On ω-Limit Sets of Zadeh’s Extension of Nonautonomous Discrete Systems on an Interval

Abstract

Let $I=[0,1]$ and $f_n$ be a sequence of continuous self-maps on I which converge uniformly to a self-map f on I. Denote by $\mathcal{F}\left(I\right)$ the set of fuzzy numbers on I, and denote by $(\mathcal{F}\left(I\right),\widehat{f})$ and $(\mathcal{F}\left(I\right),{\widehat{f}}_n)$ the Zadeh’s extensions of $(I,f)$ and $(I,f_n)$, respectively. In this paper, we study the $\omega$-limit sets of $(\mathcal{F}\left(I\right),{\widehat{f}}_n)$ and show that, if all periodic points of f are fixed points, then $\omega (A,{\widehat{f}}_n)\subset F\left(\widehat{f}\right)$ for any $A\in \mathcal{F}\left(I\right)$, where $\omega (A,{\widehat{f}}_n)$ is the $\omega$-limit set of A under $(\mathcal{F}\left(I\right),{\widehat{f}}_n)$ and $F\left(\widehat{f}\right)=\{A\in \mathcal{F}\left(I\right):\widehat{f}\left(A\right)=A\}$.

1. Introduction

Research for the dynamical properties of nonautonomous discrete systems on a metric space is very interesting (see [1,2,3,4,5,6,7,8,9,10,11]). In [12], Kempf investigated the $\omega$-limit sets of a sequence of continuous self-maps $f_n$ on I which converge uniformly to a self-map f on I and showed that, if $P\left(f\right)=F\left(f\right)$, then $\omega (x,f_n)$ is a closed subset of I with $\omega (x,f_n)\subset F\left(f\right)$ for any $x\in I$, where $F\left(f\right)$ and $P\left(f\right)$ are the set of fixed points of f and the set of periodic points of f, respectively, and $\omega (x,f_n)$ is the set of $\omega$-limit points of x under $(X,f_n)$. Further, Cánovas [13] showed that, if $f_n$ is a sequence of continuous self-maps on I which converge uniformly to a self-map f on I and $P\left(f\right)=F\left(f^{2^s}\right)$ for some $s\in \mathbb{N}$, then $\omega (x,f_n)={\cup }_{k=1}^{2^s}[p_k,q_k]\subset F\left(f^{2^s}\right)$ with $f\left([p_k,q_k]\right)=[p_{k+1},q_{k+1}]$ ($1\le k\le 2^s-1$) and $f\left([p_{2^s},q_{2^s}]\right)=[p_1,q_1]$ for any $x\in I$. In [14], we studied the $\omega$-limit sets of a sequence of continuous self-maps $f_n$ on a tree T which converge uniformly to a self-map f on T and showed that, if $P\left(f\right)=F\left(f\right)$, then $\omega (x,f_n)$ is a closed connected subset of T with $\omega (x,f_n)\subset F\left(f\right)$ for any $x\in T$.

It is well known [15] that the discrete dynamical system $(X,f)$ naturally induces a dynamical system $(\mathcal{F}\left(X\right),\widehat{f})$, where $\mathcal{F}\left(X\right)$ is the set of all fuzzy sets on a metric space X and $\widehat{f}$ is the Zadeh’s extension of continuous self-maps f on X. It is natural to ask how the dynamical properties of f is related to the dynamical properties of $\widehat{f}$. Already, there are many results for this question so far; see, e.g., References [16,17,18,19,20,21] and the related references therein, where different chaotic properties and topological entropies of Zadeh’s extensions of continuous seif-maps on metric spaces were considered. Our aim in this paper is to study the $\omega$-limit sets of Zadeh’s extensions of nonautonomous discrete systems on intervals.

