Uniqueness of the hyperspaces $C(p,X)$ in the class of trees

TL;DR

This paper investigates the structure of hyperspaces of subcontinua containing a point in trees and proves that such hyperspaces are uniquely determined by the tree and the point, within the class of trees.

Contribution

It establishes that for trees, the hyperspace $C(p,X)$ uniquely characterizes the pair $(X,p)$ among all trees, revealing a rigidity property.

Findings

Hyperspaces $C(p,X)$ are uniquely determined by the tree and point.

The main result confirms the uniqueness of $C(p,X)$ within the class of trees.

Provides topological and geometric insights into hyperspaces of trees.

Abstract

Given a continuum $X$ and $p\in X$, we will consider the hyperspace $C(p,X)$ of all subcontinua of $X$ containing $p$. Given a family of continua $\mathcal{C}$, a continuum $X\in\mathcal{C}$ and $p\in X$, we say that $(X,p)$ has unique hyperspace $C(p,X)$ relative to $\mathcal{C}$ if for each $Y\in\mathcal{C}$ and $q\in Y$ such that $C(p,X)$ and $C(q,Y)$ are homeomorphic, then there is an homeomorphism between $X$ and $Y$ sending $p$ to $q$. In this paper we study some topological and geometric properties about the structure of $C(p,X)$ when $X$ is a tree, being the main result that $(X,p)$ has unique hyperspace $C(p,X)$ relative to the class of trees.

Equations

$$2^X$$

$$C(X)$$

$$F_n(X)$$

$$C(p,X)$$

$$\left\{A\in 2^X:A\subset \bigcup _{i=1}^r{K}_i,A\cap K_i\ne \varnothing \text{ for each }i\in \left\{1,\dots ,r\right\}\right\}.$$

$$\left\{A\in C(p,X):\text{A has a neighborhood in C(p,X) homeomorphic to In}\right\},$$

$$T(X)=\bigcup \left\{e\in \text{edge(X) : e∩E(X)=∅}\right\}.$$

$$X\setminus \bigcup _{x\in E(X)}\overline{x{r}_x}.$$

$$\left|\{e\in \text{mbox}edge(X)\setminus \text{mbox}edge(T(X)):\text{z is an end point of e}\}\right|\ge 2.$$

$$X$$

$$U_Y:=\left\{Y\cup \bigcup h_e(t_e):e\in \text{mbox}edge(X)\setminus \text{mbox}edge(Y),e\cap Y\ne \varnothing ,t_e\in (0,1)\right\}.$$

$$\dim U_Y$$

$$\le |\{e\in \text{edge}(X)\setminus \text{edge}(Y^{\prime }):e\cap Y^{\prime }\ne \varnothing \}|=\dim U_{Y^{\prime }}.$$

$$\text{ord}(p,X)$$

$$|E(X)|$$

$$U_G=[G;f_1,\dots ,f_m]=\left[e_1,\dots ,e_k;f_1,\dots ,f_m\right],$$

$$U_G$$

$$U_{G^{\prime }}$$

$$\bullet$$

$$U_G$$

$$U_{G^{\prime }}$$

$$\dim (U_G)$$

$$\dim (U_{G^{\prime }})$$

$$l(p_n)$$

$$=\dim (U_{G^{\prime }})-\dim (U_G)+2-\text{ord}(p_n,T(X)),$$

$$l(p_n)$$

$$=\dim (h(U_{G^{\prime }}))-\dim (h(U_G))+2-\text{ord}(q_n,T(Y))$$

$$=\dim (U_{\overline{q{q}_n}})-\dim (U_{\overline{q{q}_{n-1}}})+2-\text{ord}(q_n,T(Y))$$

$$=\text{ord}(q_n,Y)-\text{ord}(q_n,T(Y))$$

$$=l(q_n)$$

$$=|\{q^{\prime }\in E(Y):\text{there is an edge }f=\overline{q_n{q}^{\prime }}\subset Y\}|,$$

$$H(A)=h(A\cap ((X\setminus l_p)\cup \{p\}))\cup g(f^{-1}(A\cap l_p)),\text{ for each A∈C(p,X)}$$

$$K(X)=\{C(x,X):x\in X\}.$$

Full text

††journal: Elsevier

1 Introduction

A continuum is a nonempty compact connected metric space. Given a continuum $X$, by a hyperspace of $X$ we mean a specified collection of subsets of $X$. In the literature, some of the most studied hyperspaces are the following:

[TABLE]

The collection $2^{X}$ is called the hyperspace of closed subsets of $X$ whereas that $C(X)$ is called the hyperspace of subcontinua of $X$. These hyperspaces are considered with the Hausdorff metric, see [11, p. 1].

Given a finite collection $K_{1},\dots,K_{r}$ of subsets of $X$, $\left\langle K_{1},\dots,K_{r}\right\rangle$, denotes the following subset of $2^{X}$:

[TABLE]

It is known that the family of all subsets of $2^{X}$ of the form $\left\langle K_{1},\dots,K_{r}\right\rangle$, where each $K_{i}$ is an open subset of $X$, forms a basis for a topology for $2^{X}$ (see [11, Theorem 0.11, p. 9]) called the Vietoris Topology. The Vietoris topology and the topology induced by the Hausdorff metric coincide (see [11, Theorem 0.13, p. 10]). The hyperspaces $C(X)$, $F_{n}(X)$ and $C(p,X)$ are considered as subspaces of $2^{X}$.

