this intuition we could find the stable probes of Section 5.2.1, see Fig. 5.
This notwithstanding, finding stable probes in our model was a nontrivial
task, due to the fact that we had to take into account also the presence of
hypermultiplets, which give mass to one of the vectors, and this made the
numerical work challenging.

Figure 13: Example of fluxed D6 brane potential for the spherical AdS4
solution from the $S^{7}$ truncation. The plot is for the parameters
$l_{AdS}=1/2$, $T=0.05(\text{blue}),0.11(\text{orange}),0.38(\text{green})$,
with probe charge $\kappa=0.44$. For each of the solutions we have shifted the
radial coordinate in such a way that the horizon is at $r=0$.

As a last remark, we point out that there exist also purely magnetic solutions
with a regular supersymmetric limit [86, 39, 87] and without axions. The
solution can have spherical, planar and hyperbolic horizons, and reads

$U^{2}(r)=\frac{1}{\sqrt{H_{0}H_{1}H_{2}H_{3}}}\left(2g^{2}r^{2}+c_{1}-\frac{\mu}{r}+\frac{c_{2}}{r^{2}}\right)\,,\qquad
h^{2}(r)=r^{2}\sqrt{H_{0}H_{1}H_{2}H_{3}}\,,$ (177)

with

$\qquad H_{\Lambda}=1+\frac{b_{\Lambda}}{r}\,,\qquad
F_{\theta\phi}^{\Lambda}=\frac{1}{2\sqrt{2}}P^{\Lambda}f^{\prime}(\theta)\,,$
(178)

scalars

$\tau_{1}=\sqrt{\frac{H_{0}H_{1}}{H_{2}H_{3}}}\,,\qquad\tau_{2}=\sqrt{\frac{H_{2}H_{0}}{H_{1}H_{3}}}\,,\qquad\tau_{3}=\sqrt{\frac{H_{0}H_{3}}{H_{2}H_{1}}}\,,$
(179)

with parameters

$\sum_{\Lambda=0}^{3}b_{\Lambda}=0\,,\qquad
c_{1}=\kappa+2b_{0}g^{2}(b_{1}+b_{2}+b_{3})+2b_{1}g^{2}(b_{2}+b_{3})+2b_{2}b_{3}g^{2}\,.$
(180)

Eliminating one of the scalar parameters, for example $b_{3}$, we obtain

$\displaystyle(P^{0})^{2}$ $\displaystyle=$ $\displaystyle-
b_{0}^{2}\left(2b_{1}^{2}g^{2}+2b_{1}b_{2}g^{2}+2b_{2}^{2}g^{2}+1\right)-2b_{0}^{3}g^{2}(b_{1}+b_{2})+b_{0}\mu+c_{2}\,,$
$\displaystyle(P^{1})^{2}$ $\displaystyle=$ $\displaystyle-
b_{1}^{2}\left(2b_{0}^{2}g^{2}+2b_{0}b_{2}g^{2}+2b_{2}^{2}g^{2}+1\right)-2b_{1}^{3}g^{2}(b_{0}+b_{2})+b_{1}\mu+c_{2}\,,$
$\displaystyle(P^{2})^{2}$ $\displaystyle=$ $\displaystyle-
b_{2}^{2}\left(2b_{0}^{2}g^{2}+2b_{0}b_{1}g^{2}+2b_{1}^{2}g^{2}+1\right)-2b_{2}^{3}g^{2}(b_{0}+b_{1})+b_{2}\mu+c_{2}\,,$
$\displaystyle(P^{3})^{2}$ $\displaystyle=$ $\displaystyle
2g^{2}(b_{0}+b_{1}+b_{2})^{2}(b_{0}(b_{1}+b_{2})+b_{1}b_{2})-b_{0}^{2}-2b_{0}(b_{1}+b_{2})-b_{1}^{2}+$
$\displaystyle-$ $\displaystyle
2b_{1}b_{2}-b_{2}^{2}+c_{2}-\mu(b_{0}+b_{1}+b_{2})\,,$

with supersymmetric limit obtained when

$c_{2}=\frac{c_{1}^{2}}{8g^{2}}\,.$ (182)

Scanning the parameter space to the best of our possibilities, we could not
find stable or metastable probes with these backgrounds. This result goes in
the same direction as Sec. 5.2.3, denoting the absence of stable probes on
these sort of supersymmetric magnetic backgrounds.

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# Nonblockers for hereditarily decomposable continua with the property of
Kelley

Javier Camargo and Mayra Ferreira1112010 Mathematics Subject Classification.
54B20, 54F15. Key word and phrases. Continuum, Hyperspaces of continua,
Nonblockers, Property of Kelley, Hereditarily decomposable continua.

###### Abstract

Given a continuum $X$, let $\mathcal{NB}(\mathcal{F}_{1}(X))$ be the
hyperspace of nonblockers of $\mathcal{F}_{1}(X)$. In this paper, we show that
if $X$ is hereditarily decomposable with the property of Kelley such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum, then $X$ is a simple closed
curve. Thus, we characterize the simple closed curve as the unique
hereditarily decomposable continuum with the property of Kelley $X$ such that
its hyperspace $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum.

##  1 Introduction

A continuum is a nonempty compact connected metric space. Given a continuum
$X$, $2^{X}$ denotes the hyperspace of all nonempty closed subsets of $X$
topologized with the Hausdorff metric. The set $\mathcal{F}_{1}(X)$ denotes
the family of all one-point subsets of $X$.

Let $X$ be a continuum and let $B\in 2^{X}$. We say that $B$ does not block
$\mathcal{F}_{1}(X)$ provided that for any $x\in X\setminus B$, the union of
all subcontinua of $X$ containing $x$ and contained in $X\setminus B$ is a
dense subset of $X$. The collection of all $B$ in $2^{X}$ such that $B$ does
not block $\mathcal{F}_{1}(X)$ is denoted by
$\mathcal{NB}(\mathcal{F}_{1}(X))$.

The notion of nonblocker is introduced by A. Illanes and P. Krupski in [5].
Since $\mathcal{NB}(\mathcal{F}_{1}(X))\subseteq 2^{X}$,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ has the subspace topology and, it is
natural to ask, over what conditions is the hyperspace
$\mathcal{NB}(\mathcal{F}_{1}(X))$ a continuum?

In [4, Theorem 4.4], it is proved that if $X$ is locally connected, then,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum if and only if $X$ is a
simple closed curve. Also, in [4], the authors showed that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum when $X$ is the circle of
pseudoarcs. In [2], the simple closed curve is characterized as the unique
continuum $X$ such that the hyperspace $\mathcal{NB}(\mathcal{F}_{1}(X))$ is
homeomorphic to $\mathcal{F}_{1}(X)$. In [3], the authors show examples of
continua $X$ such that $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum.
Recently, A. Illanes and B. Vejnar showed in [7] that for any completely
metrizable and separable space $X$, there exists a continuum $Y$ such that
$\mathcal{NB}(\mathcal{F}_{1}(Y))$ is homeomorphic to $X$.

The following question was taken from [4, Question 4.5].

######  Question 1.1.

For which nonlocally connected continua $X$ is
$\mathcal{NB}(\mathcal{F}_{1}(X))$ a continuum?

We note that in all the examples presented in [3], [4] or [7], if $X$ is not a
simple closed curve and $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum,
then $X$ contain an infinite number of indecomposable continua. Hence, we rise
the following question:

######  Question 1.2.

Let $X$ be an hereditarily decomposable continuum such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Then, does it follow that
$X$ is the simple closed curve?

The main result in this paper is Theorem 5.3, where we characterize the simple
closed curve as the unique hereditarily decomposable continuum with the
property of Kelley $X$, such that the hyperspace of nonblockers
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum.

The reader can see [1], [10], or [14] for additional information about
nonblockers in hyperspaces.

##  2 Preliminaries

The symbols $\mathbb{N},\mathbb{R}$ and $\mathbb{C}$ denote the set of
positive integers, reals numbers and complex numbers, respectively. In this
paper, every map will be a continuous function. Given a metric space $X$ and
$A\subseteq X$, $\mathrm{Int}(A)$, $\mathrm{Cl}(A)$ and $\mathrm{Bd}(A)$
denote the interior, the closure and the boundary of $A$, respectively. For
$x\in X$ and $r>0$, we represent $B(x,r)=\\{y\in X:d(x,y)<r\\}$ where $d$ is a
metric on $X$. A continuum is a nonempty compact connected metric space. A
subcontinuum is a continuum contained in some metric space. We say that a
continuum $X$ is irreducible between two of its points if no proper
subcontinuum of $X$ contains both points. A continuum is irreducible if it is
irreducible between two of its points. A subcontinuum $Z$ of a continuum $X$
is said to be terminal if each subcontinuum $Y$ of $X$ that intersects $Z$
satisfies either $Z\subseteq Y$ or $Y\subseteq Z$. An arc is any continuum
homeomorphic to $[0,1]$, and a simple closed curve is any continuum
homeomorphic to $S^{1}=\\{z\in\mathbb{C}:|z|=1\\}$.

A continuum is decomposable provided that it is the union of two of its proper
subcontinua. A continuum that is not decomposable is said to be
indecomposable. A continuum is hereditarily decomposable if every subcontinuum
is decomposable.

Let $X$ be a continuum and let $\mathcal{D}$ be a decomposition of $X$. We say
that $\mathcal{D}$ is an upper semicontinuous decomposition provided that
whenever $D\in\mathcal{D}$, $U$ open subset of $X$, and $D\subseteq U$, there
exists an open subset $V$ of $X$ with $D\subseteq V$ such that if
$A\in\mathcal{D}$ and $A\cap V\neq\emptyset$, then $A\subseteq U$. We say that
$\mathcal{D}$ is lower semicontinuous provided that for each $D\in\mathcal{D}$
any two points $x$ and $y$ of $D$ and each open subset $U$ of $X$ such that
$x\in U$, there exists an open subset $V$ of $X$ such that $y\in V$ and such
that if $A\in\mathcal{D}$ and $A\cap V\neq\emptyset$, then $A\cap
U\neq\emptyset$. Finally, we say that $\mathcal{D}$ is continuous if
$\mathcal{D}$ is both upper and lower semicontinuous. If $\mathcal{D}$ is a
decomposition of $X$, we denote $\rho\colon X\to\mathcal{D}$ by the natural
function such that $\rho(x)=D_{x}$ where $D_{x}$ is unique
$D_{x}\in\mathcal{D}$ with $x\in D_{x}$. The following result is not difficult
to prove.

