# Connectivity degrees of complements of closed sets in continua

Mauricio Chacón-Tirado<EMAIL_ADDRESS>César Piceno
<EMAIL_ADDRESS>Benemérita Universidad Autónoma de Puebla, Facultad de
Ciencias Físico Matemáticas, Avenida San Claudio y 18 Sur, Colonia San Manuel.
Edificio FM1-101B, Ciudad Universitaria, C.P. 72570, Puebla, Mexico.

###### Abstract

In the literature, various types of points and meager sets whose complements
are connected have been studied, such as colocally connected points, non-weak
cut points/sets, non-block points/sets, shore points/sets, etc. We extend that
study, in the following way: considering a continuum $X$ and a natural number
$n$, we investigate sets $A\in 2^{X}$ meeting the criterion that $X-A$ has at
most $n$ components, and we introduce degrees of connectivity of the
complement of $A$. When $n=1$ and $A$ is meager or singleton, these new
definitions are equivalent to the known definitions of non-cut points/sets.

###### keywords:

Continuum , hyperspace , non-cut points , shore points, non-block points ,
shore points , aposyndesis.

###### MSC:

[2020] Primary 54B20, 54F15.

††journal: a journal

##  1 Introduction

One of the main topics of interest in topology is being able to determine
whether a space is connected or not, and, when a space $X$ is connected, it is
interesting to determine how “strongly connected” $X$ is. In the case of
continua, various types of points and sets whose complements are connected
have been studied, and some “degree of connectivity” of these complements has
also been investigated. One of the most relevant works in this regard was
published by R.L. Moore [8], where the existence of non-cut points in all
continua is demonstrated. Another important result concerning the degree of
connectivity of the complement of a point in a continuum is the one obtained
by Bing [1], which states that for any point of a nondegenerate metrizable
continuum, there is a proper continuumwise connected dense subset containing
that point. Some of the articles that can be consulted on the topic are [3],
[4], [5], [6], [9], [12] and [13].

If the space is not connected, we are also interested in knowing if it is
composed of a finite number of components and how “strongly disconnected” is
the space. In a continuum, it is of particular interest to study sets that cut
the space, and if they do (or not), we are also interested in knowing the
degree of connectivity of their complements. For example, in Section 5, we
show that in a locally connected continuum $X$, if $A\in 2^{X}$ and $X-A$ has
a finite number of components, then each component is continuumwise connected.

Besides this introduction, this paper contains 4 more sections. In Section 2,
we provide the definitions that we will use throughout the paper. In
particular, we define degrees of connectivity, which we call $n$-$Q1$ to
$n$-$Q7$, and $n$-$Qo$, being $n$-$Q1$ the stronger one and $n$-$Q7$ the
weakest one. Something we wish to emphasize is the uniformity of the
definitions for classifying the degree of connectivity of a space, which makes
some results straightforward.

In Section 3, for each degree of connectivity we consider the hyperspace of
closed sets whose complements have that degree of connectivity, we explore the
relationships between those hyperspaces, whose elements we call non-$n$-cut
sets. We provide conditions to ensure when a complement that is $n$-$Q7$
implies also that is $n$-$Q1$. We offer tools to discover new non-$n$-cut sets
from others and ultimately examine the Borel classes of some hyperspaces of
non-$n$-cut sets.

In Section 4, we delve into investigating the types of functions that preserve
non-$n$-cut sets under their image or preimage.

Finally, in Section 5, we investigate the relationships between a continuum
$X$ and its hyperspaces of non-$n$-cut sets. Among other interesting results,
we provide a characterization of the arc. We discover that for irreducible
continua, certain hyperspaces of non-$n$-cut sets coincide. Additionally, we
prove that if $X$ is aposyndetic with respect to $A$ and the complement of $A$
has at most $n$ components, then for every neighborhood $U$ containing $A$,
there exists a neighborhood $V\subset U$ of $A$ such that the complement of
$V$ has at most $n$ components.

##  2 Definitions and notation

In this paper all spaces are metric. The set $\mathbb{N}$ represents the
positive integers. Given a subset $A$ of a space $X$, the closure and the
interior of $A$ are denoted by $cl_{X}(A)$ and $int_{X}(A)$, respectively, and
we omit the subindex when we feel there is no risk of confusion regarding our
space. A map is a continuous function. A continuum is a compact space with
more than one point. A continuum $X$ is aposyndetic at $p$ with respect to
$A$, where $p\in X$ and $A\subset X$, provided that there is a continuum
$B\subset X-A$ such that $p\in int_{X}(B)$. A continuum $X$ is aposyndetic
with respect to $A$ if $X$ is aposyndetic at $p$ with respect to $A$ for all
$p\in X-A$. A continuum $X$ is aposyndetic at $p$ provided that $X$ is
aposyndetic at $p$ with respect to each singleton $\\{q\\}\subset X-\\{p\\}$.
A continuum $X$ is mutually aposyndetic if for two distinct points $p,q\in X$,
there exist two subcontinua $A$ and $B$ of $X$ such that $p\in\text{int}(A)$,
$q\in\text{int}(B)$, and $A\cap B=\emptyset$.

A compact metric space $X$ is indecomposable provided that each subcontinuum
of $X$ has empty interior. A continuum $X$ is said to be irreducible about
$A\subset X$ provided that no proper subcontinuum of $X$ contains $A$. A
continuum $X$ is said to be irreducible provided that $X$ is irreducible about
$\\{p,q\\}$ for some $p,q\in X$, in which case we say $X$ is irreducible
between $p$ and $q$. A space $Y$ is continuumwise connected if any pair of
points is contained in a continuum $X\subset Y$. Let
$\mathcal{S}^{1}=\\{x\in\mathbb{R}^{2}:||x||=1\\}$.

Given a non-empty space $X$ and $n\in\mathbb{N}$, we consider the following
hyperspaces of $X$:

$2^{X}=\\{A\subset X:A\text{ is non-empty and compact}\\},$ $M(X)=\\{A\in
2^{X}:A\text{ has empty interior}\\},$ $C_{n}(X)=\\{A\in 2^{X}:A\text{ has at
most $n$ components}\\},$ $D_{0}(X)=\\{A\in 2^{X}:A\text{ has dimension
}0\\},$

and

$F_{n}(X)=\\{A\in 2^{X}:A\text{ has at most $n$ elements}\\}.$

These hyperspaces are endowed with the Hausdorff metric. We write $C(X)$
instead of $C_{1}(X)$, the elements of $C(X)$ are called _subcontinua_ of $X$.

Clearly $F_{n}(X)\subset C_{n}(X)\subset 2^{X}$ and $F_{1}(X)$ is homeomorphic
to $X$.

For a finite collection $X_{1},\ldots,X_{m}$ of subsets of $X$, we define
$\left\langle X_{1},\ldots,X_{m}\right\rangle$ as the set $\\{A\in
2^{X}:A\subset X_{1}\cup\ldots\cup X_{m}\text{ and }A\cap
X_{i}\neq\emptyset\text{ for each }i\in\\{1,\ldots,m\\}\\}$.

It is known that if $X_{1},\ldots,X_{m}$ are closed subsets of $X$, then
$\left\langle X_{1},\ldots,X_{m}\right\rangle$ is closed in $2^{X}$ and that
the collection of all subsets of the form $\left\langle
U_{1},\ldots,U_{m}\right\rangle$, where $U_{1},\ldots,U_{m}$ are open subsets
of $X$, is a base for the topology of $2^{X}$ (see [7]).

The objective of the following definition is to introduce the degree of
connectivity of a space.

######  Definition 2.1.

Given a non-empty space $X$ and $n\in\mathbb{N}$, we say that $X$ is:

  1. 1.

n-Q1 if there exists $B\in F_{n}(X)$, such that for each $x\in X$, there
exists a continuum $D\subset X$ such that $x\in int(D)$ and $B\cap
D\neq\emptyset$;

  2. 2.

n-Q2 if there exists $B\in F_{n}(X)$ such that, for each $x\in X$, there
exists a continuum $D\subset X$ such that $x\in D$ and $B\cap D\neq\emptyset$;

  3. 3.

n-Q3 if for every $x\in X$, there exists $B\in F_{n}(X)$ with $x\in B$, such
that for every non-empty open set $U$ of $X$, there exists a continuum
$D\subset X$ such that $B\cap D\neq\emptyset\neq D\cap U$;

  4. 4.

n-Qo if there exists $B\in F_{n}(X)$ such that for every non-empty open set
$U$ of $X$, there exists a continuum $D\subset X$ such that $B\cap
int(D)\neq\emptyset\neq int(D)\cap U$;

  5. 5.

n-Q4 if there exists $B\in F_{n}(X)$, such that for every non-empty open set
$U$ of $X$, there exists a continuum $D\subset X$ such that $B\cap
D\neq\emptyset\neq D\cap U$;

  6. 6.

n-Q5 if for each finite family $\mathcal{U}$ of non-empty open sets contained
in $X$, there exists $D\in C_{n}(X)$ such that $D\cap U\neq\emptyset$, for
every $U\in\mathcal{U}$;

  7. 7.

n-Q6 if for each collection of $n+1$ non-empty open sets
$U_{1},\ldots,U_{n+1}$ of $X$, there exists $D\in C_{n}(X)$ such that $D\cap
U_{i}\neq\emptyset$ for every $i\in\\{1\dots,n+1\\}$.

  8. 8.

n-Q7 if $X$ has at most $n$ components.

######  Note 2.2.

Clearly, for each $n\in\mathbb{N}$ and $m\in\\{o\\}\cup\\{1,\dots,6\\}$, being
$n$-$Qm$ implies being $(n+1)$-$Qm$, and being $n$-$Q(m+1)$, if
$m\in\\{1,\dots,5\\}$. Also, being $n$-$Q1$ implies being $n$-$Qo$, being
$n$-$Qo$ implies being $n$-$Q4$, being $n$-$Q4$ implies being $(n+1)$-$Q3$,
and being $n$-$Q6$ implies having at most $n$ components. In the following
figure, what has been stated here is represented.

$(n+1)$-$Q1$$(n+1)$-$Q2$$(n+1)$-$Q3$$(n+1)$-$Q4$$(n+1)$-$Q5$$(n+1)$-$Q6$$n$-$Q1$$n$-$Q2$$n$-$Q3$$n$-$Q4$$n$-$Q5$$n$-$Q6$$n$-$Qo$$(n+1)$-$Qo$
Figure 1: Relationships between degrees of connectivity.

Notice that in Figure 1, it is indicated that there is no relationship between
$n$-$Qo$ and $n$-$Q3$. Counterexamples to this fact are Example 3.2 for
$n$-$Qo\nRightarrow n$-$Q3$ and the Dyadic Solenoid for $n$-$Q3\nRightarrow
n$-$Qo$. On the other hand, we were unable to prove or deny $n$-$Q2\Rightarrow
n$-$Qo$. A question related to this, is Question 3.3.

The following definitions are provided to classify sets based on the degree of
connectivity of their complements. Most of them are generalizations of those
presented in [3] and [5].

######  Definition 2.3.

Given a non-degenerate compact metric space $X$ and $n\in\mathbb{N}$, an
element $A\in 2^{X}$ is said to be:

  1. 1.

set colocal connectedness of degree $n$ of $X$ provided that $A=X$ or $X-A$ is
n-Q1;

  2. 2.

not a weak cut set of degree $n$ of $X$ provided that $A=X$ or $X-A$ is n-Q2;

  3. 3.

nonblock set of degree $n$ of $X$ if $A=X$ or $X-A$ is n-Q3;

  4. 4.

a set that does not block opens of degree $n$ of $X$ provided that $A=X$ or
$X-A$ is n-Qo;

  5. 5.

weak nonblock set of degree $n$ of $X$ provided that $A=X$ or $X-A$ is n-Q4;

  6. 6.

a shore set of degree $n$ of $X$ provided that $A=X$ or $X-A$ is n-Q5;

  7. 7.

not a strong center set of degree $n$ of $X$ provided that $A=X$ or $X-A$ is
n-Q6.

