#### Abstract

We introduce and study two properties of dynamical systems: topologically transitive and topologically mixing under the set-valued setting. We prove some implications of these two properties for set-valued functions and generalize some results from a single-valued case to a set-valued case. We also show that both properties of set-valued dynamical systems are equivalence for any compact intervals.

#### 1. Introduction

In dynamical systems, one of the most important research topics is to determine the chaotic behaviour of the system. Various definitions of chaos have been introduced by mathematicians in the past (see [1–5]) as there is no universally accepted definition of chaos. Definitions of chaos are constructed based on some topological properties. One of the commonly used properties is topologically transitive. The concept of topologically transitive was introduced by Birkhoff [6] in 1920. Dynamical systems with a topologically transitive property contain at least one point which moves under iteration from one arbitrary neighborhood to any other neighborhood. This property has been studied intensively by mathematicians since it is a global characteristic in the dynamical system. Some prefer to study topologically mixing of dynamical systems as it is a notion stronger than topologically transitive.

Numerous studies related to the transitivity and mixing properties of the dynamical systems especially in a one-dimensional system have been done; see [7–12]. As we all know, in general, dynamical systems are studied in the view of a single point. However, knowing how the points of the system move is not sufficient as there are problems and applications that require one to know how the subsets of the system move. In recent years, several works and research on the topological dynamics of set-valued dynamical systems can be found (see [13–20]). However, there are many properties for the dynamics of set-valued dynamical systems yet to be discovered. Loranty and Pawlak [21] studied the connection between transitivity and a dense orbit for multifunction in generalized topological spaces. Information about topologically transitive for the single-valued dynamical systems can be found in [22–25].

In this paper, we will introduce and study the notion of topologically transitive and topologically mixing for set-valued functions. We prove some elementary results of these two properties. Some of the results are generalization from the single-valued case (e.g., [11, 26–28]). We also prove that the definitions of these two properties for a set-valued function on compact intervals are equivalence. This paper is organized as follows. In Section 2, we give some background settings and define topologically transitive and topologically mixing for set-valued functions. In Section 3, we present some elementary implication results of topologically transitive and topologically mixing. In Section 4, we prove the equivalence of topologically transitive and topologically mixing of set-valued function on an arbitrary compact interval. In Section 5, we present some conclusions.

#### 2. Preliminaries

Let be a compact metric space. We denote as the collection of all nonempty closed subsets of . We call a function as set-valued function. If , then . is said to be upper semicontinuous at if for any open subset of containing there is an open subset of containing such that for every , . is upper semicontinuous if it is upper semicontinuous at every point of . Throughout the paper, we assume the set-valued function is upper semicontinuous unless explicitly stated.

Since is compact, by [29] the hyperspace is compact. Therefore, every element of is a nonempty compact subset of since they are nonempty closed subsets of . The pair is called as set-valued dynamical system. is denoted as the identity on and for all integers . The inverse set-valued function is defined as below.

*Definition 1 [30]. *Let be a set-valued function; then, the inverse set-valued function is defined by for all .

Recall that in the single-valued dynamical system where represent a continuous function, for any point , we define the orbit of under as where and for all integers . The point is said to be a periodic point of with period provided and for all integers . If the point has period , then, it is called as a fixed point. We extend these definitions to a set-valued case.

*Definition 2 [31]. *Let be a set-valued dynamical system. For any point , an orbit of is a sequence such that and for all integers . The collection of all orbits of is called as the complete orbit of , denoted by .

*Definition 3 [30, 31]. *For a set-valued dynamical system , let and let be an orbit of . The orbit is said to be a periodic orbit if there exists such that for all integers . The point is a periodic point if it has at least one periodic orbit. The period of is the smallest number satisfying for all integers . If , then, is said to be a fixed point.

In Definitions 2 and 3, we can see that in set-valued dynamical systems, the orbits of under is no longer uniquely determined. The following example shows that the orbit is not necessarily periodic even if there exists such that .

*Example 4. *Let and let the set-valued function defined by and . Let ; then, one of the orbit of under is . We can see that but for some . Therefore, the orbit is not a periodic orbit of .