The topological structure of the hyperspaces $C(p,X)$ has been recently studied, for example, in [1], [2], [10], [13], [14], and [15], being useful to characterize classes of continua.

In general, given a hyperspace $\mathcal{H}(X)\in\{2^{X},C(X),F_{n}(X)\}$, we say that $X$* has unique hyperspace $\mathcal{H}(X)$* if for each continuum $Y$ such that $\mathcal{H}(X)$ is homeomorphic to $\mathcal{H}(Y)$, it holds that $X$ is homeomorphic to $Y$. This concept has served to characterize classes of continua through the structure of their hyperspaces, see for example [3], [4], [6] and [7].

In a similar setting, given a family of continua $\mathcal{C}$, a continuum $X\in\mathcal{C}$ and $p\in X$, we say that $(X,p)$ has unique hyperspace $C(p,X)$ in (or relative to) $\mathcal{C}$ if for each $Y\in\mathcal{C}$ and $q\in Y$ such that $C(p,X)$ is homeomorphic to $C(q,Y)$, there is an homeomorphism $\mathfrak{h}:X\to Y$ such that $\mathfrak{h}(p)=q$ (compare with the continua $\mathcal{C}$-determined given in [5, p. 12] or [9]). The main goal in this paper is to show that for each tree $X$ and $p\in X$, $(X,p)$ has unique hyperspace $C(p,X)$ in the class of trees (see Theorem 4.14).

The following examples show that the uniqueness of hyperspace $C(p,X)$ in the general setting is false, being our results the best that can be obtained in the class of trees.

Example 1.1**.**

In [13, Lemma 3.19] it was proved that a continuum $Y$ is hereditarily indecomposable if and only if for each $q\in Y$, $C(q,Y)$ is an arc. Also in [13, Theorem 3.17] it was proved that if $X$ is an arc and $p\in E(X)$, $C(p,X)$ is an arc. It follows that $(X,p)$ has no unique hyperspace in the class of continua.

Example 1.2**.**

By using [13, Observation 5.18 and Theorem 5.21], is easy to see that the Knaster continuum $Z$ with two end points $a$ and $b$ (see [8, p. 205]) satisfy that $C(a,Z)$ and $C(b,Z)$ are arcs and $C(q,Z)$ is a 2-cell for each $q\in Z\setminus\{a,b\}$. This continuum shows that if $X$ is an arc and $p\in X$, then $(X,p)$ has no unique hyperspace in the class of continua.

Example 1.3**.**

*If $X$ is a tree and $p\in X$, then there exist a continuum $Y$ and $q\in Y$, such that $Y$ is not a tree and $C(p,X)$ is homeomorphic to $C(q,Y)$.

To get such a continuum, let $Z$ be a Knaster continuum with two end points $a$ and $b$ and let $T$ be a simple triod with vertex $p\in T$. Choose an end point $x\in T$ and attach $Z$ to $T$ by identifying $x$ with $a$. Let $Y$ be the space we get, $q:=[p]\in Y$. It is not difficult to see that $C(p,T)$ is homeomorphic to $C(q,Y)$. This shows that trees have no unique hyperspace $C(p,X)$ in the class of continua.*

Example 1.4**.**

If $X$ is an arc with end points $a$ and $b$, and $Y$ is a simple closed curve, then $C(p,X)$ is homeomorphic to $C(q,Y)$ for each $p\in X\setminus\{a,b\}$ and $q\in Y$. This shows that if $p\in X\setminus\{a,b\}$ then $(X,p)$ has no unique hyperspace in the class of finite graphs.

2 Definitions and preliminaries

By a finite graph we mean a continuum $X$ which can be written as the union of finitely many arcs, any two of which are either disjoint or intersect only in one or both of their end points. A tree is a finite graph without simple closed curves. Given a positive integer $n$, a simple $n$-od is a finite graph, denoted by $T_{n}$, which is the union of $n$ arcs emanating from a single point, $v$, and otherwise disjoint from each another. The point $v$ is called the vertex of the simple $n$-od. A simple $3$-od, $T_{3}$, will be called a simple triod. A $n$-cell, $I^{n}$, is any space homeomorphic to $[0,1]^{n}$. For a subset $A$ of $X$ we denote by $|A|$ the cardinality of $A$ and we use the symbols $\textrm{int}(A)$ and $\overline{A}$ to denotes the interior and the clousure of $A$ in $X$, respectively.

Given a finite graph $X$, $p\in X$ and a positive integer $n$, we say that $p$ is of order $n$ in $X$, denoted by ord$(p,X)=n$, if $p$ has a closed neighborhood which is homeomorphic to a simple $n$-od having $p$ as the vertex. If ord$(p,X)=1$ the point $p$ is called an end point of $X$. The set of all end points of $X$ will be denoted by $E(X)$. If ord$(p,X)=2$ the point $p$ is called an ordinary point of $X$. The set of all ordinary points of $X$ will be denoted by $O(X)$. A point $p\in X$ is a ramification point of $X$ if ord$(p,X)\geq 3$. The set of all ramification points of $X$ will be denoted by $R(X)$. The vertices of a finite graph $X$ will be the end points and the ramification points of $X$, we denote by $V(X)$ the set of all vertices of $X$. An edge will be an arc joining two elements of $V(X)$ and containing no more than two points of that set; the set of edges of $X$ will be denoted by $\mbox{edge}(X)$.