We consider the following subspaces of $2^{X}$, these are called hyperspaces
of non-cut sets of degree $n$ of $X$:

  1. I.

${CC}_{n}(X)=\\{A\in 2^{X}:A\text{ is a set of colocal connectedness of degree
$n$ of $X$}\\}$;

  2. II.

${NWC}_{n}(X)=\\{A\in 2^{X}:A\text{ is not a weak cut set of degree $n$ of
}X\\}$;

  3. III.

${NB}_{n}(X)=\\{A\in 2^{X}:A\text{ is a nonblock set of degree $n$ of }X\\}$;

  4. IV.

${NBO}_{n}(X)=\\{A\in 2^{X}:A\text{ does not block opens of degree $n$ of
}X\\}$;

  5. V.

${NB}_{n}^{*}(X)=\\{A\in 2^{X}:A\text{ is a weak nonblock set of degree $n$ of
}X\\}$;

  6. VI.

$S_{n}(X)=\\{A\in 2^{X}:A\text{ is a shore set of degree $n$ of }X\\}$;

  7. VII.

${NSC}_{n}(X)=\\{A\in 2^{X}:A\text{ is not a strong center set of degree $n$
of }X\\}$;

  8. VIII.

${NC}_{n}(X)=\\{A\in 2^{X}:X-A\text{ has at most $n$ components}\\}$.

######  Note 2.4.

For a continuum $X$, according to our definitions and the definitions P1, P2,
P3, P4, P5 given in [3], $p$ is a P1 point if and only if $X-\\{p\\}$ is 1-Q1;
$p$ is a P2 point if and only if $X-\\{p\\}$ is 1-Q2; $p$ is a P3 point if and
only if $X-\\{p\\}$ is 1-Q4; $p$ is a P4 point if and only if $X-\\{p\\}$ is
1-Q5, and $p$ is a P5 point if and only if $X-\\{p\\}$ is 1-Q6.

######  Note 2.5.

The sets $NWC(X)$, $NB(F_{1}(X))$, $NB^{*}(F_{1}(X))$, $S(X)$, $NC(X)$ defined
in [5], coincide with the sets ${NWC}_{1}(X)\cap M(X)$, ${NB}_{1}(X)\cap
M(X)$, ${NB}^{*}_{1}(X)\cap M(X)$, $S_{1}(X)\cap M(X)$ and ${NC}_{1}(X)\cap
M(X)$, respectively.

##  3 General properties of non-$n$-cut sets

This section presents several key results in the context of hyperspaces of
non-$n$-cut sets of continua. These results establish relationships between
different families of non-$n$-cut sets, shedding light on their structural
properties and interconnections.

The following theorem is immediate from Note 2.2.

######  Theorem 3.1.

Given a continuum $X$ and $n\in\mathbb{N}$, the following conditions hold:

  1. 1.

If $H_{n}(X)$ represents an hyperspace of non-$n$-cut sets, then
$H_{n}(X)\subset H_{n+1}(X)$;

  2. 2.

${CC}_{n}(X)\subset{NWC}_{n}(X)\subset{NB}_{n}(X)\subset{NB}_{n}^{*}(X)\subset
S_{n}(X)\subset{NSC}_{n}(X)\subset{NC}_{n}(X);$

  3. 3.

${CC}_{n}(X)\subset{NBO}_{n}(X)\subset{NB}_{n}^{*}(X)$; and

  4. 4.

$NB_{n}^{*}(X)\subset NB_{n+1}(X)$.

The following example shows that the inclusion
${NBO}_{1}(X)\subset{NB}_{1}(X)$ is false in general.

######  Example 3.2.

The _harmonic fan_ is the set
$([0,1]\times\\{0\\})\cup(\cup\\{(x,\frac{x}{n}):x\in[0,1],n\in\mathbb{N}\\})$,
considered as a subspace of $\mathbb{R}^{2}$. Let $X$ be the harmonic fan and
let $A=\\{(\frac{1}{2},0)\\}$. Then, $A\in{NBO}_{1}(X)$ and
$A\notin{NB}_{1}(X)$.

From Example 3.2 and 3 from Theorem 3.1, we know that there are continua $X$
such that $NWC_{1}(X)\neq NBO_{1}(X)$. What we do not know is the following:

######  Question 3.3.

Does there exist a continuum $X$ such that $NWC_{1}(X)\not\subset NBO_{1}(X)$?

######  Note 3.4.

The examples ($P2\setminus P1$), ($P4\setminus P3$), ($P5\setminus P4$) and
($P6\setminus P5$) presented in [3] satisfy that
$\bigcup_{n=1}^{\infty}{CC}_{n}(X)\subsetneq{NWC}_{1}(X)$,
$\bigcup_{n=1}^{\infty}{NB^{*}}_{n}(X)\subsetneq S_{1}(X)$,
$\bigcup_{n=1}^{\infty}{S}_{n}(X)\subsetneq NSC_{1}(X)$ and
$\bigcup_{n=1}^{\infty}{NSC}_{n}(X)\subsetneq NC_{1}(X)$ respectively. While
the examples a) and b) of Remark 3.3 of [5] satisfy that
$\cup_{n=1}^{\infty}NWC_{n}(X)\subsetneq NB_{1}(X)$ and $NB_{1}(X)\subsetneq
NB_{1}^{*}(X)$ respectively. Additionally, by Example 3.2 and Theorem 3.1.3,
we have that $CC_{1}(X)\subset NBO_{1}(X)$ can also be proper. Finally, in the
case of the dyadic solenoid $X$, we have that $NBO_{1}(X)\subset
NB_{1}^{*}(X)$ is also a proper inclusion.

######  Definition 3.5.

Let $X$ be a continuum and $A\in 2^{X}$. We say that $A$ is colocally
connected of degree closed if $A=X$ or there exists a set $B\in 2^{X}$
satisfying the following conditions: $B\cap A=\emptyset$ and for every $y\in
X-A$, there exists a continuum $D\subset X-A$ such that $y\in\text{int}(D)$
and $D\cap B\neq\emptyset$.

We denote the set of all subsets of $X$ that are colocally connected of degree
closed by $CC_{2^{X}}$.

######  Theorem 3.6.

For every continuum $X$, $CC_{2^{X}}=\bigcup_{n=1}^{\infty}CC_{n}(X)$.

###### Proof.

The contention $\bigcup_{n=1}^{\infty}CC_{n}(X)\subset CC_{2^{X}}$ is clear.
For the converse contention, let $A\in CC_{2^{X}}$ and $B$ be a closed set
satisfying the conditions of Definition 3.5. Then, for each $y\in X-A$, there
exists a continuum $D_{y}\subset X-A$ such that $y\in int(D_{y})$ and
$D_{y}\cap B\neq\emptyset$. Hence, $\mathcal{D}=\\{int(D_{y}):y\in X-A\\}$ is
an open cover of $B$. Given that $B\in 2^{X}$, there exists finite subcover
$\\{int(D_{1}),\ldots,int(D_{n})\\}\subset\mathcal{D}$ of $B$. For each $i\leq
n$, choose $x_{i}\in D_{i}\cap B$. Let $B_{n}=\\{x_{1},\ldots,x_{n}\\}$.
Observe that for every $y\in X-A$, $D_{y}\subset X-A$ is such that $y\in
int(D_{y})$ and $D_{y}\cap B\neq\emptyset$. Hence, $D_{y}\cap
D_{i}\neq\emptyset$ for some $i\leq n$. Therefore, $D=D_{y}\cup D_{i}$ is a
continuum such that $D\subset X-A$, $y\in int(D)$ and $D\cap
B_{n}\neq\emptyset$. Hence, $A\in CC_{n}(X)$. In conclusion,
$CC_{2^{X}}\subset\bigcup_{n=1}^{\infty}CC_{n}(X)$. ∎

######  Lemma 3.7.

Let $X$ be a continuum and let $A\in CC_{n}(X)$, for some $n\in\mathbb{N}$. If
$X-A$ is connected, then $A\in CC_{1}(X)$.

###### Proof.

If $A=X$, we are done. Assume $A\neq X$. Let $B\in F_{n}(X)$ as in Definition
2.1.1 for $X-A$. For each $b\in B$, let $X_{b}=\\{y\in X-A:$ there exists a
continuum $D\subset X-A\text{ such that }b\in D\text{ and }y\in int(D)\\}$.
Notice that each $X_{b}$ is open and $\\{X_{b}:b\in B\\}$ is a cover of $X-A$.
Assume $b,c\in B$ satisfy $X_{b}\cap X_{c}\neq\emptyset$, and let $y\in
X_{b}\cap X_{c}$ and $z\in X_{c}$, thus there exist $D,E,F$ subcontinua of
$X-A$ such that $y\in int(D),b\in D,y\in int(E),c\in E$, and $z\in int(F),c\in
F$, thus $D\cup E\cup F$ is a subcontinuum of $X-A$ containing $z$ in its
interior and containing $b$, thus $z\in X_{b}$, therefore $X_{c}\subset
X_{b}$, and analogously $X_{b}\subset X_{c}$. In conclusion $\\{X_{b}:b\in
B\\}$ is a partition of $X-A$, since $X-A$ is connected, we have that
$X_{b}=X-A$ for some $b\in B$. Thus the set $\\{b\\}$ satisfies Definition
2.3.1 to show that $A\in CC_{1}(X)$. ∎

Examples 3.2 and 3.9 illustrate that we cannot replace the sets $CC_{n}(X)$
and $CC_{1}(X)$ in Lemma 3.7 with $NWC_{n}(X)$ and $NWC_{1}(X)$, $NB_{n}(X)$
and $NB_{1}(X)$, or $NBO_{n}(X)$ and $NBO_{1}(X)$, respectively.

######  Theorem 3.8.

Let $H_{n}(X)$ represent a hyperspace of non-cut sets of degree $n$ of a
continuum $X$. If $CC_{n}(X)=H_{n}(X)$ for some $n>1$, then
$CC_{m}(X)=H_{m}(X)$ for each $m<n$.

###### Proof.

Let $m<n$. By Theorem 3.1, $CC_{m}(X)\subset H_{m}(X)\subset
H_{n}(X)=CC_{n}(X)$. Let $A\in H_{m}(X)$, by Theorem 3.1, $A\in NC_{m}(X)$.
Let $U_{1}\ldots U_{k}$ be the components of $X-A$; given that $X-U_{i}\in
NC_{1}(X)$ for each $i\leq k$, $X-U_{i}\in CC_{1}(X)$ (Lemma 3.7). Hence, for
each $i\leq k$, there exists $x_{i}\in U_{i}$ such that for each $y\in U_{i}$,
there exists a continuum $D\subset U_{i}$ such that $y\in int(D)$ and
$x_{i}\in D$. Therefore, for $B=\\{x_{1},\ldots,x_{k}\\}$ and each $y\in X-A$,
there exists a continuum $D$ such that $y\in int(D)$, $D\cap A=\emptyset$ and
$D\cap B\neq\emptyset$. Given that $k\leq m$, $A\in CC_{m}(X)$. We conclude
that $H_{m}(X)=CC_{m}(X)$. ∎

The following example shows that the converse of Theorem 3.8 is not true.

######  Example 3.9.

For the continuum
$X=\\{(x,\sin(\frac{1}{x})):x\in[-1,0)\cup(0,1]\\}\cup(\\{0\\}\times[-1,1])$,
we have $CC_{1}(X)=NWC_{1}(X)=NB_{1}(X)=NB^{*}_{1}(X)=S_{1}(X)=NSC_{1}(X)$.
However $CC_{n}(X)\neq NWC_{n}(X)$ and $CC_{n}(X)\neq NB_{n}(X)$, for each
$n\geq 3$, and $CC_{n}(X)\neq NB^{*}_{n}(X)$, $CC_{n}(X)\neq S_{n}(X)$,
$CC_{n}(X)\neq NSC_{n}(X)$ for each $n\geq 2$.

Given the previous results, it is natural to ask the following.

