N-Expansive Property for Flows

Le Huy Tien, Le Duc Nhien

Department of Mathematics, Mechanics and Informatics, Vietnam National University at Hanoi, Hanoi, Vietnam.

DOI: 10.4236/jamp.2019.72031                

Abstract

In this paper, we discuss the dynamics of n-expansive homeomorphisms with the shadowing property defined on compact metric spaces in continuous case. For every n∈N, we exhibit an n-expansive homeomorphism but not (n-1)-expansive. Furthermore, that flow has the shadowing property and admits an infinite number of chain-recurrent classes.

Keywords

Expansive, Flow, N-Expansive, Shadowing

1. Introduction and Preliminaries

The classical terms, expansive flows on a metric space are presented by Bowen and Walters [1] which generalized the similar notion by Anosov [2]. Besides, Walters [3] investigated continuous transformations of metric spaces with discrete centralizers and unstable centralizers and proved that expansive homeomorphisms have unstable centralizers; other result was studied in [4]. In discrete case, this concept originally introduced for bijective maps by Utz [5] has been generalized to positively expansiveness in which positive orbits are considered instead [6]. Further generalizations are the pointwise expansiveness (with the above radius depending on the point [7]), the entropy-expansiveness [8], the continuum-wise expansiveness [9], the measure-expansiveness and their corresponding positive counterparts. However, as far as we know, no one has considered the generalization in which at most n companion orbits are allowed for a certain prefixed positive integer n. For simplicity we call these systems n-expansive (or positively n-expansive if positive orbits are considered instead). A generalization of the expansiveness property that has been given attention recently is the n-expansive property (see [10] [11] [12] [13] [14]).

In this paper, we introduce a notion of n-expansivity for flows which is generalization of expansivity, and show that there is an n-expansive flow but not $\left(n-1\right)$ -expansive flow. Moreover, that flow is shadowable and has infinite number of chain-recurrent classes.

Let $\left(X\mathrm{,}d\right)$ be a metric space. A flow on X is a map $\varphi \mathrm{<mo>\; :\; </mo>}X\times ℝ\to X$ satisfying $\varphi \left(x,0\right)=x$ and $\varphi \left(\varphi \left(x,s\right),t\right)=\varphi \left(x,s+t\right)$ for $x\in X$ and $t\mathrm{,}s\in ℝ$. For convenience, we will denote

$\varphi \left(x,s\right)={\varphi }_s\left(x\right)\text{and}{\varphi }_{\left(a,b\right)}\left(x\right)=\left\{{\varphi }_t\left(x\right):t\in \left(a,b\right)\right\}.$

The set ${\varphi }_{ℝ}\left(x\right)$ is called the orbit of $\varphi$ through $x\in X$ and will be denoted by ${\text{Orb}}_{\varphi }\left(x\right)$. We have the following several basis concepts (see [1] [15] [16]).

Definition 1.1. Let $\varphi$ be a flow in a metric space $\left(X\mathrm{,}d\right)$. We say that $\varphi$ is n-expansive ( $n\in \mathbb{N}$ ) if there exists $c>0$ such that for every $x\in X$ the set

$\Gamma \left(x,c\right):=\left\{y\in X;d\left({\varphi }_t\left(x\right),{\varphi }_t\left(y\right)\right)\le c,\forall t\in ℝ\right\},$

contains at most n different points of X.

We say that $\varphi$ is finite expansive if there exists $c>0$ such that for every $x\in X$ the set $\Gamma \left(x\mathrm{,}c\right)$ is finite.

Definition 1.2. Let $x\in X$. We say that x is a period point if there exists $T>0$ such that ${\varphi }_{t+T}\left(x\right)={\varphi }_t\left(x\right),\forall t\in ℝ$. Denote that $\pi \left(x\right)$ is the period of x, which is the smallest non-negative number satisfying this equation.

