# Topologically Transitive and Mixing Properties of Set-Valued Dynamical Systems

Authors: Koon Sang Wong; Zabidin Salleh
Journal: Abstract and Applied Analysis
 (2021)

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## 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.


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## Body

## 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

LetX be a compact metric space. We denote 2X as the collection of all nonempty closed subsets of X. We call a function F:X⟶2X as set-valued function. If A⊆X, then FA=y∈X:thereisapointx∈Asuchthaty∈Fx. F is said to be upper semicontinuous at x∈X if for any open subset V of X containing Fx there is an open subset U of X containing x such that for every t∈U, Ft⊆V. F is upper semicontinuous if it is upper semicontinuous at every point of X. Throughout the paper, we assume the set-valued function F is upper semicontinuous unless explicitly stated.SinceX is compact, by [29] the hyperspace 2X is compact. Therefore, every element of 2X is a nonempty compact subset of X since they are nonempty closed subsets of X. The pair X,F is called as set-valued dynamical system. F0 is denoted as the identity on X and Fn=F∘Fn−1 for all integers n>0. The inverse set-valued function is defined as below.Definition 1 [30].
LetF:X⟶2X be a set-valued function; then, the inverse set-valued function F−1:X⟶2X is defined by F−1y=x∈X:y∈Fx for all y∈X.
Recall that in the single-valued dynamical systemX,f where f:X⟶X represent a continuous function, for any point x∈X, we define the orbit of x under f as xii=0∞ where x0=x and xi+1=fxi for all integers i≥0. The point x∈X is said to be a periodic point of f with period n provided fnx=x and fjx≠x for all integers 0<j<n. If the point x has period n=1, then, it is called as a fixed point. We extend these definitions to a set-valued case.Definition 2 [31].
LetX,F be a set-valued dynamical system. For any point x∈X, an orbit of x is a sequence xii=0∞ such that x0=x and xi+1∈Fxi for all integers i≥0. The collection of all orbits of x is called as the complete orbit of x, denoted by COx.Definition 3 [30, 31].
For a set-valued dynamical systemX,F, let x∈X and let xii=0∞ be an orbit of x. The orbit is said to be a periodic orbit if there exists m∈ℕ such that xi=xi+m for all integers i≥0. The point x is a periodic point if it has at least one periodic orbit. The period of x is the smallest number m∈ℕ satisfying xi=xi+m for all integers i≥0. If m=1, then, x is said to be a fixed point.In Definitions2 and 3, we can see that in set-valued dynamical systems, the orbits of x under F is no longer uniquely determined. The following example shows that the orbit xii=0∞ is not necessarily periodic even if there exists j∈ℕ such that x=x0=xj.Example 4.
LetX=0,1 and let the set-valued function F:X⟶2X defined by F1=0,1 and F0=1. Let x=0; then, one of the orbit of x under F is 0,1,1,0,1,0,1,⋯. We can see that x=x3 but xi≠xi+3 for some i>0. Therefore, the orbit 0,1,1,0,1,0,1,⋯ is not a periodic orbit of x.
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 F×F:X×X⟶2X×X is defined by F×Fx,x′=y,y′∈X×X:y∈Fxandy′∈Fx′ for all x,x′∈X.Definition 5.
LetX,F be a set-valued dynamical system. The set-valued function F is topologically transitive if for any nonempty open subsets U and V of X, there exists m∈ℕ and x∈U with an orbit xii=0∞ such that xm∈V.Definition 6 [31].
A set-valued functionF is topologically mixing if for any nonempty open sets U and V in X, there is an M∈ℕ such that for any m>M, there is an x0∈U with an orbit xii=0∞ such that xm∈V.Definition 7.
LetF be a set-valued function of a compact metric space X. Then F is said to be
(1)
topologically bitransitive ifF2 is topologically transitive(2)
totally transitive ifFn is topologically transitive for all n∈ℕ(3)
topologically weakly mixing if the product set-valued functionF×F is topologically transitive
We end this section by recalling some basic concepts from topology. The interior ofU denoted by intU is the union of all open subsets of U. The closure of U denoted by clU is the intersection of all closed subsets of X containing U. A set U is said to be dense in X if clU=X. In other words, we can say that U is dense in X if every open subset of X contains at least a point of U (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.