A subtree $Y$ of the tree $X$ is a subcontinuum of $X$ such that if $e\in\mbox{edge}(Y)$ then $e\in\mbox{edge}(X)$, i.e., every edge in $Y$ should be an edge in $X$, it is avoid to use only a part of an edge of $X$ when constructing a subtree $Y$ of $X$.

In this paper, dimension means inductive dimension as defined in [11, (0.44), p. 21]. The symbol $\dim(X)$ will be used to denote the dimension of the space $X$.

3 About the cells in $C(p,X)$

From now on $X$ will denote a tree and $p\in X$.

Given $n\in\mathbb{N}$, we denote by $U_{n}(X)$ the following set

[TABLE]

we set $\mathcal{U}(X)=\bigcup\limits_{n\in\mathbb{N}}U_{n}(X)$, and as customary, $\pi_{0}(\mathcal{U}(X))$ denote the set of connected components of $\mathcal{U}(X)$.

Given a tree $X$, which is not an arc nor a simple $n$-od, we consider the subcontinuum

[TABLE]

For each $x\in E(X)$ there is exactly one point $r_{x}\in R(X)$ such that the arc joining $x$ with $r_{x}$, $\overline{xr_{x}}$, is an edge of $X$. With this notation the continuum $T(X)$ can be described as the closure of

[TABLE]

The key idea is to remove all arcs incident in any end point of $X$ (see the picture below).

It is clear that the following statements hold.

$T(X)$ is a subtree of $X$ and $R(T(X))\subset R(X)\subset T(X)$.

• 2.

$x\in R(X)$ is stills a ramification point of $T(X)$ if and only if there is at least three different elements $y,z,w\in R(X)$ such that the arcs $\overline{yx},\overline{zx}$ and $\overline{wx}$ are edges of $X$.

• 3.

For each $z\in E(T(X))$ holds

[TABLE]

[TABLE]

Let $\mbox{Sub}_{p}(T(X))$ the collection of all subtrees of $T(X)$ containing $p$. One of the main goals of this section is to establish a one to one correspondence between $\mbox{Sub}_{p}(T(X))$ and the components of $\mathcal{U}(X)$. In order to achieve this, let $Y\in\mbox{Sub}_{p}(T(X))$ and for each $e\in\mbox{edge}(X)\setminus\mbox{edge}(Y)$ such that $e\cap Y\neq\emptyset$, the intersection $e\cap Y$ consist of exactly one point, so we can pick a homeomorphism $h_{e}:[0,1]\rightarrow e$ in such a way that $h_{e}(0)\in Y$.

We define the following set of $C(p,X)$

[TABLE]

Observe that $U_{Y}\subset C(Y,X):=\{A\in C(X):Y\subset A\}$. On the other hand, notice that if the set $\{e\in\mbox{edge}(X)\setminus\mbox{edge}(Y),\ e\cap Y\neq\emptyset\}$ is numbered as $\{e_{1},\dots,e_{k}\}$ then $U_{Y}=\{Y\cup A\in C(X):A\in\langle\textrm{int}(e_{1}),\dots,\textrm{int}(e_{k})\rangle\}$, thus $U_{Y}$ is homeomorphic to $\prod^{k}_{i=1}\textrm{int}(e_{i})$; these two facts shows the following result.

Proposition 3.1**.**

Let $Y\in\textup{Sub}_{p}(T(X))$, $A\in U_{Y}$ and $n=|E(A)|$. Then

1. a)

$n=|\{e\in\textup{edge}(X)\setminus\textup{edge}(Y):e\cap Y\neq\emptyset\}|$* and $U_{Y}\subset U_{n}(X)$.* 2. b)

For each $B\in U_{Y}$, $Y=\bigcup\{A\in\textup{Sub}_{p}(T(X)):A\subset B\}$, consequently, $Y$ is the largest element in $\textup{Sub}_{p}(T(X))$ contained in $B$. 3. c)

If $Y^{\prime}\in\textup{Sub}_{p}(T(X))$ is such that $U_{Y}\cap U_{Y^{\prime}}\neq\emptyset$, then $Y=Y^{\prime}$. 4. d)

$U_{Y}$* is open and connected in $\mathcal{U}(X)$.*

Next we establish some consequences of the previous proposition.

According to (c) and (d), given $Y\in\textup{Sub}_{p}(T(X))$, $U_{Y}$ is a component of $\mathcal{U}(X)$, therefore we get a function $\phi_{X}:\textup{Sub}_{p}(T(X))\to\pi_{0}(\mathcal{U}(X))$, given by $\phi_{X}(Y)=U_{Y}$.

Corollary 3.2**.**

The function $\phi_{X}$ is bijective.

Proof.

By (c) from Proposition 3.1, we have that $\phi_{X}$ is one to one. In order to prove that $\phi_{X}$ is surjective, it is enough to see that for each $B\in\mathcal{U}(X)$ there exists $Y\in\textup{Sub}_{p}(T(X))$ such that $B\in U_{Y}$. Let $Y=\bigcup\{A\in\textup{Sub}_{p}(T(X)):A\subset B\}$. It is clear that $Y\in C(p,X)\cap\textup{Sub}_{p}(T(X))$ and $B\in U_{Y}$. ∎

Since $U_{Y}$ is a component of $\mathcal{U}(X)$ there exists $n\in\mathbb{N}$ such that $U_{Y}\subset U_{n}(X)$, thus $\dim(U_{Y})=n$. We have the following monotony property of $\phi_{X}$.

