# On Limit Sets of Monotone Maps on Regular Curves

Aymen Daghar and Habib Marzougui  Aymen Daghar, University of Carthage,
Faculty of Science of Bizerte, (UR17ES21), “Dynamical systems and their
applications”, Jarzouna, 7021, Bizerte, Tunisia<EMAIL_ADDRESS>Habib
Marzougui, University of Carthage, Faculty of Science of Bizerte, (UR17ES21),
“Dynamical systems and their applications”, 7021, Jarzouna, Bizerte, Tunisia
<EMAIL_ADDRESS>
###### Abstract.

We investigate the structure of $\omega$-limit (resp. $\alpha$-limit) sets for
a monotone map $f$ on a regular curve $X$. We show that for any $x\in X$
(resp. for any negative orbit $(x_{n})_{n\geq 0}$ of $x$), the $\omega$-limit
set $\omega_{f}(x)$ (resp. $\alpha$-limit set $\alpha_{f}((x_{n})_{n\geq 0})$)
is a minimal set. This also hold for $\alpha$-limit set $\alpha_{f}(x)$
whenever $x$ is not a periodic point. These results extend those of Naghmouchi
[18] established whenever $f$ is a homeomorphism on a regular curve and those
of Abdelli [1] , whenever $f$ is a monotone map on a local dendrite. Further
results related to the basin of attraction of an infinite minimal set are also
obtained.

###### Key words and phrases:

Minimal set, regular curve, $\omega$-limit set, $\alpha$-limit set, monotone
map, recurrent point.

###### 2000 Mathematics Subject Classification:

37B20, 37B45, 54H20

##  1\. Introduction

In the last two decades, the study of dynamical systems in one dimensional
continua has developed and become increasingly intensive due to the rich
dynamical properties that one dimensional maps can exhibit. Examples of
continua studied include, graphs, dendrites and local dendrites (see for
instance [1], [2], [3], [20]). Recently regular curves has given a special
attention (see for example, [4], [10], [11], [18], [19], [21]). These form a
large class of continua which includes local dendrites. The Sierpiński
triangle is a well known example of a regular curve which is not a local
dendrite. Regular curves appear in continuum theory and also in other branches
of Mathematics such as complex dynamics; for instance, the Sierpiński triangle
can be realized as the Julia set of the complex polynomial $p(z)=z^{2}+2$ (see
[7]). Also dendrites appear naturally as the Julia set of a complex polynomial
$p(z)=z^{2}+c$, then for certain values of $c$ (e.g. $c=i$), the Julia set $J$
of $p$ is a dendrite (see [6], p. 296). Recall that the Julia set of $p$ is
$J(p)=\\{z\in\mathbb{C}:(p^{n}(z))_{n\geq 1}\textrm{ is bounded}\\}$. In [21],
Seidler proved that every homeomorphism of a regular curve has zero
topological entropy (later, this result was extended by Kato in [10] to
monotone maps).

In the present paper, we deal with several questions/problems.

First, we are interested in whether every $\omega$-limit set is a minimal set.
Recall that this was proved in two cases:

$-$ for homeomorphisms on regular curves (Naghmouchi [18])

$-$ for monotone maps on local dendrites (Abdelli [1]).

In Theorem 3.1 we prove a more general result by showing that this is in fact
true for monotone maps on regular curves. We do not know whether the above
result can be extended to still a larger class of curves. However, let us
mention that already for monotone maps (resp. homeomorphism) on dendroids, it
can happen that an $\omega$-limit set is not a minimal set (Makhrova ([14],
Theorem 5), (resp. Naghmouchi [20]). The dendroids constructed therein are
rational curves. Some consequences are derived from Theorem 3.1, we prove e.g.
in Section 4 that the basin of attraction of an infinite minimal set for
monotone maps on regular curves is closed extending the one of [18] for
homeomorphisms on regular curves.

Second, we are interested to the study of accumulated points of backward
orbit, say $\alpha$-limit set. For the case of homeomorphisms on regular
curves, any $\alpha$-limit set is a minimal set as a consequence from above.
In Theorem 5.7, we prove a more general result by showing that for monotone
map $f$ on a regular curve, any $\alpha$-limit set $\alpha_{f}(x)$ of a non
periodic point $x$ is a minimal set. However, we built an example of a
monotone map $f$ on an infinite star for which $\alpha_{f}(x)$ is not a
minimal set for some periodic point $x$.

Third, we address the question of the equality between the set of recurrent
points and the set of almost periodic points. This was proved in two cases:

$-$ for homeomorphisms on regular curves (Naghmouchi [19])

$-$ for monotone maps on local dendrites (Abdelli et al. [3])

##  2\. Preliminaries

Let $X$ be a compact metric space with metric $d$ and let $f:X\longrightarrow
X$ be a continuous map. Let $\mathbb{Z},\ \mathbb{Z}_{+}$ and $\mathbb{N}$ be
the sets of integers, non-negative integers and positive integers,
respectively. For $n\in\mathbb{Z_{+}}$ denote by $f^{n}$ the $n$-th iterate of
$f$; that is, $f^{0}=\textrm{identity}$ and $f^{n}=f\circ f^{n-1}$ if
$n\in\mathbb{N}$. For any $x\in X$, the subset
$\textrm{Orb}_{f}(x)=\\{f^{n}(x):n\in\mathbb{Z}_{+}\\}$ is called the orbit of
$x$ (under $f$). A subset $A\subset X$ is called $f-$invariant (resp. strongly
$f-$invariant) if $f(A)\subset A$ (resp., $f(A)=A$); it is further called a
minimal set (under $f$) if it is closed, non-empty and does not contain any
$f-$invariant, closed proper non-empty subset of $X$. We define the
$\omega$-limit set of a point $x$ to be the set:

$\displaystyle\omega_{f}(x)$ $\displaystyle=\\{y\in
X:\liminf_{n\to+\infty}d(f^{n}(x),y)=0\\}$
$\displaystyle=\underset{n\in\mathbb{N}}{\cap}\overline{\\{f^{k}(x):k\geq
n\\}}.$

A point $x\in X$ is called:

$-$ Periodic of period $n\in\mathbb{N}$ if $f^{n}(x)=x$ and $f^{i}(x)\neq x$
for $1\leq i\leq n-1$; if $n=1$, $x$ is called a fixed point of $f$ i.e.
$f(x)=x$;

$-$ Almost periodic if for any neighborhood $U$ of $x$ there is
$N\in\mathbb{N}$ such that $\\{f^{i+k}(x):0\leq i\leq N\\}\cap
U\neq\emptyset$, for all $k\in\mathbb{N}$.

$-$Recurrent if $x\in\omega_{f}(x)$.

We denote by P$(f$), AP$(f$), R$(f$) and $\Lambda(f)$ the sets of periodic
points, almost periodic points, recurrent points and the union of all
$\omega$-limit sets of $f$, respectively. Define the space

$X_{\infty}=\displaystyle\bigcap_{n\in\mathbb{N}}f^{n}(X)$. From the
definition, we have the following inclusions:

$P(f)\subseteq\textrm{AP}(f)\subseteq R(f)\subseteq\Lambda(f)\subseteq
X_{\infty}.$

This was proved in two cases:

$-$ for homeomorphisms on regular curves (Naghmouchi [19])

$-$ for monotone maps on local dendrites (Abdelli et al. [3])

In the definitions below, we use the terminology from Nadler [17] and
Kuratowski [12].

######  Definition 2.1 ([12], p. 131).

Let $X,Y$ be two topological spaces. A continuous map $f:X\longrightarrow Y$
is said to be monotone if for any connected subset $C$ of $Y$, $f^{-1}(C)$ is
connected.

When $f$ is closed and onto, Definition 2.1 is equivalent to that the preimage
of any point by $f$ is connected (cf. [12], p. 131). Notice that $f^{n}$ is
monotone for every $n\in\mathbb{N}$ when $f$ itself is monotone.

A _continuum_ is a compact connected metric space. An _arc_ $I$ (resp. a
_circle_) is any space homeomorphic to the compact interval $[0,1]$ (resp. to
the unit circle $\mathbb{S}^{1}=\\{z\in\mathbb{C}:\ |z|=1\\}$). A space is
called degenerate if it is a single point, otherwise; it is non-degenerate. A
dendrite is a locally connected continuum which contains no circle. A local
dendrite is a continuum every point of which has a dendrite neighborhood. By
([12], Theorem 4, p. 303), a local dendrite is a locally connected continuum
containing only a finite number of circles. As a consequence every sub-
continuum of a local dendrite is a local dendrite ([12], Theorems 1 and 4, p.
303). Every graph and every dendrite is a local dendrite.

A regular curve is a continuum $X$ with the property that for every point
$x\in X$ and each open neighborhood $V$ of $x$ in $X$, there exists an open
neighborhood $U$ of $x$ in $V$ such that the boundary set $\partial U$ of $U$
is finite. Each regular curve is a $1$-dimensional locally connected
continuum. It follows that each regular curve is locally arcwise connected
(for more details see [12] and [17]). In particular every local dendrite is a
regular curve (cf. [12], p. 303). A continuum $X$ is said to be rational if
any point in $X$ has a local base of open sets with countable boundary. In
particular every regular curve is a rational curve.