2. Preliminaries

Throughout this paper, let $(X,d)$ be a metric space, write $I=[0,1]$, and denote by $\mathbb{N}$ the set of all positive integers. Let $C^0\left(X\right)$ be the set of all continuous self-maps on X. For a given $f\in C^0\left(X\right)$, let $f^{n+1}=f\circ f^n$ for any $n\in \mathbb{N}$. We call $F\left(f\right)=\{x\in X:f(x)=x\}$ the set of fixed points of f and $P\left(f\right)=\{x\in X:f^n\left(x\right)=x$ for some $n\in \mathbb{N}\}$ the set of periodic points of f.

Let $f_n\in C^0\left(X\right)(n\in \mathbb{N})$ and $F^0$ be the identity map of X, and write

$$F_n=f_n\circ f_{n-1}\circ \cdots \circ f_1\mathrm{for} \mathrm{any}n\in \mathbb{N}.$$

$y\in X$ is called the $\omega$-limit point of $x(\in X)$ under $(X,f_n)$ if there are $n_1<n_2<\cdots <n_k<\cdots$ such that

$$\underset{k⟶\infty }{lim}{F}_{n_k}\left(x\right)=y.$$

Denote by $\omega (x,f_n)$ the set of $\omega$-limit points of x under $(X,f_n)$. We write $f_n⟹f$ if $f_n$ converges uniformly to f.

Now, let us recall some definitions for fuzzy theory which are from [15].

Definition 1.

Let X be a metric space. A mapping $A:X⟶[0,1]$ is called a fuzzy set on X. For each fuzzy set A and each $\alpha \in (0,1]$, $A_{\alpha }=\{t\in X:A\left(t\right)\ge \alpha \}$ is called an α-level set of A and $A_0=\overline{\{t\in X:A(t)>0\}}$ is called the support of A, where $\overline{B}$ means the closure of subset B of X.

Definition 2.

A fuzzy set A on I is said to be a fuzzy number if it satisfies the following conditions:

(1) 

$A_1\ne \mathsf{\varnothing }$;

(2) 

A is an upper semicontinuous function;

(3) 

For any $t_1,t_2\in I$ and any $\lambda \in [0,1]$, $A(\lambda t_1+(1-\lambda )t_2)\ge min\{A\left(t_1\right),A\left(t_2\right)\}$;

(4) 

$A_0$ is compact.

Let $\mathcal{F}\left(I\right)$ denote the set of fuzzy numbers on I. It is known that $\alpha$-level set $A_{\alpha }$ of A determines the fuzzy number A and that every $A_{\alpha }$ is a closed connected subset of I. If $A\in I$, then $A\in \mathcal{F}\left(I\right)$ with $A_{\alpha }=[A,A]$ for any $\alpha \in [0,1]$.

For any $A,B\in \mathcal{F}\left(I\right)$ with $A_{\alpha }=[A_{l,\alpha },A_{r,\alpha }]$ and $B_{\alpha }=[B_{l,\alpha },B_{r,\alpha }]$ for any $\alpha \in (0,1]$, we define the metric of A and B as follows:

$$D(A,B)=\underset{\alpha \in (0,1]}{sup}max\left\{\right|A_{l,\alpha }-B_{l,\alpha }|,|A_{r,\alpha }-B_{r,\alpha }\left|\right\}.$$

Obviously, we have

$$D(A,B)=\underset{\alpha \in (0,1]}{sup}max\{\underset{x\in A_{\alpha }}{sup}d(x,B_{\alpha }),\underset{y\in B_{\alpha }}{sup}d(y,A_{\alpha })\}H(A_{\alpha },B_{\alpha }),$$

where $d(x,J)={inf}_{y\in J}d(x,y)$ for any $x\in I$ and $J\subset I$. It is known that $\left(\mathcal{F}\right(I),D)$ is a complete metric space (refer to [15]).