Corollary 3.3**.**

Let $Y,Y^{\prime}\in\textup{Sub}_{p}(T(X))$. If $Y\subset Y^{\prime}$ then $\dim(\phi_{X}(Y))\leq\dim(\phi_{X}(Y^{\prime}))$.

Proof.

Using (a) from Proposition 3.1, we get

[TABLE]

∎

Corollary 3.4**.**

The following holds

[TABLE]

Moreover $U_{\textup{ord}(p,X)}(X)=U_{\{p\}}$ and $U_{|E(X)|}(X)=U_{T(X)}$.

Proof.

The right-hand side equalities are trivial from previous corollary. The left-hand side equalities are also an easy consequence of (a) from Proposition 3.1. ∎

4 Unique hyperspace $C(p,X)$ of trees

We will denote by $\mathcal{T}$ the class of trees, by $\mathcal{I}$ the class of arcs, by $\mathcal{N}$ the class of simple $n$-ods and by $\widehat{\mathcal{T}}=\mathcal{T}\setminus(\mathcal{I}\cup\mathcal{N})$.

The following result is a slight modification of [1, Corollary 3.4, p. 45], and its proof is essentially the same.

Proposition 4.1**.**

Let $X,Y\in\mathcal{T}$ and $p\in X$, $q\in Y$. If $C(p,X)$ is homeomorphic to $C(q,Y)$, then $\textup{ord}(p,X)=\textup{ord}(q,Y)$.

Definition 4.2**.**

Let $n\in\mathbb{N}$. A continuum $Y$ is an $n$-od if there exists $K\in C(Y)$ such that $Y\setminus K$ has at least $n$ components. Further, we will say that $K$ is a core of the $n$-od. If $n=3$, $Y$ is called a triod.

Theorem 4.3**.**

If $X\in\mathcal{I}$ and $p\in X$, then $(X,p)$ has unique hyperspace in $\mathcal{T}$.

Proof.

If $X$ is an arc with end points $a$ and $b$ and $p\in X$, then $C(p,X)$ is an arc if $p\in\{a,b\}$ and $C(p,X)$ is a two cell if $p\in X\setminus\{a,b\}$ (see [13, Theorem 3.17, p. 265]). By [13, Lemma 3.15, p. 264], if $Y$ is a continuum and $q\in Y$ is such that $C(q,Y)$ is an arc or a two cell, then $q$ is not in a core of a triod in $Y$, thus in the case that $Y\in\mathcal{T}$, $Y$ must be an arc. By Proposition 4.1, is easy to see that $C(p,X)$ homeomorphic to $C(q,Y)$ implies that there exists an homeomorphism $h:X\to Y$ sending $p$ to $q$. ∎

Theorem 4.4**.**

If $X\in\mathcal{N}$ and $p\in X$, then $(X,p)$ has unique hyperspace in $\mathcal{T}$.

Proof.

Let $p\in X$ and suppose that $Y\in\mathcal{T}$ and $q\in Y$ is such that $C(p,X)$ is homeomorphic to $C(q,Y)$. We consider the following two cases.

Case 1. If $p\in R(X)$. By Proposition 4.1, $q\in R(Y)$ and $\textrm{ord}(p,X)=\textrm{ord}(q,Y)$. If $n=\textrm{ord}(p,X)$, then $C(p,X)$ and $C(q,Y)$ are $n$-cells, thus $Y$ is a simple $n$-od with vertex $q$.

Case 2. If $q\in E(X)\cup O(X)$. By Proposition 4.1, $q\in E(Y)\cup O(X)$. Consider $l_{p}$ and $l_{q}$ the edges in $X$ and $Y$, containing $p$ and $q$, respectively. Let $x\in(V(X)\setminus\{p\})\cap l_{p}$ and $y\in(V(Y)\setminus\{q\})\cap l_{q}$. By [1, Proposition 3.7, p. 45], we have that $\textrm{ord}(x,X)=\textrm{ord}(y\in Y)$. If $Y$ contains a point $r\in R(Y)\setminus\{y\}$, then $C(q,Y)$ contains an $(\textrm{ord}(q,Y)+\textrm{ord}(r,Y))$-cell, but this in impossible because $\dim(C(q,Y))=\dim(C(p,X))=\textrm{ord}(x,X)=\textrm{ord}(y,Y)$. Therefore, $Y$ is an simple $\textrm{ord}(x,X)$-od.

In both cases the existence of an homeomorphism $h:X\to Y$ sending $p$ to $q$ is trivial. ∎

Notation. Let $X\in\widehat{\mathcal{T}}$. Suppose that $G\in\textup{Sub}_{p}(T(X))$ consists of the edges $e_{1},\dots,e_{k}$ and let $f_{1},\ldots,f_{m}$ be the edges of $X\setminus G$ incident on $G$. We will use the notation

[TABLE]

to describe the component $\phi(G)$ of $\mathcal{U}(X)$.

We will need the following lemma:

Lemma 4.5**.**

Let $X\in\widehat{\mathcal{T}}$. Let $G=e_{1}\cup\ldots\cup e_{k}\subset T(X)$ be a collection of edges and let $f_{1},\ldots,f_{m}$ be the edges of $X\setminus G$ incident on $G$. Then $m\geq 2$.

Proof.

By induction on $k$. For $k=1$, let $e=\overline{p_{1}p_{2}}$, we can assume $p_{2}\neq p$. Then ord$(p_{2},X)\geq 3$, since all the vertices $T(X)$ are ramification points; in other words, there are at least two incident edges on $e=\overline{p_{1}p_{2}}$.