######  Proposition 2.1.

Let $X$ be a continuum, let $\mathcal{D}$ be a decomposition of $X$ such that
$D$ is closed for each $D\in\mathcal{D}$. Then, $\mathcal{D}$ is upper
semicontinuous (lower semicontinuous) if and only if for each sequence
$(x_{n})_{n\in\mathbb{N}}$ in $X$ such that $\lim_{n\to\infty}x_{n}=x$ for
some $x\in X$, we have that $\limsup\rho(x_{n})\subseteq\rho(x)$
($\rho(x)\subseteq\liminf\rho(x_{n})$, respectively).

Given a map between continua $f\colon X\to Y$, we say that $f$ is monotone
provided that $f^{-1}(D)$ is a continuum for each subcontinuum $D$ of $Y$. We
say that $f$ is open provided that $f(U)$ is open for every open subset $U$ of
$X$.

######  Definition 2.2.

Let $X$ be a continuum, and let $p,q\in X$. We say that $X$ is aposyndetic at
$p$ with respect to $q$ provided that there exists a subcontinuum $W$ of $X$
such that $p\in\mathrm{Int}(W)\subseteq W\subseteq X\setminus\\{q\\}$. Now,
$X$ is aposyndetic at $p$ if $X$ is aposyndetic at $p$ with respect to each
point of $X\setminus\\{p\\}$. We say that $X$ is aposyndetic provided that $X$
is aposyndetic at each of its points.

A continuum $X$ is locally connected provided that for every $p\in X$ and each
open $V$ such that $p\in V$, there exists an open connected set $U$ such that
$p\in U\subseteq V$. We say that $X$ is semi-locally connected provided that
for every $p\in X$ and each open $V$ such that $p\in V$, there exists an open
$U$ such that $p\in U\subseteq V$ and $X\setminus U$ has a finite number of
components.

Let $X$ be a continuum. Let $2^{X}=\\{A\subseteq X:A\neq\emptyset\text{ and
}\mathrm{Cl}(A)=A\\}$. It is known that the collection of sets $\langle
U_{1},U_{2},...,U_{l}\rangle$ form a base on $2^{X}$ (Vietoris topology, see
[6, p.3]), where $U_{1},U_{2},...,U_{l}$ are open sets in $X$ and

$\langle U_{1},U_{2},...,U_{l}\rangle=\\{A\in
2^{X}:A\subseteq\cup_{i=1}^{l}U_{i}\text{ and }A\cap U_{i}\neq\emptyset\text{
for each }i\\}.$

We denote $\mathcal{C}(X)=\\{A\in 2^{X}:A\text{ is connected}\\}$ and
$\mathcal{F}_{1}(X)=\\{\\{x\\}:x\in X\\}$. Since
$\mathcal{F}_{1}(X)\subseteq\mathcal{C}(X)\subseteq 2^{X}$, the sets
$\mathcal{F}_{1}(X)$ and $\mathcal{C}(X)$ are considered as subspaces of
$2^{X}$. Let $\mathcal{H}$ be defined for each $A,B\in 2^{X}$ by

$\mathcal{H}(A,B)=\inf\\{r>0:A\subseteq N(B;r)\text{ and }B\subseteq
N(A;r)\\},$

where $N(D,s)=\\{x\in X:d(x,y)<s\text{ for some }y\in D\\}$, for any $D\in
2^{X}$ and $s>0$. $\mathcal{H}$ is a metric on $2^{X}$ and it is known as the
Hausdorff metric [12, Theorem 0.2]. It is well known that the topology induced
by the Hausdorff metric and the Vietoris topology are the same [6, Theorem
3.1].

Let $A,B\in 2^{X}$ such that $A\cap B=\emptyset$. We use the notation

$\kappa_{X\setminus A}(B)=\bigcup\\{L\in\mathcal{C}(X):B\cap
L\neq\emptyset\text{ and }L\subseteq X\setminus A\\}.$

If $B=\\{b\\}$ for some $b\in X$, we write $\kappa_{X\setminus A}(b)$ instead
of $\kappa_{X\setminus A}(\\{b\\})$.

######  Lemma 2.3.

Let $X$ be a continuum and let $A,B\in 2^{X}$ such that $A\cap B=\emptyset$.
If $B$ is connected, then there exists a sequence of subcontinua
$(K_{n})_{n\in\mathbb{N}}$ of $X$ such that $\kappa_{X\setminus
A}(B)=\bigcup_{n\in\mathbb{N}}K_{n}$.

###### Proof.

Let $U_{n}=N(A,1/n)$ for each $n\in\mathbb{N}$. Since $A$ and $B$ are compact
and $A\cap B=\emptyset$, we may suppose that $U_{n}\cap B=\emptyset$ for every
$n\in\mathbb{N}$. Let $K_{n}$ be the component of $X\setminus U_{n}$ such that
$B\subseteq K_{n}$ for each $n\in\mathbb{N}$. It is not difficult to see that
$\kappa_{X\setminus A}(B)=\bigcup_{n\in\mathbb{N}}K_{n}$. ∎

A subset $W$ of a continuum $X$ is said to be connected by continua provided
for any two points $x,y\in W$ there exists a subcontinuum $L$ of $W$ such that
$\\{x,y\\}\subseteq L\subseteq W$. Observe that for $A\in 2^{X}$, $X\setminus
A$ is connected by continua if $\kappa_{X\setminus A}(x)=X\setminus A$ for
every $x\in X\setminus A$.

######  Definition 2.4.

Let $X$ be a continuum and let $A,B\in 2^{X}$ such that $A\cap B=\emptyset$.
We say that $A$ does not block $B$ provided that $\kappa_{X\setminus A}(B)$ is
a dense subset of $X$. In another way we say that $A$ blocks $B$; i.e., $A$
blocks $B$ provided that $\mathrm{Cl}(\kappa_{X\setminus A}(B))\neq X$.

In this paper, we will focus on the following set:

$\mathcal{NB}(\mathcal{F}_{1}(X))=\\{A\in 2^{X}:A\text{ does not block
}\\{x\\}\text{ for each }x\in X\setminus A\\}.$

The sets $\mathcal{F}_{1}(X),\mathcal{C}(X),\mathcal{NB}(\mathcal{F}_{1}(X))$
and $2^{X}$ are called hyperspaces of $X$. In particular,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is called the hyperspace of nonblockers of
$\mathcal{F}_{1}(X)$. In [4, Theorem 4.4], it is proved the following theorem.

######  Theorem 2.5.

Let $X$ be a locally connected continuum. Then,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum if and only if $X$ is a
simple closed curve.

Let $X$ be a continuum. We say that $X$ has the property of Kelley provided
that for any sequence $(x_{n})_{n\in\mathbb{N}}$ such that
$\lim_{n\to\infty}x_{n}=x$ for some $x\in X$, and for any $A\in\mathcal{C}(X)$
such that $x\in A$, there exists a sequence on subcontinua
$(A_{n})_{n\in\mathbb{N}}$ of $X$ such that $x_{n}\in A_{n}$ for each
$n\in\mathbb{N}$, and $\lim_{n\to\infty}A_{n}=A$.

The proof of the next result can be found in [9, Corollary 4.2].

######  Theorem 2.6.

Let $X$ be a continuum, let $A\in 2^{X}$ and let $K$ be a subcontinuum of
$X\setminus A$ such that $\mathrm{Int}(K)\neq\emptyset$. If $X$ has the
property of Kelley, then there exists a subcontinuum $L$ of $X$ such that
$K\subseteq\mathrm{Int}(L)\subseteq L\subseteq X\setminus A$.

##  3 On minimal nonblockers

In [14], it is introduced the function $\pi\colon X\to\mathcal{C}(X)$ defined
by

$\pi(x)=\begin{cases}\bigcap\\{A\in\mathcal{NB}(\mathcal{F}_{1}(X)):x\in
A\\},&\hbox{ if }x\in\bigcup\mathcal{NB}(\mathcal{F}_{1}(X));\\\ X,&\hbox{ if
}x\notin\bigcup\mathcal{NB}(\mathcal{F}_{1}(X)).\end{cases}$ (3.1)

By Proposition 3.1 and Theorem 3.3 of [14], $\pi$ is well defined and
$\pi(x)\in\mathcal{NB}(\mathcal{F}_{1}(X))$ for each $x\in X$ where
$x\in\bigcup\mathcal{NB}(\mathcal{F}_{1}(X))$. We prove in Theorem 3.6 that
$\pi$ is a map whenever $X$ is a decomposable continuum with the property of
Kelley such that $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum, giving a
partial answer to Question 3.6 of [14]. Furthermore, we show in Theorem 3.9
that $\\{\pi(x):x\in X\\}$ is a continuous decomposition of $X$.

######  Lemma 3.1.

Let $X$ be a decomposable continuum with the property of Kelley. If $L$ is a
proper subcontinuum of $X$, then there exists
$A\in\mathcal{NB}(\mathcal{F}_{1}(X))$ such that $A\subseteq X\setminus L$.

