, in h(Cone(T )), which is homeomorphic to a booklet and for distinct p, q ∈ Ram(T ), αp ∩ αq = h(νT ). This trivial observation will play a key role in the proof of Theorem 16, we introduce before some notation. Given v = {p, q} a vertex of AT , an edge-arc from v is an arc in AT joining v and one point of F1 (T ) and contains only one point of F1 (T ). A set of edge-arcs from v is called admissible if the intersection of any two of them is {v}. By the previous discussion, if h(νT ) = v, then v ∈ F2 (T ) − F1 (T ) and there exists an admissible set of k edge-arcs from v, where k = | Ram(T )|, hence OrdAT (v) ≥ | Ram(T )|. Let EA(v) = max{|K| : K is an admissible set of edge-arcs from v}. Note that | Ram(T )| ≤ EA(v).
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We proof the following lemma. Lemma 4. Let p, q ∈ Ram(T ) with p = q, and v = {p, q}. Then v is a ramification point of AT with the following properties: (1) roAT (v) = roT (p) + roT (q); (2) if p ∈ / NT (q), then | Ram(T )| ≥ roT (p) + roT (q) + 1. Proof. To show (1), let w = {a, b} ∈ NAT (v). Then {a, b} ∩ {p, q} consists of exactly one ramification point of T . Assume that p = a ∈ Ram(T ). Thus, b ∈ NT (q) and {p, b} is a ramification point of AT if b ∈ Ram(T ). Hence, {p, b} is a ramification point adjacent to v for each ramification point b adjacent to q. Similarly, assuming that q = a, {q, b} is a ramification point adjacent to v for each ramification point b adjacent to p. Since p = q, we get that roAT (v) = roT (p) + roT (q). In order to prove (2), we consider the arc α in T joining p and q. Since p and q are not adjacent, there exist p1 ∈ NT (p) and q1 ∈ NT (q) such that p1 , q1 ∈ α. In case that p1 = q1 , then there are at least roT (p) +roT (q) distinct ramification points which are different than p and q. Thus, | Ram(T )| ≥ roT (p) + roT (q) + 2. If p1 = q1 , there are at least roT (p) + roT (q) − 1 distinct ramification points which are different than p and q. Hence, | Ram(T )| ≥ roT (p) + roT (q) + 1. 2 Remark 5. Given v = {p, q} ∈ Ram(AT ), then EA(v) ≤ roAT (v) since no end point of AT different to v belongs to and edge-arc from v. 4. Three special trees In this section we introduce three special trees that will be important in order to characterize trees which are cone-embeddable in its second symmetric product. Definition 6. Let T be a tree. We say that T is a g-star, if there exists p ∈ Ram(T ) such that the closure in T of each connected component of T − {p} is a tree with at most one ramification point. Any such point p will be called a center of the g-star. The letter “g” in the term g-star comes from “generalized”. Each simple n-od is a g-star and the following figures show examples of g-stars with center p.
It is easy to see that if | Ram(T )| ≥ 3 then the center of the g-star is uniquely determined. Definition 7. Let T be a tree. We say that T is a bi-star (or a double centered star), if there are two different adjacent elements p, q ∈ Ram(T ) such that the closure in T of each connected component of T − pq is a tree with at most one ramification point. The ramification points p and q are called the centers of the bi-star.
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Remark 8. Each g-star with at least 2 ramification points is a bi-star. Not every bi-star is a g-star. Let T be a bi-star with centers p and q. If x is a vertex of T , we define Nr (x) = NT (x) ∩ (Ram(T ) − {p, q}). We will also consider the sets 1. Tp = T − {[v, w) : w ∈ Nr (q) and v ∈ NT (w) − q}, 2. Tq = T − {[v, w) : w ∈ Nr (p) and v ∈ NT (w) − p} and 3. Tp,q = (Tp )q = (Tq )p = Tq,p . Note that Tp , Tq and Tp,q are g-stars. We also observed that in Tp we have Nr (q) = ∅, in Tq we have Nr (p) = ∅ and in Tp,q we have Nr (p) = Nr (q) = ∅ and, indeed, Ram(Tp,q ) = {p, q}. For each k ∈ N, it will be very useful to introduce the following tree, denoted by RLk and called the ramified Lk , where k is its number of ramification points each one having order 3.