Next, assume the lemma holds for $k$ and consider $G=e_{1}\cup\ldots\cup e_{k}\cup e_{k+1}$, since $T(X)$ is a tree we can choose, without loss of generality, $e_{k+1}$ such that it intersects at most one of $e_{1},\ldots,e_{k}$. Let $e_{k+1}=f_{1},\ldots,f_{m}$ be the edges incident on $G$, by hypothesis $m\geq 2$. Furthermore let the point $p_{2}$ be the extreme of $e_{k+1}$ which is not in $G$, once again we have ord$(p_{2},X)\geq 3$, that is, the edges incident on $G^{\prime}=e_{1}\cup\ldots\cup e_{k}\cup e_{k+1}$ are $m-1+\textrm{ord}(p_{2},X)-1\geq m+1\geq 2$. This finishes the induction and the proof of the lemma.

∎

Lemma 4.6**.**

Let $X\in\widehat{\mathcal{T}}$. Let $G,G^{\prime}\in\textup{Sub}_{p}(T(X))$, suppose there is an $e\in\textup{edge}(G)$ such that $e\cap G^{\prime}=\emptyset$ (i.e. $e$ is not an edge of $G^{\prime}$ nor it is incident on $G^{\prime}$). Then $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}=\emptyset$.

Proof.

Let $f_{1},\ldots,f_{m}$ be the edges of $X\setminus G$ incident on $G$, similarly, let $f^{\prime}_{1},\ldots,f^{\prime}_{m^{\prime}}$ be the edges of $X\setminus G^{\prime}$ incident on $G^{\prime}$. Note that an element $K^{\prime}\in\overline{U}_{G^{\prime}}$ is of the form $K^{\prime}=G^{\prime}\cup\bigcup^{m^{\prime}}_{i=1}\{A_{i}:A_{i}\in C(f^{\prime}_{i})\}$, similarly an element $K\in\overline{U}_{G}$ is of the form $K=G\cup\bigcup^{m}_{j=1}\{A_{j}:A_{j}\in C(f_{j})\}$, therefore $e\subset K$, however there is no way $e\subset K^{\prime}$, since $e$ is not an edge of $G^{\prime}$ nor it is incident on $G^{\prime}$. Hence $\overline{U}_{G}$ and $\overline{U}_{G^{\prime}}$ cannot have elements in common. ∎

The case when all the edges of $G$ are edges of $G^{\prime}$ or are incident on $G^{\prime}$ and vice versa is covered in the following:

Proposition 4.7**.**

Let $X\in\widehat{\mathcal{T}}$. Consider $G$, $G^{\prime}\in\textup{Sub}_{p}(T(X))$ and let

$e_{1},\ldots,e_{k}$* be the edges of $G\cap G^{\prime}$.*

• 2.

$e_{k+1},\ldots,e_{k+l}$* be the edges of $G$ which are not in $G^{\prime}$.*

• 3.

$e^{\prime}_{k+1},\ldots,e^{\prime}_{k+l^{\prime}}$* be the edges of $G^{\prime}$ which are not in $G$.*

• 4.

$d_{1},\ldots,d_{n}$* be the edges of $X\setminus(G\cup G^{\prime})$ incident on $G\cap G^{\prime}$.*

• 5.

$f_{1},\ldots,f_{m}$* be the edges of $X\setminus(G\cup G^{\prime})$ incident on $G$ but not on $G^{\prime}$.*

• 6.

$f^{\prime}_{1},\ldots,f^{\prime}_{m^{\prime}}$* be the edges of $X\setminus(G\cup G^{\prime})$ incident on $G^{\prime}$ but not on $G$.*

Suppose that $e^{\prime}_{k+1},\ldots,e^{\prime}_{k+l^{\prime}}$ are incident on $G$ and $e_{k+1},\ldots,e_{k+l}$ are incident on $G^{\prime}$. Then

$\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$. 2. 2.

$\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=n$.

Proof.

Note that, $G$ consists of the edges $e_{1},\ldots,e_{k},\ldots e_{k+l}$ and the edges incident on $G$ are $d_{1},\ldots,d_{n},e^{\prime}_{k+1}\ldots e^{\prime}_{k+l^{\prime}},f^{\prime}_{1},\ldots,f^{\prime}_{m^{\prime}}$. Similarly for $G^{\prime}$, thus,

[TABLE]

The following pictures should make clear the above notation.

[TABLE]

In the case of the above subtrees, the edges are as follows:

[TABLE]

Therefore, $\dim(U_{G})$= $n+l^{\prime}+m$ and $\dim(U_{G^{\prime}})$= $n+l+m^{\prime}$. Moreover, it can be seen from this description that $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}$ consists of elements of the form $K=G\cup G^{\prime}\cup\bigcup^{n}_{k=1}\{A_{k}:A_{k}\in C(d_{k})\}$. Hence $\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=n$. ∎

As an important consequence of the description given above, we have:

Proposition 4.8**.**

Let $X\in\widehat{\mathcal{T}}$. Consider $G$, $G^{\prime}\in\textup{Sub}_{p}(T(X))$ such that $G^{\prime}=G\cup e$, for some $e\in\textup{edge}(T(X))$. Then

$\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$. 2. 2.

$\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=\dim(U_{G})-1$. 3. 3.

$\dim(U_{G^{\prime}})=\dim(U_{G})+\textup{ord}(p^{\prime},X)-2$, where $p^{\prime}$ is the vertex of $e$ which is not in $G$.