Let $f\in C^0\left(I\right)$. We define the Zadeh’s extension $\widehat{f}:\mathcal{F}\left(I\right)⟶\mathcal{F}\left(I\right)$ of f for any $x\in I$ and $A\in \mathcal{F}\left(I\right)$ by

$$\left(\widehat{f}\left(A\right)\right)\left(x\right)=\underset{f\left(y\right)=x}{sup}A\left(y\right).$$

It follows from [15] that f is continuous if and only if $\widehat{f}$ is continuous, and it follows from [22] (Lemma 2.1) that

$${\left[\widehat{f}\left(A\right)\right]}_{\alpha }=f\left(A_{\alpha }\right)$$

for any $A\in \mathcal{F}\left(I\right)$ and $\alpha \in (0,1]$. In this paper, we will show the following theorem.

Theorem 1.

Let $f_n$ be a sequence of continuous self-maps on I with $f_n⟹f$. If $P\left(f\right)=F\left(f\right)$, then $\omega (A,{\widehat{f}}_n)\subset F\left(\widehat{f}\right)$ for any $A\in \mathcal{F}\left(I\right)$.

3. Proof of the Main Result

In this section, we let $f\in C^0\left(I\right)$ and $\widehat{f}$ is the Zadeh’s extension of f. Let ${\widehat{F}}^0$ be the identity map of $\mathcal{F}\left(I\right)$ and for any $n\in \mathbb{N}$, we write

$${\widehat{F}}_n={\widehat{f}}_n\circ {\widehat{f}}_{n-1}\circ \cdots \circ {\widehat{f}}_1.$$

Lemma 1.

Assume that $f_n\in C^0\left(I\right)$ for any $n\in \mathbb{N}$ with $f_n⟹f$. Then, ${\widehat{f}}_n⟹\widehat{f}$ on $\mathcal{F}\left(I\right)$.

Proof. 

Since $f_n⟹f$ on I, it follows that, for any $\epsilon >0$, there is an $N\in \mathbb{N}$ such that, when $n\ge N$, we have

$$|f\left(x\right)-f_n\left(x\right)|<\frac{\epsilon }{2}$$

for any $x\in I$, which implies that, for any $B\subset I$, we have $d(z,f_n\left(B\right))<\epsilon /2$ for any $z\in f\left(B\right)$ and $d(z,f(B\left)\right)<\epsilon /2$ for any $z\in f_n\left(B\right)$. Thus when $n\ge N$, we have

$$D(\widehat{f}\left(A\right),{\widehat{f}}_n\left(A\right))=\underset{\alpha \in (0,1]}{sup}H(f\left(A_{\alpha }\right),f_n\left(A_{\alpha }\right))\le \frac{\epsilon }{2}<\epsilon$$

for any $A\in \mathcal{F}\left(I\right)$. Lemma 1 is proven.  □

Lemma 2.

Assume that $f_n\in C^0\left(I\right)$ for any $n\in \mathbb{N}$ with $f_n⟹f$. If $B\in \omega (A,{\widehat{f}}_n)$ for some $A\in \mathcal{F}\left(I\right)$, then $\omega (x,f_n)\cap B_{\alpha }\ne \mathsf{\varnothing }$ for any $\alpha \in (0,1]$ and $x\in A_{\alpha }$.

Proof. 

Let $B\in \omega (A,{\widehat{f}}_n)$ and $\alpha \in (0,1]$. Let $n_1<n_2<\cdots <n_k<\cdots$ such that

$$\underset{k⟶\infty }{lim}D(\widehat{F}{n}_k\left(A\right),B)=0.$$

Then,

$$\underset{k⟶\infty }{lim}H(F{n}_k\left(A_{\alpha }\right),B_{\alpha })=0.$$

Let $x\in A_{\alpha }$. By taking a subsequence, we let ${lim}_{k⟶\infty }F{n}_k\left(x\right)=y\in \omega (x,f_n)$. If $y\notin B_{\alpha }$, then $\epsilon =d(y,B_{\alpha })>0$. Since ${lim}_{k⟶\infty }H(F{n}_k\left(A_{\alpha }\right),B_{\alpha })=0$, there is an $N\in \mathbb{N}$ such that, when $n_k>N$, we have

$$H(F{n}_k\left(A_{\alpha }\right),B_{\alpha })<\frac{\epsilon }{2},$$

which implies $d(F{n}_k\left(x\right),B_{\alpha })<\epsilon /2$ and ${lim}_{k⟶\infty }F{n}_k\left(x\right)\ne y$ since $F{n}_k\left(x\right)\in F{n}_k\left(A_{\alpha }\right)$. This is a contradiction. Thus, $y\in B_{\alpha }$. Lemma 2 is proven.  □