Let $X$ be a compact metric space. We denote by $2^{X}$ (resp. $C(X)$) the set
of all non-empty compact subsets (resp. compact connected subsets) of $X$. The
Hausdorff metric $d_{H}$ on $2^{X}$ (respectively $C(X)$) is defined as
follows: $d_{H}(A,B)=\max\Big{(}\sup_{a\in A}d(a,B),\sup_{b\in
B}d(b,A)\Big{)}$, where $A,B\in 2^{X}$ (resp. $C(X)$). For $x\in X$ and $M\in
2^{X}$, $d(x,M)=\inf_{y\in M}d(x,y)$. With this distance, $(C(X),d_{H})$ and
$(2^{X},d_{H})$ are compact metric spaces. Moreover if $X$ is a continuum,
then so are $2^{X}$ and $C(X)$ (for more details, see [17]). Let
$f:X\longrightarrow X$ be a continuous map of $X$. We denote by
$2^{f}:2^{X}\longrightarrow 2^{X},A\to f(A)$, called the induced map. Then
$2^{f}$ is also a continuous self mapping of $(2^{X},d_{H})$ (cf. [17]). For a
subset $A$ of $X$, we denote by diam$(A)=\sup_{x,y\in A}d(x,y)$.

A family $(A_{i})_{i\in I}$ of subsets of $X$ is called a null family if for
any infinite sequence $(i_{n})_{n\geq 0}$ of
$I,\;\displaystyle\lim_{n\to+\infty}\textrm{diam}(A_{i_{n}})=0$. It is well
known that each pairwise disjoint family of subcontinua of a regular curve is
null (see [13]). For the convenience of the reader, we give below a proof of
it.

######  Proposition 2.2.

Let $X$ be a regular curve. Then for any $\varepsilon>0$ and for any family of
pairwise disjoint subcontinua $(A_{i})_{i\in I}$ of $X$, the set
$\\{i\in I:\textrm{diam}(A_{i})\geq\varepsilon\\}$ is finite. In particular if
$(A_{n})_{n\geq 0}$ is a sequence of pairwise disjoint continua, then
$(A_{n})_{n\geq 0}$ is a null family.

###### Proof.

Assume that for some $\varepsilon>0$, there exists an infinite family of
pairwise disjoint continua $(A_{n})_{n\geq 0}$ such that for any $n\geq 0$, we
have diam$(A_{n})>\varepsilon$. So for each $n\geq 0$, there exist
$\\{x_{n},y_{n}\\}\subset A_{n}$ such that $d(x_{n},y_{n})\geq\varepsilon$. As
$X$ is a compact set, one can assume that the two sequences $(x_{n})_{n\geq
0}$ and $(y_{n})_{n\geq 0}$ converge to $x$ and $y$, respectively. Moreover
$x\neq y$. So let $O_{x}$ be an open set with finite boundary such that $x\in
O_{x}$ and $y\notin\overline{O_{x}}$. Then there exists some $N\geq 0$ such
that for any $n\geq N$, we have $x_{n}\in O_{x}\cap A_{n}$. As
$y\notin\overline{O_{x}}$, there exists $N_{1}>N$ such that for any $n\geq
N_{1}$, we have $y_{n}\notin O_{x}$. Thus $A_{n}$ meets the finite boundary
$\partial O_{x}$, this will lead to a contradiction since the $A_{n},n\geq 0$
are disjoint. In particular, if $(A_{n})_{n\geq 0}$ is a sequence of pairwise
disjoint continua, then
$\displaystyle\lim_{n\to+\infty}\textrm{diam}(A_{n})=0$. ∎

We will use the following lemma which is well known in the literature.

######  Lemma 2.3.

$($[8]$)$ Let $X$ be a compact metric space, $f:~{}X\rightarrow X$ be a
continuous map and let $\omega_{f}(x)$ be an $\omega$-limit set of a point
$x\in X$. If $\omega_{f}(x)$ is the union of $N$ pairwise disjoint closed
subsets $A_{0},A_{1},\dots,A_{N-1}$ which are cyclicly permuted by $f$ $($i.e.
$f(A_{i})=A_{i+1(\textrm{mod})N})$, then there exists
$j\in\\{0,1,\dots,N-1\\}$ such that $\omega_{f^{N}}(x)=A_{j}.$

In [15], it was introduced the notion of totally periodic $\omega$-limit set.
Recall that an $\omega$-limit set is said to be totally periodic  if it is
composed of periodic points. A compact metric space $X$ has the $\omega$-FTP
property provided that for each self-continuous map $f$ of $X$, every totally
periodic $\omega$-limit set of $f$ is finite. In [4], it was proved that
regular continua have the $\omega$-FTP property for homeomorphisms. Recently,
the above result was relaxed to monotone maps in [16].

######  Theorem 2.4.

$($[16]$)$ Let $X$ be a regular curve and $f:X\longrightarrow X$ be a monotone
map. Then any totally periodic $\omega$-limit set of $f$ is finite.

##  3\. The $\omega$-limit sets for monotone maps on regular curves

The aim of this section is to prove the following theorem.

######  Theorem 3.1.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$. Then each
$\omega$-limit set of $f$ is a minimal set.

###### Proof.

Let $x\in X$. If $\omega_{f}(x)$ is finite, then it is a periodic orbit. Now
assume that $\omega_{f}(x)$ is infinite and $\omega_{f}(x)$ is not minimal.
Then there is a minimal set $M\subsetneq\omega_{f}(x)$. We distinguish two
cases:

Case 1. $M$ is infinite. Let $U$ be an open neighborhood of $M$ in $X$ with
finite boundary of cardinality $k\in\mathbb{N}$ and such that
$\omega_{f}(x)\setminus\overline{U}\neq\emptyset$. Then there exist $k+1$
positive integers $n_{0}<\dots<n_{k}$ such that $f^{n_{i}}(x)\in
X\setminus\overline{U}$, set $r=n_{k}-n_{0}$. Let $z\in M$, hence its orbit is
infinite. So $f^{i}(z)\neq f^{j}(z)$, for $i\neq j$ and then as $X$ is locally
arcwise connected, there exists a pairwise disjoint family $(W_{i})_{0\leq
i\leq r}$ of arcwise connected open neighborhoods of $f^{i}(z)$: $0\leq i\leq
r$. As $M\subset\omega_{f}(x)$, there exist some $m>n_{0}$ and a family of
arcs $(J_{i})_{0\leq i\leq r}$ such that for any $0\leq i\leq r,\;J_{i}\subset
W_{i}$ and $J_{i}$ join $f^{i}(z)$ and $f^{m+i}(x)$. As the arcs
$J_{0},J_{1},\dots,J_{r}$ are pairwise disjoint, then so are the
$f^{-(m-n_{0})}(J),\dots,f^{-(m-n_{0})}(J_{r})$. Set
$C_{i}:=f^{-(m-n_{0})}(J_{n_{i}-n_{0}})$, $0\leq i\leq k$. Then the $C_{i}$
are pairwise disjoint, $0\leq i\leq k$. Since $f$ is monotone, each $C_{i}$ is
connected. Moreover $C_{i}$ contains $f^{n_{i}}(x)$ and
$f^{-(m-n_{0})}(f^{n_{i}-n_{0}}(z))$. As $M=f^{p}(M)$, for any $p\geq 0$,
$u_{i}:=f^{n_{i}-n_{0}}(z)\in M$ and there is $v_{i}\in M$ such that
$f^{m-n_{0}}(v_{i})=u_{i}$. Thus $f^{-(m-n_{0})}(f^{n_{i}-n_{0}}(z))$ contains
$v_{i}$ and hence $f^{-(m-n_{0})}(f^{n_{i}-n_{0}}(z))\cap M\neq\emptyset$.
This implies that $C_{i}$ intersects $U$ and $X\backslash U$, so it intersects
the boundary $\partial U$ of $U$ for each $0\leq i\leq k$. Therefore $\partial
U$ contains at least $k+1$ points, a contradiction.

Case 2. $M$ is finite. In this case, $M=\textrm{Orb}_{f}(a)$ is a periodic
orbit.

Let $N$ be the period of $a$. By Lemma 2.3, $\omega_{f}(x)=\underset{0\leq
i\leq N}{\cup}\omega_{f^{N}}(f^{i}(x))$, then $a\in\omega_{f^{N}}(f^{i}(x))$
for some $0\leq i\leq N$. Set $g=f^{N}$ and $y=f^{i}(x)$. Then $g$ is
monotone, $g(a)=a$ and $a\in\omega_{g}(y)$. Moreover
$\\{a\\}\subsetneq\omega_{g}(y)$; indeed, otherwise, $\\{a\\}=\omega_{g}(y)$
and so $\omega_{f}(y)=\omega_{f}(x)$ is finite, a contradiction. Therefore
without loss of generality, we may assume that $M=\\{a\\}$ is a single fixed
point of $f$ with $\\{a\\}\subsetneq\omega_{f}(x)$.

Claim 1.  For each open neighborhood $V$ of $a$ in $X$ such that
$\omega_{f}(x)\setminus\overline{V}\neq\emptyset$, there is a point
$y_{V}\in\omega_{f}(x)\cap\overline{V}$ such that $y_{V}\neq\\{a\\}$ and
$y_{V}\in f^{j}(\overline{V}\cap\omega_{f}(x))$, for all $j\in\mathbb{N}$.