And vice versa, if $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$ and $\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=\dim(U_{G})-1$, then there exists an unique $e\in\textup{edge}(T(X))$ such that $G^{\prime}=G\cup e$.

Proof.

Using the notation of the previous proposition we have in this case that $G=G\cap G^{\prime}=\{e_{1},\ldots,e_{k}\}$, $e=e^{\prime}_{k+1}$ is the only edge of $G^{\prime}$ incident on $G$, $d_{1},\ldots,d_{n}$ are the edges incident on $G$, there are no $f$’s and $e,f^{\prime}_{1},\ldots,f^{\prime}_{m^{\prime}}$ are the edges incident on $p^{\prime}$. Hence,

[TABLE]

From the previous it can be seen that $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}$ consists of elements of the form $K=G^{\prime}\cup\bigcup^{n}_{i=1}\{A_{i}:A_{i}\in C(d_{i})\}$. Thus, $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$ and $\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=n$.

Moreover,

[TABLE]

This concludes the first part.

For the converse, first observe that by Lemma 4.6 and the condition $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$, all the edges of $G\setminus G^{\prime}$ are incident on $G^{\prime}$ and all the edges of $G^{\prime}\setminus G$ are incident on $G$; thus satisfying the condition of the previous proposition. Then the condition $\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=\dim(U_{G})-1$ translates to $n=n+l^{\prime}+m-1$, that is, $l^{\prime}+m=1$. We only have two possibilities $(l^{\prime},m)=(1,0)$ or $(l^{\prime},m)=(0,1)$. The first case is precisely when $e=e^{\prime}_{k+1}$ is the only edge of $G^{\prime}$ which is not in $G$, and $m=0$ implies that $G\subset G^{\prime}$. The second case, when $m=1$, is not possible due to Lemma 4.5. Therefore the only possibility is $(l^{\prime},m)=(1,0)$, so $G^{\prime}=G\cup e$. Since $G^{\prime}$ is given, then the edge $e$ is unique. ∎

Definition 4.9**.**

Let $p,p^{\prime}\in X$ be vertices in $X\in\mathcal{T}$, we will denote by $\overline{pp^{\prime}}$ the smallest connected subtree containing $p$ and $p^{\prime}$.

Proposition 4.10**.**

Let $X,Y\in\mathcal{T}$, $p\in X$ and $q\in Y$. If $h:C(p,X)\rightarrow C(q,Y)$ is an homeomorphism then for each vertex $p^{\prime}\in T(X)$ there exists a unique $q^{\prime}\in T(Y)$ such that $h(U_{\overline{pp^{\prime}}})=U_{\overline{qq^{\prime}}}$. Moreover, if $\overline{pp^{\prime}}$ consists of the edges $e_{i}=\overline{p_{i-1}p_{i}}$, $i=1,\ldots,n$, with $p=p_{0}$ and $\overline{qq^{\prime}}$ consists of the edges $f_{i}=\overline{q_{i-1}q_{i}}$, $i=1,\ldots,m$, with $q=q_{0}$, then $n=m$ and $h(U_{\overline{pp_{i}}})=U_{\overline{qq_{i}}}$ for $i=0,1,\ldots,n$.

Proof.

By induction on $n$. The case $n=0$ follows straightforward from Corollary 3.4. The case $n=1$ is covered by Proposition 4.8, more precisely, consider an edge $e=\overline{pp^{\prime}}\subset T(X)$ and let $G=\{p\}$ and $G^{\prime}=G\cup e$, then $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$ and $\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=\dim(U_{G})-1$. These conditions are preserved under the homeomorphism $h$, that is, $h(\overline{U}_{G})\cap h(\overline{U}_{G^{\prime}})\neq\emptyset$ and $\dim(h(\overline{U}_{G})\cap h(\overline{U}_{G^{\prime}}))=\dim(h(U_{G}))-1$ and we know $h(U_{G})=U_{\{q\}}$. Since $\phi_{Y}$ is bijective there exists $F^{\prime}\in\textup{Sub}_{p}(T(Y))$ such that $\phi_{Y}(F^{\prime})=U_{F^{\prime}}=h(U_{G^{\prime}})$. By the second part of Proposition 4.8 applied to $U_{\{q\}}$ and $U_{F^{\prime}}$, there exists an unique edge $f^{\prime}\in\textup{edge}(T(Y))$ such that $F^{\prime}=\{q\}\cup f^{\prime}$, in other words, there is a unique vertex $q^{\prime}\in T(X)$ and an edge $f^{\prime}=\overline{qq^{\prime}}\subset T(Y)$ such that $h(U_{\overline{pp^{\prime}}})=U_{\overline{qq^{\prime}}}$.