LetF:X⟶2X be a set-valued function. If there exists at least a point x∈X with an orbit xii=0∞∈COx such that the orbit is dense in X, then, F is topologically transitive.Proof 9.
LetU and V be open sets in X. We need to find a point x∈U and a positive integer m such that x has an orbit xii=0∞ with xm∈V. By hypothesis, there is a point in X with an orbit that is dense in X. Let y∈X be such a point with a dense orbit yii=0∞, then, we know that there exists a positive integer k such that yk∈U. Now, we try to show that there is a positive integer m such that yk+m∈V. Then, the proof is done by letting x=yk, so we have x∈U with an orbit xii=0∞ where xm=yk+m∈V.
Since the orbityii=0∞ of y is dense, V contains at least one iterate of y. Suppose that there are only finitely many iterates of y in V. Let z be any element of V such that z is not an iterate of y and let ε=mindz,yj:j=0,1,2,3,⋯ where d is the metric on X. We have ε>0 and the neighborhood Nε/2z does not contain any iterates of y. This implies that the orbit yii=0∞ of y is not dense in X, a contradiction. Therefore, V must contain infinitely many iterates of y. Since there are only finitely many positive integers less than k, there exists an integer n>k such that yn∈V. We let m=n−k and the proof is complete.With Proposition8, we obtain the following theorem.Theorem 9.
A set-valued functionF:X⟶2X is topologically transitive if and only if there exists at least a point x∈X with an orbit xii=0∞∈COx such that the orbit is dense in X.Proof 10.
Clearly by the definition of topologically transitive, we will have at least a pointx∈X with an orbit xii=0∞∈COx such that the orbit is dense in X. 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 F is topologically mixing, then it implies that F is topologically transitive. We show that the converse is not true in the following example.Example 11.
Let us consider the unit circleS1=z∈ℂ:z=1 and let Rα:S1⟶S1 be an irrational rotation of S1. We define a set-valued function R of S1 by Rz=z,Rαz. Since all the points of S1 have dense orbits, for an open subset U of S1, it will intersect with other open subset V of S1 for some iterates under R. Therefore, R is topologically transitive. But Rα is an irrational rotation, so there exists at least one further iterate of U under R that did not intersect with V. Hence, R is not topologically mixing.
Next, we discuss some connections between Definitions5, 6, and 7 in Section 2.Proposition 12.
LetX,F be a set-valued dynamical system. If the set-valued function F is topologically mixing, then F is topologically weakly mixing.Proof 13.
Assume thatF is topologically mixing. Let W1,W2 be any two nonempty open sets in X×X. There exist nonempty open sets U1,U2,V1,andV2 in X such that U1×U2⊂W1 and V1×V2⊂W2. Since F is topologically mixing, there exists N1∈ℕ such that for all n≥N1, there is a x∈U1 with an orbit xii=0∞ such that xn∈V1. Similarly, there exists N2∈ℕ such that for all n≥N2, there is a y∈U2 with an orbit yii=0∞ such that yn∈V2. Let M=maxN1,N2. Then, for all n≥M, there exists x,y∈W1 with an orbit xi,yii=0∞ such that xn,yn∈W2. Hence, we conclude that F is topologically weakly mixing.Proposition 14.
LetX,F be a set-valued dynamical system. If the set-valued function F is topologically mixing, then F is topologically bitransitive.Proof 15.
Assume thatF is topologically mixing. Then, for any two nonempty open sets U and V of X, there is a M∈ℕ such that for any positive integer m>M, there is a x∈U with an orbit xii=0∞ such that xm∈V. If we take m to be an even positive integer greater than M, i.e., m=2n>M where n is a positive integer, then, there exists a x∈U with an orbit xii=0∞ such that x2n∈V. This implies that F2 is topologically transitive and hence F is topologically bitransitive.Lemma 16.