Definition 1.3. Give $\delta \mathrm{,}T\ge 0$. We say that a sequence of pairs ${\left(x_i\mathrm{,}{t}_i\right)}_{i\in \mathbb{Z}}\subset X\times ℝ$ is a $\left(\delta \mathrm{,}T\right)$ -pseudo orbit of $\varphi$ if $t_i\ge T$ and $d\left({\varphi }_{t_i}\left(x_i\right)\mathrm{,}{x}_{i+1}\right)\le \delta \mathrm{,}\forall i\in \mathbb{Z}$.

We define

$s_i=(\begin{array}{ll}{\displaystyle \underset{j=0}{\overset{i-1}{\sum }}{t}_j},\hfill & i>0,\hfill \\ 0,\hfill & i=0,\hfill \\ -{\displaystyle \underset{j=i}{\overset{-1}{\sum }}{t}_j},\hfill & i<0,\hfill \end{array}$

and $x_0\star t={\varphi }_{t-s_i}\left(x_i\right)$ whenever $s_i\le t<s_{i+1}$.

Definition 1.4. We say that $\varphi$ is shadowing property if for each $ϵ>0$ there is $\delta >0$ such that for any $\left(\delta \mathrm{,1}\right)$ -pseudo orbit ${\left(x_i\mathrm{,}{t}_i\right)}_{i\in \mathbb{Z}}$, there exists $x\in X$ and an orientation preserving homeomorphism $h\mathrm{<mo>\; :\; </mo>}ℝ\to ℝ$ such that $h\left(0\right)=0$ and $d\left(x_0\star t\mathrm{,}{\varphi }_{h\left(t\right)}\left(x\right)\right)\le ϵ$.

Denote by Rep the set of orientation preserving homeomorphism $h\mathrm{<mo>\; :\; </mo>}ℝ\to ℝ$ such that $h\left(0\right)=0$.

Definition 1.5. Give two points p and q in X. We say p and q are $\left(\delta \mathrm{,}T\right)$ -related if there are two $\left(\delta \mathrm{,}T\right)$ -chains ${\left(x_i,t_i\right)}_{i=0}^m$ and ${\left(y_i,s_i\right)}_{i=0}^n$ such that $p=x_0=y_n$ and $q=y_0=x_m$. We say that p and q are related $\left(p\sim q\right)$ if they are $\left(\delta \mathrm{,}T\right)$ -related for every $\delta ,T>0$. The chain-recurrent class of a point $p\in X$ is the set of all points $q\in X$ such that $p\sim q$.

Theorem 1.1. For every $n\in \mathbb{N}$, there is an n-expansive flow, define in a compact metric space, that is not $\left(n-1\right)$ -expansive, has the shadowing property and admits an infinite number of chain-recurrent classes.

2. Proof of the Main Theorem

Consider a flow $\varphi$ defined in a compact metric space $\left(M\mathrm{,}{d}_0\right)$, and $\varphi$ has 1-expansive, and has the shadowing property. Further, suppose it has an infinite number of period points ${\left\{p_k\right\}}_{k\in \mathbb{N}}$, which we can suppose belong to different orbits. Let E be an infinite set, such that there exists a bijection $r\mathrm{<mo>\; :\; </mo>}ℝ\to E$. Let

$Q={\displaystyle \underset{k\in N}{\cup }}\left\{\mathrm{1,}\cdots \mathrm{,}n-1\right\}\times \left\{k\right\}\times \left[\mathrm{0,}\pi \left(p_k\right)\right)\mathrm{,}$

and note that there exists a bijection $s\mathrm{<mo>\; :\; </mo>}Q\to ℝ$. Consider the bijection $q\mathrm{<mo>\; :\; </mo>}Q\to E$ defined by

$q\left(i\mathrm{,}k\mathrm{,}j\right)=r\circ s\left(i\mathrm{,}k\mathrm{,}j\right)\mathrm{.}$

Let $X=M\cup E$. Thus, any point $x\in E$ has the form $x=q\left(i,k,j\right)$ for some $\left(i\mathrm{,}k\mathrm{,}j\right)\in Q$. Define a function $d\mathrm{<mo>\; :\; </mo>}X\times X\to {ℝ}^{+}$ by