######  Question 3.10.

For which spaces $X$ and for which hyperspaces of non-cut sets of degree 1
$H_{1}(X)$, the following implication holds: if $CC_{1}(X)=H_{1}(X)$, then
$CC_{n}(X)=H_{n}(X)$ for every $n\in\mathbb{N}$?

Partial answers to Question 3.10 are given in Theorem 3.14 and Corollary 5.2.

######  Theorem 3.11.

Let $X$ be a continuum and $A\in 2^{X}$. If $A\in CC_{n}(X)$, then for each
component $K$ of $X-A$, we have $A\in CC_{1}(A\cup K)$.

###### Proof.

If $A=X$, we are done. Assume $A\neq X$. Let $A\in CC_{n}(X)$, notice $A\in
NC_{n}(X)$. Let $B\in F_{n}(X)$ as in Definition 2.1.1 for $X-A$ and let $K$
be a component of $X-A$. Let $y\in K$, thus there exists a continuum $D\subset
X-A$ such that $B\cap D\neq\emptyset$, $y\in int(D)$. Since $K$ is a component
of $X-A$ and $y\in K$, then $D\subset K$, therefore $B\cap K\neq\emptyset$.
Let $B^{\prime}=B\cap K$, thus $B^{\prime}\in F_{n}(X)$ satisfies Definition
2.1.1 for $K$ to show that $A\in CC_{n}(A\cup K)$; by Lemma 3.7, $A\in
CC_{1}(A\cup K)$. ∎

######  Theorem 3.12.

Let $H_{n}(X)$ represent a hyperspace of non-cut sets of degree $n$ of a
continuum $X$. If
$\bigcup_{n=1}^{\infty}CC_{n}(X)=\bigcup_{n=1}^{\infty}H_{n}(X)$, then
$CC_{1}(X)=H_{1}(X)$.

###### Proof.

Let $A\in H_{1}(X)$. Therefore, $A\in NC_{1}(X)$ and $A\in CC_{m}(X)$ for some
$m\in\mathbb{N}$. Hence, Theorem 3.11 implies $A\in CC_{1}(X)$. ∎

######  Theorem 3.13.

Let $H_{n}(X)$ represent a hyperspace of non-cut sets of degree $n$ of a
continuum $X$. Let $A\in 2^{X}$ be such that $X-A$ has exactly $n$ components.
Then, $A\in H_{n}(X)$ if and only if for each component $K$ of $X-A$, $A\in
H_{1}(A\cup K)$.

###### Proof.

Clearly, $A\in NC_{n}(X)$ and for each component $K$ of $X-A$, $A\in
NC_{1}(A\cup K)$.

Let $K$ be a component of $X-A$.

Assume $A\in CC_{n}(X),NWC_{n}(X),NB_{n}^{*}(X)$ or $NBO_{n}(X)$. We prove
that $A\in CC_{1}(A\cup K),NWC_{1}(A\cup K),NB_{1}^{*}(A\cup K)$, or
$NBO_{1}(A\cup K)$, respectively. Let $B\in F_{n}(X-A)$ be a set that
satisfies the respective definitions for $X-A$. Notice that $B$ must intersect
each component of $X-A$, and the intersection of $B$ with each component
consists of exactly one point. Thus $K\cap B$ consists of only one point, and
$K\cap B$ satisfies the respective definitions for $A$ to assert that $A\in
CC_{1}(A\cup K),NWC_{1}(A\cup K),NB_{1}^{*}(A\cup K)$ or $NBO_{1}(A\cup K)$,
respectively.

Assume $A\in NB_{n}(X)$. We prove that $A\in NB_{1}(A\cup K)$. For each $x\in
K$, we can find $B_{x}\in F_{n}(X-A)$ that satisfies the Definition 2.1.3 for
$X-A$ and $x\in B_{x}$. Notice that $B_{x}$ must intersect each component of
$X-A$, and the intersection of $B_{x}$ with each component consists of exactly
one point. Thus $B_{x}\cap K=\\{x\\}$ is the set that asserts that $A\in
NB_{1}(A\cup K)$.

Assume $A\in S_{n}(X)$. We prove that $A\in S_{1}(A\cup K)$. Let $\mathcal{U}$
be a finite family of open sets of $(A\cup K)-A=K$. Let
$\mathcal{V}=\mathcal{U}\cup\\{R:R$ is a component of $X-A$ and $R\neq K\\}$.
Given that $\mathcal{V}$ is a finite family of open sets contained in $X-A$,
there exists $D\in C_{n}(X-A)$ such that $D$ intersects each element of the
family $\mathcal{V}$. Since $D$ intersects each component of $X-A$, $D$ has
exactly $n$ components. Thus $D\cap K\in C_{1}(K)$ and $D\cap K$ intersects
each element of $\mathcal{U}$. Therefore, $A\in S_{1}(A\cup K)$.

Assume $A\in NSC_{n}(X)$. Let $U,V$ be two open sets such that $U\cup V\subset
K$. Let $\mathcal{V}=\\{U,V\\}\cup\\{R:R$ is a component of $X-A$ and $R\neq
K\\}$. Given that $\mathcal{V}$ is a family of $n+1$ open sets contained in
$X-A$, there exists $D\in C_{n}(X-A)$ such that $D$ intersects each element of
the family $\mathcal{V}$. Since $D$ intersects each component of $X-A$, $D$
has exactly $n$ components. Thus $D\cap K\in C_{1}(K)$ and $D\cap K$
intersects $U$ and $V$. Therefore, $A\in NSC_{1}(A\cup K)$.

Let $K_{1},\ldots K_{n}$ be the components of $X-A$. Assume that for each
$K_{i}$, $A\in CC_{1}(A\cup K_{i})$, so $K_{i}$ is a $1$-$Q1$ space. For each
$i\in\\{1,\dots,n\\}$, choose $x_{i}\in K_{i}$ such that $\\{x_{i}\\}$
witnesses that $K_{i}$ is a $1$-$Q1$ space. Then, $B=\\{x_{1},\ldots x_{n}\\}$
witnesses $\bigcup_{i=1}^{n}K_{i}$ is a $n$-$Q1$ space, so $A\in CC_{n}(X)$.
Analogously, if $A\in NWC_{1}(A\cup K_{i})$ for each $i\in\\{1,\dots n\\}$, or
$A\in NB_{1}^{*}(A\cup K_{i})$ for each $i\in\\{1,\dots n\\}$, or $A\in
NBO_{1}(A\cup K_{i})$ for each $i\in\\{1,\dots n\\}$, then $A\in NWC_{n}(X),$
or $A\in NB_{n}^{*}(X)$, or $A\in NBO_{n}(X)$, respectively.

Assume that for each $i\in\\{1,\dots,n\\}$, $A\in NB_{1}(A\cup K_{i})$, so
$K_{i}$ is a 1-$Q3$ space. For each $i\in\\{1,\dots,n\\}$, choose $x_{i}\in
K_{i}$. For $x\in X-A$, let $j$ be such that $x\in K_{j}$. Notice that
$B=\\{x\\}\cup(\\{x_{1},\ldots x_{n}\\}-\\{x_{j}\\})$ witnesses
$\bigcup_{i=1}^{n}K_{i}$ is a $n$-$Q3$ space, so $A\in NB_{n}(X)$.

Assume that for each $i\in\\{1,\dots,n\\}$, $A\in S_{1}(A\cup K_{i})$, so
$K_{i}$ is a 1-$Q5$ space. Let $\mathcal{U}$ be a finite family of non-empty
open sets contained in $\bigcup_{i=1}^{n}K_{i}$. For each
$i\in\\{1,\dots,n\\}$, let $\mathcal{U}_{i}=\\{K_{i}\\}\cup\\{U\cap
K_{i}:U\in\mathcal{U}\text{ and }U\cap K_{i}\neq\emptyset\\}$. Since $K_{i}$
is a $1$-$Q5$ space, there exists $D_{i}\in C(K_{i})$ such that $D_{i}\cap
U\neq\emptyset$ for every $U\in\mathcal{U}_{i}$. Hence,
$D=\cup_{i=1}^{n}D_{i}\in C_{n}(\bigcup_{i=1}^{n}K_{i})$ satisfies $D\cap
U\neq\emptyset$ for each $U\in\mathcal{U}$. Therefore $\bigcup_{i=1}^{n}K_{i}$
is a $n$-$Q5$ space, so $A\in S_{n}(X)$.

Assume that for each $i\in\\{1,\dots,n\\}$, $A\in NSC_{1}(A\cup K_{i})$, so
$K_{i}$ is a 1-$Q6$ space. Let $\mathcal{U}=\\{U_{1},\ldots,U_{n+1}\\}$ be a
family of $n+1$ non-empty open sets contained in $\bigcup_{i=1}^{n}K_{i}$. Let
$K$ be a component of $X-A$ such that $K\cap U_{i}\neq\emptyset\neq K\cap
U_{j}$ for some $i\neq j$. Since $K$ is a $1$-$Q6$ space, there exists
$D_{K}\in C(X)$ such that $D_{K}\cap U_{i}\neq\emptyset\neq\emptyset\neq
D_{K}\cap U_{j}$. Now, for each $m\in\\{1,\dots,n+1\\}$, choose $x_{m}\in
U_{m}$. Hence, $D=D_{K}\cup\\{x_{m}:m\in\\{1,\dots,n+1\\}-\\{i,j\\}\\}\in
C_{n}(X)$ satisfies $D\cap U\neq\emptyset$ for each $U\in\mathcal{U}$.
Therefore $\bigcup_{i=1}^{n}K_{i}$ is a $n$-$Q6$ space, so $A\in NSC_{n}(X)$.
∎

In Example 3.9, if $A=\\{(-1,0)\\}$, the set $X-A$ has $1$ component. However,
although $A$ belongs to $NWC_{3}(X),NB_{3}(X),NB^{*}_{2}(X),S_{2}(X)$ and
$NSC_{2}(X)$, it does not belong to $NWC_{1}(X),$ $NB_{1}(X),$
$NB^{*}_{1}(X),$ $S_{1}(X),$ or $NSC_{1}(X)$. This implies that the result in
Theorem 3.13 is false if we remove the condition that $X-A$ has exactly $n$
components.

######  Theorem 3.14.

If $H_{1}(X)$ is a hyperspace of non-cut sets such that $H_{1}(X)=NC_{1}(X)$,
then $H_{n}(X)=NC_{n}(X)$ for each $n\in\mathbb{N}$.

###### Proof.

By Theorem 3.1, $H_{n}(X)\subset NC_{n}(X)$. Let $A\in NC_{n}(X)$. If $A=X$,
then $A\in H_{n}(X)$. Suppose that $X-A$ has exactly $m$ components, for some
$m\in\\{1,\dots,n\\}$. Observe that for each component $K$ of $X-A$, $X-K\in
NC_{1}(X)=H_{1}(X)$. Since $X-(X-K)=(A\cup K)-A$, we have $A\in H_{1}(A\cup
K)$. By Theorem 3.13, $A\in H_{m}(X)\subset H_{n}(X)$. Therefore
$H_{n}(X)=NC_{n}(X)$. ∎

By Theorems 3.6, 3.12 and 3.14, we obtain the following result.

######  Corollary 3.15.

The following conditions are equivalent:

  * 1.

$CC_{1}(X)=NC_{1}(X)$;

  * 2.

$CC_{n}(X)=NC_{n}(X)$ for some $n\in\mathbb{N}$;

  * 3.

$CC_{n}(X)=NC_{n}(X)$ for each $n\in\mathbb{N}$;

  * 4.

$\bigcup_{n=1}^{\infty}CC_{n}(X)=\bigcup_{n=1}^{\infty}NC_{n}(X)$;

  * 5.

$CC_{2^{X}}=\bigcup_{n=1}^{\infty}NC_{n}(X)$.

Recognizing or obtaining new non-cut sets of degree $n$ from those already
known is of great interest; Theorems 3.16 and 3.20 address this.