###### Proof.

Let $L$ be a proper subcontinuum of $X$ and let $M$ and $N$ be proper
subcontinua of $X$ such that $X=M\cup N$. We consider two cases:

1. $\mathrm{Int}(L)=\emptyset$. Note that either $M\cap L\neq\emptyset$ or $N\cap L\neq\emptyset$. Suppose that $M\cap L\neq\emptyset$. Since $X\setminus M$ is a nonempty open subset of $X$ and $X\setminus M\nsubseteq L$, we have that $M\cup L\neq X$. Let $L_{1}=M\cup L$. Since $\mathrm{Int}(M)\neq\emptyset$, $\mathrm{Int}(L_{1})\neq\emptyset$. By Theorem 2.6, there exists a proper subcontinuum $R$ of $X$ such that $L_{1}\subseteq\mathrm{Int}(R)$. By [6, Theorem 14.6], there exists an order arc in $\mathcal{C}(X)$ from $R$ to $X$. Let $\alpha$ be an order arc from $R$ to $X$ and let $f\colon[0,1]\to\mathcal{C}(X)$ be an embedding such that $f(0)=R,f(1)=X$ and $f([0,1])=\alpha$. Since $f$ is a map, there exists $t\in(0,1)$ such that $\mathcal{H}(f(t),X)<\frac{1}{2}$. Let $L_{2}=f(t)$. Note that $L_{1}\subseteq\mathrm{Int}(L_{2})$. Hence inductively, we may construct an increasing sequence of continua $(L_{n})_{n\in\mathbb{N}}$ such that $L_{n}\subseteq\mathrm{Int}(L_{n+1})$, $L_{n}\neq X$ and $\mathcal{H}(L_{n},X)<\frac{1}{n}$, for each $n\in\mathbb{N}$. Let $A=\bigcap_{n\in\mathbb{N}}(X\setminus\mathrm{Int}(L_{n}))$. Since $\mathrm{Int}(L_{n})\subseteq\mathrm{Int}(L_{n+1})$, $X\setminus\mathrm{Int}(L_{n+1})\subseteq X\setminus\mathrm{Int}(L_{n})$, for each $n\in\mathbb{N}$. Furthermore, $X\setminus\mathrm{Int}(L_{n})\neq\emptyset$ for each $n\in\mathbb{N}$. Thus, $A$ is a nonempty compact subset of $X$. Observe that $X\setminus A=\bigcup_{n\in\mathbb{N}}\mathrm{Int}(L_{n})$ and, since $\mathrm{Int}(L_{n})\subseteq L_{n+1}$, $X\setminus A=\bigcup_{n\in\mathbb{N}}L_{n}$. Also, $X\setminus A$ is dense, because $\mathcal{H}(L_{n},X)<\frac{1}{n}$ for each $n\in\mathbb{N}$. Therefore, $A\in\mathcal{NB}(\mathcal{F}_{1}(X))$ and $A\subseteq X\setminus L$. Similarly, we construct $A$ if $N\cap L\neq\emptyset$.

2. $\mathrm{Int}(L)\neq\emptyset$. We define $L_{1}=L$ and continue the same argument as in case 1 to construct $A$ and complete the proof. ∎

With the next theorem, we have that $\pi(x)\neq X$ for each $x\in X$ when $X$
is a decomposable continuum with the property of Kelley such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum.

######  Theorem 3.2.

Let $X$ be a decomposable continuum with the property of Kelley. If
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum, then
$X=\bigcup\mathcal{NB}(\mathcal{F}_{1}(X))$.

###### Proof.

By [14, Theorem 3.3],
$\mathcal{NB}(\mathcal{F}_{1}(X))\cap\mathcal{C}(X)\neq\emptyset$. Since
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum,
$\bigcup\mathcal{NB}(\mathcal{F}_{1}(X))$ is a subcontinuum of
$\mathcal{C}(X)$, by [6, Exercise 15.9 (2)]. Suppose that
$\bigcup\mathcal{NB}(\mathcal{F}_{1}(X))$ is a proper subcontinuum of
$\mathcal{C}(X)$. Thus, there exists $A\in\mathcal{NB}(\mathcal{F}_{1}(X))$
such that $A\subseteq X\setminus\bigcup\mathcal{NB}(\mathcal{F}_{1}(X))$, by
Lemma 3.1. A contradiction. Therefore,
$X=\bigcup\mathcal{NB}(\mathcal{F}_{1}(X))$. ∎

In this section we use the following notation:

$B(x)=\\{y\in X:x\text{ blocks }y\\}=\\{y\in
X:\mathrm{Cl}(\kappa_{X\setminus\\{x\\}}(y))\neq X\\},\text{ for }x\in X.$

Observe that $X\setminus\pi(x)\subseteq X\setminus B(x)$ for each $x\in X$.
Thus, $B(x)\subseteq\pi(x)$ for each $x\in X$. The following remark is not
difficult to prove.

######  Remark 3.3.

Let $X$ be a continuum and let $x\in X$. If $L\in\mathcal{C}(X)$ is such that
$L\cap B(x)\neq\emptyset$ and $L\cap(X\setminus B(x))\neq\emptyset$, then
$x\in L$.

######  Proposition 3.4.

Let $X$ be a decomposable continuum such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. If $X$ has the property of
Kelley, then $\pi(x)=B(x)$ for each $x\in X$.

###### Proof.

Let $x\in X$. Note that there exists $A\in\mathcal{NB}(\mathcal{F}_{1}(X))$
such that $x\in A$, by Theorem 3.2. Thus, $\pi(x)\neq X$. Since
$B(x)\subseteq\pi(x)$, we need to show that $\pi(x)\subseteq B(x)$. Observe
that if $B(x)$ is closed, then $B(x)\in\mathcal{NB}(\mathcal{F}_{1}(X))$, and
hence, $B(x)\in\\{A\in\mathcal{NB}(\mathcal{F}_{1}(X)):x\in A\\}$ and
$\pi(x)\subseteq B(x)$ (see (3.1)).

We see that $B(x)$ is closed. Let $(w_{n})_{n\in\mathbb{N}}$ be a sequence in
$B(x)$ such that $\lim_{n\to\infty}w_{n}=w$ for some $w\in X$. Suppose that
$w\in\mathrm{Cl}(B(x))\setminus B(x)$. Since $B(x)\subseteq\pi(x)$, $\pi(x)$
is closed and $\mathrm{Int}(\pi(x))=\emptyset$, we have that there exists a
nonempty open subset $U$ of $X$ such that $U\cap\mathrm{Cl}(B(x))=\emptyset$.
Furthermore, there exists $L\in\mathcal{C}(X)$ such that $w\in L$, $x\notin L$
and $L\cap U\neq\emptyset$, because $w\notin B(x)$. Since $X$ has the property
of Kelley, there exists a sequence $(L_{n})_{n\in\mathbb{N}}$ of
$\mathcal{C}(X)$ such that $w_{n}\in L_{n}$ for each $n\in\mathbb{N}$, and
$\lim_{n\to\infty}L_{n}=L$. Note that $L\in\langle
X\setminus\\{x\\},U\rangle$. Thus, there exists a positive number
$k\in\mathbb{N}$ such that $L_{i}\in\langle X\setminus\\{x\\},U\rangle$ for
each $i\geq k$. Let $k_{0}>k$. Note that $L_{k_{0}}\cap B(x)\neq\emptyset$,
$L_{k_{0}}\cap U\neq\emptyset$ and $x\notin L_{k_{0}}$. This contradicts
Remark 3.3. Therefore, $B(x)$ is closed and $\pi(x)=B(x)$. ∎

######  Question 3.5.

Let $X$ be a decomposable continuum. If $X$ has the property of Kelley, then
does it follow that $B(x)$ is closed for each $x\in X$?

It is easy to verify that $B(x)$ is one composant of $X$, whenever $X$ is an
hereditarily indecomposable continuum. Thus, if $X$ is hereditarily
indecomposable, then $X$ has the property of Kelley and $B(x)$ is not closed
(see [6, 20.6, p.168]).

######  Theorem 3.6.

Let $X$ be a decomposable continuum with the property of Kelley. If
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum, then $\pi$ is a map.

###### Proof.

Let $(x_{n})_{n\in\mathbb{N}}$ be a sequence in $X$ such that
$\lim_{n\to\infty}x_{n}=x$ for some $x\in X$. We prove that
$\lim_{n\to\infty}\pi(x_{n})=\pi(x)$. To see this, we show that
$\limsup\pi(x_{n})\subseteq\pi(x)$ and $\pi(x)\subseteq\liminf\pi(x_{n})$.

We see that $\limsup\pi(x_{n})\subseteq\pi(x)$. Let $z\in\limsup\pi(x_{n})$.
Let $(n_{k})_{k\in\mathbb{N}}$ be an increasing sequence of positive numbers
such that, for each $k\in\mathbb{N}$, there exists
$z_{n_{k}}\in\pi(x_{n_{k}})$ where $\lim_{k\to\infty}z_{n_{k}}=z$.
Furthermore, since $\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact and
$\pi(x_{n_{k}})\in\mathcal{NB}(\mathcal{F}_{1}(X))$ for all $k$, without loss
of generality we may suppose that $\lim_{k\to\infty}\pi(x_{n_{k}})=D$ for some
$D\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Hence, $\mathrm{Int}(D)=\emptyset$.
Note that $z\in D$. Suppose that $z\in X\setminus\pi(x)$. Let $U$ and $V$ be
nonempty open subsets of $X$ such that $D\subseteq U$ and $U\cap V=\emptyset$.
Since $D\in\langle U\rangle$, there exists $m\in\mathbb{N}$ such that
$\pi(x_{n_{k}})\in\langle U\rangle$ for each $k\geq m$; i.e.,
$\pi(x_{n_{k}})\subseteq U$ for each $k\geq m$. Also, since $z\notin\pi(x)$,
there exists a subcontinuum $L$ of $X$ such that $z\in L,x\notin L$ and $L\cap
V\neq\emptyset$. Observe that $\langle X\setminus\\{x\\},V\rangle$ is an open
subset of $2^{X}$ such that $L\in\langle X\setminus\\{x\\},V\rangle$. Since
$X$ has the property of Kelley, there exists a sequence of subcontinua
$(L_{n_{k}})_{k\in\mathbb{N}}$ of $X$ such that $z_{n_{k}}\in L_{n_{k}}$ for
each $k\in\mathbb{N}$, and $\lim_{k\to\infty}L_{n_{k}}=L$. Thus, there is
$l\in\mathbb{N}$ such that $L_{n_{k}}\in\langle X\setminus\\{x\\},V\rangle$
for each $k\geq l$. Let $j>\max\\{m,l\\}$. Since $\pi(x_{n_{j}})\subseteq U$,
$L_{n_{j}}\cap V\neq\emptyset$ and $U\cap V=\emptyset$, we have that
$L_{n_{j}}\cap X\setminus\pi(x_{n_{j}})\neq\emptyset$. Furthermore,
$L_{n_{j}}\cap\pi(x_{n_{j}})\neq\emptyset$, because $z_{n_{j}}\in
L_{n_{j}}\cap\pi(x_{n_{j}})$. By Proposition 3.4,
$\pi(x_{n_{j}})=B(x_{n_{j}})$. Hence, $x_{n_{j}}\in L_{n_{j}}$, by Remark 3.3.
Since $j$ was an arbitrary integer greater than $\max\\{m,l\\}$, we have that
$x_{n_{j}}\in L_{n_{j}}$ for each $j\geq\max\\{m,l\\}$. Thus, $x\in L$. A
contradiction. Therefore, $z\in\pi(x)$ and $\limsup\pi(x_{n})\subseteq\pi(x)$.