###### Proof.

Since $a\in\omega_{f}(x)$ and
$\omega_{f}(x)\cap(X\setminus\overline{V})\neq\emptyset$, there are an
increasing sequence of positive integers $(n_{i})_{i\geq 0}$ and a sequence of
integers $(t_{i})_{i\geq 0}$ with the following properties:
(i) for each $i\in\mathbb{Z}_{+}$, $t_{i}\geq i$ and $f^{n_{i}+j}(x)\in V$ for
$0\leq j\leq t_{i}$
(ii) $f^{n_{i}+t_{i}+1}(x)\notin V$.
Since $X$ is a compact space, we may assume that the sequence
$(f^{n_{i}+t_{i}}(x))_{i\geq 0}$ converges to a point $y_{V}$. Clearly,
$y_{V}\in\omega_{f}(x)\cap\overline{V}$. Now let $j\in\mathbb{N}$. For every
$i\geq j$, we have $0\leq n_{i}+t_{i}-j\leq n_{i}+t_{i}$ and so
$f^{n_{i}+t_{i}}(x)=f^{j}(f^{n_{i}+t_{i}-j}(x))$. Let $y_{V,j}$ be some limit
point of $(f^{n_{i}+t_{i}-j}(x))_{i\geq 0}$. Clearly
$y_{V,j}\in\overline{V}\cap\omega_{f}(x)$ and $f^{j}(y_{V,j})=y_{V}$.
Therefore for all $j\in\mathbb{N}$, we have $y_{V}\in
f^{j}(\overline{V}\cap\omega_{f}(x))$.
On the other hand, if $y_{V}=a$, then $f(y_{V})=a$. As
$(f^{n_{i}+t_{i}}(x))_{i\geq 0}$ converges to $y_{V}$, then
$(f^{n_{i}+t_{i}+1}(x))_{i\geq 0}$ converges to $f(y_{V})=a$. Thus for $i$
large enough, we have $f^{n_{i}+t_{i}+1}(x)\in V$, which is a contradiction.
This ends the proof of the claim. ∎

Since $X$ is a regular curve, there is an open neighborhood $U$ of $a$ in $X$
with finite boundary of cardinality $k\in\mathbb{N}$ and such that
$\omega_{f}(x)\setminus\overline{U}\neq\emptyset$. Then there exist $k+1$
positive integers $n_{0}<\dots<n_{k}$ such that $f^{n_{i}}(x)\in
X\setminus\overline{U}$. Set $r=n_{k}-n_{0}$. We distinguish two sub-cases.

Case 2.1. For some open neighborhood $V$ of $a$ in $X$ such that
$\overline{V}\subset U$, there exists a point $y:=y_{V}$ defined as in Claim 1
which is not periodic i.e. $y\notin P(f)$. Then the $f^{-i}(\\{y\\})$ ($0\leq
i\leq r$) are pairwise disjoint non-empty connected compact subsets. Let
$(W_{i})_{0\leq i\leq r}$ be a disjoint family of arcwise connected open
neighborhoods of $f^{-i}(\\{y\\})$, $0\leq i\leq r$. Since
$y\in\omega_{f}(x)$, there exists $b_{r}\in\omega_{f}(x)\cap f^{-r}(y)$. Then
for any $0\leq i\leq r,\;f^{i}(b_{r})\in
f^{-(r-i)}(\\{y\\})\cap\omega_{f}(x)\subset W_{r-i}\cap\omega_{f}(x)$. Hence
there exists an integer $m>n_{k}$ such that $f^{m+i}(x)\in W_{r-i}$. Indeed,
there exists $m_{j}\to+\infty$ so that
$b_{r}=\underset{j\to+\infty}{\lim}f^{m_{j}}(x)$. Then
$f^{i}(b_{r})=\underset{j\to+\infty}{\lim}f^{i+m_{j}}(x)$, for any $0\leq
i\leq r$. It follows that there exists $j_{i}>n_{k}$ such that for all $j\geq
j_{i}$, $f^{i+m_{j}}(x)\in W_{r-i}$. It suffices to take $m=m_{N}$, where
$N=\underset{0\leq i\leq r}{\max}j_{i}$, to get that $f^{i+m}(x)\in W_{r-i}$.
For any $0\leq i\leq r$, denote by $C_{i}=[f^{i}(b_{r}),f^{m+i}(x)]\cup
f^{-(r-i)}(\\{y\\})$, where $[f^{i}(b_{r}),f^{m+i}(x)]$ is an arc in $W_{r-i}$
joining $f^{i}(b_{r})$ and $f^{m+i}(x)$. One has $C_{i}\subset W_{r-i}$. Since
$f$ is monotone, $C_{i}$ is connected, for any $0\leq i\leq r$. Moreover the
$C_{i}$ ($0\leq i\leq r$) are pairwise disjoint connected subsets, and among
them $C_{n_{i}-n_{0}}$ ($0\leq i\leq k$). Set
$K_{i}=f^{-(m-n_{0})}(C_{n_{i}-n_{0}})$. Again by monotonicity of $f$, the
$K_{i}$ ($0\leq i\leq k$) are pairwise disjoint connected sets. Observe that
$K_{i}$ contains $f^{n_{i}}(x)$. On the other hand, $K_{i}$ contains
$f^{-(m-n_{0})}(f^{-(r-(n_{i}-n_{0}))}(y))=f^{-(m-n_{i}+r)}(y)$. As $y\in
f^{j}(\overline{V})$, for all $j\in\mathbb{N}$, so there is
$z_{i}\in\overline{V}$ such that $y=f^{m-n_{i}+r}(z_{i})$, since $m>n_{k}$.
Thus $f^{-(m-n_{0})}(f^{-(r-(n_{i}-n_{0}))})(y)$ contains $z_{i}$. Therefore
$K_{i}\cap\overline{V}\neq\emptyset$. In result, $K_{i}$ intersects
$X\setminus\overline{U}$ and $U$, so it intersects the boundary $\partial U$,
for $0\leq i\leq k$. This implies that $\partial U$ contains at least $k+1$
points. A contradiction.

Case 2.2. For each open neighborhood $V$ of $a$ in $X$ such that
$\overline{V}\subset U$, each point $y_{V}$ defined as in Claim 1 is periodic.

For each $n\in\mathbb{N}$, let $V_{n}=B(a,\frac{1}{n})$ be the open ball
centered at $a$ of radius $\dfrac{1}{n}$. One can assume that
$\overline{V_{n}}\subset U$ and
$\omega_{f}(x)\setminus\overline{V_{n}}\neq\emptyset$. By Claim 1, there
exists an infinite sequence $(y_{n})_{n\in\mathbb{N}}$ such that for any
$n\in\mathbb{N}$, $y_{n}\in\omega_{f}(x)\cap\overline{V_{n}}$. Moreover,
$y_{n}\neq a$ and $y_{n}\in f^{j}(\overline{V_{n}}\cap\omega_{f}(x))$, for all
$n,j\in\mathbb{N}$. Clearly the sequence $(y_{n})_{n\in\mathbb{N}}$ converges
to $a$ and moreover for any $n\geq 0$, $y_{n}\in P(f)$.

For each $n\in\mathbb{N}$, denote by $p_{n}$ the period of $y_{n}$.

Claim 2: The sequence $(p_{n})_{n\geq 0}$ is bounded.

###### Proof.

On the contrary, there exists an increasing subsequence $(p_{n_{i}})_{i\geq
0}$ of $(p_{n})_{n\in\mathbb{N}}$. Since $f^{j}(y_{n})$ converges to $a$, so
for every $0\leq j\leq r$, there is $N>0$ such that $p_{N}>r$ and
$\\{y_{N},f(y_{N}),\dots,f^{r}(y_{N})\\}\subset U$. Moreover
$(f^{i}(y_{N}))_{0\leq i\leq r}\subset\omega_{f}(x)$ are pairwise distinct.
Choose disjoint open arcwise connected neighborhoods $(W_{i})_{0\leq i\leq r}$
of $f^{i}(y_{N})$, $0\leq i\leq r$. Since $y_{N}\in\omega_{f}(x)$, there exist
an integer $m>n_{k}$ such that $f^{m+i}(x)\in W_{i}$. Denote by
$I_{i}=[f^{i}(y_{N}),f^{m+i}(x)]\subset W_{i}$ an arc joining $f^{i}(y_{N})$
and $f^{m+i}(x)$, $0\leq i\leq r$. We see that the $I_{i}$ ($0\leq i\leq r$)
are pairwise disjoint connected subsets, and among them $I_{n_{i}-n_{0}}:0\leq
i\leq k$. Set $K_{i}=f^{-(m-n_{0})}(I_{n_{i}-n_{0}})$. By monotonicity of $f$,
the $K_{i}$ ($0\leq i\leq r$) are pairwise disjoint connected sets. Observe
that $K_{i}$ contains $f^{n_{i}}(x)$. On the other hand, $K_{i}$ contains
$f^{-(m-n_{0})}(f^{n_{i}-n_{0}}(y_{N}))$. As $m-n_{0}>n_{i}-n_{0}$, we have