######  Theorem 3.16.

Let $X$ be a continuum and $n\in\mathbb{N}$. Let $A\in 2^{X}$ and let $C\in
2^{X}$ be such that $int(A)\subset C\subset A$.

  1. 1.

If $A\in NBO_{n}(X)$, then $C\in NBO_{n}(X)$;

  2. 2.

if $A\in NB_{n}^{*}(X)$, then $C\in NB_{n}^{*}(X)$;

  3. 3.

if $A\in S_{n}(X)$, then $C\in S_{n}(X)$;

  4. 4.

if $A\in NSC_{n}(X)$, then $C\in NSC_{n}(X)$;

  5. 5.

if $A\in NC_{n}(X)$, then $C\in NC_{n}(X)$; and

###### Proof.

  1. 1.

Let $A\in NBO_{n}(X)$ and let $B\in F_{n}(X-A)$ witnessing that $X-A$ if a
$n$-$Qo$ space. Let $V$ be a non-empty open set of $X-C$. Given that
$int(A)\subset C\subset A$, $V_{A}=V-A$ is a non-empty open set of $X-A$.
Therefore, there exists a continuum $D\subset X-A$ such that $int(D)\cap
V_{A}\neq\emptyset\neq int(D)\cap B$, which implies that $int(D)\cap
V\neq\emptyset\neq int(D)\cap B$. Hence, $B$ witnesses that $X-C$ is a
$n$-$Qo$ space, so $C\in NBO_{n}(X)$.

  2. 2.

Let $A\in NB_{n}^{*}(X)$, and let $B\in F_{n}(X-A)$ witnessing that $X-A$ is a
$n$-$Q4$ space. Given that $int(A)\subset C\subset A$, for every non-empty
open set $V$ of $X-C$, $V_{A}=V-A\neq\emptyset$ is an open set of $X-A$.
Therefore, there exists a continuum $D\subset X-A\subset X-C$ such that $D\cap
V_{A}\neq\emptyset\neq D\cap B$, which implies that $D\cap V\neq\emptyset$.
Hence, Hence, $B$ witnesses that $X-C$ is a $n$-$Q4$ space, so $C\in
NB_{n}^{*}(X)$.

  3. 3.

Let $A\in S_{n}(X)$ and let $\mathcal{U}$ be a finite family of non-empty open
sets of $X-C$. Given that $int(A)\subset C\subset A$, the family
$\mathcal{V}=\\{U-A:U\in\mathcal{U}\\}$ is a finite family of non-empty open
sets of $X-A$. Therefore, there exists $D\in C_{n}(X-A)$ such that $D\cap
V\neq\emptyset$ for each $V\in\mathcal{V}$, which implies that $D\cap
U\neq\emptyset$ for each $U\in\mathcal{U}$. Hence, $X-C$ is a $n$-$Q5$ space,
so $C\in S_{n}(X)$.

  4. 4.

Let $A\in NSC_{n}(X)$ and let $U_{1},\ldots,U_{n+1}$ be a collection of $n+1$
non-empty open sets of $X-C$. For each $i\in\\{1,\dots,n+1\\}$, let
$V_{i}=U_{i}-A$. We have $V_{1},\ldots,V_{n+1}$ is a collection of $n+1$ non-
empty open sets of $X-A$. Therefore, there exists $D\in C_{n}(X-A)$ such that
$D\cap V_{i}\neq\emptyset$ for each $i\in\\{1,\dots,n+1\\}$. Hence, $X-C$ is a
$n$-$Q6$ space, so $C\in NSC_{n}(X)$.

  5. 5.

Notice that $X-A\subset X-C\subset X-int(A)=cl(X-A)$. Hence, $X-C$ has at most
the same number of components as $X-A$.

∎

Example 3.17 shows that Theorem 3.16 cannot be extended to the sets
$CC_{n}(X)$, $NWC_{n}(X)$, and $NB_{n}(X)$.

######  Example 3.17.

Let $X$ be the circle of pseudo-arcs and let $f:X\to\mathcal{S}^{1}$ be the
quotient map from $X$ onto the circle described on [2]. Then $f$ is an onto,
monotone and open map. Hence, for every $x\in\mathcal{S}^{1}$, $f^{-1}(x)\in
CC_{1}(X)$ (see Proposition 4.5 and Proposition 4.7), and no proper subset of
$f^{-1}(x)$ is an element of $CC_{1}(X),NWC_{1}(X)$ neither $NB_{1}(X)$.

The following lemma gives us a characterization of the elements in
$NBO_{n}(X)$.

######  Proposition 3.18.

Let $X$ be a continuum and $A\in 2^{X}$. Then, $A\in NBO_{n}(X)$ if and only
if for each non-empty finite family $\mathcal{U}$ of non-empty open sets
contained in $X-A$, there exists $D\in C_{n}(X-A)$ such that $int(D)\cap
U\neq\emptyset$ for all $U\in\mathcal{U}$.

###### Proof.

Let $A\in NBO_{n}(X)$ and let $B\in F_{n}(X-A)$ witnessing that $X-A$ is a
$n$-$Qo$ space. Let $\mathcal{U}=\\{U_{1},\ldots U_{m}\\}$ be a non-empty
finite family of non-empty open sets contained in $X-A$. Then, for each
$i\in\\{1,\dots,m\\}$, there exists $D_{i}\in C(X-A)$, such that
$int(D_{i})\cap U_{i}\neq\emptyset\neq int(D_{i})\cap B$. Therefore,
$D=\cup_{i=1}^{m}D_{i}\in C_{n}(X-A)$ satisfies $int(D)\cap U\neq\emptyset$
for each $U\in\mathcal{U}$.

Now suppose that $A\in 2^{X}$ satisfies that for each non-empty finite family
$\mathcal{U}$ of non-empty open sets of $X-A$, there exists $D\in C_{n}(X-A)$
such that $int(D)\cap U\neq\emptyset$ for each $U\in\mathcal{U}$. If $A=X$, we
have $A\in NBO_{n}(X)$.

Assume $A\neq X$. For $D\in C_{n}(X-A)$, define $\alpha(D)=\bigcup\\{K\in
C(X-A):K\cap D\neq\emptyset\\}$.

Claim: There exists $D\in C_{n}(X-A)$ such that $\alpha(D)$ is dense in $X-A$
and each component of $D$ has non-empty interior.

Proof of Claim:

Let $\mathcal{F}=\\{\\{K_{1},K_{2},\dots,K_{m}\\}:m\in\mathbb{N},$ for each
$i,j\in\\{1,\dots,m\\}$, $K_{i}\in C(X-A)$, $int(K_{i})\neq\emptyset$, and if
$i\neq j,\alpha(K_{i})\cap\alpha(K_{j})=\emptyset\\}$.

By the properties of $A$, if $\\{K_{1},\dots,K_{m}\\}\in\mathcal{F}$, then
$m\leq n$. Let $M$ be the maximum number of elements of the members of
$\mathcal{F}$ and let $\\{K_{1},\dots,K_{M}\\}\in\mathcal{F}$.

Notice that $\bigcup_{i=1}^{M}\alpha(K_{i})$ is dense in $X-A$. Let
$D=\bigcup_{i=1}^{M}K_{i}$, notice that $D\in C_{n}(X-A)$ and $\alpha(D)$ is
dense in $X-A$. The claim is proved.

Now, take $D$ as in the claim and let $B\in F_{n}(X)$ containing exactly one
point in the interior of each component of $D$. If $U$ is any non-empty open
set of $X-A$, then there exist a continuum $F$ such that $int(F)\cap
U\neq\emptyset$ and a continuum $K\subset X-A$ such that $K\cap
D\neq\emptyset\neq K\cap(U\cap int(F))$.

Let $E$ be a component of $D$ that intersects $K$. Therefore, $E\cup K\cup
F\in C(X-A)$, $B\cap int(E\cup K\cup F)\neq\emptyset$ and $int(E\cup K\cup
F)\cap U\neq\emptyset$. Thus $B$ witnesses that $X-A$ is a $n$-$Qo$ space, so
$A\in NBO_{n}(X)$. ∎

The following lemma gives us a characterization of the elements in
$CC_{n}(X)$.

######  Lemma 3.19.

Let $X$ be a continuum, $A\in 2^{X}$ and $n\in\mathbb{N}$. Then, $A\in
CC_{n}(X)-\\{X\\}$ if and only if for each open set $U$ with $A\subset U$,
there exists an open set $V$ such that $A\subset V\subset U$ and $X-V\in
C_{n}(X)$.

###### Proof.

Let $A\in CC_{n}(X)-\\{X\\}$ and let $U$ be an open set such that $A\subset
U$. Let $B\in F_{n}(X-A)$ witnessing that $X-A$ is a $n$-$Q1$ space. For each
$y\in X-U$, let $D_{y}$ be a continuum such that $y\in int(D_{y})$, $D_{y}\cap
B\neq\emptyset$ and $D_{y}\subset X-A$. Since $X-U$ is compact and
$\\{int(D_{y}):y\in X-U\\}$ is an open cover of $X-U$, there exists a finite
subcover $\\{int(D_{1}),\ldots,int(D_{k})\\}$ of $X-U$. Let
$V=(X-\cup_{i=1}^{k}D_{i})\subset U$, observe that $V$ is an open set,
$A\subset V\subset U$ and $X-V=\cup_{i=1}^{k}D_{i}\in C_{n}(X)$.

Now suppose that $A\in 2^{X}$ is such that for each open set $U$ such that
$A\subset U$, there exists an open set $V$ such that $A\subset V\subset U$ and
$X-V\in C_{n}(X)$. Observe that $X-A\neq\emptyset$ must have at most $n$
components. Choose $B\in F_{n}(X)$ such that $B$ intersects each component of
$X-A$. Let $y\in X-A$, let $V_{y}\subset X-A$ be a closed neighborhood of $y$
and let $U=X-(V_{y}\cup B)$. Since $A\subset U$, there exists $V$ open such
that $A\subset V\subset U$ and $X-V\in C_{n}(X)$, which implies that the
component $D$ of $X-V\subset X-A$ containing $y$ is a continuum such that
$B\cap D\neq\emptyset$ and $y\in int(D)$. ∎

######  Theorem 3.20.

Let $X$ be a continuum and $n\in\mathbb{N}$. Let $A,C\in 2^{X}$ such that $C$
is union of some components of $A$.

  1. 1.

If $A\in CC_{n}(X)$, then $C\in CC_{n}(X)$;

  2. 2.

if $A\in NWC_{n}(X)$, then $C\in NWC_{n}(X)$; and

  3. 3.

if $A\in NC_{n}(X)$, then $C\in NC_{n}(X)$.

###### Proof.

If $A=X$, the result is trivial. Assume $A\neq X$.

  1. 1.

Let $A\in CC_{n}(X)$ and $B\in F_{n}(X-A)$ witnessing that $X-A$ is a $n$-$Q1$
space. Let $x\in X-C$. If $x\in X-A$, there exists a continuum $G$ containing
$x$ in its interior such that $G\cap B\neq\emptyset$ and $B\subset X-A$. Now,
assume $x\in A-C$, and let $D$ be the component of $x$ in $A-C$. Since $A$ is
compact, there exist two open sets $U$ and $V$ such that $C\subset U$,
$D\subset V$, $U\cap V=\emptyset$, $A\subset U\cup V$; moreover, by Lemma
3.19, we may assume that $E=X-(U\cup V)$ has at most $n$ components.
Therefore, $E\cup V=X-U\in C_{n}(X)$ (Theorem 5.6 from [10]) and contains $x$
in its interior. Since the component of $E\cup V$ containing $x$ has a non-
empty interior in $X-A$, there exists a continuum $G$ containing $x$ in its
interior such that $G\cap B\neq\emptyset$ and $B\subset X-C$. Hence, $C\in
CC_{n}(X)$.

  2. 2.