$\begin{array}{l}d\left(x,y\right)\\ =(\begin{array}{ll}0,\hfill & x=y,\hfill \\ d_0\left(x,y\right),\hfill & x,y\in M,\hfill \\ \frac{1}{k}+d_0\left(y,{\varphi }_j\left(p_k\right)\right),\hfill & x=q\left(i,k,j\right),y\in M,\hfill \\ \frac{1}{k}+d_0\left(x,{\varphi }_j\left(p_k\right)\right),\hfill & x\in M,y=q\left(i,k,j\right),\hfill \\ \frac{1}{k},\hfill & x=q\left(i,k,j\right),y=q\left(l,k,j\right),i\ne l,\hfill \\ \frac{1}{k}+\frac{1}{m}+d_0\left({\varphi }_t\left(p_k\right),{\varphi }_r\left(p_m\right)\right),\hfill & x=q\left(i,k,j\right),y=q\left(l,m,r\right),k\ne m\text{or}j\ne r.\hfill \end{array}\end{array}$

Now we prove that function d is a metric in X. Indeed, we see that $d\left(x,y\right)=0$ iff $x=y$, and that $d\left(x,y\right)=d\left(y,x\right)$ for any pair $\left(x\mathrm{,}y\right)\in X\times X$. We shall prove that the triangle inequality $d\left(x\mathrm{,}z\right)\le d\left(x\mathrm{,}y\right)+d\left(y\mathrm{,}z\right)$ for any triple $\left(x\mathrm{,}y\mathrm{,}z\right)\in X\times X\times X$. When $\left(x\mathrm{,}y\mathrm{,}z\right)\in M\times M\times M$ we have that $d_{|M\times M}=d_0$, and $d_0$ is a metric in M. When $\left(x\mathrm{,}y\mathrm{,}z\right)\in M\times M\times E$ then $z=q\left(i,k,j\right)$ and

$d\left(x\mathrm{,}z\right)=\frac{1}{k}+d_0\left(x\mathrm{,}{\varphi }_j\left(p_k\right)\right)\le d_0\left(x\mathrm{,}y\right)+\frac{1}{k}+d_0\left(y\mathrm{,}{\varphi }_j\left(p_k\right)\right)=d\left(x\mathrm{,}y\right)+d\left(y\mathrm{,}z\right)\mathrm{.}$

Therefore, when $\left(x\mathrm{,}y\mathrm{,}z\right)\in E\times M\times M$, changing the role of x and z in the previous case, we obtain this result. When $\left(x\mathrm{,}y\mathrm{,}z\right)\in M\times E\times M$, we have $y=q\left(i,k,j\right)$ and

$d\left(x\mathrm{,}z\right)=d_0\left(x\mathrm{,}z\right)\le \frac{2}{k}+d_0\left(x\mathrm{,}{\varphi }_j\left(p_k\right)\right)+d_0\left(z\mathrm{,}{\varphi }_j\left(p_k\right)\right)=d_0\left(x\mathrm{,}y\right)+d_0\left(y\mathrm{,}z\right)\mathrm{.}$

When $\left(x\mathrm{,}y\mathrm{,}z\right)\in M\times E\times E$, we have $y=q\left(i,k,j\right)$ and $z=\left(l,m,r\right)$. If $k\ne m$ or $j\ne r$ then

$\begin{array}{c}d\left(x,z\right)=\frac{1}{m}+d_0\left(x,{\varphi }_r\left(p_m\right)\right)\\ <\frac{2}{k}+\frac{1}{m}+d_0\left(x,{\varphi }_j\left(p_k\right)\right)+d_0\left({\varphi }_j\left(p_k\right),{\varphi }_r\left(p_m\right)\right)\\ =d\left(x,y\right)+d\left(y,z\right).\end{array}$