$\displaystyle f^{-(m-n_{0})}(f^{n_{i}-n_{0}}(y_{N}))$
$\displaystyle=f^{-((m-n_{0})-(n_{i}-n_{0}))}(f^{-(n_{i}-n_{0})}(f^{n_{i}-n_{0}}(y_{N})))$
$\displaystyle\supseteq
f^{-((m-n_{0})-(n_{i}-n_{0}))}(y_{N})=f^{-(m-n_{i})}(y_{N}).$

Recall that $f^{-j}(y_{N})\cap\overline{V_{N}}\neq\emptyset$, for all
$j\in\mathbb{N}$, hence
$f^{-(m-n_{i})}(y_{N}))\cap\overline{V_{N}}\neq\emptyset$, for all $0\leq
i\leq k$. Therefore $K_{i}\cap\overline{V_{N}}\neq\emptyset$. In result, for
$0\leq i\leq k$, $I_{i}$ intersects $X\setminus\overline{U}$ and $U$, so it
intersects the boundary $\partial U$. This implies that $\partial U$ contains
at least $k+1$ points. A contradiction. ∎

Let $k\in\mathbb{N}$ such that $p_{n}\leq k$ for all $n$. Set
$Y=\\{y_{n}:n\in\mathbb{N}\\}$. Then $Y$ is infinite with
$Y\subset\textrm{Fix}(f^{k!})$. By considering $g=f^{k!}$ instead of $f$, we
can assume that $Y\subset\textrm{Fix}(f)\cap\omega_{f}(x)$. Since
$\omega_{f}(x)$ is infinite, so by Theorem 2.4, $\omega_{f}(x)\nsubseteq
P(f)$. Hence there is a point $z\in\omega_{f}(x)\setminus P(f)$ and thus
$(f^{-k}(z))_{k\geq 0}$ is pairwise disjoint. Fix a negative orbit
$(z_{n})_{n}$ of $z$: $z_{0}=z$ and for any $n\geq 0$ $f(z_{n+1})=z_{n}$ with
$z_{n}\in\omega_{f}(x)$ and denote by $Z$ the set of limit points of the
sequence $(z_{n})_{n\geq 0}$. Clearly $Z\subset\omega_{f}(x)$.

Claim 3: $Y\nsubseteq Z$.

###### Proof.

Assume that $Y\subset Z$. Since $y_{1}\neq a$, there are disjoint open
neighborhoods $O_{a}$ (resp. $O_{1}$) of $a$ (resp. $y_{1}$) of $X$ such that
the boundary of $O_{a}$ is finite; set $s=\textrm{card}(\partial(O_{a}))$. As
$(y_{n})_{n\geq 0}$ converges to $a$, there exists some $N>1$ such that
$\\{y_{N},\dots y_{N+1},\dots y_{N+s}\\}\subset O_{a}$. As $X$ is locally
arcwise connected and $Y\subset Z$, we can find $m_{0}<m_{1}<\dots<m_{s}$ and
a family $(I_{j})_{0\leq j\leq s}$ of pairwise disjoint arcs in $O_{a}$
joining $y_{N+j}$ and $z_{m_{j}}$. As $y_{1}\in Z\cap\textrm{Fix}(f)$, then
there is $p>0$ such that for any $0\leq j\leq s$, $z_{m_{j}+p}\in O_{1}$. By
monotonicity of $f$, the $f^{-p}(I_{j})$ ($0\leq j\leq s$) are pairwise
disjoint continua, each of them contains $y_{N+j}\in O_{a}$ and
$z_{m_{j}+p}\in O_{1}$. This implies that the boundary $\partial O_{a}$
contains at least $s+1$ points. A contradiction. ∎

End of the proof of Theorem 3.1. Let $y:=y_{N}\in Y\setminus Z$, for some
$N>0$. We can find an open neighborhood $O_{y}$ of $y$ in $X$ with finite
boundary such that $\overline{\\{z_{n}:n\geq 0\\}}\cap O_{y}=\emptyset$. Set
$s=\textrm{card}(\partial O_{y})$. As $y\in\omega_{f}(x)\cap\textrm{Fix}(f)$
we can find $m>0$ such that $\\{f^{m}(x),f^{m+1}(x),\dots,f^{m+s}(x)\\}\subset
O_{y}$. The negative orbit $(z_{n})_{n\geq 0}$ of $z$ is infinite, so
$(z_{s},z_{s-1},\dots,z)$ is family of a pairwise distinct points of
$\omega_{f}(x)$. We can find an integer $p_{1}>m$ and a family of pairwise
disjoint arcs $(I_{j})_{0\leq j\leq s}$ each of which joins $z_{s-j}$ to
$f^{p_{1}+j}(x)$. It follows that the $(f^{-(p_{1}-m)}(I_{j}))_{0\leq j\leq
s}$ is family of pairwise disjoint continua each of which meets $O_{y}$ in at
least $f^{m+j}(x)$ and meets $X\setminus O_{y}$ in at least $z_{s-j+p_{1}-m}$.
Hence they meet the boundary $\partial O_{a}$ in at least $s+1$ points. A
contradiction. In result, $\omega_{f}(x)$ is itself minimal. This ends the
proof of Theorem 3.1. ∎

######  Corollary 3.2 ([1], [18]).

If $f$ is a monotone map on a local dendrite (resp. $f$ is a homeomorphism on
a regular curve), then each $\omega$-limit set of $f$ is a minimal set.

######  Corollary 3.3.

Let $f$ be a monotone map on a regular curve $X$. Then
$\Lambda(f)=R(f)=\textrm{AP}(f)$.

##  4\. The basin of attraction of an infinite minimal set

Let $f$ be a monotone map on a regular curve $X$. If $M$ is an infinite
minimal set of $f$, we call $A(M)=\\{x\in X:\ \omega_{f}(x)=M\\}$ the basin of
attraction of $M$.

######  Theorem 4.1.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$. If $M$ is
an infinite minimal set of $f$, then $A(M)$ is $f$-invariant and closed in
$X$.

###### Proof.

From the equality $\omega_{f}(f(x))=\omega_{f}(x)$ for any $x\in X$, it
follows that $A(M)$ is $f$-invariant. Let $(x_{n})_{n\geq 0}$ be a sequence of
$A(M)$ that converges to some $x\in X$. Assume that $x\notin A(M)$ that is
$\omega_{f}(x)\neq M$. Then by Theorem 3.1, $\omega_{f}(x)\cap M=\emptyset$.
Set $L=\overline{O_{f}(x)}$. Then there exists an open neighborhood $U_{L}$ of
$L$ with finite boundary such that $M\cap U_{L}=\emptyset$. We let
$s=\textrm{card}(\partial U_{L})$. As $f$ is continuous, we can find $N\geq 0$
such that $\\{x_{N},f(x_{N}),\dots,f^{s}(x_{N})\\}\subset U_{L}$. Let $y\in
M$, as $M$ is an infinite minimal set, then $\\{y,f(y),\dots f^{s}(y)\\}$ are
pairwise distinct, so we may pick $s+1$ pairwise disjoint arcwise connected
neighborhoods $(V_{f^{i}(y)})_{0\leq i\leq s}$ of $(f^{i}(y))_{0\leq i\leq
s}$. Recall that $\omega_{f}(x_{N})=M$, then we can find $k>s$ and a family of
pairwise disjoint arcs $(I_{j})_{0\leq j\leq s}$ in $V_{f^{i}(y)}$, each of
which with endpoints $f^{k+j}(x_{N})$ and $f^{j}(y)$. By monotonicity of $f$,
the $(f^{-k}(I_{j}))_{0\leq j\leq s}$ is a family of pairwise disjoint
continua each of which meets $U_{L}$ in at least $f^{j}(x_{N})$ and meets
$X\setminus U_{L}$ in at least a point of $M$ in $f^{-k}(f^{j}(y))$. Hence
these continua meet the boundary of $U_{L}$ in at least $s+1$ points, which is
a contradiction.
This finish the proof of Theorem 4.1. ∎

######  Remark 4.2.

(1) In addition to that $A(M)$ is $f$-invariant, we also have that
$f^{-1}(A(M))\subset A(M)$.
(2) Theorem 4.1 is not true in general when $M$ is finite (see [18]).

######  Corollary 4.3.

Let $X$ be a regular curve, $f$ a monotone self mapping of $X$ and let $M$ be
an infinite minimal set of $f$. If $M$ is connected then so is $A(M)$.

###### Proof.

Suppose that $A(M)$ is not connected then there exist at least two connected
components $C_{1}$ and $C_{2}$ of $A(M)$. Let $x\in C_{1}$, $y\in C_{2}$.

Claim. There exists an open set $U$ of $X$ with finite boundary such that
$A(M)\subset U$ and for any sub-continuum $C$ of $X$ that contains $x,y$ we
have $C\nsubseteq U$.