Remark 9. 1. If T is a bi-star and RLk is a subgraph of T then k ≤ 4. 2. If T is a g-star and RLk is a subgraph of T then k ≤ 3. Lemma 10. Suppose that the tree T contains a homeomorphic copy of RL4 , let p ∈ Ram(T ) − Ext(T ) and q ∈ Ext(T ). Then there exist a copy of RL4 with p and q as vertices. Proof. Suppose first that p and q are adjacent. In this case, note that p is the only ramification point adjacent to q. Since p is not an exterior point, roT (p) ≥ 2. If roT (x) = 1 for each x ∈ NT (p) − {q}, then T is a g-star with p as the center. Thus, there is no copy of RL4 in T , which is a contradiction. We conclude that there exists x ∈ NT (p) − {q} such that roT (x) ≥ 2. Then, there exists a ramification point y ∈ NT (x) − {p}. We obtain a copy of RL4 by using the ramification points q, p, x and y. In case that p and q are not adjacent, there is a ramification point x in the arc pq such that x = p and x = q. Since p is not an exterior point, roT (p) ≥ 2 and there exists a ramification point y ∈ NT (p) such that y ∈ / pq. Thus, there is a copy of RL4 with vertices q, x, p and y. 2 Note that if T contains a homeomorphic copy of RL4 and p, q ∈ Ram(T ) − Ext(T ) are such that p = q, then there exists a homeomorphic copy L of RL4 with p and q as vertices of L. If p, q ∈ Ext(T ) and p = q,
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then it is not possible to guarantee the existence of a homeomorphic copy L of RL4 with p and q as vertices of L. 5. The case F2 (T ) To characterize cone-embeddable trees T in F2 (T ), we will introduce some notation. For the particular case when the tree is the arc pq, fixing a homeomorphism f = fpq : [0, 1] → pq such that f (0) = p and f (1) = q, a good embedding h = hpq : Cone(pq) → F2 (pq) can be defined as follows: given x ∈ pq, let tx = f −1 (x); define h(x, t) = {f ((1 − t)tx ), f ((1 − t)tx + t)}. It is not difficult to see that h is a good embedding such that αp = {{p, x} : x ∈ pq} = {p}, pq 2 , αq = {{q, x} : x ∈ pq} = {q}, pq 2 and h(νpq ) = {p, q}. Recall that νpq denotes the vertex of Cone(pq). It is now easy to describe inductively how to extend this map if the tree T is a simple n-od with q as the ramification point:
We only need to write down the extension to pz with z ∈ E(T ) and q ∈ pz, described by hzpq : Cone(pz) −→ F2 (pz) such that hzpq (x, 0) = {x}, for each x ∈ pz; hzpq |Cone(pq) = hpq ; hzpq ({z} × [0, 1]) = {{x, z} : x ∈ pz} ∪ {{p, y} : y ∈ qz} = {z}, pz 2 ∪ {p}, qz 2 . For each x ∈ (q, z), hzpq ({x} × [0, 1]) is an arc in F2 (pz) from {x} to {p, q} and contained in {{a, b} ∈ F2 (pz) : {a, b} ∩ qz = ∅}, as shown in the figure below. 5. hzpq (qz × [0, 1]) = {{a, b} ∈ F2 (pz) : {a, b} ∩ qz = ∅}.
1. 2. 3. 4.
Doing this for each vertex which is adjacent to q and distinct from p, we get an good embedding hq : Cone(T ) → F2 (T ) defined as hq (x, t) = hzpq (x, t) if x ∈ pz and z is a vertex which is adjacent to q and distinct to p. Since Cone(pz1 ) ∩ Cone(pz2 ) = Cone(pq) and hzpq1 |Cone(pq) = hpq = hzpq2 |Cone(pq) for distinct z1 , z2 ∈ N (q) − {p}, by [4, Theorem 9.4, p. 83], hq is a continuous function. Moreover, it is not difficult to see that hq is a good embedding with the following properties: 1. hq (x, 0) = {x} for each x ∈ T ; 2. hq |Cone(pq) = hpq ; 3. hq |Cone(pz) = hzpq for each z ∈ NT (q) − {p}. Lemma 11. If T is a g-star with at most two ramification points, then T is cone-embeddable in F2 (T ).
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Proof. For the previous discussion, we only need to prove the case in which T has exactly two ramification points p and q. Such a g-star has the following form:
In this case, T can be written, in a natural way, as the union of two simple n-ods Rp and Rq having p and q as ramification points, respectively, and such that Rp ∩ Rq = pq. The preceding discussion give us two good embeddings hp : Cone(Rp ) → F2 (Tp ) and hq : Cone(Rq ) → F2 (Tq ) satisfying the properties above. Define h : Cone(T ) → F2 (T ) as h(x, t) =
hp (x, t), if (x, t) ∈ Cone(Rp ); hq (x, t), if (x, t) ∈ Cone(Rq ).