Let $A\in NWC_{n}(X)$ and $B\in F_{n}(X-A)$ witnessing that $X-A$ is an
$n$-$Q2$ space. Let $x\in X-C$. If $x\in X-A$, there exists a continuum $F$
containing $x$ such that $F\subset X-A$, and $F\cap B\neq\emptyset$. Now,
assume $x\in A-C$. Let $D$ be the component of $A$ containing $x$. Since $A$
is compact, there exist two open sets $U$ and $V$ such that $C\subset U$,
$D\subset V$, $U\cap V=\emptyset$, $A\subset U\cup V$. Thus, there exists a
continuum $G$ such that $D\subsetneq G\subset U$ (Corollary 5.5 of [10]).
Choose $r\in G-A$. Since $r\in X-A$, there exists a continuum $F$ containing
$r$ such that $F\subset X-A$ and $F\cap B\neq\emptyset$. Therefore, $F\cup G$
is a continuum containing $x$ such that $F\subset X-C$, and $(F\cup G)\cap
B\neq\emptyset$. In conclusion, $B$ witnesses that $X-C$ is an $n$-$Q2$ space,
thus $C\in NWC_{n}(X)$.

  3. 3.

Let $A\in NC_{n}(X)$. Let $x\in X-C$. If $x\in X-A$, the component of $x$ in
$X-A$ is contained in the component of $x$ in $X-C$. Now, assume $x\in A-C$.
Let $D$ be the component of $A$ containing $x$. Since $A$ is compact, there
exist two open sets $U$ and $V$ such that $C\subset U$, $D\subset V$, $U\cap
V=\emptyset$, $A\subset U\cup V$. Thus, there exists a continuum $G$ such that
$D\subsetneq G\subset U$ (Corollary 5.5 of [10]). Choose $r\in G-A$. Since
$r\in X-A$, the component of $r$ in $X-C$ contains $G$ and contains the
component of $r$ in $X-A$, and the component of $x$ in $X-C$ is the same as
the component of $r$ in $X-C$. Thus, each component of $X-C$ contains some
component of $X-A$, so $C\in NC_{n}(X)$.

∎

The following examples show that we can not generalize the previous theorem to
the hyperspaces ${NB}_{n}(X)$, ${NBO}_{n}(X)$, ${NB}_{n}^{*}(X)$, ${S}_{n}(X)$
and ${NSC}_{n}(X)$.

######  Example 3.21.

Let $Y$ be the dyadic solenoid, let $S\subset Y$ be an arc and let
$h:I\times\\{0\\}\to S$ be an homeomorphism. Define $X=Y\cup_{h}(I\times I)$,
$A=\\{p\\}\cup(I\times I)$ where $p$ is a point of $X$ not in the composant of
$I\times I$, and let $C=\\{p\\}$. Observe that $A\in NB_{1}(X)$ and $C\notin
NB_{1}(X)$.

######  Example 3.22.

Let $Y$ be a compactification of the ray with remainder $\mathcal{S}^{1}$, let
$S\subset\mathcal{S}^{1}$ be an arc and $h:I\times\\{0\\}\to S$ an
homeomorphism. Define $X=Y\cup_{h}(I\times I)$, $A=\\{p\\}\cup(I\times I)$
where $p\in\mathcal{S}^{1}-S$, and $C=\\{p\\}$. Observe that $A\in
NBO_{1}(X)$, $C$ is a component of $A$ and $C\notin NSC_{1}(X)$.

In Proposition 2.4 of [3], the authors studied what is the Borel type class of
the sets $CC_{1}(X)\cap F_{1}(X)$, $S_{1}(X)\cap F_{1}(X)$, $NSC_{1}(X)\cap
F_{1}(X)$, and $NC_{1}(X)\cap F_{1}(X)$. We extend the analysis for some
hyperspaces of non-cut sets of degree $n$.

######  Proposition 3.23.

Let $X$ be a continuum and $n\in\mathbb{N}$. The following is true.

  * (i)

$CC_{n}(X)$ is of type $G_{\delta}$,

  * (ii)

$NBO_{n}(X)$ is of type $G_{\delta}$,

  * (iii)

$S_{n}(X)$ is of type $G_{\delta}$, and

  * (iv)

$NSC_{n}(X)$ is of type $G_{\delta}$.

###### Proof.

  * (i)

For each $k\in\mathbb{N}$, define $CC_{n}^{k}$ as the union of all the sets
$\left\langle U_{1},\ldots,U_{m}\right\rangle\subset 2^{X}$ with
$m\in\mathbb{N}$, $U_{i}$ non-empty and open in $X$, $diam(U_{i})<\frac{1}{k}$
for each $i$, such that there exists $B\in F_{n}(X)$ satisfying that for every
$z\in X-\bigcup_{i=1}^{m}U_{i}$, there exists a continuum $D\subset
X-\bigcup_{i=1}^{m}U_{i}$ such that $z\in{int}(D)$ and $B\cap D\neq\emptyset$.
It holds that $CC_{n}(X)=\bigcap_{k=1}^{\infty}CC_{n}^{k}$.

  * (ii)

Let $\mathcal{B}$ be a countable base with $\emptyset\notin\mathcal{B}$. For
each $\mathcal{U}\subset\mathcal{B}$ finite, define the set
$NBO_{\mathcal{U}}$ as follows:

$NBO_{\mathcal{U}}=\bigcup\\{\langle X-K\rangle:K\in C_{n}(X),\forall
U\in\mathcal{U},int(K)\cap
U\neq\emptyset\\}\cup(\bigcup_{U\in\mathcal{U}}\langle X,U\rangle).$

Let $\mathcal{C}=\\{\mathcal{U}:\mathcal{U}\text{ is a non-empty finite subset
of }\mathcal{B}\\}$. It holds that
$NBO_{n}(X)=\bigcap_{\mathcal{U}\in\mathcal{C}}NBO_{\mathcal{U}}$.

  * (iii)

Let $\mathcal{B}$ and $\mathcal{C}$ as in (ii). For each
$\mathcal{U}\subset\mathcal{B}$ finite, define the set $S_{\mathcal{U}}$ as
follows:

$S_{\mathcal{U}}=\bigcup\\{\langle X-K\rangle:K\in C_{n}(X),\forall
U\in\mathcal{U},K\cap U\neq\emptyset\\}\cup(\bigcup_{U\in\mathcal{U}}\langle
X,U\rangle).$

It holds that $S_{n}(X)=\bigcap_{\mathcal{U}\in\mathcal{C}}S_{\mathcal{U}}$.

  * (iv)

Let $\mathcal{B}$ and $S_{\mathcal{U}}$ as in (ii). It holds that
$NSC_{n}(X)=\bigcap\\{S_{\mathcal{U}}:\mathcal{U}\subset\mathcal{B}$ non-empty
with at most $n+1\text{ elements}\\}$.

∎

In Example 2.5 from [3], the authors show a continuum $X$ where
$NWC_{1}(X)\cap F_{1}(X)$ is not Borel, consequently, $NWC_{1}(X)$ itself is
not Borel. In (ii), they show that the sets $CC_{1}(X)\cap F_{1}(X)$,
$S_{1}(X)\cap F_{1}(X)$, $NSC_{1}(X)\cap F_{1}(X)$, and $NC_{1}(X)\cap
F_{1}(X)$ do not necesarilly fall into the category of $F_{\sigma}$, implying
the same for $CC_{1}(X)$, $S_{1}(X)$, $NSC_{1}(X)$, and $NC_{1}(X)$. Finally,
in (iii), the authors furnish an example where $NC_{1}(X)\cap F_{1}(X)$ lacks
the $G_{\delta}$ property, hence $NC_{1}(X)$ is not $G_{\delta}$.

######  Theorem 3.24.

If $X$ is a compact metric space, then $M(X)$ is a $G_{\delta}$ set in
$2^{X}$.

###### Proof.

Let $\\{U_{n}:n\in\mathbb{N}\\}$ be a countable base of $X$. For each
$n\in\mathbb{N}$, we define

$\mathcal{U}_{n}=\\{A\in 2^{X}:U_{n}\subset A\\}.$

It is clear that each $\mathcal{U}_{n}$ is closed and
$2^{X}-M(X)=\cup_{n}\mathcal{U}_{n}$ which is an $F_{\sigma}$ set. Hence,
$M(X)$ is a $G_{\delta}$ set. ∎

As a consequence of the following corollary, we can specify the Borel type
class of some sets mentioned in Note 2.5.

######  Corollary 3.25.

Let $X$ be a continuum and $n\in\mathbb{N}$. The following is true.

  * (i)

$CC_{n}(X)\cap M(X)$ is of type $G_{\delta}$,

  * (ii)

$NBO_{n}(X)\cap M(X)$ is of type $G_{\delta}$,

  * (iii)

$S_{n}(X)\cap M(X)$ is of type $G_{\delta}$, and

  * (iv)

$NSC_{n}(X)\cap M(X)$ is of type $G_{\delta}$.

##  4 Properties of non-$n$-cut sets preserved by continuous functions

The results presented herein highlight how different types of mappings, such
as onto mappings, open mappings, and monotone mappings, preserve some
properties of non-$n$-cut sets.

We start with a lemma.

######  Lemma 4.1.

Let $f:X\to Y$ be an onto mapping between continua. If $y\in Y$, and $D\in
2^{X}$ are such that $f^{-1}(y)\subset int_{X}(D)$, then $y\in int_{Y}(f(D))$.

###### Proof.

Let $y\in Y$, $D\in 2^{X}$ and assume $f^{-1}(y)\subset int_{X}(D)$. Notice
that $y\in Y-f(X-int_{X}(D))\subset f(D)$ and $Y-f(X-int_{X}(D))$ is open in
$Y$. Thus $y\in int_{Y}(f(D))$. ∎

######  Proposition 4.2.

Let $f:X\to Y$ be an onto mapping between continua, let $A\in 2^{Y}$ and let
$n\in\mathbb{N}$. The following statements hold:

  1. 1.

If $f^{-1}(A)\in CC_{n}(X)$, then $A\in CC_{n}(Y)$;

  2. 2.

if $f^{-1}(A)\in NWC_{n}(X)$, then $A\in NWC_{n}(Y)$;

  3. 3.

if $f^{-1}(A)\in NB_{n}(X)$, then $A\in NB_{n}(Y)$;

  4. 4.

if $f^{-1}(A)\in NB_{n}^{*}(X)$, then $A\in NB_{n}^{*}(Y)$;

  5. 5.

if $f^{-1}(A)\in S_{n}(X)$, then $A\in S_{n}(Y)$;

  6. 6.

if $f^{-1}(A)\in NSC_{n}(X)$, then $A\in NSC_{n}(Y)$; and

  7. 7.

if $f^{-1}(A)\in NC_{n}(X)$, then $A\in NC_{n}(Y)$.

###### Proof.

Observe that the statements are true if $A=Y$. Suppose that $A\neq Y$.

  1. 1.

Assume that $f^{-1}(A)\in CC_{n}(X)$. Let $B\in F_{n}(X-f^{-1}(A))$ witnessing
that $X-f^{-1}(A)$ is a $n$-$Q1$ space. Let $y\in Y-A$. For each $x\in
f^{-1}(y)$, there exists a continuum $D_{x}$ contained in $X-f^{-1}(A)$ such
that $x\in int_{X}(D_{x})$ and $B\cap D_{x}\neq\emptyset$. By compactness of
$f^{-1}(y)$, we can find a finite number of elements $x_{1},\dots,x_{n}\in
f^{-1}(y)$ such that $f^{-1}(y)\subset\bigcup_{i=1}^{m}D_{x_{i}}$. Notice that
$\bigcup_{i=1}^{m}D_{x_{i}}\subset X-f^{-1}(A)$ is in $C_{n}(X)$ and contains
$f^{-1}(y)$ in its interior. By Lemma 4.1,
$f(\bigcup_{i=1}^{m}D_{x_{i}})\subset Y-A$ is a subcontinuum of $Y$ such that
$f(B)\cap f(\bigcup_{i=1}^{n}D_{a_{i}})\neq\emptyset$ and has $y$ in its
interior, so $f(B)$ witnesses that $Y-A$ is a $n$-$Q1$ space, therefore $A\in
CC_{n}(Y)$.