Indeed, by Theorem 4.1, $A(M)$ is a compact subset of $X$. So we can find a
decreasing sequence of open sets $(O_{n})_{n\geq 0}$ with finite boundaries
such that $A(M)=\displaystyle\bigcap_{n\geq 0}O_{n}$. If for infinitely many
$n\geq 0$, there exists a sub-continuum $K_{n}$ of $X$ containing $x,y$ such
that $K_{n}\subset O_{n}$, then there is a sub-continuum $K\subset A(M)$
containing $x$ and $y$ (it suffices to select a suitable converging
subsequence of $(K_{n})_{n\geq 0}$), a contradiction with the fact that either
$K\subset C_{1}$ or $K\subset C_{2}$. Hence for some $N$ and for any sub-
continuum $K$ of $X$ containing $x,y$, we have $K\nsubseteq O_{n}$ for any
$n\geq N$. Choose then $U=O_{N}$. As $M$ is a sub-continuum of $X$, we can
find a decreasing sequence of arcwise connected open sets $(U_{n})_{n\geq 0}$
such that $M=\displaystyle\bigcap_{n\geq 0}U_{n}$. Since
$\omega_{f}(x)=\omega_{f}(y)=M\subset U_{n}$ for any $n\geq 0$, there exists
$k_{n}\geq 0$ such that for any $k\geq k_{n}$ we have
$\\{f^{k}(x),f^{k}(y)\\}\subset U_{n}$. Then we can find an increasing
sequence of integers $m_{n}$ such that for any $n\geq 0$ we have
$\\{f^{m_{n}}(x),f^{m_{n}}(y)\\}\subset U_{n}$. Let $I_{n}$ be an arc in
$U_{n}$ joining $f^{m_{n}}(x)$ and $f^{m_{n}}(y)$. Then $f^{-m_{n}}(I_{n})$ is
a sub-continuum of $X$ that contains $x$ and $y$. By the claim above,
$f^{-m_{n}}(I_{n})\nsubseteq U$. As $x,y\in U$, $f^{-m_{n}}(I_{n})$ meets the
finite boundary $\partial U$. Therefore there exists $z\in\partial U$ such
that $z\in f^{-m_{n}}(I_{n})$ for infinitely many $n$. Hence for infinitely
many $n$, we have $f^{m_{n}}(z)\in I_{n}\subset U_{n}$ and so
$\omega_{f}(z)=M$. It follows that $z\in A(M)\subset U$, this contradicts the
fact that $z\in\partial U$. ∎

In [19], Naghmouchi proved that homeomorphisms on regular curves without
periodic points, have a unique minimal set. In [9], Daghar showed that this
later is either a circle or homeomorphic to a Cantor set. In this setting,
does this last result remains true for monotone maps on regular curves without
periodic points?

A partial answer to the above question is given by the following.

######  Corollary 4.4.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$, If
$P(f)=\emptyset$, then there exist one or uncountably many minimal sets.
Moreover at most a finite number of them are connected.

###### Proof.

Let $(M_{i})_{i\in I}$ be a family of minimal sets of $(X,f)$. As
$P(f)=\emptyset$, then for any $i\in I,\;M_{i}$ is an infinite minimal set and
by Theorem 4.1, $A(M_{i})$ is a closed set of $X$. As the $(A(M_{i}))_{i\in
I}$ are disjoint and $X=\displaystyle\bigcup_{i\in I}A(M_{i})$, so by ([12],
Theorem 6, p. 173), $I$ is either uncountable or reduced to a point. For the
rest, assume that there is infinitely many connected minimal sets
$(M_{n})_{n>0}$. Then pick a sequence $(M_{n})_{n>0}$ of them converging (in
the Hausdorff metric) to a continuum $C$ of $X$. Then by Proposition 2.2, we
have $\displaystyle\lim_{n\to+\infty}\textrm{diam}(M_{n})=0$. Hence $C$ is
reduced to a fixed point, which is a contradiction. ∎

##  5\. The $\alpha$-limit sets for monotone maps on regular curves

The aim of this section is to investigate the $\alpha$-limit sets. First
recall the definition of the $\alpha$-limit set. Let $X$ be a compact metric
space, $f:X\longrightarrow X$ a continuous map and $x\in X$. The set
$\alpha_{f}(x)=\displaystyle\bigcap_{k\geq
0}\displaystyle\overline{\bigcup_{n\geq k}f^{-n}(x)}$ is called the
$\alpha$-limit set of $x$. Equivalently a point $y\in\alpha_{f}(x)$ if and
only if there exist an increasing sequence of positive integers
$(n_{k})_{k\in\mathbb{N}}$ and a sequence of points $(x_{k})_{k\geq 0}$ such
that $f^{n_{k}}(x_{k})=x$ and
$\displaystyle\lim_{k\rightarrow+\infty}x_{k}=y.$ It is much harder to deal
with $\alpha$-limit sets since there are many choices for points in a backward
orbit. When $f$ is a homeomorphism, $\alpha_{f}(x)=\omega_{f^{-1}}(x)$.

Balibrea et al. [5] considered exactly one branch of the backward orbit as
follows.

######  Definition 5.1.

Let $x\in X$. A negative orbit of $x$ is a sequence $(x_{n})_{n}$ of points in
$X$ such that $x_{0}=x$ and $f(x_{n+1})=x_{n}$, for every $n\geq 0$. The
$\alpha$-limit set of $(x_{n})_{n\geq 0}$ denoted by $\alpha_{f}((x_{n})_{n})$
is the set of all limit points of $(x_{n})_{n\geq 0}$.

It is clear that $\alpha_{f}((x_{n})_{n})\subset\alpha_{f}(x)$. The inclusion
can be strict;

Balibrea et al. [5] gave an example of a pair ($X,\sigma$), where $\sigma$ is
the shift map on $\\{0,1\\}^{\mathbb{N}}$, for any point $x$,
$\alpha_{f}(x)=X$ and there are two negative orbits $(x_{n})_{n}$ and
$(y_{n})_{n}$ of $x$ under $f$ such that $\alpha_{f}((x_{n})_{n})$ consists of
a single point and $\alpha_{f}((y_{n})_{n})=X$. In the present paper, we show
that the above inclusions can also be strict for onto monotone maps on regular
curves (see Example 5.10). Notice that we have the following equivalence:

(i) $\alpha_{f}(x)\neq\emptyset$, (ii) $x\in X_{\infty}$. In particular, if
$f$ is onto then $\alpha_{f}(x)\neq\emptyset$ for every $x\in X$.

###  5.1. The limsup of a particular sequence of compact sets

Let $f:X\longrightarrow X$ be a continuous map of a compact space $X$ and let
$(G_{n})_{n\geq 0}$ be a sequence of compact subsets of $X$ satisfying
$f(G_{n+1})\subset G_{n}$, for any $n\geq 0$. We denote by $\limsup
G_{n}=\displaystyle\bigcap_{k\geq 0}\displaystyle\overline{\bigcup_{n\geq
k}G_{n}}$. A point $y\in\limsup G_{n}$ if and only if there exists an
increasing sequence of positive integers $(n_{k})_{k\in\mathbb{N}}$ and a
sequence of points $(x_{k})_{k\geq 0}$ such that $x_{k}\in G_{n_{k}}$ and
$\displaystyle\lim_{k\rightarrow+\infty}x_{k}=y$.

In particular:

(1) If $x\in X_{\infty}$ and $G_{n}=f^{-n}(x),n\geq 0$, then
$f(G_{n+1})\subset G_{n}$, for every $n\geq 0$ and we have $\limsup
G_{n}=\alpha_{f}(x)$.

(2) If $(x_{n})_{n\geq 0}$ is a negative orbit of $x$ and $G_{n}=\\{x_{n}\\}$,
$n\geq 0$, then $f(G_{n+1})=G_{n}$ and we have $\limsup
G_{n}=\alpha((x_{n})_{n})$.

First, we establish the following lemma which will be useful for most of the
proofs in this section.

######  Lemma 5.2.

Let $(G_{n})_{n\geq 0}$ be a null sequence of compact subsets of $X$
satisfying $f(G_{n+1})\subset G_{n}$, for every $n\geq 0$. Then:

  * (i)

If $A\in 2^{X}$ and for some sequence of positive integers $(n_{i})_{i\geq 0}$
we have $\displaystyle\lim_{i\to+\infty}d_{H}(A,G_{n_{i}})=0$, then $A$ is
degenerate.

  * (ii)

$x\in\limsup G_{n}$ if and only if there exists a sequence of positive
integers $(n_{i})_{i\geq 0}$ such that
$\displaystyle\lim_{i\to+\infty}d_{H}(x,G_{n_{i}})=0$.

  * (iii)

If $(G_{n_{i}})_{i\geq 0}$ converges to $\\{x\\}$, then for any sequence of
non-empty compact subsets $(A_{n_{i}})_{i\geq 0}$ of $X$ such that
$A_{n_{i}}\subset G_{n_{i}}$ for any $i\geq 0$, the sequence
$(A_{n_{i}})_{i\geq 0}$ also converges to $\\{x\\}$.