Now suppose the proposition holds for $n$ and consider the subtree $\overline{pp^{\prime}}$ having edges $e_{i}=\overline{p_{i-1}p_{i}}$, $i=1,2,\ldots,n+1$. Once again we rely completely on Proposition 4.8, let $G=e_{1}\cup\ldots\cup e_{n}$, $G^{\prime}=G\cup e_{n+1}$, then $\overline{U}_{G}\cap\overline{U}_{G^{\prime}}\neq\emptyset$ and $\dim(\overline{U}_{G}\cap\overline{U}_{G^{\prime}})=\dim(U_{G})-1$. These conditions are preserved under the homeomorphism $h$, that is, $h(\overline{U}_{G})\cap h(\overline{U}_{G^{\prime}})\neq\emptyset$ and $\dim(h(\overline{U}_{G})\cap h(\overline{U}_{G^{\prime}}))=\dim(h(U_{G}))-1$ and we know by hypothesis of induction that $h(U_{\overline{pp_{i}}})=U_{\overline{qq_{i}}}$, for $i=0,1,\ldots,n$; in particular, $h(U_{G})=U_{\overline{qq_{n}}}$. As above, let $F^{\prime}\in\textup{Sub}_{p}(T(Y))$ such that $U_{F^{\prime}}=h(U_{G^{\prime}})$, by Proposition 4.8 there exists an edge $f^{\prime}\in\textup{edge}(T(Y))$ such that $F^{\prime}=\overline{qq_{n}}\cup f^{\prime}$, in other words, there is a unique vertex $q^{\prime}=q_{n+1}\in T(X)$ and an edge $f^{\prime}=\overline{q_{j}q_{n+1}}\subset T(Y)$ (for some $j=0,1,\ldots,n$) such that $h(U_{\overline{pp^{\prime}}})=U_{\overline{qq^{\prime}}}$. Now, it is left to prove that $j=n$. This is a consequence of Lemma 4.6, indeed, for $i=0,1,2,\ldots,n-1$, let $G_{i}=\{p_{0}\}\cup e_{1}\cup\ldots\cup e_{i}$ and $G^{\prime}=\{p_{0}\}\cup e_{1}\cup\ldots\cup e_{n}\cup e_{n+1}$. Note that $e_{n+1}$ is not incident on $G_{i}$, then $\overline{U}_{G_{i}}\cap\overline{U}_{G^{\prime}}=\emptyset$, this condition is preserved under $h$, that is, $\overline{U}_{\overline{qq_{i}}}\cap h(\overline{U}_{G^{\prime}})=h(\overline{U}_{G_{i}})\cap h(\overline{U}_{G^{\prime}})=\emptyset$ (for $i=0,1,\ldots,n-1$) this means that the edge $f^{\prime}=\overline{q_{j}q_{n+1}}$ is not incident on any of $q_{0},\ldots,q_{n-1}$ and therefore it must be incident on $q_{n}$ and then $h(U_{\overline{pp_{i}}})=U_{\overline{qq_{i}}}$, for each $i=0,1,\ldots,n+1$. ∎

Finally we put all the pieces together to prove one of the main theorems of this paper.

Theorem 4.11**.**

Let $X,Y\in\mathcal{T}$, $p\in R(X)$ and $q\in Y$. If $h:C(p,X)\rightarrow C(q,Y)$ is an homeomorphism then there exists an homeomorphism $\mathfrak{h}:X\rightarrow Y$ with $\mathfrak{h}(p)=q$.

Proof.

We will prove the following stronger statement: there is an isomorphism $\mathfrak{h}:X\rightarrow Y$ such that $\mathfrak{h}(p)=q$ (this implies the existence of the desired homeomorphism). The Proposition 4.10 allows us to define a function $\mathfrak{h}:V(T(X))\rightarrow V(T(Y))$ given by $\mathfrak{h}(p^{\prime})=q^{\prime}$ if $h(U_{\overline{pp^{\prime}}})=U_{\overline{qq^{\prime}}}$. The same construction applied to $h^{-1}$ provides us with its inverse, therefore $\mathfrak{h}:V(T(X))\rightarrow V(T(Y))$ is a bijection. It is left to prove that it preserves edges. Formally, consider an edge $e=\overline{p^{\prime}p^{\prime\prime}}\subset T(X)$ and without lost of generality assume that the path $\overline{pp^{\prime\prime}}$ consists of edges $e_{i}=\overline{p_{i-1}p_{i}}$, $i=1,\ldots,n$, with $p=p_{0}$, $p^{\prime}=p_{n-1}$ and $p^{\prime\prime}=p_{n}$, then by Proposition 4.10, we know that the path $\overline{qq^{\prime\prime}}$ consists of the edges $f_{i}=\overline{q_{i-1}q_{i}}$, $i=1,\ldots,n$, with $q_{i}=\mathfrak{h}(p_{i})$ and $q=q_{0}$, $q^{\prime}=q_{n-1}$ and $q^{\prime\prime}=q_{n}$. Hence, there is an edge between $q^{\prime}=\mathfrak{h}(p^{\prime})$ and $q^{\prime\prime}=\mathfrak{h}(p^{\prime\prime})$.

Finally we need to prove that $\mathfrak{h}:V(T(X))\rightarrow V(T(Y))$ can be extended to all of $V(X)$. Indeed, consider a point $p^{\prime}\in E(X)$ then there is an edge $e=\overline{p_{n}p^{\prime}}$ with $p_{n}\in T(X)$. Let $l(p_{n})=|\{p^{\prime}\in E(X):\textrm{there is an edge }e=\overline{p_{n}p^{\prime}}\subset X\}|$, we will prove $l(p_{n})=l(\mathfrak{h}(p_{n}))$. Consider the path $\overline{pp_{n}}$ consisting of edges $e_{i}=\overline{p_{i-1}p_{i}}$, $i=1,\ldots,n$, with $p=p_{0}$. We can count $l(p_{n})$ as follows:

[TABLE]

where in the second line we have used part 3 of Proposition 4.8 for $G=\overline{pp_{n-1}}$ and $G^{\prime}=\overline{pp_{n}}$. Next, it is already known that $\mathfrak{h}:V(T(X))\rightarrow V(T(Y))$ is a bijection preserving edges, which, together with $h$ being a homeomorphism and preserving dimensions implies that