Next, we define topologically transitive of set-valued functions. For topologically mixing of set-valued functions, we adopt the definition which has been defined by [31]. Note that the product set-valued function is defined by for all .

*Definition 5. *Let be a set-valued dynamical system. The set-valued function is topologically transitive if for any nonempty open subsets and of , there exists and with an orbit such that .

*Definition 6 [31]. *A set-valued function is topologically mixing if for any nonempty open sets and in , there is an such that for any , there is an with an orbit such that .

*Definition 7. *Let be a set-valued function of a compact metric space . Then is said to be
(1)topologically bitransitive if is topologically transitive(2)totally transitive if is topologically transitive for all (3)topologically weakly mixing if the product set-valued function is topologically transitiveWe end this section by recalling some basic concepts from topology. The interior of denoted by is the union of all open subsets of . The closure of denoted by is the intersection of all closed subsets of containing . A set is said to be dense in if . In other words, we can say that is dense in if every open subset of contains at least a point of (see [32, 33]).

#### 3. Topologically Transitive and Mixing of Set-Valued Functions

In a single-valued case, there are two commonly used definitions for topologically transitive: one is defined by using open sets and another one is defined by using points with a dense orbit (see [25]). Block [26] showed that both definitions coincide when the space is compact. But in general, both characterizations of transitivity are not equivalent as shown in [34, 35]. On a compact metric space with a set-valued function, we show that if there is a point with a dense orbit, then it will imply the transitivity of the set-valued function.

Proposition 8. *Let be a set-valued function. If there exists at least a point with an orbit such that the orbit is dense in , then, is topologically transitive.*

*Proof 9. *Let and be open sets in . We need to find a point and a positive integer such that has an orbit with . By hypothesis, there is a point in with an orbit that is dense in . Let be such a point with a dense orbit , then, we know that there exists a positive integer such that . Now, we try to show that there is a positive integer such that . Then, the proof is done by letting , so we have with an orbit where .

Since the orbit of is dense, contains at least one iterate of . Suppose that there are only finitely many iterates of in . Let be any element of such that is not an iterate of and let where is the metric on . We have and the neighborhood does not contain any iterates of . This implies that the orbit of is not dense in , a contradiction. Therefore, must contain infinitely many iterates of . Since there are only finitely many positive integers less than , there exists an integer such that . We let and the proof is complete.

With Proposition 8, we obtain the following theorem.

Theorem 9. *A set-valued function is topologically transitive if and only if there exists at least a point with an orbit such that the orbit is dense in .*

*Proof 10. *Clearly by the definition of topologically transitive, we will have at least a point with an orbit such that the orbit is dense in . For the converse part, we have proved in Proposition 8.

Similar to the single-valued case (see [36]), it is easy to see that by the definition, if the set-valued function is topologically mixing, then it implies that is topologically transitive. We show that the converse is not true in the following example.

*Example 11. *Let us consider the unit circle and let be an irrational rotation of . We define a set-valued function of by . Since all the points of have dense orbits, for an open subset of , it will intersect with other open subset of for some iterates under . Therefore, is topologically transitive. But is an irrational rotation, so there exists at least one further iterate of under that did not intersect with . Hence, is not topologically mixing.

Next, we discuss some connections between Definitions 5, 6, and 7 in Section 2.

Proposition 12. *Let be a set-valued dynamical system. If the set-valued function is topologically mixing, then is topologically weakly mixing.*

*Proof 13. *Assume that is topologically mixing. Let be any two nonempty open sets in . There exist nonempty open sets in such that and . Since is topologically mixing, there exists such that for all , there is a with an orbit such that . Similarly, there exists such that for all , there is a with an orbit such that . Let . Then, for all , there exists with an orbit such that . Hence, we conclude that is topologically weakly mixing.

Proposition 14. *Let be a set-valued dynamical system. If the set-valued function is topologically mixing, then is topologically bitransitive.*

*Proof 15. *Assume that is topologically mixing. Then, for any two nonempty open sets and of , there is a such that for any positive integer , there is a with an orbit such that . If we take to be an even positive integer greater than , i.e., where is a positive integer, then, there exists a with an orbit such that . This implies that is topologically transitive and hence is topologically bitransitive.