###### Proof.

(i) Let $a,b\in A$. As $\displaystyle\lim_{i\to+\infty}d_{H}(A,G_{n_{i}})=0$,
so there exist $a_{i},b_{i}\in G_{n_{i}}$ such that
$\displaystyle\lim_{i\to+\infty}d(a_{i},a)=0$ and
$\displaystyle\lim_{i\to+\infty}d(b_{i},b)=0$. Since $(G_{n})_{n\geq 0}$ is a
null family, $\displaystyle\lim_{i\to+\infty}\textrm{diam}(G_{n_{i}})=0$. For
any $i\geq 0$, we have $d(a_{i},b_{i})\leq\textrm{diam}(G_{n_{i}})$, thus
$d(a,b)=0$ and so $a=b$. In result $A$ is degenerate.
(ii) Let $x\in\limsup G_{n}$, then there exists a sequence of positive
integers $(n_{i})_{i\geq 0}$ such that $x_{i}$ converges to $x$ with $x_{i}\in
G_{n_{i}}$. We have

$d_{H}(x,G_{n_{i}})\leq d_{H}(x,x_{i})+d_{H}(x_{i},G_{n_{i}})\leq
d_{H}(x,x_{i})+\textrm{diam}(G_{n_{i}}).$

Since $\displaystyle\lim_{i\to+\infty}\textrm{diam}(G_{n_{i}})=0$,
$\displaystyle\lim_{i\to+\infty}d_{H}(x,G_{n_{i}})=0$. The converse is
trivial.
(iii) We have

$d_{H}(x,A_{n_{i}})\leq d_{H}(x,G_{n_{i}})+d_{H}(G_{n_{i}},A_{n_{i}})\leq
d_{H}(x,G_{n_{i}})+\textrm{diam}(G_{n_{i}}).$

As $\displaystyle\lim_{i\to+\infty}\textrm{diam}(G_{n_{i}})=0$ and
$\displaystyle\lim_{i\to+\infty}d_{H}(x,G_{n_{i}})=0$, thus
$\displaystyle\lim_{i\to+\infty}d_{H}(x,A_{n_{i}})=0$. ∎

######  Lemma 5.3.

Let $f$ be a continuous self mapping of a compact space $X$ and
$(G_{n})_{n\geq 0}$ be a null sequence of compact subsets satisfying
$f(G_{n+1})\subset G_{n}$, for every $n\geq 0$. Then $\limsup G_{n}$ is non-
empty, closed and strongly $f$-invariant.

###### Proof.

By compactness of $X$, it is obvious that $\limsup G_{n}$ is non-empty. Let
$x\in\limsup G_{n}$. Then by Lemma 5.2, there exists a sequence of positive
integers $(n_{i})_{i\geq 0}$ such that $G_{n_{i}}$ converges to $\\{x\\}$.
Consider now the sequence $(G_{n_{i}+1})_{i\geq 0}$. Since $(G_{n})_{n\geq 0}$
is a null family, we can assume that $(G_{n_{i}+1})_{i\geq 0}$ converges to
some point $\\{z\\}$. By Lemma 5.2, $z\in\limsup G_{n}$. By continuity of
$2^{f}$, we have $(f(G_{n_{i}+1}))_{i\geq 0}$ converges to $\\{f(z)\\}$. As
$f(G_{n_{i}+1})\subset G_{n_{i}}$ then by Lemma 5.2, $f(z)=x$ and thus
$\limsup G_{n}\subset f(\limsup G_{n})$. The other inclusion can be proved by
the same argument since $f(G_{n_{i}})\subset G_{n_{i}-1}$. ∎

######  Corollary 5.4.

Let $f:X\longrightarrow X$ be a continuous self mapping of a compact space
$X$. Let $x\in X_{\infty}$. Then:

  * (i)

$\alpha_{f}(x)$ is non-empty, closed and $f$-invariant. In addition, it is
strongly $f$-invariant whenever
$\displaystyle\lim_{n\to+\infty}\textrm{diam}(f^{-n}(x))=0$.

  * (ii)

$\alpha_{f}((x_{n})_{n\geq 0})$ is non-empty, closed and strongly
$f$-invariant, for any negative orbit $(x_{n})_{n\geq 0}$ of $x$.

###  5.2. Case of monotone maps on regular curves

The aim of this subsection is to prove the following theorem.

######  Theorem 5.5.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$. Let
$(G_{n})_{n\geq 0}$ be a null family of compact sets of $X$ such that
$f(G_{n+1})\subset G_{n}$, for every $n\geq 0$. Then $\limsup G_{n}$ is a
minimal set.

First we establish the following proposition.

######  Proposition 5.6.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$. If $x\in
X_{\infty}$ and $\omega_{f}(x)$ is infinite then
$\omega_{f}(x)=\alpha_{f}(x)$.

###### Proof.

Set $M=\omega_{f}(x)$. By Theorem 4.1 and Remark 4.2, $A(M)$ is a closed and
invariant under $f$ and $f^{-1}$. As $x\in A(M)$, then $\alpha_{f}(x)\subset
A(M)$. As $\alpha_{f}(x)$ is non-empty and $f$-invariant (Corollary 5.4) then
$M\subset\alpha_{f}(x)$. Now assume that $M\subsetneq\alpha_{f}(x)$, then
there exists some $a\in\alpha_{f}(x)\setminus M$. Let $O_{M}$ be an open
neighborhood of $M$ with finite boundary of cardinality $s$ such that
$a\notin\overline{O_{M}}$. Then there exist $n_{s}<n_{1}<\dots<n_{0}$ such
that $\\{x_{n_{j}}:\;0\leq j\leq s\\}\cap O_{M}=\emptyset$ with
$f^{n_{j}}(x_{n_{j}})=x$. Set $r=n_{0}-n_{s}$. Let $t\in M$. As $M$ is
infinite, we can find an integer $k\geq n_{0}$ such that $f^{k+j}(x)\in O_{M}$
for all $0\leq j\leq r$ and a family of pairwise disjoint arcs $(I_{j})_{0\leq
j\leq r}\subset O_{M}$, each of which joins $f^{k+j}(x)$ and $f^{j}(t)$.
Consider now the pairwise disjoint family $(f^{-(n_{0}+k)}(I_{j}))_{0\leq
j\leq r}$, each member of it meets $M$ and contains $x_{n_{0}-j}$ (since
$x_{n_{0}-j}\in f^{-(n_{0}+k)}(f^{k+j}(x))$). On the other hand, for
$j=n_{0}-n_{i},;\ 0\leq i\leq s$, $f^{-(n_{0}+k)}(I_{j})$ contains
$x_{n_{i}}$, hence it intersects $X\setminus O_{M}$. Therefore for
$j=n_{0}-n_{i},;\ 0\leq i\leq s,\;f^{-(n_{0}+k)}(I_{j})$ meets the boundary
$\partial O_{M}$. This contradicts card($\partial O_{M})=s$. ∎

###### Proof of Theorem 5.5.

We know by Lemma 5.3 that $\limsup G_{n}$ contains a minimal set. Assume that
$\limsup G_{n}$ is not a minimal set, thus for any minimal set $M$, we have
$M\neq\limsup G_{n}$.

Claim 1. Any minimal set $M\subsetneq\limsup G_{n}$ is finite.

Proof. On the contrary, there exists some infinite minimal set
$M\subsetneq\limsup G_{n}$.

$\bullet$ First, let us prove that for any $t\in\limsup G_{n}$, we have
$\omega_{f}(t)=M$:

Assume that there is $t\in\limsup G_{n}$ such that $\omega_{f}(t)\neq M$. So
$\omega_{f}(t)$ is a minimal set by Theorem 3.1. Let $U$ be an open
neighborhood of $M$ in $X$ with finite boundary of cardinality $k$ such that
$\omega_{f}(t)\cap\overline{U}=\emptyset$ and let $z\in M$. Since the orbit
$O_{f}(z)$ is infinite and $X$ is locally connected, there exists a connected
closed neighborhood $K_{j}$ of $f^{j}(z)$ such that $(K_{j})$ $(0\leq j\leq
k)$ are pairwise disjoint. Thus we can find $m>k$ such that $G_{m-j}\subset
K_{j}$ for any $0\leq j\leq k$. Now as $\omega_{f}(t)$ is a minimal set in
$\limsup G_{n}\setminus U$, we can find $n>m$ such that $G_{n-j}\cap
U=\emptyset$, for any $0\leq j\leq k$. Therefore $f^{-(n-m)}(K_{j})$ ($0\leq
j\leq k$) is a family of pairwise disjoint continua each of which meets $U$ at
some point in $f^{-(n-m)}(f^{j}(z))\cap M$ and meets $X\setminus U$ at
$G_{n-j}$ since $G_{n-j}\subset f^{-(n-m)}(G_{m-j})$. Hence
$f^{-(n-m)}(K_{j})$ meets the boundary $\partial U$ for any $0\leq j\leq k$. A
contradiction. We conclude that for any $t\in\limsup G_{n}$, we have
$\omega_{f}(t)=M$.

$\bullet$ Second, let $t\in\limsup G_{n}\backslash M$. By above we have
$\omega_{f}(t)=M$. As $\limsup G_{n}$ is strongly invariant (Lemma 5.3), so
$t\in X_{\infty}$. Now by Proposition 5.6, we have $\alpha_{f}(t)=M$. Let
$V_{t}$ and $V_{M}$ be two disjoint open neighborhoods of $t$ and $M$
respectively with finite boundary. Since $M\subset V_{M}$ there exists some
$N>0$ such that $f^{-n}(t)\subset V_{M}$ for any $n\geq N$. As $X$ is locally
connected, let $V_{n}$ be an open connected neighborhood of $t$ such that
$V_{n}\subset B(t,\frac{1}{n})$. Clearly $(\overline{V_{n}})_{n\geq 0}$ is a
sequence of continua which converges to $\\{t\\}$ (with respect to the
Hausdorff metric). As $t\in\limsup G_{n}$, then for any $n\geq 0$, we can find
two increasing sequences of positive integers $(m_{n})_{n\geq 0}$ and
$(s_{n})_{n\geq 0}$ such that $G_{m_{n}}\subset V_{n}$ and
$G_{m_{n}+s_{n}}\subset V_{t}$. We can assume that $s_{n}>N$ for any $n\geq
0$. Then $f^{-s_{n}}(V_{n})$ meets $V_{M}$ at some point of $f^{-s_{n}}(t)$
and $V_{t}$ at $G_{m_{n}+s_{n}}$ since $G_{m_{n}+s_{n}}\subset
f^{-s_{n}}(G_{m_{n}})$. Hence $f^{-s_{n}}(V_{n})$ meets the finite boundary
$\partial V_{t}$, and then it meets one point $b\in\partial V_{t}$ for
infinitely many $n$. Thus $f^{s_{n}}(b)\in V_{n}$ for infinitely many $n$ and
then $t\in\omega_{f}(b)$. Since $\omega_{f}(b)$ is minimal, it follows that
$\omega_{f}(t)=\omega_{f}(b)=M$. So $t\in M$, a contradiction. This ends the
proof of Claim $1$.