Note that Cone(T ) = Cone(Rp ) ∪ Cone(Rq ). Since Rp ∩ Rq = pq, then Cone(Rp ) ∩ Cone(Rq ) = Cone(pq). Moreover, hp |Cone(pq) = hpq = hq |Cone(pq) . Hence, h is well defined and, by [4, Theorem 9.4, p. 83], it is continuous. Since, hp and hq are one-to-one functions, we conclude that h is a good embedding. 2 Corollary 12. If T is a bi-star, then Tp,q is cone-embeddable in F2 (Tp,q ). Proposition 13. Let T be a bi-star with centers p and q. Assume that Nr (p) = ∅, Nr (q) = ∅ and, for each w ∈ Nr (p), we have OrdT (w) ≤ OrdT (q). Then T is cone-embeddable in F2 (T ). Proof. Since Tp,q is a g-star and Ram(Tp.q ) = {p, q}, by Lemma 11, we have that Tp,q is cone-embeddable in F2 (Tp,q ).
Let hp,q : Cone(Tp,q ) → F2 (Tp,q ) be the good embedding given in the proof of Lemma 11. Let w ∈ Nr (p). Since OrdT (w) ≤ OrdT (q), there exists an one-to-one function fw : (NT (w) − {p}) → (NT (q) − {p}). For each v ∈ NT (w) − {p}, there is a natural embedding hv : Cone(vfw (v)) → F2 (vfw (v)) such that hv |Cone(wfw (v)) = hp,q |Cone(wfw (v)) , {v}, vfw (v) 2 ⊂ hv ({v} × [0, 1]) and hv ({v} × [0, 1]) ∩ {fw (v)}, vfw (v) 2 = {v, fw (v)}. The diagram below describes the image of hv in F2 (vfw (v)), and the images of some arcs of the form {x} × [0, 1], for some points x ∈ vfw (v), in the hyperspace F2 (vfw (v)).
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Note that if v1 , v2 ∈ NT (w) − {p} and v1 = v2 , then hv1 (Cone(v1 fw (v1 ))) ∩ hv2 (Cone(v2 fw (v2 ))) = hp,q (Cone(wq)). Define h : Cone(T ) → F2 (T ) as h(x, t) =
hp,q (x, t), if x ∈ Tp,q ; if x ∈ vfw (v) for some w ∈ Nr (p) and v ∈ NT (w) − {p}. hv (x, t),
It is easy to see that h is the required good embedding. 2 Corollary 14. If T is a g-star with | Ram(T )| ≥ 3, then T is cone-embeddable in F2 (T ). Theorem 15. Let T be a bi-star with centers p and q. Suppose that the following two conditions are satisfied: 1) Nr (p) = ∅ and, for each w ∈ Nr (p), we have OrdT (w) ≤ OrdT (q); 2) Nr (q) = ∅ and, for each v ∈ Nr (p), we have OrdT (v) ≤ OrdT (p). Then T is cone-embeddable in F2 (T ). Proof. Let hp,q : Cone(Tp,q ) → F2 (Tp,q ) be a good embedding. Since both conditions 1) and 2) are satisfied, by Proposition 13 applied to Tp and Tq , there exist good embeddings hp : Cone(Tp ) → F2 (Tp ) and hq : Cone(Tq ) → F2 (Tq ) such that hp |Cone(Tp,q ) = hq |Cone(Tp,q ) = hp,q . Defining h : Cone(T ) → F2 (T ) as h(x, t) = hp (x, t) if x ∈ Tp and h(x, t) = hq (x, t) if x ∈ Tq , it follows, by [4, Theorem 9.4, p. 83], that h is a well defined continuous function and, then, a good embedding. Hence, T is cone-embeddable in F2 (T ). 2 The following results state necessary conditions for a good embedding for a tree containing RL4 . Theorem 16. Let T be a tree containing RL4 . If h : Cone(T ) → F2 (T ) is a good embedding, then h(νT ) = {p, q}, where p, q ∈ Ram(T ) − Ext(T ) and p ∈ NT (q). Proof. Let h(νT ) = v = {p, q}. Because of Lemma 3 we know that at least one of p and q is a ramification point of T . Suppose that p is a ramification point. Since there are at least four ramification points in T , we have that OrdA (v) ≥ 4. Thus, q is neither an end point of T nor an ordinary point of T . Hence q ∈ Ram(T ) (note that only three ramification points are needed for this conclusion). If p, q ∈ Ext(T ), then roT (p) = roT (q) = 1. Hence, by Lemma 4, | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + roT (q) = 2, which is a contradiction. Suppose that q ∈ Ext(T ) and p ∈ Ram(T ) − Ext(T ), then roT (q) = 1. Thus, by Lemma 4, | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + 1. Moreover, by Lemma 10 there exists a homeomorphic copy of RL4 having p and q as vertices.