If $k=m$, $j=r$ and $i\ne l$ then

$d\left(x,z\right)=\frac{1}{m}+d_0\left(x,{\varphi }_r\left(p_m\right)\right)<\frac{1}{k}+\frac{1}{m}+d_0\left(x,{\varphi }_j\left(p_k\right)\right)=d\left(x,y\right)+d\left(y,z\right).$

So if $\left(x\mathrm{,}y\mathrm{,}z\right)\in E\times E\times M$, change the role of x and z in previous case, and we get the result. If $\left(x\mathrm{,}y\mathrm{,}z\right)\in E\times M\times E$ then $x=q\left(i,k,j\right)$ and $z=q\left(l,m,r\right)$. Hence,

$d\left(x,y\right)+d\left(y,z\right)=\frac{1}{k}+\frac{1}{m}+d_0\left(y,{\varphi }_j\left(p_k\right)\right)+d_0\left(y,{\varphi }_r\left(p_m\right)\right)$

and

$d\left(x,z\right)=(\begin{array}{ll}\frac{1}{k}+\frac{1}{m}+d_0\left({\varphi }_j\left(p_k\right),{\varphi }_r\left(p_m\right)\right)\hfill & \text{if}k\ne m\text{or}j\ne r,\hfill \\ \frac{1}{k}\hfill & \text{if}k=m,j=r\text{and}i\ne l.\hfill \end{array}$

Thus, we always get the result $d\left(x,z\right)<d\left(x,y\right)+d\left(y,z\right)$ for both of 2 cases. When $\left(x\mathrm{,}y\mathrm{,}z\right)\in E\times E\times E$, we let

$x=q\left(i_1,k_1,j_1\right),y=q\left(i_2,k_2,j_2\right),z=q\left(i_3,k_3,j_3\right)$.

Case 1. If $k_1=k_3$ and $j_1=j_3$ we have $d\left(x,z\right)=\frac{1}{k_1}$, and

$\begin{array}{l}d\left(x\mathrm{,}y\right)+d\left(y\mathrm{,}z\right)\\ =(\begin{array}{ll}\frac{2}{k_1}\mathrm{,}\hfill & k_1=k_2=k_3\text{and}{j}_1=j_2=j_3\mathrm{,}\hfill \\ \frac{2}{k_1}+\frac{2}{k_2}+d_0\left({\varphi }_{j_1}\left(k_1\right)\mathrm{,}{\varphi }_{j_2}\left(k_2\right)\right)+d_0\left({\varphi }_{j_2}\left(k_2\right)\mathrm{,}{\varphi }_{j_3}\left(k_3\right)\right)\mathrm{,}\hfill & k_1=k_3\ne k_2\text{or}{j}_1=j_3\ne j_2\mathrm{.}\hfill \end{array}\end{array}$

It means that $d\left(x,z\right)<d\left(x,y\right)+d\left(y,z\right)$ for both of 2 cases.

Case 2. If $k_1\ne k_3$ or $j_1\ne j_3$, we have

$d\left(x\mathrm{,}z\right)=\frac{1}{k_1}+\frac{1}{k_3}+d_0\left({\varphi }_{j_1}\left(k_1\right)\mathrm{,}{\varphi }_{j_3}\left(k_3\right)\right)\mathrm{,}$

and

$\begin{array}{l}d\left(x\mathrm{,}y\right)+d\left(y\mathrm{,}z\right)\\ =(\begin{array}{ll}\frac{2}{k_1}+\frac{1}{k_3}+d_0\left({\varphi }_{j_2}\left(k_2\right)\mathrm{,}{\varphi }_{j_3}\left(k_3\right)\right)\mathrm{,}\hfill & k_1=k_2\text{and}{j}_1=j_2\mathrm{,}\hfill \\ \frac{1}{k_1}+\frac{2}{k_3}+d_0\left({\varphi }_{j_1}\left(k_1\right)\mathrm{,}{\varphi }_{j_2}\left(k_2\right)\right)\mathrm{,}\hfill & k_2=k_3\text{and}{j}_2=j_3\mathrm{,}\hfill \\ \frac{1}{k_1}+\frac{2}{k_2}+\frac{1}{k_3}+d_0\left({\varphi }_{j_1}\left(k_1\right)\mathrm{,}{\varphi }_{j_2}\left(k_2\right)\right)+d_0\left({\varphi }_{j_2}\left(k_2\right)\mathrm{,}{\varphi }_{j_3}\left(k_3\right)\right)\mathrm{,}\hfill & k_1\ne k_2\ne k_3\text{or}{j}_1\ne j_2\ne j_3\mathrm{.}\hfill \end{array}\end{array}$

Hence, $d\left(x,z\right)<d\left(x,y\right)+d\left(y,z\right)$.