  2. 2.

Assume that $f^{-1}(A)\in NWC_{n}(X)$. Let $B\in F_{n}(X)$ witnessing that
$X-f^{-1}(A)$ is a $n$-$Q2$ space, let $y\in Y-A$ and $x\in f^{-1}(y)$. Since
$X-f^{-1}(A)$ is $n$-$Q2$, there exists a continuum $D\subset X$ such that
$D\cap f^{-1}(A)=\emptyset$, $w\in D$ and $D\cap B\neq\emptyset$. Therefore
$f(D)\subset Y-A$ is a continuum, $x\in f(D)$ and $f(B)\cap
f(D)\neq\emptyset$, so $f(B)$ witnesses that $Y-A$ is a $n$-$Q2$ space,
therefore $A\in NWC_{n}(Y)$.

  3. 3.

Assume that $f^{-1}(A)\in NB_{n}(X)$. Let $y\in Y-A$ and $x\in f^{-1}(y)$. Let
$B\in F_{n}(X)$ witnessing that $X-f^{-1}(A)$ is a $n$-$Q3$ space and $x\in
B$. Let $U\subset Y-A$ be and non-empty open set. Notice that
$f^{-1}(U)\subset X-f^{-1}(A)$ is a non-empty open set of $X$; since
$X-f^{-1}(A)$ is $n$-$Q3$, there exists a continuum $D\subset X-f^{-1}(A)$
such that $D\cap B\neq\emptyset$ and $D\cap f^{-1}(U)\neq\emptyset$, thus
$f(D)\subset Y-A$ is a continuum such that $f(D)\cap f(B)\neq\emptyset$,
$f(D)\cap U\neq\emptyset$ and $y\in f(B)$, so $f(B)$ witnesses that $Y-A$ is a
$n$-$Q3$ space, therefore $A\in NB_{n}(Y)$.

  4. 4.

Assume that $f^{-1}(A)\in NB_{n}^{*}(X)$. Let $B\in F_{n}(X)$ witnessing that
$X-f^{-1}(A)$ is a $n$-$Q4$ space. Let $U\subset Y-A$ be and non-empty open
set. Notice that $f^{-1}(U)\subset X-f^{-1}(A)$ is a non-empty open set of
$X$; since $X-f^{-1}(A)$ is $n$-$Q4$, there exists a continuum $D\subset
X-f^{-1}(A)$ such that $D\cap B\neq\emptyset$ and $D\cap
f^{-1}(U)\neq\emptyset$, thus $f(D)\subset Y-A$ is a continuum such that
$f(D)\cap f(B)\neq\emptyset$ and $f(D)\cap U\neq\emptyset$, so $f(B)$
witnesses that $Y-A$ is a $n$-$Q4$ space, therefore $A\in NB_{n}^{*}(Y)$.

  5. 5.

Assume that $f^{-1}(A)\in S_{n}(X)$. Let $U_{1},\dots,U_{m}$ be a finite
number of non-empty open sets contained in $Y-A$. Since
$f^{-1}(U_{1}),\dots,f^{-1}(U_{m})$ is a finite number of non-empty open sets
contained in $X-f^{-1}(A)$ and $X-f^{-1}(A)$ is a $n$-$Q5$ space, there exists
an element $D\in C_{n}(X)$ such that $D\subset X-f^{-1}(A)$, $D\cap
f^{-1}(U_{i})\neq\emptyset$ for each $i\in\\{1,\dots,m\\}$, hence $f(D)\subset
Y-A$ is an element of $C_{n}(Y)$ such that $f(D)\cap U_{i}\neq\emptyset$ for
each $i\in\\{1,\dots,m\\}$. So $Y-A$ is a $n$-$Q5$ space, therefore $A\in
S_{n}(Y)$.

  6. 6.

Assume that $f^{-1}(A)\in NSC_{n}(X)$. Let $U_{1},\dots,U_{n+1}$ be non-empty
open sets contained in $Y-A$. Since $f^{-1}(U_{1}),\dots,f^{-1}(U_{n+1})$ are
non-empty open sets contained in $X-f^{-1}(A)$ and $X-f^{-1}(A)$ is a $n$-$Q6$
space, there exists an element $D\in C_{n}(X)$ such that $D\subset
X-f^{-1}(A)$, $D\cap f^{-1}(U_{i})\neq\emptyset$ for each
$i\in\\{1,\dots,n+1\\}$, hence $f(D)\subset Y-A$ is an element of $C_{n}(Y)$
such that $f(D)\cap U_{i}\neq\emptyset$ for each $i\in\\{1,\dots,n+1\\}$. So
$Y-A$ is a $n$-$Q6$ space, therefore $A\in NSC_{n}(Y)$.

  7. 7.

Assume that $f^{-1}(A)\in NC_{n}(X)$. Since $Y-A=f(X-f^{-1}(A))$ and
$X-f^{-1}(A)$ has at most $n$ components, then $Y-A$ has at most $n$
components. Therefore $A\in NC_{n}(X)$.

∎

In the following example, we show that it is not possible to extend
Proposition 4.2 to the hyperspace $NBO_{n}(X)$. A weaker result is presented
in Proposition 4.4.

######  Example 4.3.

Let $Y$ be the Knaster buckethandle continuum, $p\in Y$ be the endpoint of
$Y$, and let $\alpha:[0,1)\rightarrow K$ be an onto injective mapping such
that $\alpha(0)=p$, where $K$ is the composant of $Y$ containing $p$. Define
$X=(Y\times\\{0\\})\cup\\{(\alpha(t),1-t):t\in[0,1)\\}\subset Y\times[0,1]$
and let $f:X\to Y$ be the projection onto $Y$. Then $f$ is an onto mapping.
Note that for $z\in Y-K$, $f^{-1}(z)=\\{(z,0)\\}\in{NBO}_{1}(X)$, and
$\\{z\\}\notin\bigcup_{n=1}^{\infty}{NBO}_{n}(Y)$.

######  Proposition 4.4.

Let $f:X\to Y$ be an onto and open mapping between continua, let $A\in 2^{Y}$
and let $n\in\mathbb{N}$. If $f^{-1}(A)\in NBO_{n}(X)$, then $A\in
NBO_{n}(Y)$.

###### Proof.

Assume that $f^{-1}(A)\in NBO_{n}(X)$. If $A=Y$, then $A\in NBO_{n}(X)$.
Assume $A\neq Y$. Let $B\in F_{n}(X)$ witnessing that $X-f^{-1}(A)$ is a
$n$-$Qo$ space. Let $U$ be an open set of $Y-A$. Then, there exists $D\in
C(X)$ such that $D\subset X-f^{-1}(A)$ and $B\cap D\neq\emptyset\neq
int(D)\cap f^{-1}(U)$. Hence, $f(D)\subset Y-A$ and $f(B)\cap
f(D)\neq\emptyset$, and since $f$ is open, $int(f(D))\cap U\neq\emptyset$,
which implies that $Y-A$ is a $n$-$Qo$ space, therefore $A\in NBO_{n}(Y)$. ∎

######  Proposition 4.5.

Let $f:X\to Y$ be an onto monotone mapping between continua, let $A\in 2^{Y}$
and let $n\in\mathbb{N}$. The following statements hold:

  1. 1.

If $A\in CC_{n}(Y)$, then $f^{-1}(A)\in CC_{n}(X)$;

  2. 2.

if $A\in NWC_{n}(Y)$, then $f^{-1}(A)\in NWC_{n}(X)$; and

  3. 3.

if $A\in NC_{n}(Y)$, then $f^{-1}(A)\in NC_{n}(X)$.

###### Proof.

Observe that the statements are true if $A=Y$. Suppose that $A\neq Y$.

  1. 1.

Assume $A\in CC_{n}(Y)$. Let $B\in F_{n}(Y-A)$ witnessing that $Y-A$ is a
$n$-$Q1$ space and $B^{\prime}\in F_{n}(X-f^{-1}(A))$ such that
$f(B^{\prime})=B$. Let $x\in X-f^{-1}(A)$. Choose $D\in C(Y)$ such that
$D\subset Y-A$, $B\cap D\neq\emptyset$ and $f(x)\in int_{Y}(D)$. Hence
$f^{-1}(D)\subset X-f^{-1}(A)$ is a continuum with $B^{\prime}\cap
f^{-1}(D)\neq\emptyset$ and $x\in int_{X}(f^{-1}(D))$. So $f^{-1}(A)\in
CC_{n}(X)$.

  2. 2.

Assume $A\in CC_{n}(Y)$. Let $B\in F_{n}(Y-A)$ witnessing that $Y-A$ is a
$n$-$Q2$ space and $B^{\prime}\in F_{n}(X-f^{-1}(A))$ such that
$f(B^{\prime})=B$. Let $x\in X-f^{-1}(A)$. By the properties of $B$, there
exists a continuum $D\subset Y-A$ such that $f(x)\in D$ and $B\cap
D\neq\emptyset$. Hence, $f^{-1}(D)\subset X-f^{-1}(A)$ is a continuum with
$x\in f^{-1}(D)$ and $f^{-1}(D)\cap B^{\prime}\neq\emptyset$. So $f^{-1}(A)\in
NWC_{n}(X)$.

  3. 3.

Let $A\in NC_{n}(Y)$. Observe that $f^{-1}(Y-A)=X-f^{-1}(A)$ has at most $n$
components, therefore $f^{-1}(A)\in NC_{n}(X)$.

∎

The following example shows that Theorem 4.5 does not hold for the hyperspaces
$NB_{n}(Y),NB_{n}^{*}(Y),S_{n}(Y)$ and $NSC_{n}(Y)$. We do not know if
Proposition 4.5 holds for $NBO_{n}(X)$.

######  Example 4.6.

Let $Y$ be the dyadic solenoid, let $S\subset Y$ be an arc and let
$h:I\times\\{0\\}\to S$ be a homeomorphism. Take $X=Y\cup_{h}(I\times I)$ and
let $f:X\to Y$ be defined as $f(x)=x$, if $x\in Y$, and $f(x)=h(x_{1},0)$ if
$x=(x_{1},x_{2})\in I\times I$. Observe that $f:X\to Y$ is an onto, monotone
mapping. However, for $A=\\{h(\frac{1}{2},0)\\}\in NB_{1}(Y)$ and
$f^{-1}(A)\notin NSC_{1}(X)$.

However, if we add to the conditions of Theorem 4.5 that the map $f:X\to Y$ is
open, Theorem 4.5 holds for the remaining hyperspaces of non-cut sets.

######  Proposition 4.7.

Let $f:X\to Y$ be an onto, open and monotone mapping between continua, let
$A\in 2^{Y}$ and let $n\in\mathbb{N}$. The following statements hold:

  1. 1.

If $A\in NB_{n}(Y)$, then $f^{-1}(A)\in NB_{n}(X)$;

  2. 2.

if $A\in NBO_{n}(Y)$, then $f^{-1}_{n}(A)\in NBO_{n}(X)$;

  3. 3.

if $A\in NB_{n}^{*}(Y)$, then $f^{-1}(A)\in NB_{n}^{*}(X)$;

  4. 4.

if $A\in S_{n}(Y)$, then $f^{-1}(A)\in S_{n}(X)$; and

  5. 5.

if $A\in NSC_{n}(Y)$, then $f^{-1}(A)\in NSC_{n}(X)$.

###### Proof.

Observe that the statements are true if $A=Y$. Suppose that $A\neq Y$.

  1. 1.

Assume $A\in NB_{n}(Y)$. Let $x\in X-f^{-1}(A)$. Since $Y-A$ is a $n$-$Q3$
space, let $B\in F_{n}(Y-A)$ satisfying that $f(x)\in B$ and for each non-
empty open set $U\subset Y-A$, there exists a continuum $D\subset Y-A$ such
that $D\cap B\neq\emptyset\neq D\cap U$.