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If p ∈ NT (q), there are at least roT (p) − 1 ramification points pairwise distinct and different from p and q. On the other hand, since p and q are vertices of a copy of RL4 and q is exterior, there exists a ramification point x ∈ / NT (p) such that x ∈ / {p, q}. Thus, there are at least roT (p) + 2 ramification points of T . Hence | Ram(T )| ≤ roT (p) + 1 < roT (p) + 2 ≤ | Ram(T )|, which is a contradiction. Assume that p and q are not adjacent, then, by Lemma 4, | Ram(T )| ≥ roT (p) + roT (q) + 1 = roT (p) + 2. We conclude again that | Ram(T )| ≤ roT (p) + 1 < roT (p) + 2 ≤ | Ram(T )|. This contradiction proves that q∈ / Ext(T ). We have proved that p, q ∈ Ram(T ) −Ext(T ). If p ∈ / NT (q), by Lemma 4, | Ram(T )| ≥ roT (p) +roT (q) +1. On the other hand, we have that | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + roT (q) < roT (p) + roT (q) + 1 ≤ | Ram(T )|. Which is absurd. 2 Lemma 17. Let T be a bi-star with centers p, q and suppose that T is not a g-star. If h : Cone(T ) → F2 (T ) is a good embedding, then 1. αw = w, wp 2 ∪ w, pq 2 ∪ q, wp 2 for each w ∈ Nr (p); and 2. αz = z, zq 2 ∪ z, pq 2 ∪ p, zq 2 for each z ∈ Nr (q). Proof. Note that, since T is no a g-star, p and q are the only two non exterior ramification points of T . Then, by Theorem 16, h(νT ) = v = {p, q}. The proof will be given in three steps. Claim 1. For each w ∈ Nr (p), there exists tw ∈ (0, 1) such that h(w, tw ) = {w, p}. Moreover, w, wp 2 = h({w} × [0, tw ]). Since w ∈ Ram(T ), by Lemma 3, h({w} × [0, 1]) ⊂ A. Observe that the vertices of A adjacent to {w} are of the form {w, a} for some a ∈ NT (w) ∩ E(T ) or {w, p}. Inasmuch as {w, a} is an end point of A, it does not belong to an edge-arc from {v}. This proves Claim 1. Note that similar claim can be proved for z ∈ Nr (q). Claim 2. For each w ∈ Nr (p) there exists sw ∈ (tw , 1) such that h(w, sw ) = {w, q}. Moreover, w, pq 2 = h({w} × [tw , sw ]). Note that the vertices of A adjacent to {w, p} are one of the following: 1. 2. 3. 4.
{p, a}, for some a ∈ NT (w) ∩ E(T ); {w, b}, for some b ∈ Nr (p); {w, e}, for some e ∈ NT (p) ∩ E(T ); or {w, q}.
Since vertices in 1 and 3 are end points of A, they do not belong to an edge-arc from v. Thus, there are two possibilities: w, pb 2 ⊂ h({w × [tw , 1]}) or w, pq 2 ⊂ h({w × [tw , 1]}). Suppose that w, pb 2 ⊂ h({w × [tw , 1]}) for some b ∈ Nr (p). That is, assume that there exists s ∈ (tw , 1) such that h(w, s ) = {w, b}; thus h({w} × [tw , s ]) = w, pb 2 . We already know that w, wp 2 = h({w × [0, tw ]}). Then, b = w. Note that the vertices adjacent to {w, b} are the following: 1. 2. 3. 4.
{w, c}, for some c ∈ NT (b) ∩ E(T ); {b, a}, for some a ∈ NT (w) ∩ E(T ); {p, b}; or, {p, w}.