It implies d is a metric in X.

Next, we prove that $\left(X\mathrm{,}d\right)$ is a compact metric space. Let any sequences ${\left(x_n\right)}_{n\in \mathbb{N}}\in X$. We prove that this sequence has a convergent subsequence. If ${\left(x_n\right)}_{n\in \mathbb{N}}$ has infinite elements in M, then the compactness of M and the fact $d_{|M\times M}=d_0$, so ${\left(x_n\right)}_{n\in \mathbb{N}}$ has a convergent subsequence. We consider ${\left(x_n\right)}_{n\in \mathbb{N}}$ has finite elements in M; therefore, it has infinite elements in E. We can assume that ${\left(x_n\right)}_{n\in \mathbb{N}}\subset E$ then $x_n=q\left(i_n,k_n,j_n\right)$. If there is $N\in \mathbb{N}$ such that $k_n<N,\forall n\in \mathbb{N}$ then the set $\left\{x_n\mathrm{<mo>\; ;\; </mo>}n\in \mathbb{N}\right\}$ is finite, so at least one point of ${\left(x_n\right)}_{n\in \mathbb{N}}$ appears infinite times, forming a convergent subsequence. Now suppose ${\left(k_n\right)}_{n\in \mathbb{N}}$ is unbounded, therefore, $\underset{n\to \infty }{\lim }{k}_n=\infty$. We choose $y_n={\varphi }_{j_n}\left(p_{k_n}\right)$,

so ${\left(y_n\right)}_{n\in \mathbb{N}}\subset M$ and $d\left(x_n,y_n\right)=\frac{1}{k_n},\forall n\in \mathbb{N}$. Since ${\left(y_n\right)}_{n\in \mathbb{N}}$ is a subset of

compact set M, ${\left(y_n\right)}_{n\in \mathbb{N}}$ has a subsequence ${\left(y_{n_l}\right)}_{l\in \mathbb{N}}$ converging to $y\in M$. Thus, we have

$d\left(x_{n_l}\mathrm{,}y\right)<d\left(x_{n_l}\mathrm{,}{y}_{n_l}\right)+d\left(y_{n_l}\mathrm{,}y\right)=\frac{1}{k_{n_l}}+d\left(y_{n_l}\mathrm{,}y\right)\to 0\text{when}l\to \infty \mathrm{.}$

It implies that ${\left(x_n\right)}_{n\in \mathbb{N}}$ has a subsequence ${\left(x_{n_l}\right)}_{l\in \mathbb{N}}$ which converges to y. Thus, $\left(X\mathrm{,}d\right)$ is a compact metric space.

For all $t\in ℝ$, we define a map ${\psi }_t$ by

${\psi }_t\left(x\right)=(\begin{array}{ll}{\varphi }_t\left(x\right)\hfill & \text{if}x\in M\mathrm{,}\hfill \\ q\left(i,k,\left(j+t\right)\mathrm{mod}\pi \left(p_k\right)\right)\hfill & \text{if}x=q\left(i\mathrm{,}k\mathrm{,}j\right)\mathrm{.}\hfill \end{array}$

We can see that j, t, $j+t$ cannot be in $\mathbb{N}$, but we can define a real number: $t\mathrm{mod}\pi \left(p_k\right):=r$, when