We show that $\pi(x)\subseteq\liminf\pi(x_{n})$. Suppose that there exists
$y\in\pi(x)$ such that $y\in X\setminus\liminf\pi(x_{n})$. Then there exist an
open subset $U$ of $X$ such that $y\in U$, and an increasing sequence of
positive integers $(n_{k})_{k\in\mathbb{N}}$ such that $\pi(x_{n_{k}})\cap
U=\emptyset$ for each $k\in\mathbb{N}$. Since
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact and
$(\pi(x_{n_{k}}))_{k\in\mathbb{N}}$ is a sequence in
$\mathcal{NB}(\mathcal{F}_{1}(X))$, we may suppose that
$\lim_{k\to\infty}\pi(x_{n_{k}})=D$ for some
$D\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Observe that $D\cap U=\emptyset$ and
$x\in D$. This contradicts that $y\in\pi(x)$,
$D\in\mathcal{NB}(\mathcal{F}_{1}(X))$ and $y\notin D$. Therefore,
$\pi(x)\subseteq\liminf\pi(x_{n})$. ∎

######  Lemma 3.7.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Let $x,y\in X$. If
$\pi(x)\cap\pi(y)\neq\emptyset$, then $\pi(x)\subseteq\pi(y)$ or
$\pi(y)\subseteq\pi(x)$.

###### Proof.

Let $x,y\in X$ such that $\pi(x)\cap\pi(y)\neq\emptyset$. Suppose that
$\pi(x)\setminus\pi(y)\neq\emptyset$. Observe that $\pi(y)=B(y)$, by
Proposition 3.4. Furthermore, $\pi(x)\in\mathcal{C}(X)$ such that $\pi(x)\cap
B(y)\neq\emptyset$ and $\pi(x)\cap X\setminus B(y)\neq\emptyset$. Thus,
$y\in\pi(x)$, by Remark 3.3. Therefore, $\pi(y)\subseteq\pi(x)$, by (3.1). ∎

Given a decomposable continuum with the property of Kelley $X$ such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Let

$\mathcal{L}=\\{\pi(x):x\in X\\}.$ (3.2)

Note that $\mathcal{L}=\pi(X)$. Hence, $\mathcal{L}$ is a subcontinuum of
$\mathcal{NB}(\mathcal{F}_{1}(X))$, by Theorem 3.6. Since $\mathcal{L}$ is
compact, there are maximal and minimal elements of $\mathcal{L}$ that we
represent:

$\mathcal{M}=\\{\pi(x):\pi(x)\text{ is maximal in }\mathcal{L}\\};\text{ and
}\mathcal{N}=\\{\pi(x):\pi(x)\text{ is minimal in }\mathcal{L}\\}.$ (3.3)

######  Proposition 3.8.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Let
$\mathcal{L},\mathcal{M}$ and $\mathcal{N}$ be as (3.2) and (3.3). Then,

  1. 1.

$\mathcal{M}$ is an upper semicontinuous decomposition of $X$.

  2. 2.

$\pi(x)$ is terminal, for each $\pi(x)\in\mathcal{N}$.

  3. 3.

$\mathcal{N}$ is compact.

  4. 4.

$\bigcup\mathcal{N}$ is a subcontinuum of $X$.

###### Proof.

We see that $X=\bigcup\mathcal{M}$. Let $x\in X$. By (3.1), $x\in\pi(x)$.
Furthermore, since $\mathcal{L}$ is compact, there exists $z\in X$ such that
$\pi(x)\subseteq\pi(z)$ and $\pi(z)\in\mathcal{M}$. Therefore,
$x\in\bigcup\mathcal{M}$ and $X=\bigcup\mathcal{M}$. Note that if
$\pi(x)\cap\pi(y)\neq\emptyset$ and $\pi(x),\pi(y)\in\mathcal{M}$, then
$\pi(x)=\pi(y)$, by Lemma 3.7. Thus, $\mathcal{M}$ is a decomposition of $X$.
We show that $\mathcal{M}$ is an upper semicontinuous decomposition. Let
$\rho\colon X\to\mathcal{M}$ be the quotient map. Let
$(x_{n})_{n\in\mathbb{N}}$ be a sequence of $X$ such that
$\lim_{n\to\infty}x_{n}=x$ for some $x\in X$. We see that
$\limsup\rho(x_{n})\subseteq\rho(x)$. Let $z\in\limsup\rho(x_{n})$. Since
$(\rho(x_{n}))_{n\in\mathbb{N}}$ is a sequence in $\mathcal{L}$ and
$\mathcal{L}$ is compact, there exists a subsequence
$(\rho(x_{n_{k}}))_{k\in\mathbb{N}}$ of $(\rho(x_{n}))_{n\in\mathbb{N}}$, such
that $\lim_{k\to\infty}\rho(x_{n_{k}})=D$ for some $D\in\mathcal{L}$ where
$z\in D.$ Let $w\in D$ be such that $D=\pi(w)$. Since
$x_{n_{k}}\in\rho(x_{n_{k}})$ for each $k\in\mathbb{N}$, and
$\lim_{k\to\infty}x_{n_{k}}=x$, we have that $x\in\pi(w)$. Hence,
$\rho(x)\cap\pi(z)\neq\emptyset$. Since $\rho(x)\in\mathcal{M}$,
$\pi(z)\subseteq\rho(x)$. Thus, $z\in\rho(x)$ and
$\limsup\rho(x_{n})\subseteq\rho(x)$. Therefore, $\mathcal{M}$ is an upper
semicontinuous decomposition of $X$, by Proposition 2.1. We have proved 1.

Let $x\in X$ be such that $\pi(x)\in\mathcal{N}$. We prove that $\pi(x)$ is
terminal. Since $\pi(x)\in\mathcal{NB}(\mathcal{F}_{1}(X))$,
$\pi(y)\subseteq\pi(x)$ for every $y\in\pi(x)$, by (3.1). Since $\pi(x)$ is
minimal, $\pi(y)=\pi(x)$ for all $y\in\pi(x)$. Let $L$ be a subcontinuum of
$X$ such that $\pi(x)\cap L\neq\emptyset$. Suppose that $\pi(x)\setminus
L\neq\emptyset$. Let $y\in\pi(x)\setminus L$. Note that $\pi(y)\cap
L\neq\emptyset$. We know that $\pi(y)=B(y)$, by Proposition 3.4. Thus,
$L\subseteq\pi(y)$, by Remark 3.3. Since $\pi(y)=\pi(x)$, $L\subseteq\pi(x)$.
Therefore, $\pi(x)$ is terminal. The proof of 2 is completed.

We see 3. Suppose that $\mathcal{N}$ is not compact, that is, there exists a
sequence $(\pi(x_{n}))_{n\in\mathbb{N}}$ in $\mathcal{N}$ such that
$\lim_{n\to\infty}\pi(x_{n})=D$ where $D\notin\mathcal{N}$. Since
$\mathcal{L}$ is compact, $D\in\mathcal{L}$. Let $z\in D$ be such that
$D=\pi(z)$. Since $D\notin\mathcal{N}$, there exists $w\in\pi(z)$ such that
$\pi(w)\subseteq\pi(z)$ and $\pi(w)\neq\pi(z)$. Let
$p\in\pi(z)\setminus\pi(w)$. Since $\mathrm{Int}(\pi(x_{n}))=\emptyset$ for
each $n\in\mathbb{N}$, $\mathrm{Int}(\pi(z))=\emptyset$ and
$\lim_{n\to\infty}\pi(x_{n})=\pi(z)$, we have that
$K=(\bigcup_{n\in\mathbb{N}}\pi(x_{n}))\cup\pi(z)$ is a compact such that
$\mathrm{Int}(K)=\emptyset$ (see [17, Corollary 25.4]). Let $V$ be a nonempty
open subsets of $X$ such that $V\cap K=\emptyset$. Since $p\notin\pi(w)$,
there exists a continuum $L$ such that $p\in L$, $L\cap\pi(w)=\emptyset$ and
$L\cap V\neq\emptyset$. Observe that $\langle X,V\rangle$ is an open subset of
$2^{X}$ such that $L\in\langle X,V\rangle$. Since
$\lim_{n\to\infty}\pi(x_{n})=\pi(z)$ and $p\in\pi(z)$, there exists a sequence
$(p_{n})_{n\in\mathbb{N}}$ such that $p_{n}\in\pi(x_{n})$ for each
$n\in\mathbb{N}$, and $\lim_{n\to\infty}p_{n}=p$. Since $X$ has the property
of Kelley, there exists a sequence $(L_{n})_{n\in\mathbb{N}}$ in
$\mathcal{C}(X)$ such that $p_{n}\in L_{n}$ for each $n$, and
$\lim_{n\to\infty}L_{n}=L$. Hence, there is $m\in\mathbb{N}$ such that
$L_{n}\in\langle X,V\rangle$ for each $n\geq m$. Note that we have both
$L_{n}\cap\pi(x_{n})\neq\emptyset$ and
$L_{n}\cap(X\setminus\pi(x_{n}))\neq\emptyset$, for each $n\geq m$. Thus,
$\pi(x_{n})\subseteq L_{n}$, because $\pi(x_{n})$ is terminal, for each $n\geq
m$ (see 2 in this proposition). Therefore, $\pi(z)\subseteq L$. This
contradicts the facts that $L\cap\pi(w)=\emptyset$ and
$\pi(w)\subseteq\pi(z)$. We finish the proof of 3.