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Vertices in 1 and 2 are end points of A, thus they do not belong to an edge-arc from v. By applying Claim 1 to b ∈ Nr (p), there exists tb < 1 such that h(b, tb ) = {b, p}. Thus, {p, b} ∈ / h({w} × [0, 1]). On the other hand, if h(w, s ) = {p, w} = h(w, s ) for some s > s , this contradicts that h is one-to-one. This contradiction shows Claim 2. Similar claim can be proved for z ∈ Nr (q). Claim 3. For each w ∈ Nr (p), q, wp 2 = h({w} × [sw , 1]). Observe that the vertices of A adjacent to {w, q} are one of the following: 1. 2. 3. 4.
{q, a}, for some a ∈ NT (w) ∩ E(T ); {w, b}, for some b ∈ Nr (q); {w, e}, for some e ∈ NT (q) ∩ E(T ); or {p, q}.
Note that vertices in 1 and 3 are end points of A, then they do not belong to an edge-arc from v. By assuming that h(w, s ) = {w, b} for some s ∈ (sw , 1) and b ∈ Nr (q), we can proceed as above to prove that h(w, s ) = {p, b} for some s ∈ (s , 1). Thus, by applying Claims 1 and 2 to the vertex b, we have that h(b, sb ) = {p, b} = h(b, s ) for some sb ∈ (0, 1), which contradicts the injectivity of h. This proves Claim 3. The lemma follows by using similar arguments for z ∈ Nr (q). 2 In order to prove the converse of Theorem 15, and gives the complete characterization of trees T which are cone-embeddable in F2 (T ), we will need the following notion used in [2]. Let Z be a continuum and W an open subset of Z. For each open subset U of Z, let c(U, W) = |{C ⊂ Z : C is component of U ∩W}|, if this number is finite and c(U, W) = ∞, otherwise. For each p ∈ clZ (W), define σ(p, W) = min({m ∈ N : p has a basis of neighborhoods β in Z such that c(U, W) = m for each U ∈ β} ∪ {∞}). The following lemma is easy to prove. Lemma 18. Let h : Z → Y be a homeomorphism and W an open subset of Z. If p ∈ clZ (W), then σ(p, W) = σ(h(p), h(W)). Theorem 19. Let T be a bi-star which is not a g-star and let p and q the centers of T . If T is cone-embeddable in F2 (T ) then deg(v) ≤ deg(p)
for all v ∈ Nr (q),
deg(w) ≤ deg(q)
for all w ∈ Nr (p).
Proof. Suppose to the contrary that deg(w) > deg(q) for some w ∈ Nr (p). Let h : Cone(T ) → F2 (T ) be a good embedding. Let W = Cone(T ) − {{x} × [0, 1] : x ∈ Ram(T )} ⊂ Cone(T ). Note that W is an open subset of Cone(T ). Let m = deg(w). For each t ∈ (0, 1), (w, t) has a basis of neighborhoods β whose elements are m-booklets. For each U ∈ β, U ∩ W has exactly m components and (w, t) ∈ clCone(T ) (C) for each component C of U ∩ W. Thus, by [2, Lemma 5.2], σ((w, t), W) = m, for each t ∈ (0, 1). Since h is an embedding, we have that σ(h(w, t), h(W)) = m for each t ∈ (0, 1), in h(Cone(T )). Thus, it is not difficult to see that σ(h(w, t), F2 (T ) − A) ≥ m for each t ∈ (0, 1). On the other hand, take t ∈ (0, 1) such that h(w, t) = {x, q} for some x ∈ wp, which exists by Lemma 17. By [2, Lemma 5.4 (b)], σ(h(w, t), F2 (T ) − A) = deg(q) < m ≤ σ(h(w, t), F2 (T ) − A). This contradiction completes the proof. 2