$t=m\pi \left(p_k\right)+r,m\in \mathbb{Z},0\le r<\pi \left(p_k\right).$

By definition of flow, it's easy to see that $\psi$ is a flow of X. Indeed, we can prove that ${\psi }_{t+s}={\psi }_t\circ {\psi }_s,\forall t,s\in ℝ$. If $x\in M$, we get

${\psi }_{t+s}\left(x\right)={\varphi }_{t+s}\left(x\right)={\varphi }_t\circ {\varphi }_s\left(x\right)={\psi }_t\circ {\psi }_s\left(x\right),\forall t,s\in ℝ.$

If $x=q\left(i,k,j\right)$, we have

${\psi }_{t+s}\left(x\right)=q\left(i,k,\left(j+t+s\right)\mathrm{mod}\pi \left(p_k\right)\right)={\psi }_t\circ {\psi }_s\left(x\right).$

Therefore, $\psi$ is the flow with the previous properties.

In order to prove that $\psi$ is n-expansive, first we see that $\varphi$ is 1-expansive; so there is $a>0$ such that if $d\left({\varphi }_t\left(x\right)\mathrm{,}{\varphi }_t\left(y\right)\right)\le a\mathrm{,}\forall t\in \mathbb{N}$, then $x=y$. Suppose that $\left\{x_1\mathrm{,}\cdots \mathrm{,}{x}_{n+1}\right\}$ are $n+1$ different points of X satisfying

$d\left({\psi }_t\left(x_i\right)\mathrm{,}{\psi }_t\left(x_j\right)\right)\le a\mathrm{,}\forall t\in ℝ\mathrm{,}\forall \left(i\mathrm{,}j\right)\in \left\{\mathrm{1,}\cdots \mathrm{,}n+1\right\}\times \left\{\mathrm{1,}\cdots \mathrm{,}n+1\right\}\mathrm{.}$

Hence, at most one of these points belong to M. Consequently, at least n of them belong to E. Without loss of generality, we get $x_m=q\left(i_m,k_m,j_m\right),m\in \left\{1,\cdots ,n\right\}$. Because $i_m\in \left\{\mathrm{1,}\cdots \mathrm{,}n-1\right\}$ and we have n number $i_m$ ; thus, by Pigeonhole principle, at least two of these points are of the form $q\left(i\mathrm{,}k\mathrm{,}j\right)$ and $q\left(i\mathrm{,}m\mathrm{,}r\right)$. We prove that $k\ne m$. Indeed, if $k=m$, we have 2 points are $q\left(i\mathrm{,}k\mathrm{,}j\right)$ and $q\left(i\mathrm{,}k\mathrm{,}r\right)$ with $j\ne r$ (because all of $n+1$ points are different). For each $s\in ℝ$ we have

$\begin{array}{l}d\left({\varphi }_s\left({\varphi }_j\left(p_k\right)\right)\right)\mathrm{,}d\left({\varphi }_s\left({\varphi }_r\left(p_k\right)\right)\right)\\ =d\left({\psi }_s\left(q\left(i\mathrm{,}k\mathrm{,}j\right)\mathrm{,}{\psi }_s\left(q\left(i\mathrm{,}k\mathrm{,}r\right)\right)\right)\right)-\frac{2}{k}\\ <d\left({\psi }_s\left(q\left(i,k,j\right)\right),{\psi }_s\left(q\left(i,k,r\right)\right)\right)<a.\end{array}$

This implies that ${\varphi }_j\left(p_k\right)={\varphi }_r\left(p_k\right)$ (by the Proposition of 1-expansive of $\varphi$ ), which implies that $j=r$ and we obtain a contradiction. Therefore, $k\ne m$.