We prove 4. Since $\mathcal{N}$ is compact, $\bigcup\mathcal{N}$ is compact,
by [6, Exercise 11.5]. We suppose that $\bigcup\mathcal{N}$ is not connected,
that is, $\bigcup\mathcal{N}=A\cup B$ where $A$ and $B$ are nonempty compact
disjoint subsets of $X$. Observe that if $x\in\bigcup\mathcal{N}$, then there
exists $\pi(z)\in\mathcal{N}$ such that $x\in\pi(z)$. Hence,
$\pi(x)\subseteq\pi(z)$ and, since $\pi(z)$ is minimal, $\pi(x)=\pi(z)$. We
have that $\pi(x)\subseteq\bigcup\mathcal{N}$ for each
$x\in\bigcup\mathcal{N}$. Furthermore, $\pi(x)$ is connected for every
$x\in\bigcup\mathcal{N}$. Thus, $\pi(x)\subseteq A$ for each $x\in A$, and
$\pi(y)\subseteq B$ for each $y\in B$.

Note that $\mathcal{L}$ is a continuum and $\langle X\setminus
B\rangle\cap\mathcal{L}$ is a proper open subset of $\mathcal{L}$. Hence,
there exists a sequence $(\pi(x_{n}))_{n\in\mathbb{N}}$ in $\langle X\setminus
B\rangle\cap\mathcal{L}$ such that $\lim_{n\to\infty}\pi(x_{n})=\pi(x)$ for
some $\pi(x)\in\mathcal{L}\setminus\langle X\setminus B\rangle$. Let
$y\in\pi(x)\cap B$. Observe that $\pi(y)\subseteq B$. Let
$(y_{n})_{n\in\mathbb{N}}$ be a sequence such that $y_{n}\in\pi(x_{n})$ for
each $n\in\mathbb{N}$, and $\lim_{n\to\infty}y_{n}=y$. Since
$\pi(y_{n})\subseteq\pi(x_{n})\subseteq X\setminus B$, $\pi(y_{n})\cap
A\neq\emptyset$ for each $n\in\mathbb{N}$. By Theorem 3.6, $\pi$ is a map and
$\lim_{k\to\infty}\pi(y_{n})=\pi(y)$. Thus, $\pi(y)\cap A\neq\emptyset$. This
contradicts that $\pi(y)\subseteq B$ and $A\cap B=\emptyset$. Therefore,
$\bigcup\mathcal{N}$ is a subcontinuum of $X$. ∎

######  Theorem 3.9.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Let
$\mathcal{L},\mathcal{M}$ and $\mathcal{N}$ be as (3.2) and (3.3). Then,
$\mathcal{N}=\mathcal{L}$ and $\mathcal{N}$ is a continuous decomposition of
$X$.

###### Proof.

In order to prove that $\mathcal{N}=\mathcal{L}$, we show first that
$\bigcup\mathcal{N}=X$. We know that $\bigcup\mathcal{N}$ is a continuum, by
Proposition 3.8 (4). Suppose that $\bigcup\mathcal{N}\neq X$. By Lemma 3.1,
there exists $B\in\mathcal{NB}(\mathcal{F}_{1}(X))$ such that
$B\cap\bigcup\mathcal{N}=\emptyset$. Note that $\pi(x)\subseteq B$ for each
$x\in B$. Hence, there exists $z\in X$ such that $\pi(z)\in\mathcal{N}$ and
$\pi(z)\subseteq B$. A contradiction. Therefore, $\bigcup\mathcal{N}=X$.

We see that $\mathcal{N}=\mathcal{L}$. Let $x\in X$. Since
$\bigcup\mathcal{N}=X$, there exists $z\in X$ such that $\pi(z)\in\mathcal{N}$
and $x\in\pi(z)$. Hence, $\pi(x)\subseteq\pi(z)$ and, since $\pi(z)$ is
minimal, $\pi(x)=\pi(z)$. Thus, $\pi(x)\in\mathcal{N}$ and
$\mathcal{N}=\mathcal{L}$.

Observe that if every $\pi(x)$ is minimal, then every $\pi(x)$ is maximal;
i.e., $\mathcal{N}=\mathcal{M}$. Thus, $\mathcal{N}$ is a decomposition, by
Proposition 3.8 (1). Finally, we have that $\mathcal{N}$ is a continuous
decomposition because $\pi$ is a map (see Proposition 2.1 and Theorem 3.6). ∎

We know that $\mathcal{N}$ is a continuous decomposition of $X$, and
$\mathcal{N}\subseteq\mathcal{C}(X)$. Let $T(\mathcal{N})$ denote the
decomposition topology on $\mathcal{N}$ induced by the surjective map $q\colon
X\to\mathcal{N}$ defined by $q(x)=\pi(x)$ for each $x\in X$, and let
$T_{V}(\mathcal{N})$ denote the topology on $\mathcal{N}$ as a subspace of
$\mathcal{C}(X)$. Then, by [11, Theorem 13.10], $q$ is an open map, and
$T(\mathcal{N})=T_{V}(\mathcal{N})$. Furthermore, since
$q(x)\in\mathcal{C}(X)$ for each $x\in X$, $q$ is a monotone map.

######  Theorem 3.10.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Let
$\mathcal{N}=\\{\pi(x):x\in X\\}$ be the continuous decomposition of $X$ and
let $q\colon X\to\mathcal{N}$ be the quotient map. Then,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is homeomorphic to
$\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$, and
$\mathcal{F}_{1}(\mathcal{N})\subseteq\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$.

###### Proof.

Let $D\in\mathcal{NB}(\mathcal{F}_{1}(X))$. By (3.1), $\pi(x)\subseteq D$ for
each $x\in D$. Hence, $D=\bigcup\\{\pi(x):x\in D\\}$, for each
$D\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Thus, we have that $D=q^{-1}(q(D))$
for each $D\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Since $q$ is an open monotone
map, by [4, Lemma 5.1], we have that

$q(D)\in\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))\text{ for each
}D\in\mathcal{NB}(\mathcal{F}_{1}(X)).$ (3.4)

Let
$g\colon\mathcal{NB}(\mathcal{F}_{1}(X))\to\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$
be defined by $g(D)=q(D)$ for each $D\in\mathcal{NB}(\mathcal{F}_{1}(X))$.
Note that $g$ is well defined, by (3.4). Furthermore,
$g=2^{q}|_{\mathcal{NB}(\mathcal{F}_{1}(X))}$ where $2^{q}\colon 2^{X}\to
2^{\mathcal{N}}$ is the induced map given by $2^{q}(A)=q(A)$ for each $A\in
2^{X}$. Since $2^{q}$ is a map, $g$ is a map (see [6, p.188]).

We see that $g$ is one-to-one. Let $A,B\in\mathcal{NB}(\mathcal{F}_{1}(X))$ be
such that $A\neq B$. Without loss of generality, we may suppose that there
exists $x\in A\setminus B$. Hence, $\pi(x)\in q(A)$. If $\pi(x)\in q(B)$, then
there exists $z\in B$ such that $\pi(z)=\pi(x)$. Since $\pi(z)\subseteq B$,
$x\in B$. A contradiction. Thus, $\pi(x)\notin q(B)$. Therefore, $g(A)\neq
g(B)$ and $g$ is one-to-one.

Let $B\in\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$. By [4, Lemma 5.1],
$q^{-1}(B)\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Since $q$ is surjective,
$q(q^{-1}(B))=B$. Thus, $g(q^{-1}(B))=B$ and $g$ is surjective. Since
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact and
$\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$ is a Hausdorff space, $g$ is a
homeomorphism (see [17, Theorem 17.14]). Therefore,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is homeomorphic to
$\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$.

Finally, if $D\in\mathcal{F}_{1}(\mathcal{N})$, then there exists $x\in X$
such that $D=\\{\pi(x)\\}$. Since $\pi(x)\in\mathcal{NB}(\mathcal{F}_{1}(X))$,
$g(\pi(x))\in\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$. Therefore,
$D\in\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$ and
$\mathcal{F}_{1}(\mathcal{N})\subseteq\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$.
∎

##  4 More about the property of Kelley

In this section, the show some extra properties related to a continuum with
the property of Kelley and its hyperspace of nonblockers. The main result is
Theorem 4.8 when we show that if $X$ is decomposable with the property of
Kelley, $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum and
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$, then $X$ is a
aposyndetic.

######  Lemma 4.1.

Let $X$ be a decomposable continuum with the property of Kelley. Let
$A\in\mathcal{C}(X)$ and let $x,y\in X\setminus A$ be such that
$\kappa_{X\setminus A}(x)\cap\kappa_{X\setminus A}(y)=\emptyset$. If
$\kappa_{X\setminus A}(x)\cap\mathrm{Cl}(\kappa_{X\setminus
A}(y))\neq\emptyset$, then $\kappa_{X\setminus
A}(x)\subseteq\mathrm{Cl}(\kappa_{X\setminus A}(y))$.

###### Proof.

Let $A\in\mathcal{C}(X)$ and let $x,y\in X\setminus A$ be such that
$\kappa_{X\setminus A}(x)\cap\kappa_{X\setminus A}(y)=\emptyset$ and
$\kappa_{X\setminus A}(x)\cap\mathrm{Cl}(\kappa_{X\setminus
A}(y))\neq\emptyset$. Let $z\in\kappa_{X\setminus
A}(x)\cap\mathrm{Cl}(\kappa_{X\setminus A}(y))$. Suppose that
$\kappa_{X\setminus A}(x)\setminus\mathrm{Cl}(\kappa_{X\setminus
A}(y))\neq\emptyset$. Let $w\in\kappa_{X\setminus
A}(x)\setminus\mathrm{Cl}(\kappa_{X\setminus A}(y))$. Since
$\\{z,w\\}\subseteq\kappa_{X\setminus A}(x)$, there exists a continuum $L$
such that $\\{z,w\\}\subseteq L$ and $L\cap A=\emptyset$. Let $U$ be an open
subset of $X$ such that $w\in U$ and $U\cap\mathrm{Cl}(\kappa_{X\setminus
A}(y))=\emptyset$. Note that $L\in\langle X\setminus A,U\rangle$ where
$\langle X\setminus A,U\rangle$ is an open subset of $2^{X}$. Since
$z\in\mathrm{Cl}(\kappa_{X\setminus A}(y))\setminus\kappa_{X\setminus A}(y)$,
there exists a sequence $(z_{n})_{n\in\mathbb{N}}$ in $\kappa_{X\setminus
A}(y)$ such that $\lim_{n\to\infty}z_{n}=z$. Since $X$ has the property of
Kelley, there exists a sequence of subcontinua $(L_{n})_{n\in\mathbb{N}}$ of
$X$ such that $z_{n}\in L_{n}$ for each $n\in\mathbb{N}$, and
$\lim_{n\to\infty}L_{n}=L$. Hence, there exists $k\in\mathbb{N}$ such that
$L_{i}\in\langle X\setminus A,U\rangle$ for each $i\geq k$. Let $m>k$. Hence,
$L_{m}\cap\kappa_{X\setminus A}(y)\neq\emptyset$ and $L_{m}\cap A=\emptyset$.
Thus, $L_{m}\subseteq\kappa_{X\setminus A}(y)$. This contradicts that
$L_{m}\cap U\neq\emptyset$ and $U\cap\mathrm{Cl}(\kappa_{X\setminus
A}(y))=\emptyset$. Therefore, $\kappa_{X\setminus
A}(x)\subseteq\mathrm{Cl}(\kappa_{X\setminus A}(y))$. ∎

######  Lemma 4.2.

Let $X$ be a continuum and let $(A_{n})_{n\in\mathbb{N}}$ be a sequence of
pairwise disjoint subcontinua of $X$ such that $\lim_{n\to\infty}A_{n}=A$ for
some $A\in\mathcal{C}(X)$. Then, there exists a countable set $D$ such that
$D\subseteq\bigcup_{n\in\mathbb{N}}A_{n}$, $|D\cap A_{i}|=1$ for each
$i\in\mathbb{N}$, and $A\subseteq\mathrm{Cl}(D)$.