Lemma 16. *Let be a set-valued dynamical system, and the iterations of any open set are an open set of . If the set-valued function is topologically weakly mixing, then, the product set-valued dynamical system is topologically transitive for all integers .*

*Proof 17. *For all open sets , in , we define

Let , and be nonempty open sets in . Since is topologically weakly mixing, there exists a natural number such that there is a point with an orbit such that . This means that there is a point with orbit such that and there is a point with orbit such that . We can say that for any nonempty open sets in , .

Now, we are going to show that there exist nonempty open sets in such that . Let us define the open sets and as follow:
We have already shown that these sets are not empty. Let . This integer exists and satisfies the condition of there exists a with an orbit such that and . This means that there is an with an orbit such that , . Then, we can deduce that there is a with an orbit such that and there is a with an orbit such that . Therefore, we obtain and and this implies that . By using the principle of mathematical induction, we were able to see that for all nonempty open sets in , there exist nonempty open sets in such that
Hence, we conclude that is topologically transitive.

Theorem 18. *Let be a set-valued dynamical system and the iterations of any open set is an open set of . If the set-valued function is topologically weakly mixing, then, is totally transitive.*

*Proof 19. *Assume that is topologically weakly mixing. Let be an arbitrary fixed positive integer and be nonempty open sets in . We define two open sets in as follows:
By Lemma 16, is topologically transitive. So, there exists a positive integer such that there is a with an orbit such that
This implies that for all , there is a with an orbit such that and and there is a with an orbit such that and . Since and for all , there exists such that and such that for all . So, we can write that for all , there is a with an orbit such that and and there is a with an orbit such that and .

Now, we choose an such that where is a positive integer. We deduce that there is a point with an orbit under such that . This shows that is topologically weakly mixing which implies that is topologically transitive. Since is chosen arbitrary, is topologically transitive for all , and hence, is totally transitive.

Proposition 17. *Let be a set-valued dynamical system. If the set-valued function is totally transitive, then, is topologically bitransitive.*

*Proof 18. *Let be any two nonempty open sets of . Since is totally transitive, is topologically transitive for all positive integers . For each , there exists an such that there is an with an orbit such that . When we take , there exists an such that there is an with an orbit such that . This implies that is topologically transitive, hence is topologically bitransitive.

The implications between the various conditions of topologically transitive and mixing in this section are summarized as follows:

#### 4. Transitivity and Mixing Properties of Set-Valued Functions on Compact Intervals

In this section, we investigate the properties of topologically transitive and topologically mixing of set-valued functions on arbitrary compact interval , where such that . Note that for a single-valued case, the set of periodic points is dense in if the function is topologically transitive [26]. First, we present a lemma that will be used in the following proposition.

Lemma 19. *Let be a set-valued dynamical system and let be a subinterval of which contains no periodic point of . Suppose that, has an orbit such that for some integers and has an orbit such that for some integers . If , then , and if , then, .*

*Proof 20. *Without loss of generality, suppose that . Let ; then, the subinterval of contains no periodic point of . If for some , then, as there is no fixed point of inside the interval . Clearly by mathematical induction, for all . So, in particular, we have .

Assume that . With similar argument as the above (the order of inequality is reversed), we yield . Consequently, there exists a point lying in between and such that . This leads to a contradiction as contains no periodic point of . Therefore, we conclude that .

Proposition 21. *If the set-valued function is topologically transitive, then, the set of periodic points of is dense in .*

*Proof 22. *We will prove this by contradiction. Suppose that the set of periodic points of is not dense in . There exist two points where such that the open interval contains no any periodic points of . Since is topologically transitive, by Theorem 9, there is a point with a dense orbit in . Hence, there exist integers and such that and . Let . Since , has an orbit such that . Then, we obtain the following inequality:
But this is impossible as Lemma 19 is applied to the open interval . Thus, we conclude that the set of periodic points of is dense in .

Next, we prove an elementary property of topologically mixing for set-valued function on compact interval. Recall that an interval is degenerate [28] if it is either empty or reduced to a single point and it is nondegenerate otherwise. We refer the notation from [28] to denote as the smallest interval containing , that is, if and if .