[TABLE]

where $q_{n-1}=\mathfrak{h}(p_{n-1})$ and $q_{n}=\mathfrak{h}(p_{n})$. Thus we have established the existence of a bijection between the sets $\{p^{\prime}\in E(X):\textrm{there is an edge }e=\overline{p_{n}p^{\prime}}\}$ and $\{q^{\prime}\in E(Y):\textrm{there is an edge }f=\overline{q_{n}q^{\prime}}\}$, by the construction this bijection preserves incidence relations. Therefore we get a full bijection $\mathfrak{h}:V(X)\rightarrow V(Y)$ preserving the edge relations of the graphs and with $\mathfrak{h}(p)=q$. ∎

To complete our discussion now we will extend the previous result to the case when $p\in O(X)\cup E(X)$. In order to do so we need the following lemma.

Lemma 4.12**.**

Let $X,Y\in\mathcal{T}$, $p\in R(X)$ and $q\in R(Y)$. If there exists edges $l_{p}$ in $X$ and $l_{q}$ in $Y$, containing $p$ and $q$, respectively, such that $l_{p}\cap E(X)\neq\emptyset$, $l_{q}\cap E(Y)\neq\emptyset$ and $C(p,(X\setminus l_{p})\cup\{p\})$ is homeomorphic to $C(q,(Y\setminus l_{q})\cup\{q\})$, then $C(p,X)$ is homeomorphic to $C(q,Y)$.

Proof.

Let $h:C(p,(X\setminus l_{p})\cup\{p\})\to C(q,(Y\setminus l_{q})\cup\{q\})$ be a homeomorphism. Take $f:[0,1]\to l_{p}$ and $g:[0,1]\to l_{q}$ two homeomorphisms such that $f(0)=p$ and $g(0)=q$. It is easy to see that the map $H:C(p,X)\to C(q,Y)$ given by

[TABLE]

is a homeomorphism. ∎

Theorem 4.13**.**

Let $X\in\mathcal{T}$ and $p\in O(X)\cup E(X)$. Then $(X,p)$ has unique hyperspace $C(p,X)$ in $\mathcal{T}$.

Proof.

By Theorems 4.3 and 4.4, we can suppose that $X\in\widehat{\mathcal{T}}$. Let $p\in O(X)$ and suppose that $Y\in\mathcal{T}$ and $q\in Y$ is such that $C(p,X)$ is homeomorphic to $C(q,Y)$. By Proposition 4.1, $\textrm{ord}(X,p)=\textrm{ord}(Y,q)$. Since $X$ can be embedded into the euclidean space $\mathbb{R}^{3}$, suppose that $X$ is a subspace of $\mathbb{R}^{4}$ such that $X\subset\mathbb{R}^{3}\times\{0\}$ and $p=(0,0,0,0)$. Similarly for $Y$. Consider $X^{\prime}=X\cup((0,0,0)\times[0,1])$ and $Y^{\prime}=Y\cup((0,0,0)\times[0,1])$. By Lemma 4.12, $C(p,X^{\prime})$ is homeomorphic to $C(q,Y^{\prime})$ and using Theorem 4.11, we obtain that there exists an homeomorphism between $X^{\prime}$ and $Y^{\prime}$ sending $p$ to $q$, thus $X$ and $Y$ are homeomorphic.

The case when $p\in E(X)$ reduces to the case when $p\in O(X)$ by attaching one arc at $p$ and one arc at $q$ becoming these ordinary points. ∎

Finally we state the main result of this paper which summarizes Theorems 4.11 and 4.13.

Theorem 4.14**.**

Let $X\in\mathcal{T}$ and $p\in X$. Then $(X,p)$ has unique hyperspace $C(p,X)$ in the class $\mathcal{T}$.

The next is an application of the previous theorem.

Given a continuum $X$, we consider the hyperspace of $C(C(X))$,

[TABLE]

In [1], it defines in $K(X)$ the following equivalence relation, $C(p,X)\sim C(q,X)$ if and only if $C(p,X)$ is homeomorphic to $C(q,X)$. Given a positive integer $n$ and a continuum $X$, it is say that $K(X)$* has size $n$* if the quotient $K(X)/\sim$ has cardinality $n$. In [1], the authors shown that in the class of graphs $X$, the size of $K(X)$ can be different of the homogeneity degree of $X$. As a concequence of Theorem 4.14, in the following result, we obtain a partial solution of [1, Problem 5.4, p. 49].

Corollary 4.15**.**

Let $X$ be a tree. Then the size of $K(X)$ is equal to the homogeneity degree of $X$.

To finish this paper we present some open problems.

Question 4.16**.**

If $X$ is a dendrite and $p\in X$, has $(X,p)$ unique hyperspace $C(p,X)$ in the class of dendrites?

Question 4.17**.**

If $X$ is a dendroid and $p\in X$, has $(X,p)$ unique hyperspace $C(p,X)$ in the class of dendroids?

Question 4.18**.**

Regarding Example 1.3, are there a continuum $X$ and $p\in X$ such that $(X,p)$ has unique hyperspace in the class of continua?

Question 4.19**.**

Are there a continuum $X$ and two points $p,q\in X$ such that $(X,p)$ has unique hyperspace $C(p,X)$, whereas $(X,q)$ does not has it?

5 Acknowledgements

The authors wish to thank to Hugo Villanueva, Eli Vanney Roblero and Sergio Guzmán for the fruitful discussions about the topic of this paper.