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To complete the characterization for n = 2, by Theorem 15, Theorem 19 and Remark 9, it remains only to work the case where T has RLk as a subgraph, for k ≥ 5. Theorem 20. If T is a tree containing RLk , k ≥ 5, as a subgraph then it is not cone-embeddable in F2 (T ). Proof. Suppose to the contrary that there exists a good embedding h : Cone(T ) → F2 (T ). By Theorem 16, h(νT ) = v = {p, q} where p, q are adjacent non exterior ramification points. Then, by Lemma 4 and Remark 5, | Ram(T )| ≤ EA(v) ≤ roA (v) = roT (p) + roT (q). On the other hand, since p and q are adjacent, there are roT (p) + roT (q) ramification points, all pairwise distinct and adjacent to p or q. Since T contains a copy of RL5 , there is a ramification point r ∈ / NT (p) ∪ NT (q). Then, T has at least roT (p) + roT (q) + 1 ramification points. Thus, | Ram(T )| ≤ roT (p) + roT (q) < | Ram(T )|, which is a contradiction. This completes the proof. 2 6. The case n ≥ 3 In this section, we will prove that each tree T is cone-embeddable in Fn (T ) for n ≥ 3. Since F3 (T ) ⊂ Fn (T ) for each n ≥ 3, it is suffices to prove this claim for n = 3. Given two points x, y ∈ T , let length(xy) denote the length of the arc xy ⊂ T . We assume that length(J) = 1 for each edge J of T . We may suppose that the metric of T is given by d(x, y) = length(xy). It is clear that if z ∈ xy, then d(x, y) = d(x, z) + d(z, y). A continuum X is said to be unicoherent if whenever A and B are subcontinua of X such that X = A ∪B, then A ∩ B is connected. A continuum X is hereditarily unicoherent if each subcontinua of X is unicoherent. A dendroid is a hereditarily unicoherent and arcwise connected continuum. A dendroid X is said to be smooth at p ∈ X if whenever (xn ) is a sequence in X convergent to x ∈ X, then the sequence of arcs (pxn ) converges to the arc px. A dendrite is a locally connected dendroid. It is known that every dendrite, and so, any tree, is smooth at each of its points ([3, Corollary 4]). Let p ∈ E(T ) and q ∈ NT (p). We label the vertices of T as follows. Let v10 = q. Let {v11 , v21 , . . . , vn1 1 } = NT (q) − {p} be the adjacent vertices to q, distinct to p. Note that d(vi1 , q) = 1 for i ∈ {1, . . . , n1 }. Let {v12 , v22 , . . . , vn2 2 } be the set {v ∈ V (T ) : d(v, q) = 2}. Note that each vi2 is adjacent to vj1 , for some j ∈ {1, . . . , n1 }. Similarly, let {v13 , . . . , vn3 3 } = {v ∈ T : d(v, q) = 3} and note that each vi3 is adjacent to vj2 for some j ∈ {1, . . . , n2 }. Following this process, we label all vertices of T .
Thus, for each x ∈ T − [p, q), there exist a non negative integer m such that x ∈ vim vjm+1 , for some i ∈ {1, . . . , nm } and some j ∈ {1, . . . , nm+1 }. Theorem 21. If T is a tree, then T is cone-embeddable in Fn (T ) for n ≥ 3. Proof. We will show that there exists a good embedding h : Cone(T ) → F3 (T ). Since F3 (T ) ⊂ Fn (T ) for n ≥ 3, the theorem will follow.
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We will define three functions, g1 , g2 , g3 : pq × [0, 1] → T by g1 (x, t) = x1 and g2 (x, t) = g3 (x, t) = x2 , where x1 , x2 ∈ pq are such that d(p, x1 ) = (1 − t)d(p, x) and d(q, x2 ) = (1 − t)d(q, x). Note that gi (x, t) ∈ pq for i ∈ {1, 2, 3}. We will show that g1 is continuous. Take a sequence ((yn , tn ))∞ n=1 in pq × [0, 1] and 1 (y, t) ∈ pq × [0, 1] such that lim(yn , tn ) = (y, t). Let yn = g1 (yn , tn ) ∈ pyn , then d(p, yn1 ) = (1 − tn )d(p, yn ). By