###### Proof.

Let $E=\\{e_{n}:n\in\mathbb{N}\\}$ be a dense subset of $A$. For each
$n\in\mathbb{N}$, let $(x_{i}^{n})_{i\in\mathbb{N}}$ be a sequence such that
$x_{i}^{n}\in A_{i}$ for each $i\in\mathbb{N}$, and
$\lim_{i\to\infty}x_{i}^{n}=e_{n}$ for each $n\in\mathbb{N}$. Let
$\mathcal{N}_{1},\mathcal{N}_{2}\ldots$ be a sequence of pairwise disjoint
subsets of $\mathbb{N}$ such that $\mathcal{N}_{n}$ is infinite for each
$n\in\mathbb{N}$ and $\mathbb{N}=\bigcup_{n\in\mathbb{N}}\mathcal{N}_{n}$. Let
$D=\bigcup_{n\in\mathbb{N}}D_{n}$ where
$D_{n}=\\{x_{i}^{n}:i\in\mathcal{N}_{n}\\}$. It is not difficult to see that
$D$ satisfies all the conditions in our lemma. ∎

######  Lemma 4.3.

Let $X$ be a continuum and let $A$ be a proper subcontinuum of $X$ such that
$X\setminus A$ is not connected by continua. Let $B\in
2^{X}\setminus\mathcal{C}(X)$ be such that $\mathrm{Int}(B)=\emptyset$ and
$B\subseteq X\setminus A$. Suppose that $B$ satisfies the following
conditions:

  1. 1.

$D\in\mathcal{NB}(\mathcal{F}_{1}(X))$ for every component $D$ of $B$;

  2. 2.

$\mathrm{Cl}(\kappa_{X\setminus A}(x))\cap\kappa_{X\setminus
A}(x^{\prime})=\emptyset$ for every $x,x^{\prime}\in X\setminus A$ such that
$\kappa_{X\setminus A}(x)\cap\kappa_{X\setminus A}(x^{\prime})=\emptyset$
$\mathrm{Cl}(\kappa_{X\setminus A}(x))\cap B\neq\emptyset$ and
$\mathrm{Cl}(\kappa_{X\setminus A}(x^{\prime}))\cap B\neq\emptyset$; and

  3. 3.

$B\cap\kappa_{X\setminus A}(x)$ is a component of $B$ for each $x\in
X\setminus A$ such that $\kappa_{X\setminus A}(x)\cap B\neq\emptyset$.

Then, $B\in\mathcal{NB}(\mathcal{F}_{1}(X))$.

###### Proof.

We prove that $X\setminus B$ is connected by continua. To this end, we show
that for every $x\in X\setminus B$, there exists a subcontinuum $L_{x}$ of
$X\setminus B$ such that $x\in L_{x}$ and $L_{x}\cap A\neq\emptyset$. Let
$x\in X\setminus B$. If $x\in A$, then we define $L_{x}=A$ and clearly $x\in
L_{x}$, $L_{x}\cap B=\emptyset$ and $L_{x}\cap A\neq\emptyset$. Hence, we
suppose that $x\in X\setminus A$. Note that if $\mathrm{Cl}(\kappa_{X\setminus
A}(x))\cap B=\emptyset$, then $L_{x}=\mathrm{Cl}(\kappa_{X\setminus A}(x))$ is
a continuum satisfying the conditions. Thus, assume that
$\mathrm{Cl}(\kappa_{X\setminus A}(x))\cap B\neq\emptyset$. Let
$z\in\mathrm{Cl}(\kappa_{X\setminus A}(x))\cap B$. It is clear that
$\mathrm{Cl}(\kappa_{X\setminus A}(z))\cap B\neq\emptyset$. Thus, by 2,
$\kappa_{X\setminus A}(x)\cap\kappa_{X\setminus A}(z)\neq\emptyset$ and
$\kappa_{X\setminus A}(x)=\kappa_{X\setminus A}(z)$. Observe that we showed
that

$\mathrm{Cl}(\kappa_{X\setminus A}(x))\cap B=\kappa_{X\setminus A}(x)\cap B.$
(4.1)

Furthermore, $\kappa_{X\setminus A}(x)\cap B$ is a component of $B$, by 3.
Since $B$ is not connected, there exists $b\in B$ such that $b\in
X\setminus\mathrm{Cl}(\kappa_{X\setminus A}(x))$ (see (4.1)). Hence,
$X\setminus\mathrm{Cl}(\kappa_{X\setminus A}(x))\neq\emptyset$. Let
$U=X\setminus\mathrm{Cl}(\kappa_{X\setminus A}(x))$ and let
$D=\kappa_{X\setminus A}(x)\cap B$. We know that
$D\in\mathcal{NB}(\mathcal{F}_{1}(X))$, by 1. Since $x\notin D$, there exists
a subcontinuum $H$ of $X\setminus D$ such that $x\in H$ and $H\cap
U\neq\emptyset$. Note that $H\cap A\neq\emptyset$. Let $L_{x}$ be the
component of $H\cap\mathrm{Cl}(\kappa_{X\setminus A}(x))$ such that $x\in
L_{x}$. If $L_{x}\cap A=\emptyset$, then there exists a subcontinuum $J$ of
$H\setminus A$ such that $L_{x}\subseteq J$ and $J\neq L_{x}$, by [11,
Corollary 5.5]. Thus, $J\subseteq\kappa_{X\setminus A}(x)$ contradicting that
$L_{x}$ is a component of $H\cap\mathrm{Cl}(\kappa_{X\setminus A}(x))$; i.e.,
$L_{x}\cap A\neq\emptyset$. Since $H\cap D=\emptyset$ and $L_{x}\subseteq H$,
$L_{x}\cap D=\emptyset$. Furthermore,
$L_{x}\subseteq\mathrm{Cl}(\kappa_{X\setminus A}(x))$ and hence, $L_{x}\cap
B=\emptyset$, by (4.1). Therefore, for every $x\in X\setminus B$, there exists
a subcontinuum $L_{x}$ of $X\setminus B$ such that $x\in L_{x}$ and $L_{x}\cap
A\neq\emptyset$.

If $x,y\in X\setminus B$, then there exist subcontinua $L_{x}$ and $L_{y}$ of
$X\setminus B$ such that $x\in L_{x},y\in L_{y}$, $L_{x}\cap A\neq\emptyset$
and $L_{y}\cap A\neq\emptyset$. Thus, $L_{x}\cup A\cup L_{y}$ is a proper
subcontinuum of $X\setminus B$ containing both $x$ and $y$. Therefore,
$X\setminus B$ is connected by continua and
$B\in\mathcal{NB}(\mathcal{F}_{1}(X))$. ∎

######  Theorem 4.4.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$ and
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact. If $A\in\mathcal{C}(X)$ is such
that $\kappa_{X\setminus A}(w)$ is dense for some $w\in X\setminus A$, then
there exists $L\in\mathcal{NB}(\mathcal{F}_{1}(X))$ such that $A\subseteq L$.

###### Proof.

Let $A\in\mathcal{C}(X)$ and let $w\in X\setminus A$ be such that
$\kappa_{X\setminus A}(w)$ is dense in $X$. If
$A\in\mathcal{NB}(\mathcal{F}_{1}(X))$, then $L=A$. Hence, suppose that
$A\in\mathcal{C}(X)\setminus\mathcal{NB}(\mathcal{F}_{1}(X))$. Let $y\in
X\setminus A$ be such that $\mathrm{Cl}(\kappa_{X\setminus A}(y))\neq X$. We
define

$L=A\cup\\{y\in X\setminus A:\mathrm{Cl}(\kappa_{X\setminus A}(y))\neq X\\}.$
(4.2)

We see that $\mathrm{Cl}(\kappa_{X\setminus A}(y))\subseteq L$ for each $y\in
L\setminus A$. Let $y\in L\setminus A$ and let
$z\in\mathrm{Cl}(\kappa_{X\setminus A}(y))$. Note that if
$z\in\mathrm{Cl}(\kappa_{X\setminus A}(y))\setminus A$, then
$\kappa_{X\setminus A}(z)\subseteq\mathrm{Cl}(\kappa_{X\setminus A}(y))$, by
Lemma 4.1. Hence, $\mathrm{Cl}(\kappa_{X\setminus A}(z))\neq X$ and $z\in L$.
Therefore, $\mathrm{Cl}(\kappa_{X\setminus A}(y))\subseteq L$ for each $y\in
L\setminus A$.