Let $B^{\prime}\in F_{n}(X-f^{-1}(A))$ be such that $x\in B^{\prime}$ and
$f(B^{\prime})=B$. If $V\subset X-f^{-1}(A)$ is a non-empty open set of $X$,
then $f(V)$ is open in $Y-A$, so there exists a continuum $D$ such that $B\cap
D\neq\emptyset\neq D\cap f(V)$. Hence, $f^{-1}(D)\subset X-f^{-1}(A)$ is a
continuum and $f^{-1}(D)\cap V\neq\emptyset\neq f^{-1}(D)\cap B^{\prime}$. So,
$X-f^{-1}(A)$ is a $n$-$Q3$ space and $f^{-1}(A)\in NB_{n}(X)$.

  2. 2.

Assume $A\in NBO_{n}(Y)$. Let $B\in F_{n}(Y-A)$ witnessing that $Y-A$ is a
$n$-$Qo$ space. Let $B^{\prime}\in F_{n}(X-f^{-1}(A))$ be such that
$f(B^{\prime})=B$. Consider a non-empty open set $U\subset X-f^{-1}(A)$. Since
$f(U)\subset Y-A$ is non-empty and open, there exists a continuum $D\subset
Y-A$ such that $B\cap int(D)\neq\emptyset\neq int(D)\cap f(U)$. Hence
$f^{-1}(D)\subset X-f^{-1}(A)$ is a continuum and $B^{\prime}\cap
int(f^{-1}(D))\neq\emptyset\neq int(f^{-1}(D))\cap U$. So $X-f^{-1}(A)$ is a
$n$-$Qo$ space and $f^{-1}(A)\in NBO_{n}(X)$.

  3. 3.

Assume $A\in NB_{n}^{*}(Y)$. Let $B\in F_{n}(Y-A)$ witnessing that $Y-A$ is a
$n$-$Q4$ space. Let $B^{\prime}\in F_{n}(X-f^{-1}(A))$ be such that
$f(B^{\prime})=B$. Consider a non-empty open set $U\subset X-f^{-1}(A)$. Since
$f(U)\subset Y-A$ is non-empty and open, there exists a continuum $D\subset
Y-A$ such that $B\cap D\neq\emptyset\neq D\cap f(U)$. Hence $f^{-1}(D)\subset
X-f^{-1}(A)$ is a continuum and $B^{\prime}\cap f^{-1}(D)\neq\emptyset\neq
f^{-1}(D)\cap U$. So $X-f^{-1}(A)$ is a $n$-$Q4$ space and $f^{-1}(A)\in
NB_{n}^{*}(X)$.

  4. 4.

Assume $A\in S_{n}(Y)$. Let $U_{1},\dots,U_{m}$ be a finite number of non-
empty open sets contained in $X-f^{-1}(A)$. Since $f(U_{1}),\dots,f(U_{m})$ is
a finite number of non-empty open sets contained in $Y-A$ and $A\in S_{n}(Y)$,
there exists an element $D\in C_{n}(Y-A)$ such that $D\cap
f(U_{i})\neq\emptyset$ for each $i\in\\{1,\dots,m\\}$. Hence $f^{-1}(D)\in
C_{n}(X-f^{-1}(A))$, satisfies $f^{-1}(D)\cap U_{i}\neq\emptyset$ for each
$i\in\\{1,\dots,m\\}$. So, $f^{-1}(A)\in S_{n}(X)$.

  5. 5.

Assume $A\in NSC_{n}(Y)$. Let $U_{1},\dots,U_{n+1}$ be a finite number of non-
empty open sets contained in $X-f^{-1}(A)$. Since $f(U_{1}),\dots,f(U_{n+1})$
is a collection of $n+1$ non-empty open sets contained in $Y-A$ and $A\in
NSC_{n}(Y)$, there exists an element $D\in C_{n}(Y)$ such that $D\subset Y-A$
and $D\cap f(U_{i})\neq\emptyset$ for each $i\in\\{1,\dots,n+1\\}$. Hence
$f^{-1}(D)\in C_{n}(X)$, $f^{-1}(D)\subset X-f^{-1}(A)$ and $f^{-1}(D)\cap
U_{i}\neq\emptyset$ for each $i\in\\{1,\dots,n+1\\}$. So, $f^{-1}(A)\in
NSC_{n}(X)$.

∎

##  5 Relations between $X$ and the hyperspaces of non-cut sets of degree $n$
of $X$

The first result we present provides an initial insight into the relationship
between hyperspaces of non-$n$-cut sets, when the original space exhibits
certain specific characteristics. We will explore properties of the original
space that may lead to the coincidence of some of its hyperspace of
non-$n$-cut sets. On the other hand, while the fact that a set is a
non-$n$-cut set may not be particularly relevant by itself, studying the
hyperspace of these sets can lead to interesting conclusions.

######  Theorem 5.1.

If X is a locally connected continuum, then ${NC}_{1}(X)={CC}_{1}(X)$.

###### Proof.

We only have to prove that ${NC}_{1}(X)\subset{CC}_{1}(X)$. Let
$A\in{NC}_{1}(X)$ and let $x,y\in X-A$. Let $U_{x}$ and $U_{y}$ be open
connected neighborhoods of $x$ and $y$, respectively, such that $cl(U_{x})\cap
A=\emptyset=cl(U_{y})\cap A$. Since $X-A$ is an open connected set, $X-A$ is
arcwise connected (Theorem 8.26 of [10]), so let $Y$ be an arc in $X-A$
joining $x$ to $y$. Observe that $D=Y\cup cl(U_{x})\cup cl(U_{y})$ is a
continuum avoiding $A$ with $x,y\in int(D)$. In conclusion, $A\in{CC}_{1}(X)$.
∎

######  Corollary 5.2.

If X is a locally connected continuum, then ${CC}_{n}(X)={NC}_{n}(X)$ for each
$n\in\mathbb{N}$.

###### Proof.

See Corollary 3.15. ∎

The following example shows that Theorem 5.1 does not hold if we replace the
locally connected continuum condition with the mutually aposyndetic continuum
condition or the Kelley continuum condition.

######  Example 5.3.

Let $X$ the suspension over the Cantor set, so $X$ is mutually aposyndetic and
Kelley. Let $p$ and $q$ the vertices of $X$, and let $A$ be a set consisting
of two points in the same arc-component of $X-\\{p,q\\}$. Then, $A\in
NC_{1}(X),$ $A\notin NB_{1}(X)$ and $A\notin CC_{1}(X)$. Notice that
$F_{1}(X)=NC_{1}(X)\cap F_{1}(X)=CC_{1}(X)\cap F_{1}(X)$.

We do not know a non-locally connected continuum $X$ where
$NC_{1}(X)=NWC_{1}(X)$. Hence, we find the following question interesting:

######  Question 5.4.

If ${CC}_{1}(X)={NC}_{1}(X)$ or ${NWC}_{1}(X)={NC}_{1}(X)$, is $X$ locally
connected?

######  Note 5.5.

When $X$ is an indecomposable continuum, we have
${CC}_{n}(X)={NWC}_{n}(X)={NBO}_{n}(X)=\\{X\\}$ and
$C_{n}(X)\subset{NB}_{n}^{*}(X)$, for each $n\in\mathbb{N}$.

With the following example, we show that $\\{X\\}={CC}_{n}(X)={NWC}_{n}(X)$
$={NBO}_{n}(X)$ does not imply that the space is indecomposable.

######  Example 5.6.

For a space $X=A\cup B$, where $A$ and $B$ are two indecomposable continua and
$A\cap B$ consists only of one point, we have
$CC_{n}(X)=NWC_{n}(X)=NBO_{n}(X)=\\{X\\}$, for each $n\in\mathbb{N}$.

As mentioned earlier, the condition of $X$ being locally connected implies
that $NC_{n}(X)=CC_{n}(X)$ for each $n\in\mathbb{N}$. As part of the research
related to Question 5.4, for a continuum $X$ and a set $A\in 2^{X}$, we are
exploring certain conditions under which $A\in NC_{n}(X)$ implies $A\in
CC_{n}(X)$. With that goal, we generalize the definition of semi-local
connectivity given by Whyburn in [13] from points to sets, and we define a
continuum $X$ to be semi-locally connected at a set $A$ provided that if $U$
is an open subset of $X$ containing $A$, there is an open subset $V$ of $X$
lying in $U$ and containing $A$ such that $X-V$ has a finite number of
components.

######  Proposition 5.7.

Let $X$ be a continuum. If $A\in\bigcup_{i=1}^{\infty}CC_{i}(X)$, then $X$ is
semi-locally connected at $A$.

###### Proof.

It follows from Lemma 3.19. ∎

While it may seem intuitive that the converse of Theorem 5.7 holds, this is
not the case. In the example 5.8, the vertex is a point of semi-local
connectivity but does not belong to
$\bigcup_{i=1}^{\infty}CC_{i}(F_{\omega})$.

######  Example 5.8.

Define
$F_{\omega}=\bigcup_{n=1}^{\infty}\\{(x,\frac{x}{n}):x\in[0,\frac{1}{n}]\\}$,
considered as a subspace of $\mathbb{R}^{2}$. Let $A=\\{(0,0)\\}$. Then,
$F_{\omega}$ is semi-locally connected at $A$ but
$A\notin\bigcup_{i=1}^{\infty}CC_{i}(F_{\omega})$.

The following theorem is based on results (6.1) and (6.21) of [13].

######  Theorem 5.9.

Let $X$ be a continuum and $A\in 2^{X}$. If $X$ is semi-locally connected at
$A$, then each component of $X-A$ is a continuumwise connected open set.

###### Proof.

Let $D$ be a component of $X-A$. Let $\\{U_{n}\\}_{n=1}^{\infty}$ be a
sequence of open sets such that for each $n$, $A\subset U_{n}$,
$cl(U_{n+1})\subset U_{n}$, $A=\bigcap_{n=1}^{\infty}U_{n}$ and $X-U_{n}$ has
a finite number of components. For each $a\in D$, let $C_{a}=\bigcup\\{K:K$ is
the component of $X-U_{n}$ containing $a$, for some $n\in\mathbb{N}\\}$.
Notice that for each $a,b\in D$, $a\in C_{a}$, $C_{a}$ is continuumwise
connected and $C_{a}\cap C_{b}\neq\emptyset$ implies $C_{a}=C_{b}$.

Now we prove that $C_{a}$ is open for each $a\in D$. If $x\in C_{a}$, then $x$
is an element the component of $X-U_{n}$ containing $a$, for some
$n\in\mathbb{N}$. Therefore $x\notin cl(U_{n+1})$, so $x\in int(K)$, where $K$
is the component of $X-U_{n+1}$. Hence $x\in int(C_{a})$.

Since, $D-C_{a}=\cup\\{C_{b}:b\in D-C_{a}\\}$ is an open set and $D$ is
connected, $C_{a}$ must be equal to $D$. ∎

The following theorem is useful to understand better the characteristics of
the sets $A$, for which $X$ is semi-locally connected at $A$.

######  Theorem 5.10.

Let $X$ be a continuum and $A\in 2^{X}$. Then, $X$ is aposyndetic with respect
to $A$ if and only if it is semi-locally connected at $A$.

###### Proof.

Assume that $X$ is aposyndetic with respect to $A$ and let $U$ be an open set
such that $A\subset U$. Then, for each $p\in X-A$, there exists a subcontinuum
$C_{p}$ of $X$ such that $p\in int(C_{p})$ and $C_{p}\cap A=\emptyset$. Hence,
$\\{int(C_{p}):p\in X-A\\}$ is an open cover of $X-U$. Therefore, there exists
a finite set $\\{p_{1},\ldots,p_{n}\\}\subset X-A$ such that
$X-U\subset\bigcup_{i=1}^{n}C_{p_{i}}$. Notice that
$V=X-\bigcup_{i=1}^{n}C_{p_{i}}\subset U$ and $X-V$ has at most $n$
components. In conclusion, $X$ is semi-locally connected at $A$.