Proposition 1.

Assume that $f_n\in C^0\left(I\right)$ for any $n\in \mathbb{N}$ with $f_n⟹f$ and $P\left(f\right)=F\left(f\right)$. Then, the following statements hold:

(1) 

If $B\in \omega (A,{\widehat{f}}_n)$ for some $A\in \mathcal{F}\left(I\right)$, then $\mathsf{\varnothing }\ne f\left(B_{\alpha }\right)\cap B_{\alpha }\cap \omega (x,f_n)\subset F\left(f\right)$ for any $\alpha \in (0,1]$ and $x\in A_{\alpha }$.

(2) 

If $B\in \omega (A,{\widehat{f}}_n)$ for some $A\in \mathcal{F}\left(I\right)$, then ${\cup }_{B\in \omega (A,{\widehat{f}}_n)}{B}_{\alpha }\cup \omega (x,f_n)$ is a connected subset of I for any $\alpha \in (0,1]$ and $x\in A_{\alpha }$.

Proof. 

It follows from Theorem 1 and Lemma 2.  □

Lemma 3

(See [14] (Lemma 2)). Assume that $f\in C^0\left(I\right)$ with $F\left(f\right)=P\left(f\right)$. Then, for any $x\in I$ and $n\in \mathbb{N}$, $f^n\left(x\right)>x$ if $f\left(x\right)>x$ and $f^n\left(x\right)<x$ if $f\left(x\right)<x$.

Now, we show the main result of this paper.

Proof of Theorem 1.

Let $B\in \omega (A,{\widehat{f}}_n)$. For any $\alpha \in (0,1]$, write $B_{\alpha }=[a_{\alpha },b_{\alpha }]$ and $f\left(B_{\alpha }\right)=[c_{\alpha },d_{\alpha }]$. Let $n_1<n_2<\cdots <n_k<\cdots$ such that

$$\underset{k⟶\infty }{lim}D(\widehat{F}{n}_k\left(A\right),B)=0.$$

(1)

By $\widehat{f}\in C^0\left(\mathcal{F}\left(I\right)\right)$, we see that, for any $\epsilon >0$, there is an $\delta =\delta \left(\epsilon \right)>0$ such that, if $D(B,C)<\delta$ with $C\in \mathcal{F}\left(I\right)$, then

$$D(\widehat{f}\left(B\right),\widehat{f}\left(C\right))<\frac{\epsilon }{3}.$$

By Lemma 1, we see that there is an $N=N\left(\epsilon \right)\in \mathbb{N}$ such that, when $n\ge N$, we have

$$D(\widehat{f}\left(W\right),{\widehat{f}}_n\left(W\right))\le \frac{\epsilon }{3}$$

(2)

for any $W\in \mathcal{F}\left(I\right)$. Take $r=n_k\ge N$ such that $D({\widehat{F}}_r\left(A\right),B)<\delta$. Thus,

$$D(\widehat{f}\left(B\right),{\widehat{F}}_{r+1}\left(A\right))\le D(\widehat{f}\left(B\right),\widehat{f}\left({\widehat{F}}_r\left(A\right)\right))+D(\widehat{f}\left({\widehat{F}}_r\left(A\right)\right),{\widehat{F}}_{r+1}\left(A\right))\le \frac{2\epsilon }{3}.$$

(3)

In the following, we show that $a_{\alpha }=c_{\alpha }$ and $b_{\alpha }=d_{\alpha }$. For convenience, write $a_{\alpha }=a,b_{\alpha }=b$, $c_{\alpha }=c$, and $d_{\alpha }=d$.