LetX,F be a set-valued dynamical system, and the iterations of any open set are an open set of X. If the set-valued function F is topologically weakly mixing, then, the product set-valued dynamical system Xn,F×⋯×F is topologically transitive for all integers n≥1.Proof 17.
For all open setsU, V in X, we defineNU,V=n≥1:thereexistsx∈Uwithanorbitxii=0∞suchthatxn∈V.
LetU1,U2,V1, and V2 be nonempty open sets in X. Since F is topologically weakly mixing, there exists a natural number m such that there is a point x,y∈U1×U2 with an orbit xi,yii=0∞ such that xm,ym∈V1×V2. This means that there is a point x∈U1 with orbit xii=0∞ such that xm∈V1 and there is a point y∈U2 with orbit yii=0∞ such that ym∈V2. We can say that for any G,H nonempty open sets in X, NG,H≠∅.
Now, we are going to show that there exist nonempty open setsU,V in X such that NU,V⊆NU1,V1∩NU2,V2. Let us define the open sets U=U1∩FnU2 and V=V1∩FnV2 as follow:
(1)U≔x:x∈U1withorbitxii=0∞suchthatx0=xandxn∈U2,V≔x:x∈V1withorbitxii=0∞suchthatx0=xandxn∈V2.
We have already shown that these sets are not empty. Letk∈NU,V. This integer exists and satisfies the condition of there exists a x∈U with an orbit xii=0∞ such that x0=x and xk∈V. This means that there is an x∈U with an orbit xii=0∞ such that x0=x∈U1, xn∈U2,xk∈V1andxk+n∈V2. Then, we can deduce that there is a x∈U1 with an orbit xii=0∞ such that xk∈V1 and there is a x∈U2 with an orbit xii=0∞ such that xk∈V2. Therefore, we obtain k∈NU1,V1 and k∈NU2,V2 and this implies that NU,V⊆NU1,V1∩NU2,V2. By using the principle of mathematical induction, we were able to see that for all nonempty open sets U1,⋯,Un,V1,⋯,Vn in X, there exist nonempty open sets U,V in X such that
(2)NU,V⊆NU1,V1∩NU2,V2∩⋯∩NUn,Vn.
Hence, we conclude thatXn,F×⋯×F is topologically transitive.Theorem 18.
LetX,F be a set-valued dynamical system and the iterations of any open set is an open set of X. If the set-valued function F is topologically weakly mixing, then, F is totally transitive.Proof 19.
Assume thatF is topologically weakly mixing. Let n≥1 be an arbitrary fixed positive integer and U,U′,V,V′ be nonempty open sets in X. We define two open sets W,W′ in X2n as follows:
(3)W=x0,x1,⋯,xn−1,y0,y1,⋯,yn−1:xi∈FiU,yi∈FiVforalli=0,1,⋯,n−1,W′=x0′,x1′,⋯,xn−1′,y0′,y1′,⋯,yn−1′:xi′∈U′,yi′∈V′foralli=0,1,⋯,n−1.
By Lemma16, X2n,F×F×⋯×F is topologically transitive. So, there exists a positive integer m such that there is a x0,x1,⋯,xn−1,y0,y1,⋯,yn−1∈W with an orbit x0j,x1j,⋯,xn−1j,y0j,y1j,⋯,yn−1jj=0∞ such that
(4)x0,x1,⋯,xn−1,y0,y1,⋯,yn−1=x00,x10,⋯,xn−10,y00,y10,⋯,yn−10∈W,x0m,x1m,⋯,xn−1m,y0m,y1m,⋯,yn−1m∈W′.