We prove that $L$ is a continuum. Since $\mathrm{Cl}(\kappa_{X\setminus
A}(y))\cap A\neq\emptyset$ for each $y\in L\setminus A$, and
$L=A\cup(\bigcup\\{\mathrm{Cl}(\kappa_{X\setminus A}(y)):y\in L\setminus
A\\})$, we have that $L$ is connected. We show that $L$ is closed. Let
$(x_{n})_{n\in\mathbb{N}}$ be a sequence of $L$ such that
$\lim_{n\to\infty}x_{n}=x$ for some $x\in X$. We see that $x\in L$. Observe
that if there exist $y_{1},y_{2},\ldots,y_{k}$ in $L\setminus A$ such that

$\mathcal{N}=\left\\{n\in\mathbb{N}:x_{n}\in
A\cup\mathrm{Cl}(\kappa_{X\setminus
A}(y_{1}))\cup\mathrm{Cl}(\kappa_{X\setminus
A}(y_{2}))\cup\cdots\cup\mathrm{Cl}(\kappa_{X\setminus A}(y_{k}))\right\\}$

is infinite, then $(x_{n})_{n\in\mathcal{N}}$ converges to some point of
$A\cup\mathrm{Cl}(\kappa_{X\setminus
A}(y_{1}))\cup\mathrm{Cl}(\kappa_{X\setminus
A}(y_{2}))\cup\cdots\cup\mathrm{Cl}(\kappa_{X\setminus A}(y_{k}))$. Hence,
$x\in L$. Thus, by Lemma 4.1, we may assume that there exists a sequence
$(y_{n})_{n\in\mathbb{N}}$ of $L\setminus A$ such that
$x_{n}\in\kappa_{X\setminus A}(y_{n})$ for each $n\in\mathbb{N}$, and

$\mathrm{Cl}(\kappa_{X\setminus A}(y_{i}))\cap\kappa_{X\setminus
A}(y_{j})=\emptyset\text{ whenever that }i\neq j.$ (4.3)

Suppose that $x\notin L$. We will have a contradiction, showing that
$X\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Let $U_{1},\ldots,U_{m}$ be open
subsets of $X$ such that $X\in\langle U_{1},\ldots,U_{m}\rangle$. Since
$x\notin L$, $\kappa_{X\setminus A}(x)$ is dense. Thus, there exists a
subcontinuum $M$ of $X\setminus A$ such that $x\in M$ and $M\in\langle
U_{1},\ldots,U_{m}\rangle$. Notice that $M\in\langle
U_{1},\ldots,U_{m}\rangle\cap\langle X\setminus A\rangle$. Since $X$ has the
property of Kelley, there exists a sequence of subcontinua
$(M_{n})_{n\in\mathbb{N}}$ of $X$ such that $x_{n}\in M_{n}$ for each
$n\in\mathbb{N}$, and $\lim_{n\to\infty}M_{n}=M$. Hence, there exists
$k\in\mathbb{N}$ such that $M_{n}\in\langle
U_{1},\ldots,U_{m}\rangle\cap\langle X\setminus A\rangle$ for each $n\geq k$.
Since $M_{n}\cap A=\emptyset$ and $x_{n}\in M_{n}$, we have that
$M_{n}\subseteq\kappa_{X\setminus A}(y_{n})$ for each $n\geq k$. Thus,
$(M_{n})_{n\geq k}$ is a sequence of pairwise disjoint subcontinua of $X$, by
(4.3). By Lemma 4.2, there exists a countable set $D\subseteq\bigcup_{n\geq
k}M_{n}$ such that $|D\cap M_{n}|=1$ for each $n\geq k$ and
$M\subseteq\mathrm{Cl}(D)$. We know that
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$. If $F$ is a
finite subset of $D$, then $F\in\mathcal{NB}(\mathcal{F}_{1}(X))$, by Lemma
4.3. Let $R_{m}=(\bigcup_{k\leq i\leq m}M_{i})\cap D$, for each $m\geq k$.
Note that $R_{n}\in\mathcal{NB}(\mathcal{F}_{1}(X))$ for each $n\geq m$.
Furthermore, since $\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact, we may
suppose that $\lim_{m\to\infty}R_{m}=Q$ for some
$Q\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Since $M\subseteq\mathrm{Cl}(D)$ and
$\mathrm{Cl}(D)\subseteq Q$, $M\subseteq Q$. Thus, $Q\in\langle
U_{1},\ldots,U_{m}\rangle$. Therefore, $X$ is a limit point of
$\mathcal{NB}(\mathcal{F}_{1}(X))$. This contradicts the compactness of
$\mathcal{NB}(\mathcal{F}_{1}(X))$ and the fact that
$X\notin\mathcal{NB}(\mathcal{F}_{1}(X))$. We have that $L$ is a subcontinuum
of $X$.

Finally, $\kappa_{X\setminus A}(x)$ is dense and $L\cap\kappa_{X\setminus
A}(x)=\emptyset$. Hence, $\mathrm{Int}(L)=\emptyset$. Furthermore,
$\kappa_{X\setminus A}(z)$ is dense, for every $z\in X\setminus L$ (see
definition of $L$ in (4.2)). Therefore, $L\in\mathcal{NB}(\mathcal{F}_{1}(X))$
and $A\subseteq L$. ∎

######  Theorem 4.5.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$ and
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact. If
$A\in\mathcal{NB}(\mathcal{F}_{1}(X))$, then $X\setminus A$ is connected by
continua.

###### Proof.

Let $x\in X\setminus A$. We prove that $X\setminus A=\kappa_{X\setminus
A}(x)$. Suppose that there exists $y\in X\setminus A$ such that
$\kappa_{X\setminus A}(y)\cap\kappa_{X\setminus A}(x)=\emptyset$. Note that
$\kappa_{X\setminus A}(y)$ is dense, because
$A\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Since $\kappa_{X\setminus A}(x)$ is
dense, there exists a sequence of subcontinua $(L_{n})_{n\in\mathbb{N}}$ of
$\kappa_{X\setminus A}(x)$ such that $L_{n}\subseteq L_{n+1}$ for each
$n\in\mathbb{N}$, and $\lim_{n\to\infty}L_{n}=X$ [4, Proposition 2.2]. Since
$\kappa_{X\setminus A}(y)\cap L_{n}=\emptyset$ and $\kappa_{X\setminus A}(y)$
is dense, we have that $\kappa_{X\setminus A}(y)\subseteq\kappa_{X\setminus
L_{n}}(y)$; i.e., $\kappa_{X\setminus L_{n}}(y)$ is dense for each
$n\in\mathbb{N}$. Hence, there exists a continuum
$K_{n}\in\mathcal{NB}(\mathcal{F}_{1}(X))$ such that $L_{n}\subseteq K_{n}$
for each $n\in\mathbb{N}$, by Theorem 4.4. Since $\lim_{n\to\infty}L_{n}=X$,
$\lim_{n\to\infty}K_{n}=X$. Thus, $X\in\mathcal{NB}(\mathcal{F}_{1}(X))$, a
contradiction. Therefore, $X\setminus A=\kappa_{X\setminus A}(x)$ and
$X\setminus A$ is connected by continua. ∎

######  Theorem 4.6.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$ and
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. If $W$ is a proper
subcontinuum of $X$, then each component of $X\setminus W$ is open.

###### Proof.

Suppose to the contrary that there exist a component $C$ of $X\setminus W$ and
$p\in C$ such that $p\in\mathrm{Cl}(X\setminus C)$.

######  Claim 4.7.

$\mathrm{Cl}(\kappa_{X\setminus W}(p))\subseteq\mathrm{Cl}(X\setminus C)$.

To prove the Claim 4.7, we need to show that $\kappa_{X\setminus
W}(p)\subseteq\mathrm{Cl}(X\setminus C)$. Let $q\in\kappa_{X\setminus W}(p)$.
Hence, there exists a subcontinuum $P$ of $X\setminus W$ such that
$\\{p,q\\}\subseteq P$. Let $(p_{n})_{n\in\mathbb{N}}$ be a sequence of
$X\setminus C$ such that $\lim_{n\to\infty}p_{n}=p$. Since $X$ is compact, we
may suppose that $(p_{n})_{n\in\mathbb{N}}$ is a sequence in $X\setminus(W\cup
C)$. Since $X$ has the property of Kelley, there exists a sequence of
subcontinua $(P_{n})_{n\in\mathbb{N}}$ of $X$ such that $p_{n}\in P_{n}$ for
all $n$, and $\lim_{n\to\infty}P_{n}=P$. Notice that $\langle X\setminus
W\rangle$ is an open subset of $2^{X}$, and $P\in\langle X\setminus W\rangle$.
Hence, there exists $k\in\mathbb{N}$ such that $P_{n}\in\langle X\setminus
W\rangle$ for each $n\geq k$. Since $p_{n}$ belongs to a different component
to $C$ in $X\setminus W$, and $P_{n}\cap W=\emptyset$ for each $n\geq k$, we
have that $P_{n}\cap C=\emptyset$ for all $n\geq k$. Hence, $P_{n}\subseteq
X\setminus(W\cup C)$ for each $n\geq k$. Thus,
$P\subseteq\mathrm{Cl}(X\setminus(W\cup C))\subseteq\mathrm{Cl}(X\setminus
C)$. Therefore, $q\in\mathrm{Cl}(X\setminus C)$ and $\kappa_{X\setminus
W}(p)\subseteq\mathrm{Cl}(X\setminus C)$. We have proved the Claim 4.7.

We prove that $2^{\mathrm{Cl}(\kappa_{X\setminus
W}(p))}\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$. Let
$F=\\{w_{1},\ldots,w_{m}\\}$ be a finite subset of $\kappa_{X\setminus W}(p)$.
By Claim 4.7, for each $j\in\\{1,\ldots,m\\}$, there exists a sequence
$(z^{j}_{n})_{n\in\mathbb{N}}$ in $X\setminus C$ such that
$\lim_{n\to\infty}z^{j}_{n}=w_{j}$. Since $W$ is compact, we may suppose that
$(z^{j}_{n})_{n\in\mathbb{N}}$ is a sequence in $X\setminus(C\cup W)$ for each
$j\in\\{1,\ldots,m\\}$. Furthermore, observe that if $D$ is a component of
$X\setminus W$ and $D\neq C$, then $\\{n\in\mathbb{N}:z^{j}_{n}\in D\\}$ is a
finite set for each $j\in\\{1,\ldots,m\\}$. Hence, if necessary, we can take a
subsequence $(z^{j}_{n_{k}})_{k\in\mathbb{N}}$ of
$(z^{j}_{n})_{n\in\mathbb{N}}$ for each $j\in\\{1,\ldots,m\\}$, such that
$|\\{z_{n_{k}}^{j}:k\in\mathbb{N},j\in\\{1,\ldots,m\\}\\}\cap D|\leq 1$ for
each component $D$ of $X\setminus W$. Let
$F_{i}=\\{z^{1}_{n_{i}},\ldots,z^{m}_{n_{i}}\\}$ for each $i\in\mathbb{N}$.
Since $\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$,
$F_{i}\in\mathcal{NB}(\mathcal{F}_{1}(X))$ for each $i\in\mathbb{N}$, by Lemma
4.3. It is not difficult to see that $\lim_{i\to\infty}F_{i}=F$. Since
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact,
$F\in\mathcal{NB}(\mathcal{F}_{1}(X))$. We have that
$F\in\mathcal{NB}(\mathcal{F}_{1}(X))$ for every finite subset $F$ of
$\kappa_{X\setminus W}(p)$. It is easy to see that the collection of all
finite subsets of $\kappa_{X\setminus W}(p)$ is dense of
$2^{\mathrm{Cl}(\kappa_{X\setminus W}(p))}$. Hence, since
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is compact,
$2^{\mathrm{Cl}(\kappa_{X\setminus
W}(p))}\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$.