(i) We will show $c\le a$. Assume on the contrary that $c>a$. Then, by Proposition 1, we see $c\le b$.

We claim that there is an $u\in (a,1]$ such that $f\left(u\right)=a$. Indeed, if $f([a,1])\subset (a,1]$, then let $\epsilon =min\left\{d\right(a,f([a,1])),c-a\}>0$. By Equation (3), we see $F_{r+1}\left(A_{\alpha }\right)\subset [a+\epsilon /3,1]$. It follows from Equation (2) that

$$H(f\left(F_{r+1}\left(A_{\alpha }\right)\right),F_{r+2}\left(A_{\alpha }\right))\le \frac{\epsilon }{3}.$$

Thus, $F_{r+2}\left(A_{\alpha }\right)\subset [a+\epsilon /3,1]$. Continuing in this fashion, we have that $F_n\left(A_{\alpha }\right)\subset [a+\epsilon /3,1]$ for any $n\ge r+1$, which contradicts Equation (1). The claim is proven.

Let $u=min\{x\in (a,1]:f(x)=a\}$. Then, $u>b$ since $f([a,b])=[c,d]$ and $u>d$ (Otherwise, if $b<u\le d$, then there exists an $u_1\in [a,b]$ and $u_2\in [u_1,u]$ such that $u_1=f\left(u_2\right)<u_2\le d=f\left(u_1\right)$. This contradicts Lemma 3.). By Lemma 3, we see $f([a,u])\subset [a,u)$. Write

$$\begin{array}{ccc}\hfill p& =& max\{b,d,maxf([a,u]\left)\right\},\hfill \\ \hfill {\epsilon }_1& =& (u-p)/2,\hfill \\ \hfill q& =& min\{c,minf\left([a,p+{\epsilon }_1]\right)\},\hfill \\ \hfill \epsilon & =& min\{(q-a)/2,{\epsilon }_1\}.\hfill \end{array}$$

By Equation (3), we see $F_{r+1}\left(A_{\alpha }\right)\subset [q-\epsilon ,p+{\epsilon }_1]$. It follows from Equation (2) that

$$H(f\left(F_{r+1}\left(A_{\alpha }\right)\right),F_{r+2}\left(A_{\alpha }\right))\le \frac{\epsilon }{3}.$$

Thus, $F_{r+2}\left(A_{\alpha }\right)\subset [q-\epsilon ,p+{\epsilon }_1]$. Continuing in this fashion, we have that $F_n\left(A_{\alpha }\right)\subset [q-\epsilon ,p+{\epsilon }_1]$ for any $n\ge r+1$, which contradicts Equation (1).

(ii) In similar fashion, we can show $d\ge b$.

(iii) We will show that, if $c=a$, then $d=b$. Assume on the contrary that $d>b$. Let $u=max\{z\in [a,b]:f(z)=d\}$ and $e=min\{z\in [a,b]:f(z)=a\}$. Then, we have $e<u$ (Otherwise, if $e>u$, then there is an $w\in [u,e]$ satisfying $u=f\left(w\right)<w<d=f^2\left(w\right)$. This contradicts Lemma 3.) and $f\left(a\right)<u$.