This implies that for alli∈0,1,⋯,n−1, there is a xi∈FiU with an orbit xijj=0∞ such that xi0∈FiU and xim∈U′ and there is a yi∈FiV with an orbit yijj=0∞ such that yi0∈FiVand xim∈V′. Since xi∈FiU and yi∈FiV for all i∈0,1,⋯,n−1, there exists u∈U such that xi∈Fiu and v∈V such that yi∈Fiv for all i∈0,1,⋯,n−1. So, we can write that for all i∈0,1,⋯,n−1, there is a u∈U with an orbit ujj=0∞ such that u=u0∈U and ui+m∈U′ and there is a v∈V with an orbit vjj=0∞ such that v0∈V and vi+m∈V′.
Now, we choose ani∈0,1,⋯,n−1 such that m+i=np where p is a positive integer. We deduce that there is a point u,v∈U×V with an orbit uj,vjj=0∞ under Fn such that unp,vnp∈U′×V′. This shows that Fn is topologically weakly mixing which implies that Fn is topologically transitive. Since n is chosen arbitrary, Fn is topologically transitive for all n≥1, and hence, F is totally transitive.Proposition 17.
LetX,F be a set-valued dynamical system. If the set-valued function F is totally transitive, then, F is topologically bitransitive.Proof 18.
LetU,V be any two nonempty open sets of X. Since F is totally transitive, Fn is topologically transitive for all positive integers n≥1. For each n≥1, there exists an m∈ℕ such that there is an x∈U with an orbit xii=0∞ such that xmn∈V. When we take n=2, there exists an m∈ℕ such that there is an x∈U with an orbit xii=0∞ such that x2m∈V. This implies that F2 is topologically transitive, hence F is topologically bitransitive.
The implications between the various conditions of topologically transitive and mixing in this section are summarized as follows:(5)mixing⇒weaklymixing⇒totalytransitive⇒bitransitive⇒transitive.


## 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 intervalI=a,b, where a,b∈ℝ such that a<b. Note that for a single-valued case, the set of periodic points is dense in I if the function is topologically transitive [26]. First, we present a lemma that will be used in the following proposition.Lemma 19.
LetI,F be a set-valued dynamical system and let J be a subinterval of I which contains no periodic point of F. Suppose that, x∈J has an orbit xii=0∞ such that xm∈J for some integers m>0 and y∈J has an orbit yii=0∞ such that yn∈J for some integers n>0. If x<xm, then y<yn, and if x>xm, then, y>yn.Proof 20.
Without loss of generality, suppose thatx<xm. Let G=Fm; then, the subinterval x,xm of J contains no periodic point of G. If xkm>x for some k≥1, then, xk+1m>xm as there is no fixed point of Gk inside the interval x,xm. Clearly by mathematical induction, xkm>x for all k≥1. So, in particular, we have xmn>x.
Assume thaty>yn. With similar argument as the above (the order of inequality is reversed), we yield y>ymn. Consequently, there exists a point z∈J lying in between x and y such that zmn=z. This leads to a contradiction as J contains no periodic point of F. Therefore, we conclude that y<yn.Proposition 21.
If the set-valued functionF:I⟶2I is topologically transitive, then, the set of periodic points of F is dense in I.Proof 22.
We will prove this by contradiction. Suppose that the set of periodic points ofF is not dense in I. There exist two points x,y∈I where x<y such that the open interval x,y contains no any periodic points of F. Since F is topologically transitive, by Theorem 9, there is a point u∈x,y with a dense orbit uii=0∞ in I. Hence, there exist integers m>0 and 0<p<q such that u<um<y and x<uq<up<u. Let v=up. Since uq∈Fq−pv, v has an orbit vii=0∞ such that vq−p=uq. Then, we obtain the following inequality:
(6)x<vq−p<v<u<um<y.
But this is impossible as Lemma19 is applied to the open interval x,y. Thus, we conclude that the set of periodic points of F is dense in I.
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 a,b as the smallest interval containing a,b, that is, a,b=a,b if a≤b and a,b=b,a if a≥b.Proposition 23.
LetI=a,b and F:I⟶2I be a set-valued function. Then, F is topologically mixing if and only if for all nondegenerate subintervals J⊆I and any pair c,d∈intI, there exists a positive integer M such that c,d⊂FnJ for all n≥M.Proof 24.