Let $Q=W\cap\mathrm{Cl}(\kappa_{X\setminus W}(p))$. Since $Q\in
2^{\mathrm{Cl}(\kappa_{X\setminus W}(p))}$,
$Q\in\mathcal{NB}(\mathcal{F}_{1}(X))$. Note that $\kappa_{X\setminus
W}(p)\subseteq C$ and $C$ is a component of $X\setminus W$. Hence,
$\mathrm{Cl}(\kappa_{X\setminus W}(p))\setminus\kappa_{X\setminus
W}(p)\subseteq W$. Thus, $p\notin Q$ and $\kappa_{X\setminus
Q}(p)\subseteq\kappa_{X\setminus W}(p)\subseteq C$. This clearly is a
contradiction, because $\kappa_{X\setminus Q}(p)$ has to be dense and $C$ is
not dense. Therefore, each component of $X\setminus W$ is open. ∎

######  Theorem 4.8.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$. If
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum, then $X$ is aposyndetic.

###### Proof.

Let $p\in X$. We show that $X$ is aposyndetic at $p$ with respect to each
point of $X\setminus\\{p\\}$ (see Definition 2.2). Let $q\in
X\setminus\\{p\\}$. Since $\\{q\\}\in\mathcal{NB}(\mathcal{F}_{1}(X))$,
$X\setminus\\{q\\}$ is connected by continua, by Theorem 4.5. Furthermore,
there exists a sequence of subcontinua $(L_{n})_{n\in\mathbb{N}}$ of
$X\setminus\\{q\\}$ such that
$X\setminus\\{q\\}=\bigcup_{n\in\mathbb{N}}L_{n}$, by Lemma 2.3. Since $X$ is
a Baire space, there exists $k\in\mathbb{N}$ such that
$\mathrm{Int}(L_{k})\neq\emptyset$ [17, Corollary 25.4]. Since
$X\setminus\\{q\\}$ is connected by continua, there exists a subcontinuum $R$
of $X\setminus\\{q\\}$ such that $p\in R$ and $R\cap L_{k}\neq\emptyset$.
Since $X$ has the property of Kelley, there exists a continuum $W$ such that
$L_{k}\cup R\subseteq\mathrm{Int}(W)\subseteq W\subseteq X\setminus\\{q\\}$,
by Theorem 2.6. Thus, $p\in\mathrm{Int}(W)$ and $q\notin W$. Therefore, $X$ is
aposyndetic at $p$ with respect to $q$ and hence, $X$ is aposyndetic. ∎

##  5 The simple closed curve

In this section, we present our main results. We show two characterizations in
theorems 5.1 and 5.3 of the simple closed curve.

######  Theorem 5.1.

Let $X$ be a decomposable continuum with the property of Kelley such that
$\mathcal{F}_{1}(X)\subseteq\mathcal{NB}(\mathcal{F}_{1}(X))$. Then,
$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum if and only if $X$ is a
simple closed curve.

###### Proof.

Suppose that $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. Then, by
Theorem 4.8, $X$ is aposyndetic. Hence, $X$ is semi-locally connected, by [8,
Theorem 1.7.17]. We prove that $X$ is locally connected. Let $p\in X$ and let
$V$ be an open subset of $X$ such that $p\in V$. Since $X$ is semi-locally
connected, there exists an open subset $U$ of $X$ such that $p\in U\subseteq
V$ and $X\setminus U$ has a finite number of components. Let
$L_{1},L_{2},\ldots,L_{k}$ be the components of $X\setminus U$. Since
$\\{p\\}\in\mathcal{NB}(\mathcal{F}_{1}(X))$, $X\setminus\\{p\\}$ is connected
by continua, by Theorem 4.5. Thus, for each $i\in\\{2,\ldots,k\\}$, there
exists a subcontinuum $M_{i}$ of $X\setminus\\{p\\}$ such that $M_{i}\cap
L_{1}\neq\emptyset$ and $M_{i}\cap L_{i}\neq\emptyset$.

Let

$W=(L_{1}\cup L_{2}\cup\cdots\cup L_{k})\cup(M_{2}\cup M_{3}\cup\cdots\cup
M_{k}).$

Observe that $W$ is a continuum and $p\notin W$. Let $J$ be the component of
$X\setminus W$ such that $p\in J$. By Theorem 4.6, $J$ is open. Furthermore,
$J\subseteq X\setminus W\subseteq U\subseteq V$. Thus, $X$ is locally
connected at $p$. Therefore, $X$ is a simple closed curve, by Theorem 2.5.

Conversely, $\mathcal{NB}(\mathcal{F}_{1}(S^{1}))=\mathcal{F}_{1}(S^{1})$.
Therefore, $\mathcal{NB}(\mathcal{F}_{1}(S^{1}))$ is a continuum. ∎

######  Theorem 5.2.

Let $X$ be a decomposable continuum with the property of Kelley.
If$\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum, then there exists a map
$f\colon X\to S^{1}$ such that $f$ is open, monotone and $f^{-1}(z)$ is a
terminal subcontinuum of $X$ for every $z\in S^{1}$.

###### Proof.

We know that $\mathcal{N}=\\{\pi(x):x\in X\\}$ is a continuous decomposition
of $X$ such that $\pi(x)$ is a terminal subcontinuum of $X$ for every $x\in
X$, by Proposition 3.8 and Theorem 3.9. Let $\rho\colon X\to\mathcal{N}$ be
the quotient map. Since $\rho$ is monotone, $\mathcal{N}$ has the property of
Kelley, by [15, Theorem 4.3]. It is not difficult to see that $\mathcal{N}$ is
decomposable. Furthermore, $\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$ is a
continuum and
$\mathcal{F}_{1}(\mathcal{N})\subseteq\mathcal{NB}(\mathcal{F}_{1}(\mathcal{N}))$,
by Theorem 3.10. Thus, $\mathcal{N}$ is a simple closed curve, by Theorem 5.1.
Let $g\colon\mathcal{N}\to S^{1}$ be a homeomorphism. It is clear that
$f=g\circ\rho$ is a map from $X$ to $S^{1}$ monotone, open and $f^{-1}(z)$ is
terminal for each $z\in S^{1}$. ∎

The following result give us a partial answer to Question 1.2.

######  Theorem 5.3.

Let $X$ be an hereditarily decomposable continuum with the property of Kelley.
Then, $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum if and only if $X$ is
a simple closed curve.

###### Proof.

We assume that $\mathcal{NB}(\mathcal{F}_{1}(X))$ is a continuum. By Theorem
5.2, there exists a map $f\colon X\to S^{1}$ such that $f$ is open, monotone
and $f^{-1}(z)$ is a terminal subcontinuum of $X$ for each $z\in S^{1}$. We
show that $|f^{-1}(z)|=1$ for every $z\in S^{1}$. Let $z_{0}\in S^{1}$ and let
$\alpha$ be an arc in $S^{1}$ such that $z_{0}\in\alpha$. Let
$L=f^{-1}(\alpha)$. Since $f$ is monotone, $L$ is a continuum. Let $p,q\in L$
be such that $\alpha$ is irreducible between $f(p)$ and $f(q)$. We see that
$L$ is irreducible between $p$ and $q$. Let $K$ be a subcontinuum of $L$ such
that $p,q\in K$. Hence, $f(K)=\alpha$. Since $f^{-1}(z)$ is terminal for each
$z\in\alpha$ and $f^{-1}(z)\cap K\neq\emptyset$, we have that
$f^{-1}(z)\subseteq K$ for each $z\in\alpha$. Thus, $K=f^{-1}(\alpha)$ and
$K=L$. Therefore, $L$ is irreducible.

Let $g=f|_{L}\colon L\to\alpha$. Since $f$ is open, $g$ is open, by [16,
Theorem 4.31]. Furthermore, $g$ monotone and $g^{-1}(z)$ is terminal for each
$z\in\alpha$. Thus, $D=\\{x\in L:g^{-1}(g(x))=x\\}$ is a dense subset of $L$,
by [13, Corollary 9, p.130]. Since $g$ is open, $\\{g^{-1}(z):z\in\alpha\\}$
is a continuous decomposition of $L$ (see [16, Theorem 4.32, p.130]). Thus,
$g^{-1}(g(x))=x$ for each $x\in L$. In particular, $|f^{-1}(z_{0})|=1$.
Therefore, $f$ is injective, and $f$ is a homeomorphism.

It is clear that $\mathcal{NB}(\mathcal{F}_{1}(S^{1}))$ is a continuum.
Therefore, we complete the proof of our theorem. ∎

The first author thanks La Vicerrectoría de Investigación y Extensión de la
Universidad Industrial de Santander y su Programa de Movilidad for financial
support.

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Escuela de Matemáticas, Facultad de Ciencias, Universidad Industrial de
Santander, Ciudad Universitaria, Carrera 27 Calle 9, Bucaramanga, Santander,
A.A. 678, COLOMBIA.

E-mail address<EMAIL_ADDRESS>
Escuela de Matemáticas, Facultad de Ciencias, Universidad Industrial de
Santander, Ciudad Universitaria, Carrera 27 Calle 9, Bucaramanga, Santander,
A.A. 678, COLOMBIA.

E-mail address<EMAIL_ADDRESS>