Claim 2. Let $M$ be a finite minimal set in $\limsup G_{n}$. Then for each
open neighborhood $V_{M}$ of $M$ in $X$ such that $\limsup
G_{n}\setminus\overline{V_{M}}\neq\emptyset$, there is a point
$y_{V}\in\limsup G_{n}\cap(\overline{V_{M}}\setminus M)$ such that $y_{V}\in
f^{j}(\overline{V_{M}}\cap\limsup G_{n})$, for all $j\in\mathbb{N}$.

Proof. Let $V_{M}$ be open neighborhood of $M$ in $X$ such that $\limsup
G_{n}\setminus\overline{V_{M}}\neq\emptyset$. Let $m\in M$ and $t\in\limsup
G_{n}\setminus V_{M}$. There exist two increasing sequences of integers
$(n_{i})_{i\geq 0}$ and $(p_{i})_{i\geq 0}$ such that:

(1) $(G_{n_{i}})_{i\geq 0}$ converges $\\{m\\}$.

(2) $(G_{p_{i}})_{i\geq 0}$ converges to $\\{t\\}$.

(3) $n_{i}>p_{i}$ for any $i\geq 0$.

Denote by $A_{i,j}=\displaystyle\bigcup_{s=0}^{j}G_{n_{i}-s}$. Since $f(M)=M$
and $(G_{n})_{n\geq 0}$ is a null family, we can easily prove that for any
$j\geq 0$, $(A_{i,j})_{i\geq 0}$ converges to $L\subset M$ (with respect to
the Hausdorff metric). Thus for any $j\geq 0$, we can find $i_{j}\geq j$ such
that $A_{i_{j},j}\subset V_{M}$. Since $(G_{n})_{n\geq 0}$ is a null family,
so by (2), we can assume that $G_{p_{i_{j}}}\cap V_{M}=\emptyset$ and by (3),
$n_{i_{j}}>p_{i_{j}}$. Therefore there exists $t_{i_{j}}\geq i_{j}$ such that:
(i) For any $0\leq s\leq t_{i_{j}},\;G_{n_{i}-s}\cap V_{M}\neq\emptyset$;
(ii) $G_{n_{i_{j}}-t_{i_{j}}-1}\cap V_{M}=\emptyset$;
(iii) $n_{i_{j}}-t_{i_{j}}-1\geq p_{i_{j}}$.

By (iii), we have
$\displaystyle\lim_{j\to+\infty}n_{i_{j}}-t_{i_{j}}=+\infty$. So we can assume
that $G_{n_{i_{j}}-t_{i_{j}}}$ converges (with respect to the Hausdorff
metric). As $(G_{n})_{n\geq 0}$ is a null family, so by Lemma 5.2,
$\displaystyle\lim_{j\to+\infty}G_{n_{i_{j}}-t_{i_{j}}}=\displaystyle\lim_{j\to+\infty}\overline{G_{n_{i_{j}}-t_{i_{j}}}\cap
V_{M}}=\\{y_{V}\\}$. Hence $y_{V}\in\limsup G_{n}\cap\overline{V_{M}}$. Let
$k\geq 0$ and let $\\{y_{V,k}\\}$ be some limit point of
$(G_{n_{i_{j}}-t_{i_{j}}+k})_{j\geq 0}$. Let us prove that
$f^{k}(y_{V,k})=y_{V}$ and $y_{V,k}\in\limsup G_{n}\cap\overline{V_{M}}$.
For $j$ large enough and from (i), we have $G_{n_{i_{j}}-t_{i_{j}}+k}\cap
V_{M}\neq\emptyset$. Again by Lemma 5.2 and the fact that $(G_{n})_{n\geq 0}$
is a null family, we have $y_{V,k}\in\limsup G_{n}\cap\overline{V_{M}}$.
Furthermore, $f^{k}(G_{n_{i_{j}}-t_{i_{j}}+k})\subset
G_{n_{i_{j}}-t_{i_{j}}}$. Since $2^{f}$ is continuous,
$\\{f^{k}(y_{V,k})\\}=\displaystyle\lim_{j\to+\infty}f^{k}(G_{n_{i_{j}}-t_{i_{j}}+k})\subset
G_{n_{i_{j}}-t_{i_{j}}}$. As $G_{n_{i_{j}}-t_{i_{j}}}$ converges to
$\\{y_{V}\\}$, we have $f^{k}(y_{V,k})=y_{V}$. In addition, $y_{V}\in
f^{k}\Big{(}\overline{V_{M}}\cap\limsup G_{n}\Big{)}$, for all
$k\in\mathbb{N}$.
It remains to show that $y_{V}\notin M$. Otherwise, $y_{V}\in M$ and then
$f(y_{V})\in M$. Thus for $j$ large enough, $G_{n_{i_{j}}-t_{i_{j}}-1}\cap
V_{M}\neq\emptyset$, which contradicts (ii). This ends the proof of Claim 2.

Claim 3. There is an open neighborhood $V$ of $M$ such that
$\limsup G_{n}\setminus\overline{V}\neq\emptyset$ and $y_{V}\notin P(f)$.

Proof. Otherwise, for any open neighborhood $V$ of $M$ such that $\limsup
G_{n}\setminus\overline{V}\neq\emptyset$, we have $y_{V}\in P(f)$. Let
$(V_{n})_{n\geq 0}$ be a decreasing sequence of neighborhoods of $M$ such that
$\limsup G_{n}\setminus\overline{V_{n}}\neq\emptyset$. So let
$y_{n}=y_{V_{n}}$ given by Claim 2. Since for any $n\geq 0,\;y_{n}\notin M$,
we can assume that the sequence $(y_{n})_{n\geq 0}$ is pairwise distinct. Let
$W_{M}$ be some open neighborhood of $M$ with finite boundary of cardinality
$k$ such that $O_{f}(y_{0})\cap\overline{W_{M}}=\emptyset$. In $W_{M}$ we can
find infinitely many $(\overline{V_{n}})_{n\geq 0}$, so let $(O_{j})_{0\leq
j\leq k}$ be a family of pairwise disjoint open connected neighborhoods of
$y_{p_{j}}$. Recall that for any $0\leq j\leq k$, we have $y_{p_{j}}\in\limsup
G_{n}$. Hence we can find $k+1$ positive integer $m_{k}<m_{k-1}<\dots<m_{0}$,
such that $G_{m_{j}}\subset O_{j}$, for any $0\leq j\leq k$. Let
$r=m_{0}-m_{k}$. Now as $O_{f}(y_{0})\subset\limsup G_{n}$, we can find
$p>m_{0}$ such that $\displaystyle\bigcup_{s=0}^{r}G_{p-s}\cap
W_{M}=\emptyset$. Therefore, for $i:=m_{0}-m_{j},\;0\leq j\leq
k,\;f^{-(p-m_{0})}(O_{i})$ meets $\overline{V_{y_{n_{j}}}}\subset W_{M}$ at
some point of $f^{-(p-m_{0})}(y_{p_{j}})$ and $X\setminus W_{M}$ at least in
$G_{p-(m_{0}-m_{j})}$, thus it meets $\partial W_{M}$. A contradiction with
$k=\textrm{card}(\partial W_{M})$.