We claim that there is an $v\in (u,1]$ such that $f\left(v\right)=u$. Indeed, if $p=minf\left(\right[u,1\left]\right)>u$, then let $\epsilon =min\left\{\right(d-b)/2,d(u,p\left)\right\}>0$. By Equation (3), we see

$$D(\widehat{f}\left(B\right),{\widehat{F}}_{r+1}\left(A\right))\le \frac{2\epsilon }{3}.$$

If $F_n\left(A_{\alpha }\right)\cap [a,u]\ne \mathsf{\varnothing }$ for any $n\ge r+1$, then we have $d-2\epsilon /3\in F_n\left(A_{\alpha }\right)$ for any $n\ge r+1$, which contradicts Equation (1). If $F_n\left(A_{\alpha }\right)\cap [a,u]=\mathsf{\varnothing }$ for some $n\ge r+1$, then let $m=min\{F_n\left(A_{\alpha }\right)\cap [a,u]=\mathsf{\varnothing }:n\ge r+1\}$. Thus, $F_m\left(A_{\alpha }\right)\subset (u,1]$, and it follows from Equation (2) that

$$H(f\left(F_m\left(A_{\alpha }\right)\right),F_{m+1}\left(A_{\alpha }\right))\le \frac{\epsilon }{3},$$

which implies $F_{m+1}\left(A_{\alpha }\right)\subset [p-\epsilon /3,1]\subset (u,1]$. Continuing in this fashion, we obtain that $F_n\left(A_{\alpha }\right)\subset [p-\epsilon /3,1]\subset (u,1]$ for any $n\ge m$, which contradicts Equation (1). The claim is proven.

Let $v=min\{x\in (u,1]:f(x)=u\}$. Then, by Lemma 3, we see $d<v$ and $p=maxf\left(\right[u,v\left]\right)<v$.

If there is an $w\in [0,a)$ satisfying $f\left(w\right)=u$, then let $w=max\{x\in [0,a):f(x)=u\}$. By Lemma 3, we see $q=minf\left(\right[w,a\left]\right)>w$ and $f\left(\right[w,v\left]\right)=[q,p]$. Write

$$\begin{array}{ccc}\hfill {\epsilon }_1& =& (v-p)/2,\hfill \\ \hfill z& =& minf\left([u,p+{\epsilon }_1]\right)>u,\hfill \\ \hfill \epsilon & =& min\{(d-b)/2,(q-w)/2,(z-u)/2,{\epsilon }_1\}.\hfill \end{array}$$

By Equation (3), we see

$$D(\widehat{f}\left(B\right),{\widehat{F}}_{r+1}\left(A\right))\le \frac{2\epsilon }{3}.$$

This implies $F_{r+1}\left(A_{\alpha }\right)\subset [q-\epsilon ,p+{\epsilon }_1]$. If $F_n\left(A_{\alpha }\right)\cap [a,u]\ne \mathsf{\varnothing }$ for any $n\ge r+1$, then we have $d-2\epsilon /3\in F_n\left(A_{\alpha }\right)$ for any $n\ge r+1$, which contradicts Equation (1). If $F_n\left(A_{\alpha }\right)\cap [a,u]=\mathsf{\varnothing }$ for some $n\ge r+1$, then let $m=min\{F_n\left(A_{\alpha }\right)\cap [a,u]=\mathsf{\varnothing }:n\ge r+1\}$. Thus, $F_m\left(A_{\alpha }\right)\subset (u,p+{\epsilon }_1]$, and it follows from Equation (2) that

$$H(f\left(F_m\left(A_{\alpha }\right)\right),F_{m+1}\left(A_{\alpha }\right))\le \frac{\epsilon }{3},$$

which implies $F_{m+1}\left(A_{\alpha }\right)\subset [z-\epsilon /3,p+{\epsilon }_1]\subset (u,p+{\epsilon }_1]$. Continuing in this fashion, we obtain that $F_n\left(A_{\alpha }\right)\subset [z-\epsilon /3,p+{\epsilon }_1]\subset (u,p+{\epsilon }_1]$ for any $n\ge m$, which contradicts Equation (1).

If $maxf\left(\right[0,a\left]\right)<u$, then $f\left(\right[0,v\left]\right)=[0,p]$. Using the similar arguments as ones developed in the above given proof, we also obtain a conclusion which contradicts Equation (1).