Proposition 23. *Let and be a set-valued function. Then, is topologically mixing if and only if for all nondegenerate subintervals and any pair , there exists a positive integer such that for all .*

*Proof 24. *Suppose that is topologically mixing. Let and . If is a nonempty open subinterval of , then, there exists such that there is an with an orbit such that for all . Similarly, there exists such that there is a with an orbit such that for all . Let . Then, we have an with an orbit such that and we have a with an orbit such that for all . This implies that by connectedness of . If is a nondegenerate subinterval, the same result holds by considering the nonempty open interval .

Conversely, suppose that for any pair and a nondegenerate subinterval , there is a positive integer such that for all . Let be two nonempty open subsets of . Choose two nonempty open subintervals such that and neither nor is an endpoint of . There exists a pair such that . By assumption, there exists a positive integer such that for all . Therefore, we have for all and this implies that there is an with an orbit such that for all . We conclude that is topologically mixing.

When topologically mixing is replaced with topologically bitransitive for set-valued functions, a weaker version of Proposition 23 can be obtained with the help of the following lemma.

Lemma 25. *Let be a set-valued function. Let be a subinterval of which contains a fixed point of and a periodic point of with a least period . Let represent the iterates of in the periodic orbit where . Then, one of the following holds:
*(1)* for all *(2)* is even, sets and lie on the opposite sides of , and for all *

*Proof 26. *Let and be two points in the periodic orbits of such that where and both are contained in the same periodic orbit of . Let such that and . If both point and lie on the same side of the fixed point for some , then, we have four possible cases:
For case 1, we have . For case 4, we have . For cases 2 and 3, the set contains the compact interval. Thus, we have . In both cases, since , we obtain and, since , we have for all .

Otherwise, if and lie on the opposite sides of for all , clearly is an even integer. Let be the even integer from the set such that is the point which stays on the same side of as and lies most far from . Then, contains the set for all . Therefore, we conclude that for all .

Theorem 27. *Let and be a set-valued function. If is topologically bitransitive, then, for any nondegenerate subinterval and any pair such that , there exists a positive integer such that for all .*

*Proof 28. *Let be a fixed point of and be any nondegenerate subinterval of . Without the loss of generality, we may assume that for all . Since is topologically bitransitive, it means that is topologically transitive and by Theorem 9, there exists a point with an orbit with respect to which is dense in .

Let be a compact subinterval in where and . Since the orbit with respect to is dense in , for some positive integers , we obtain
By Proposition 21, we know that the set of periodic points is dense in . Since , there exists a periodic point with period which is close to point and the set contains all points of the periodic orbit of which is contained in . Then, we have
Consequently, there is a positive integer where the interval contains the fixed point and the periodic point with period . For the orbit , we can see that the even iterates are distributed on both sides of . Hence, by Lemma 25, we have for all . Therefore, for all and the proof is complete.

The following theorem gives an overview on the relation between topologically transitive and topologically mixing for set-valued functions of compact intervals. In fact, for compact intervals, both definitions are equivalent, which is similar to the results in the single-valued case (see [26]).

Theorem 29. *Let and be a set-valued function. If is topologically transitive, then the following statements are equivalent:
*(1)* is topologically bitransitive*(2)* is totally transitive*(3)* is topologically weakly mixing*(4)* is topologically mixing*(5)*For any nondegenerate subinterval and any pair such that , there exists a positive integer such that for all *

*Proof 30. *. It follows from Theorem 27 that . By Proposition 23, we have for the case . Finally, by the implication diagram at the end of Section 3, we have .

#### 5. Conclusion

In this paper, we studied the two properties of dynamical systems which are topologically transitive and topologically mixing under the setting of a set-valued case. An implication diagram to show the connection between various conditions of transitivity and mixing is provided in Section 3. We also investigated transitivity and mixing properties of set-valued functions for compact intervals and showed that both definitions are equivalent. Some results are similar to a single-valued case, but it will serve as a stepping stone for future research in a set-valued case.

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#### Conflicts of Interest

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