Now, suppose that $X$ is semi-locally connected at $A$ and let $x\in X-A$. Let
$U$ be an open set such that $A\subset U\subset cl(U)\subset X-\\{x\\}$.
Hence, there exists an open set $V$ such that $A\subset V\subset U$ and $X-V$
has a finite number of components. Therefore, the component of $X-V$
containing $x$ is a continuum with $x$ in its interior. We conclude that $X$
is aposyndetic with respect to $A$. ∎

######  Corollary 5.11.

Let $X$ be a continuum and let $A\in 2^{X}$ such that $X$ is semi-locally
connected at $A$. If $A\in NC_{1}(X)$, then $A\in CC_{1}(X)$.

###### Proof.

Let $x,y$ two different points in $X-A$, then by Theorem 5.9, there exists a
continuum $K$ such that $\\{x,y\\}\subset K$ and $K\cap A=\emptyset$. By
Theorem 5.10, there exist two continua $K_{x}$ and $K_{y}$ such that $x\in
int(K_{x})$, $y\in int(K_{y})$ and $K_{x}\cap A=\emptyset=K_{y}\cap A$. Let
$D=K\cup K_{x}\cup K_{y}$, observe that $\\{x,y\\}\in int(D)$ and $D\cap
A=\emptyset$. Hence, $A\in CC_{1}(X)$. ∎

######  Corollary 5.12.

Let $X$ be a continuum, $n\in\mathbb{N}$ and $A\subset X$ such that $X$ is
semi-locally connected at $A$. If $A\in{NC}_{n}(X)$, then $A\in{CC}_{n}(X)$.

###### Proof.

Assumme $A\in NC_{n}(X)$. Using the same arguments as in the proof of 5.10, we
obtain that for each component $K$ of $X-A$, $K\in CC_{1}(A\cup K)$. Hence, by
Theorem 3.13, $A\in CC_{n}(X)$. ∎

By Theorem 3.1, Proposition 5.7, Theorem 5.10 and Corollary 5.12, we obtain
the following result, which is a characterization of the elements of
$CC_{n}(X)$.

######  Corollary 5.13.

Let $X$ be a continuum and $n\in\mathbb{N}$. Then, $A\in CC_{n}(X)$ if and
only if $X$ is aposyndetic respect to $A$ and $A\in NC_{n}(X)$.

As an application of Corollary 5.13 we present a continuum $X$ in which
$NC_{1}(X)\cap D_{0}(X)=CC_{1}(X)\cap D_{0}(X)$ and $X$ is not aposyndetic
with respect to some $A\in D_{0}(X)$.

######  Example 5.14.

Let $Y$ be the harmonic fan with vertex $v$ and let
$X=Y\times[0,1]/(\\{v\\}\times[0,1])$. Notice that $X$ is not aposyndetic with
respect to $\\{[(v,0)]\\}$ and $NC_{1}(X)\cap D_{0}(X)=\\{A\in
D_{0}(X):[(v,0)]\notin A\\}=CC_{1}(X)\cap D_{0}(X)$

######  Proposition 5.15.

Let $X$ be a continuum and $A\in NSC_{1}(X)$. If $X$ is aposyndetic at $p$
with respect to $A$, for some $p\in X-A$, then $A\in NB_{1}^{*}(X)$.

###### Proof.

Assume that $X$ is aposyndetic at $p$ with respect to $A$, let $C\subset X-A$
be a continuum containing $p$ in its interior. Let $B=\\{p\\}$. Take $U$ an
open set of $X-A$. Since $X-A$ is a $1$-$Q6$ space, for $U$ and $V=int(C)$,
there exists a continuum $D\subset X-A$ such that $D\cap U\neq\emptyset\neq
D\cap V$. Then $E=C\cup D\subset X-A$ is a continuum such that $b\in int(E)$
and $E\cap U\neq\emptyset$. Hence, $A\in NB_{1}^{*}(X)$. ∎

In [3, Proposition 2.2] the authors proved that $NC_{1}(X)\cap
F_{1}(X)=CC_{1}(X)\cap F_{1}(X)$ when $X$ is aposyndetic. It is natural to ask
if the converse is also true. We see in the next example that the answer is
negative.

######  Example 5.16.

Let $\mathcal{C}$ be the Cantor set, $Z=\mathcal{C}\times\mathcal{S}^{1}$ and
$X=Z/(\mathcal{C}\times\\{q\\})$, where $q$ is a point in $\mathcal{S}^{1}$.
Notice that $X$ is not aposyndetic at the point $[\mathcal{C}\times\\{q\\}]$
and $NC_{1}(X)\cap F_{1}(X)=\\{\\{p\\}:p\in
X-\\{[\mathcal{C}\times\\{q\\}]\\}\\}=CC_{1}(X)\cap F_{1}(X)$.

For irreducible continua, the converse of [3, Proposition 2.2] is true, we
give a stronger result in the following theorem.

######  Theorem 5.17.

Let $X$ be an irreducible continuum. If $NC_{1}(X)\cap F_{1}(X)=NWC_{1}(X)\cap
F_{1}(X)$ then $X$ is an arc.

###### Proof.

By Corollary 2 of [1], let $p,q\in X$ be distinct such that
$\\{p\\},\\{q\\}\in NC_{1}(X)$. Hence, $\\{p\\},\\{q\\}\in NWC_{1}(X)$. Given
that $\\{p\\},\\{q\\}\in NWC_{1}(X)$, $X$ must be irreducible only between $p$
and $q$. Since $NC_{1}(X)\cap F_{1}(X)=NWC_{1}(X)\cap F_{1}(X)$ and $X$ is
irreducible between $p$ and $q$, if $z\in X-\\{p,q\\}$, then $\\{z\\}\notin
NWC_{1}(X)$, so $\\{z\\}\notin NC_{1}(X)$. Therefore, by Section 3 of [1], $X$
must be an arc. ∎

With Theorem 5.17, we can provide a partial answer to Question 5.4.

######  Corollary 5.18.

Let $X$ be an irreducible continuum. Then $NC_{1}(X)=NWC_{1}(X)$ if and only
if $X$ is an arc.

The following example shows that Theorem 5.17 does not hold if we replace the
hyperspace $NWC_{1}(X)$ by a weaker one.

######  Example 5.19.

If $X$ is a dyadic solenoid, then $X$ is irreducible, and $NC_{1}(X)\cap
F_{1}(X)=F_{1}(X)=NB_{1}(X)\cap F_{1}(X)$.

Also, for irreducible continua, we have the following results.

######  Theorem 5.20.

Let $X$ be an irreducible continuum. Then $NWC_{1}(X)=CC_{1}(X)$.

###### Proof.

The contention $CC_{1}(X)\subset NWC_{1}(X)$ follows from Theorem 3.1.

We prove $NWC_{1}(X)\subset CC_{1}(X)$. Assume that $X$ is irreducible between
$a$ and $b$ and let $A\in NWC_{1}(X)$.

Claim 1: If $B$ is a component of $A$, then $a\in B$ or $b\in B$.

Proof of Claim 1. Assume $B$ is a component of $A$; by Theorem 3.20, $B\in
NWC_{1}(X)$. If $a\notin B$ and $b\notin B$, then there exists a continuum
$D\subset X-B$ such that $a,b\in D$, which is a contradiction to the
irreducibility of $X$. Thus $a\in B$ or $b\in B$.

Claim 2: The set $A$ has at most two components.

Proof of Claim 2. By Claim 1, each component of $A$ contains $a$ or contains
$b$, so $A$ has at most two components.

Claim 3: If $A$ is connected, then $A\in CC_{1}(X)$.

Proof of Claim 3. Without loss of generality, assume $a\in A$. If $b\in A$,
then $B=X\in CC_{1}(X)$. Assume $b\notin A$, let $x\in X-A$ and let $E\subset
X-A$ be a continuum with $b,x\in E$. Let $V$ be an open set such that $x\in V$
and $cl(V)\cap A=\emptyset$. Let $K$ be the component of $X-V$ that contains
$A$, and observe that $b\notin K$. Let $k\in K-A$; since $A\in NWC_{1}(X)$,
there exists a continuum $D\subset X-A$ such that $b,k\in D$. As $X$ is
irreducible between $a,b$, we have $D\cup K=X$, $b\notin K$ and $K\cap
V=\emptyset$, so $V\subset D$, which implies $x\in int(D)$. So $E\cup D$ is a
continuum with $b\in E\cup D$ and $x\in int(E\cup D)$, so $A\in CC_{1}(X)$.

Claim 4: If $A$ is not connected, then $A\in CC_{1}(X)$.

Proof of Claim 4. Assume $A$ is not connected. By Claim 2, $A$ has two
components $E$ and $F$. By Claim 1, without loss of generality assume $a\in E$
and $b\in F$. Let $p\in X-A$, let $x\in X-A$ and $V$ be an open set such that
$x\in V\subset cl(V)\subset X-A$. Let $K_{E}$ and $K_{F}$ be the components of
$X-V$ containing $E$ and $F$ respectively. Since $K_{E}\cup K_{F}\neq X$,
$K_{E}\cap K_{F}=\emptyset$. Take $e\in K_{E}-A$ and $f\in K_{F}-A$. Since
$A\in NWC_{1}(X)$, there exists a continuum $D\subset X-A$ containing
$\\{e,f\\}$. Therefore $D\cup K_{E}\cup K_{F}$ is a continuum containing $a$
and $b$. Since $X$ is irreducible between $a$ and $b$, $D\cup K_{E}\cup
K_{F}=X$. Hence $V\subset D$. This implies $A\in CC_{1}(X)$.

From Claims 1,2,3,4 and 5, we conclude that $A\in NWC_{1}(X)$ implies $A\in
CC_{1}(X)$. ∎

The following example shows that the previous result does not hold if we
replace $1$ by $n>1$.

######  Example 5.21.

Let $X=\\{(x,sin(\frac{1}{x})):x\in(0,1]\\}\cup\\{(0,x):x\in[-1,1]\\}$. Notice
that $X$ is irreducible and $\\{(0,-1)\\}\in NWC_{2}(X)-CC_{2}(X)$.

######  Theorem 5.22.

Let $X$ be a decomposable continuum and let $C\in NB_{1}(X)$. If $A,B\in
C(X)-\\{X\\}$ are such that $A\cup B=X$ and $C\cap B=\emptyset$, then $C\in
NWC_{1}(X)$.

###### Proof.

Let $p\in X-C$, given that $C\in NB_{1}(X)$ and $int(B)\neq\emptyset$, there
exists a continuum $K\subset X-C$, such that $p\in K$ and $K\cap
B\neq\emptyset$. Therefore, $C\in NWC_{1}(X)$. ∎

######  Corollary 5.23.

Let $X$ be a decomposable, irreducible continuum and let $C\in NB_{1}(X)$. If
$A,B\in C(X)-\\{X\\}$ are such that $A\cup B=X$ and $C\cap B=\emptyset$, then
$C\in CC_{1}(X)$.

###### Proof.

Apply Theorem 5.20 and Theorem 5.22. ∎

######  Corollary 5.24.

Let $X$ be a decomposable continuum and irreducible between $p$ and $q$. If
$\\{p\\}\in NB_{1}(X)$, then $\\{p\\}\in CC_{1}(X)$.

The next example shows that in Corollary 5.24, we can not replace the
condition irreducible between $p$ and $q$ by the condition irreducible about
$A\in 2^{X}$.

######  Example 5.25.

Let $X=Y\cup I$, where $Y$ is the dyadic solenoid and $I$ is an arc such that
$Y\cap I=\\{p\\}$. In this case, $X$ is irreducible about $I\cup\\{x\\}$,
where $x\in Y$ and $p$ are in distinc composants of $Y$. Notice $I\in
NB_{1}(X)$ but $I\notin NWC_{1}(X)$.

By Theorem 5.24 and Lemma 3.10 from [3], we obtain the following result.

######  Corollary 5.26.

Let $X$ be a decomposable continuum and irreducible between $p$ and $q$. If
$\\{p\\}\in NB_{1}(X)$, then $X$ is locally connected at $p$.

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