(iv) We will show $c=a$. Assume on the contrary that $c<a$. Then by claim (ii), we see $b\le d$. Using the similar arguments as ones developed in the proof of claim (iii), we can obtain $b<d$. Let $\epsilon =min\left\{\right(a-c)/2,(d-b)/2\}$. By Equation (3), we see $F_{r+1}\left(A_{\alpha }\right)\supset [a-\epsilon ,b+\epsilon ]$. It follows from Equation (2) that

$$H(f\left(F_{r+1}\left(A_{\alpha }\right)\right),F_{r+2}\left(A_{\alpha }\right))\le \frac{\epsilon }{3}.$$

Thus, $F_{r+2}\left(A_{\alpha }\right)\supset [a-\epsilon ,b+\epsilon ]$. Continuing in this fashion, we have that $F_n\left(A_{\alpha }\right)\supset [a-\epsilon ,b+\epsilon ]$ for any $n\ge r+1$, which contradicts Equation (1).

By claims (iii) and (iv), we see $f\left(B_{\alpha }\right)=B_{\alpha }$ for any $\alpha \in (0,1]$, which implies $f\left(B\right)=B$. Theorem 1 is proven. □

Using the similar arguments as ones developed in the proofs of Proposition 1.4 of [13] and Theorem 1, we may show the following result.

Corollary 1.

Let $f_n\in C^0\left(I\right)$ for any $n\in \mathbb{N}$ with $f_n⟹f$. If $P\left(f\right)=F\left(f^{2^s}\right)$ for some $s\in \mathbb{N}$, then $\omega (A,{\widehat{f}}_n)\subset F(\widehat{f}$${}^{2^s})$ for any $A\in \mathcal{F}\left(I\right)$.

The following example illustrates that there are $f_n\in C^0\left(I\right)$ for any $n\in \mathbb{N}$ such that $f_n⟹f$ with $P\left(f\right)=F\left(f\right)$ and $\omega (A,{\widehat{f}}_n)=\mathsf{\varnothing }$ for some $A\in \mathcal{F}\left(I\right)$.

Example 1.

Let $f\in C^0\left(I\right)$ with $f\left(1\right)=1>0=f\left(0\right)$ and $x<f\left(x\right)$ for any $x\in (0,1)$ and $f_n\equiv f$ for any $n\in \mathbb{N}$. Thus, $f_n⟹f$. We define $A\in \mathcal{F}\left(I\right)$ for any $x\in I$ by

$$A\left(x\right)=-x+1.$$

By calculation, we have $A_{\alpha }=[0,1-\alpha ]$ for any $\alpha \in (0,1]$ and $f^n\left(A_1\right)=\left\{0\right\}$ for any $n\in \mathbb{N}$. In the following, we assume that $\alpha \in (0,1)$ and let $f^n\left(A_{\alpha }\right)=[a_n\left(\alpha \right),b_n\left(\alpha \right)].$ Then, $a_n\left(\alpha \right)=0$ for any $n\in \mathbb{N}$ and $b_n\left(\alpha \right)\le b_{n+1}\left(\alpha \right)\le 1$ for any $n\in \mathbb{N}$ and $b_n\left(\alpha \right)⟶1.$ Since $b\left(\alpha \right)=1$ is not left continuous at $\alpha =1$, by Theorem 2.1 of [23], there is not a $B\in \mathcal{F}\left(I\right)$ such that $B_{\alpha }=[a\left(\alpha \right),b\left(\alpha \right)]=[0,1]$ for any $\alpha \in (0,1]$. Thus, $\omega (A,\widehat{f})=\mathsf{\varnothing }$.

4. Conclusions

In this paper, we investigated the $\omega$-limit sets of Zadeh’s extensions of a nonautonomous discrete system $f_n$ on an interval which converges uniformly to a map f and show that, if $P\left(f\right)=F\left(f\right)$, then $\omega (A,{\widehat{f}}_n)\subset F\left(\widehat{f}\right)$ for any $A\in \mathcal{F}\left(I\right)$.