Now we implies that: for every $s\in ℝ$ we have:

$\begin{array}{l}d\left({\varphi }_s\left({\varphi }_j\left(p_k\right)\right)\right),d\left({\varphi }_s\left({\varphi }_r\left(p_m\right)\right)\right)\\ =d\left({\psi }_s\left(q\left(i,k,j\right)\right),{\psi }_s\left(q\left(i,m,r\right)\right)\right)-\frac{1}{k}-\frac{1}{m}\\ <d\left({\psi }_s\left(q\left(i,k,j\right)\right),{\psi }_s\left(q\left(i,m,r\right)\right)\right)<a.\end{array}$

So similarly, we have ${\varphi }_j\left(p_k\right)={\varphi }_r\left(p_m\right)$ ; hence, $p_m=p_k$, which is contradiction with the fact that $k\ne m$. Thus, we cannot choose $n+1$ points satisfy this proposition; it means $\psi$ is n-expansive in X.

Next, we prove that $\psi$ is not $\left(n-1\right)$ -expansive. For any $a>0$, we choose $k\in \mathbb{N}$ such that $\frac{1}{k}<a$, so we have

$d\left({\varphi }_j\left(p_k\right),q\left(i,k,j\right)\right)=\frac{1}{k}<a,\forall j\in ℝ,\forall i\in \left\{1,\cdots ,n-1\right\}$. So $\Gamma \left(p_k\mathrm{,}a\right)$ contain at

least n points $\left\{p_k\mathrm{,}q\left(\mathrm{1,}k\mathrm{,0}\right)\mathrm{,}\cdots \mathrm{,}q\left(n-\mathrm{1,}k\mathrm{,0}\right)\right\}$ and that $\psi$ is not $\left(n-1\right)$ -expansive, because there is not $a>0$ satisfies this define about $\left(n-1\right)$ -expansive.

Now we prove that $\psi$ has the shadowing property. Since $\varphi$ has the shadowing property, for each $ϵ>0$, we can consider ${\delta }_{\varphi }>0$, so for any $\left({\delta }_{\varphi }\mathrm{,1}\right)$ -pseudo-orbit in M we have the $\frac{ϵ}{2}$ -shadowing. Now consider ${\left(x_n\mathrm{,}{t}_n\right)}_{n\in \mathbb{Z}}$ has

the $\left(\delta \mathrm{,1}\right)$ -pseudo-orbit by $\psi$ in X. We assume that $\delta <\frac{{\delta }_{\varphi }}{3}<\frac{ϵ}{3}$. So we have

$d\left({\psi }_{t_n}\left(x_n\right),x_{n+1}\right)<\delta$. Let N is a smallest integer number such that $\frac{1}{N}<\delta$, and we consider $\left(x_n\mathrm{,}{x}_{n+1}\right)$ in 3 cases.

Case 1. If $\left(x_n\mathrm{,}{x}_{n+1}\right)\in E\times M$, we have $x_n=q\left(i,k,j\right)$ and $d\left({\psi }_{t_n}\left(x_n\right),x_{n+1}\right)=\frac{1}{k}+d_0\left(x_{n+1},{\varphi }_{j+t_n}\left(p_k\right)\right)$, so $\frac{1}{k}<\delta$ ; hence, $k\ge N$.

Case 2. If $\left(x_n\mathrm{,}{x}_{n+1}\right)\in M\times E$, we obtain $x_{n+1}=q\left(i,k,j\right)$ and $d\left({\psi }_{t_n}\left(x_n\right),x_{n+1}\right)=\frac{1}{k}+d_0\left({\varphi }_j\left(p_k\right),{\varphi }_j\left(x_n\right)\right)$, so $\frac{1}{k}<\delta$ ; hence, $k\ge N$.

Case 3. If $\left(x_n\mathrm{,}{x}_{n+1}\right)\in E\times E$, we have $x_n=q\left(i,k,j\right)$ and $x_{n+1}=q\left(l,m,r\right)$. So ${\psi }_{t_n}\left(x_n\right)=q\left(i,k,j+t_n\right)$. Thus, if we want $d\left({\psi }_{t_n}\left(x_n\right),x_{n+1}\right)<\delta$, we have either if $k\ge N$, so $m\ge N$ (by similarly) or if $k<N$, we have $x_{n+1}={\psi }_{t_n}\left(x_n\right)$, such that $x_{n+1}=q\left(i,k,j+t_n\right)$. When ${\left(x_n\mathrm{,}{t}_n\right)}_{n\in \mathbb{Z}}$ is one of orbit ${\left\{q\left(l\mathrm{,}k\mathrm{,}{j}_n\right)\right\}}_{n\in \mathbb{Z}}$, and $j_{n+1}=j_n+t_n,\forall n\in \mathbb{Z}$. So one obtain $s_n=j_n-j_0$, thus,