End of the proof of Theorem 5.5. Let $V$ and $y_{V}\notin P(f)$ be as in Claim
3. Since $\omega_{f}(y_{V})\subset\limsup G_{n}$ is a minimal set, so by Claim
1, it is finite. Since $y_{V}\notin P(f)$, $y_{V}\notin\omega_{f}(y_{V})$.
Also $(f^{-k}(y_{V}))_{k\geq 0}$ is pairwise disjoint, so by Proposition 2.2,
it is a null family. By Claim 2, $f^{-k}(y_{V})\cap\overline{V}\neq\emptyset$,
for any $k\geq 0$. Therefore $\alpha_{f}(y_{V})\subset\overline{V}$. Recall
that $\limsup G_{n}\nsubseteq\overline{V}$, so let $t\in\limsup
G_{n}\setminus\alpha_{f}(y_{V})$.
We can find a sequence of positive integers $(n_{i})_{i\geq 0}$ such that
$G_{n_{i}}$ converges to $y_{V}$. As $X$ is locally arcwise connected, we can
find a decreasing sequence $(O_{i})_{i\geq 0}$ of connected open neighborhoods
of $y_{V}$. Clearly $(\overline{O_{i}})_{i\geq 0}$ is a sequence of continua
that converges to $\\{y_{V}\\}$ and we can assume that $G_{n_{i}}\subset
O_{i}$ for any $i\geq 0$. Now as $t\notin\alpha_{f}(y_{V})$, we can find two
disjoint open neighborhoods with finite boundary $V_{t}$ and
$V_{\alpha_{f}(y_{V})}$. Clearly there exists some $N\geq 0$ such that
$f^{-k}(y_{V})\subset V_{\alpha_{f}(y_{V})}$ for any $k\geq N$. Recall that
$t\in\limsup G_{n}$. Then we can find infinitely many $n\geq 0$ such that
$G_{n}\subset V_{t}$ and so we can find an increasing sequence $(s_{i})_{i\geq
0}$ such that $G_{n_{i}+s_{i}}\subset V_{t}$ for any $i\geq 0$. Clearly we can
assume that for any $i\geq 0$ we have $s_{i}>N$. It follow that for any $i\geq
0$, $f^{-s_{i}}(O_{i})$ meets $V_{t}$ and $V_{\alpha_{f}(y_{V})}$, hence it
meets the finite boundary $\partial V_{t}$. Thus there exists $b\in\partial
V_{t}$ such that $f^{s_{i}}(b)\in O_{i}$ for infinitely many $i$. Therefore
$y_{V}\in\omega_{f}(b)$ and hence $\omega_{f}(b)=\omega_{f}(y_{V})$ by
minimality of $\omega_{f}(b)$. In result, $y_{V}\in\omega_{f}(y_{V})$. A
contradiction. ∎

As a consequence, we obtain the following theorem.

######  Theorem 5.7.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$. Then we
have:

  * (i)

For any $x\in X_{\infty}\setminus P(f),\;\alpha_{f}(x)$ is a minimal set.

  * (ii)

For any $x\in X_{\infty}$, if $(x_{n})_{n\geq 0}$ is a negative orbit of $x$,
then $\alpha((x_{n})_{n})$ is a minimal set.

###### Proof.

(i) If $x\in X_{\infty}\setminus P(f)$ and $G_{n}=f^{-n}(x)$, then $\limsup
G_{n}=\alpha_{f}(x)$. Moreover $(G_{n})_{n\geq 0}$ is a family of pairwise
disjoint sub-continua of $X$ and so by Proposition 2.2, it is a null family.
Furthermore we have $f(G_{n+1})\subset G_{n}$, for every $n\geq 0$. We deduce
from Theorem 5.5 that $\alpha_{f}(x)$ is a minimal set.

(ii) Set $G_{n}=\\{x_{n}\\}$, $n\geq 0$. In this case, $\limsup
G_{n}=\alpha((x_{n})_{n\geq 0})$. Clearly $(G_{n})_{n\geq 0}$ is a null family
of compact sets and $f(G_{n+1})=G_{n}$. Thus from Theorem 5.5,
$\alpha((x_{n})_{n\geq 0})$ is a minimal set. ∎

######  Corollary 5.8.

Let $X$ be a regular curve and $f$ a monotone self mapping of $X$ and $x\in
X$. If $x\notin P(f)$ and ($x_{n})_{n\geq 0}$ is a negative orbit of $x$, then
$\alpha_{f}(x)=\alpha_{f}((x_{n})_{n\geq 0})$.

###### Proof.

As $\alpha_{f}((x_{n})_{n\geq 0})\subset\alpha_{f}(x)$, and $\alpha_{f}(x)$ is
a minimal set (Theorem 5.7), so Corollary 5.8 follows. ∎

######  Remark 5.9.

(i) The Corollary 5.8 is false in general whenever $x\in P(f)$ (see Example
5.10).

(ii) Contrarily to homeomorphisms (see [18], Corollaries 2.5 and 2.6), it may
happen, for a monotone map $f$ on a regular curve, that $\omega_{f}(x)$ is
finite but $\alpha_{f}(x)$ infinite (see Example 5.10).

######  Example 5.10.

We shall construct a monotone map $f$ on an infinite star $X$ centered at a
point $z_{0}\in\mathbb{R}^{2}$ and beam $I_{n},\;n\geq 0$ with endpoints
$z_{0}$ and $z_{n}$ satisfying the following properties.

  * (i)

For any $x\in X\setminus\\{z_{n}:n\geq 0\\}$ we have
$\omega_{f}(x)=\\{z_{0}\\}$ and $\alpha_{f}(x)=\\{z_{n_{x}}\\}$, where
$n_{x}\geq 1$ such that $x\in I_{n_{x}}$.

  * (ii)

For any $n\geq 1$ we have $\alpha_{f}(z_{n})=\omega_{f}(z_{n})=\\{z_{n}\\}$.

  * (iii)

$\omega_{f}(z_{0})=\\{z_{0}\\},\;\alpha_{f}(z_{0})=X$.

Figure 1. The map $f$ on the infinite star $X$

###### Proof.

The space $X$ is a countable union $X=\displaystyle\bigcup_{n\geq 0}I_{n}$ of
arcs $I_{n}$ in $\mathbb{R}^{2}$ for each of which $z_{0}$ is one of its
endpoints, $z_{n}$ is the other endpoint of $I_{n}$ distinct from $z_{0}$,
with $n\geq 1$, such that for any $n\neq m$, $I_{n}\cap I_{m}=\\{z_{0}\\}$ and
$\underset{n\to+\infty}{\lim}\textrm{diam}(I_{n})=0$. The space $X$, known as
the $\omega$-star, is then a regular curve. The $I_{n},\;n\geq 1$ are the
branches of $X$ and $z_{0}$ is its central point. Denote by
$I_{0}=\\{z_{0}\\}$. Let the map
$g:[0,1]\to[0,1],\;x\longrightarrow\max\\{0,2x-1\\}$. We extend $g$ into a map
of $X$ such that the restriction of $f$ on each branch is $g$, where the
central point $\\{z_{0}\\}$ is playing the role of $0$.

It is plain that the map $f$ is continuous monotone and onto. Moreover, it
satisfies the properties (i),(ii) and (iii) above. Indeed :

(i) We have $\displaystyle\bigcup_{n\geq
0}f^{-n}(z_{0})=X\setminus\\{z_{n}:n\geq 1\\}$, therefore
$\omega_{f}(x)=\\{z_{0}\\}$. Notice that $x$ has only one negative orbit
$(x_{k})_{k\geq 0}$ such that for any $k\geq 0,\;x_{k}\in I_{n_{x}}$. Hence
$\alpha_{f}(x)=\\{z_{n_{x}}\\}$.

(ii) For any $n\geq 1$, we have
$\\{f(z_{n})\\}=\\{f^{-1}(z_{n})\\}=\\{z_{n}\\}$. So (ii) follows.
(iii) $\omega_{f}(z_{0})=\\{z_{0}\\}$. Indeed, $z_{0}$ is a fixed point and
$\alpha_{f}(z_{0})=\overline{\displaystyle\bigcup_{n\geq 0}f^{-n}(z_{0})}=X$.
Notice that any negative orbit $(x_{k})_{k\geq 0}$ of $z_{0}$ is either the
constant sequence $(x_{k}=z_{0})_{k\geq 0}$ or it coincides with the negative
orbit of some $x\in X\setminus\\{z_{n}:n\geq 0\\}$, for $k$ large enough. ∎

In [16], Mchaalia proved for a monotone regular curve self-mapping $f$ of $X$,
that every totally periodic $\omega$-limit set is finite, that is a periodic
orbit. The following corollary shows analogously that every totally periodic
$\alpha$-limit set is a periodic orbit.

######  Corollary 5.11.

Let $X$ be a regular curve, $f$ a monotone self mapping of $X$ and $x\in
X_{\infty}$. If $\alpha_{f}(x)\subset P(f)$, then $\alpha_{f}(x)$ is a
periodic orbit.

###### Proof.

Assume that $\alpha_{f}(x)\subset P(f)$. If $x\in X_{\infty}\setminus P(f)$,
then by Theorem 5.7, $\alpha_{f}(x)$ is minimal, and hence it is a periodic
orbit. If $x\in P(f)$, then
$\alpha_{f}(x)=\overline{\displaystyle\bigcup_{n\geq 0}f^{-n}(x)}$. Moreover,
we have $\displaystyle\bigcup_{n\geq 0}f^{-n}(x)=O_{f}(x)$; indeed, if
$y\in\displaystyle\bigcup_{n\geq 0}f^{-n}(x)$, there is $n\geq 0$ such that
$f^{n}(y)=x$ and moreover $y\in P(f)$. As $f^{n}(y)=f^{n}(f^{sp-n})(x)$, where
$p$ is the period of $x$ and $s\geq 1$ an integer such that $sp-n\geq 0$, then
$y=f^{sp-n}(x)\in O_{f}(x)$ (since the restriction $f_{|P(f)}$ is injective).
We conclude that $\alpha_{f}(x)=O_{f}(x)$ a periodic orbit. ∎

Conflict of interest

The authors declare that there is no conflict of interests regarding the
publication of this article.

Data availability statement

All data analysed in this study are included in this article.

Acknowledgements

This work was supported by the research unit: “Dynamical systems and their
applications”, (UR17ES21), Ministry of Higher Education and Scientific
Research, Faculty of Science of Bizerte, Bizerte, Tunisia.

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