compactness of T , we may assume that there exists y 1 ∈ T such that lim yn1 = y 1 . By the smoothness of T at p, lim pyn = py, thus y 1 ∈ py. By continuity of the distance function, d(p, y 1 ) = lim d(p, yn1 ) = lim(1 −tn )d(p, yn ) = (1 −t)d(p, y). Hence, g1 (y, t) = y 1 and g1 is continuous. Similarly, g2 = g3 is continuous. Note that g1 (x, 0) = g2 (x, 0) = g3 (x, 0) = {x}, g1 (x, 1) = {p} and g2 (x, 1) = g3 (x, 1) = {q} for each x ∈ T. Moreover, g1 (q, t) = q1 ∈ pq such that d(p, q1 ) = (1 − t)d(p, q), and g2 (q, t) = g3 (q, t) = {q} for each t ∈ [0, 1]. Take s < t. Let g1 (x, s) = x1 and g1 (x, t) = y1 . Then, d(p, x1 ) = (1 − s)d(p, x) > (1 − t)d(p, x) = d(p, y1 ). Thus, y1 ∈ px1 − {x1 }. In a similar way, we have that g2 (x, t) ∈ qg2 (x, s) − {g2 (x, s)}. Also, if x2 = g2 (x, t) ∈ xq, then d(p, x1 ) ≤ d(p, x2 ) ≤ d(p, q). Thus, x2 ∈ x1 q and x1 ∈ px2 . Suppose that x ∈ T − [p, q), there exist a non negative integer m such that x ∈ vim vjm+1 , for some i ∈ {1, . . . , nm } and some j ∈ {1, . . . , nm+1 }. Define the function f : T −[p, q) → R by f (x) = m +d(x, vim )j , if x = q and f (q) = 0. Note that, if m = 0, then vim ∈ vim vjm+1 for each vjm+1 ∈ NT (vim ) and vim ∈ vim vkm−1 for each vkm−1 ∈ NT (vim ). Since d(vim , vkm−1 ) = 1, then f (x) = m +d(vim , vim ) = m = (m −1) +1 = (m −1) + d(vim , vkm−1 )i , thus f is well defined. Observe that, if x ∈ qvj1 = v10 vj1 , then f (x) = d(x, v10 )j = d(x, q)j . Hence f is continuous at q. Moreover, it is not difficult to show that f is continuous in each vim vjm+1 and so continuous in T − [p, q). Define h1 , h2 , h3 : (T −[p, q)) ×[0, 1] → T by h1 (x, t) = x1 , h2 (x, t) = x2 and h3 (x, t) = x3 , where x1 ∈ px, x2 ∈ qx and x3 ∈ x1 x2 are such that d(p, x1 ) =(1 − t)d(p, x), d(q, x2 ) =(1 − t)d(q, x), d(x1 , x3 ) =tf (x) d(x1 , x2 ) =tf (x) d(h1 (x, t), h2 (x, t)), for t = 0, x3 =x1 if t = 0 and x = q, and x3 =q if x = q. It can be proved, in a similar way that for the functions gi, that h1 and h2 are continuous. Note that h1 (x, 0) = h2 (x, 0) = {x} for each x ∈ T − [p, q). Thus, h3 (x, 0) = {x} for each x ∈ T − [p, q). Moreover, by the smoothness of T at p and since the distance function and f are continuous, then h3 is continuous. Observe that h1 (q, t) = q1 ∈ pq such that d(p, q1 ) = (1 −t)d(p, q), and h2 (q, t) = q. Thus, h1 (q, t) = g1 (q, t) and h2 (q, t) = g2 (q, t). Moreover, h3 (q, t) = g3 (q, t). For t = 1, h1 (x, 1) = p, h2 (x, 1) = q and h3 (x, 1) = x3 ∈ pq such that d(p, x3 ) = d(p, q), this implies that h3 (x, 1) = q. Let x1 = h1 (x, t) and x2 = h2 (x, t). We will prove that d(p, x1 ) ≤ d(p, x2 ). Since x1 ∈ px and x2 ∈ xq ⊂ xp, then d(p, x1 ) = (1 − t)d(p, x) = (1 − t)d(p, q) + (1 − t)d(q, x) = (1 − t)d(p, q) + d(q, x2 ) ≤ d(p, q) + d(q, x2 ) = d(p, x2 ). It follows from x3 = h3 (x, t) ∈ x1 x2 , that d(p, x1 ) ≤ d(p, x3 ) ≤ d(p, x2 ).
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Moreover, take s < t and let y2 = h2 (x, s). Then d(q, x2 ) = (1 − t)d(q, x) < (1 − s)d(q, x) = d(q, y2 ). Thus, x2 ∈ qx − {y2 }. Hence, h2 (x, t) ∈ / pq for t < 1. Define the function h : T × [0, 1] → F3 (T ) by h(x, t) =
if x ∈ pq; {g1 (x, t), g2 (x, t), g3 (x, t)}, {h1 (x, t), h2 (x, t), h3 (x, t)}, if x ∈ T − [p, q).