$d\left({\psi }_{t-s_n}\left(x_n\right),{\psi }_t\left(x_0\right)\right)=d\left(q\left(l,k,t-s_n+j_n\right),q\left(l,k,t+j_0\right)\right)=0,s_n\le t<s_{n+1}.$

Therefore, the shadowing property is proved.

When $x_i=q\left(l,k,j\right)$, then $k>N$. Define a sequence ${\left(y_n\mathrm{,}{t}_n\right)}_{n\in \mathbb{Z}}\subset M$ by

$y_n=(\begin{array}{ll}{x}_n\hfill & \text{if}{x}_n\in M\mathrm{,}\hfill \\ {\varphi }_j\left(p_k\right)\hfill & \text{if}{x}_n=q\left(l\mathrm{,}k\mathrm{,}j\right)\mathrm{.}\hfill \end{array}$

Then ${\left(y_n\mathrm{,}{t}_n\right)}_{n\in \mathbb{Z}}$ is ${\delta }_{\varphi }$ -pseudo-orbit in M since

$\begin{array}{l}d\left({\varphi }_{t_n}\left(y_n\right),y_{n+1}\right)=d\left({\psi }_{t_n}\left(y_n\right),y_{n+1}\right)\\ \le d\left({\psi }_{t_n}\left(y_n\right),{\psi }_{t_n}\left(x_n\right)\right)+d\left({\psi }_{t_n}\left(x_n\right),x_{n+1}\right)+d\left(x_{n+1},y_{n+1}\right)\\ <\frac{1}{N}+\delta +\frac{1}{N}<{\delta }_{\varphi }.\end{array}$

Hence, there exists $y\in M$ and a function $h\in Rep$ such that

$d\left({\varphi }_{t-s_n}\left(y_n\right),{\varphi }_{h\left(t\right)}\left(y\right)\right)<\frac{ϵ}{2},\forall s_n\le t<s_{n+1}.$

So

$\begin{array}{c}d\left({\varphi }_{t-s_n}\left(x_n\right),{\varphi }_{h\left(t\right)}\left(y\right)\right)<d\left({\varphi }_{t-s_n}\left(x_n\right),{\varphi }_{t-s_n}\left(y_n\right)\right)+d\left({\varphi }_{t-s_n}\left(y_n\right),{\varphi }_{h\left(t\right)}\left(y\right)\right)\\ <\frac{1}{N}+\frac{ϵ}{2}<ϵ.\end{array}$

Therefore, ${\left(x_n\mathrm{,}{t}_n\right)}_{n\in \mathbb{Z}}$ is $ϵ$ -shadowing. Hence, $\psi$ has the shadowing property.

Finally, we have $\psi$ admits an infinite number of chain-recurrent classes. Indeed, if we have $q\left(i\mathrm{,}k\mathrm{,}l\right)\in E$ then

$d\left(q\left(i,k,j\right),x\right)\ge \frac{1}{k},\forall x\in X\backslash \left\{q\left(i,k,j\right)\right\}.$

So if $0<ϵ<\frac{1}{k}$ then the orbit of $q\left(i\mathrm{,}k\mathrm{,}j\right)$ cannot be connected by $ϵ$ -pseudo orbits with any other point of X. This proves that the chain-recurrent classes of $q\left(i\mathrm{,}k\mathrm{,}j\right)$ contains only its orbit. Therefore different periodic orbits in E belong to different chain-recurrent classes and we conclude the proof.

Acknowledgements

The first author was supported in part by the VNU Project of Vietnam National University No. QG101-15.

Open Questions

How are the properties of the local stable (unstable) sets of n-expansive flows?

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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