Since gi and hi are continuous for i ∈ {1, 2, 3} and gi (q, t) = hi (q, t) for i ∈ {1, 2, 3} and t ∈ [0, 1], then h is continuous. We will show that h is one-to-one in T × [0, 1). Take x, y ∈ T and s, t ∈ [0, 1) such that h(x, t) = h(y, s). Note that h(x, t) ⊂ pq if and only if x ∈ pq. Thus, x, y ∈ pq or x, y ∈ / pq. Suppose that x, y ∈ pq. Let xi = gi (x, t), yi = gi (y, s) for i ∈ {1, 2, 3}. If h(x, t) = h(y, s), since d(p, x1 ) ≤ d(p, x2 ) and d(p, y1 ) ≤ d(p, y2 ), then xi = yi for i ∈ {1, 2, 3}. Thus, (1 − t)d(p, x) = d(p, x1 ) = d(p, y1 ) = (1 − s)d(p, y) and (1 − t)d(q, x) = d(q, x2 ) = d(q, y2 ) = (1 − s)d(q, y). Note that d(q, x) = d(p, q) − d(p, x). Hence, (1 − t)d(q, x) = (1 − t)d(p, q) − (1 − t)d(p, x) = (1 − t)d(p, q) − d(p, x1 ) and (1 − s)d(q, y) = (1 − s)d(p, q) − d(p, y1 ). We conclude that (1 − t)d(p, q) = (1 − s)d(p, q) and, then, s = t. Since (1 − t)d(p, x) = (1 − s)d(p, y), it follows from t < 1, that d(p, x) = d(p, y). Thus, x = y. Assume now that x, y ∈ / pq. Let xi = hi (x, t) and yi = hi (y, s) for i ∈ {1, 2, 3}. If h(x, t) = h(y, s), since d(p, x1 ) ≤ d(p, x3 ) ≤ d(p, x2 ) and d(p, y1 ) ≤ d(p, y3 ) ≤ d(p, y2 ), then xi = yi for i ∈ {1, 2, 3}. Then (1 − t)d(p, x) = d(p, x1 ) = d(p, y1 ) = (1 − s)d(p, y), (1 − t)d(q, x) = d(q, x2 ) = d(q, y2 ) = (1 − s)d(q, y). Hence, d(p, x) − td(p, x) = d(p, y) − sd(p, y), d(q, x) − td(q, x) = d(q, y) − sd(q, y). Since d(p, x) = d(p, q) + d(q, x), we have that d(p, q) + d(q, x) − td(p, q) − td(q, x) = d(p, q) + d(q, y) − sd(p, q) − sd(q, y). Thus, td(p, q) = sd(p, q). Then, s = t. Since (1 − t)d(q, x) = (1 − s)d(q, y), it follows from t < 1, that d(q, x) = d(q, y). Hence, there exist a non negative integer m such that x ∈ vim vjm+1 and y ∈ vkm vlm+1 for some i, k ∈ {1, . . . , nm } and some j, l ∈ {1, . . . , nm+1 }. Moreover, note that d(x, vim ) = d(y, vkm ). Since xi = yi for i ∈ {1, 2, 3}, tf (x) d(x1 , x2 ) = d(x1 , x3 ) = d(y1 , y3 ) = tf (y) d(y1 , y2 ). Thus, tf (x) = tf (y) . Since h(x, t) ∈ F1 (T ) if and only if t = 0, we may assume that t = 0, since, otherwise we have that {x} = h(x, 0) = h(y, 0) = {y} and x = y. Hence, m + d(x, vim )j = f (x) = f (y) = m + d(y, vkm )l . Thus,
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d(x, vim )j = d(y, vkm )l . Since d(x, vim ) = d(y, vkm ), we conclude that j = l and then i = k. It follows that x = y. We have proved that h is one-to-one in T × [0, 1). Note that h(x, 1) = {p, q} for each x ∈ T . Consider the quotient map π : T × [0, 1] → Cone(T ). Note that h is constant on each fiber of π. By Transgression Theorem ([4, Theorem 3.2, p. 123]), h = h ◦ π −1 : Cone(T ) → F3 (T ) is continuous. It is not difficult to show that h is one-to-one. Then h is a one-to-one map from a compact space into a Hausdorff space. It follows that h is an embedding. Moreover, h (x, 0) = h(x, 0) = {x} for each x ∈ T . We conclude that h is a good embedding. This completes the proof. 2 The following questions remain open for dendroids. Question 22. For which dendrites X, X is cone-embeddable in F2 (X)? Question 23. If X is a dendrite, is it true that X is cone-embeddable in Fn (X) for n ≥ 3? Question 24. Let X be a dendroid, if X is cone-embeddable in Fn (X) for some positive integer n, is it true that X is smooth? Acknowledgements The authors wish to thank to the referee for the valuable comments and suggestions that improve the paper. The third author was partially supported by Apoyo a la Incorporación de Nuevos PTC, PRODEP–SEP, DSA/103.5/14/10906. The authors thank the support given by FCFM–PROFOCIES–SEP–2014. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
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