Suppose thatF is topologically mixing. Let U=a,c and V=d,b. If J is a nonempty open subinterval of I, then, there exists N1∈ℕ such that there is an x∈J with an orbit xii=0∞ such that xn∈U for all n≥N1. Similarly, there exists N2∈ℕ such that there is a y∈J with an orbit yii=0∞ such that yn∈V for all n≥N2. Let M=maxN1,N2. Then, we have an x∈J with an orbit xii=0∞ such that xn∈U and we have a y∈J with an orbit yii=0∞ such that yn∈V for all n≥M. This implies that c,d⊂FnJ by connectedness of c,d. If J is a nondegenerate subinterval, the same result holds by considering the nonempty open interval intJ.
Conversely, suppose that for any pairc,d∈intI and a nondegenerate subinterval J⊆I, there is a positive integer M such that c,d⊂FnJ for all n≥M. Let U,V be two nonempty open subsets of I. Choose two nonempty open subintervals K,L such that K⊂U,L⊂V and neither a nor b is an endpoint of L. There exists a pair c,d∈intI such that L⊂c,d. By assumption, there exists a positive integer M such that c,d⊂FnK for all n≥M. Therefore, we have L⊂c,d⊂FnK for all n≥M and this implies that there is an x∈U with an orbit xii=0∞ such that xn∈V for all n≥M. We conclude that F is topologically mixing.
When topologically mixing is replaced with topologically bitransitive for set-valued functions, a weaker version of Proposition23 can be obtained with the help of the following lemma.Lemma 25.
LetF:I⟶2I be a set-valued function. Let J be a subinterval of I which contains a fixed point p of F and a periodic point q of F with a least period m≥3. Let qi represent the iterates of q in the periodic orbit where i∈0,1,⋯,m−1. Then, one of the following holds:
(1)
FnJ⊃q,q1,⋯,qm−1 for all n≥2m−2(2)
m is even, sets q,q2,⋯,qm−2 and q1,q3,⋯,qm−1 lie on the opposite sides of p, and F2nJ⊃q,q2,⋯,qm−2 for all n≥m/2−1Proof 26.
Letu and v be two points in the periodic orbits of q such that v′≤u<v≤u′ where v′∈Fv,u′∈Fu and both u′,v′ are contained in the same periodic orbit of q. Let i,j∈0,1,⋯,m−1 such that u=qi and v=qj. If both point qk and qk+1 lie on the same side of the fixed point p for some k∈0,1,⋯,m−1, then, we have four possible cases:
(7)qi=u<v<p,qk<u<p<v,u<p<v<qk,p<u<v=qj.
For case 1, we haveFiJ⊃u,v. For case 4, we have FjJ⊃u,v. For cases 2 and 3, the set FkJ contains the compact intervalp,qk. Thus, we have Fk+1J=FFkJ⊃u,v. In both cases, since Fu,v⊃u,v, we obtain FmJ⊃u,v and, since Fm−2u,v⊃q,q1,⋯,qm−1, we have FnJ⊃q,q1,⋯,qm−1 for all n≥2m−2.
Otherwise, ifqi and qi+1 lie on the opposite sides of p for all i∈0,1,⋯,m−1, clearly m is an even integer. Let r be the even integer from the set 0,1,⋯,m−1 such that qr is the point which stays on the same side of p as q and lies most far from p. Then, Fr+2jJ contains the set qm=q,q2,q4,⋯,qm−2 for all j≥0. Therefore, we conclude that F2nJ⊃q,q2,⋯,qm−2 for all n≥m/2−1.Theorem 27.
LetI=a,b and F:I⟶2I be a set-valued function. If F is topologically bitransitive, then, for any nondegenerate subinterval J⊆I and any pair c,d∈intI such that a<c<d<b, there exists a positive integer M such that c,d⊂FnJ for all n≥M.Proof 28.
Letz be a fixed point of F and J be any nondegenerate subinterval of I. Without the loss of generality, we may assume that z<x for all x∈J. Since F is topologically bitransitive, it means that F2 is topologically transitive and by Theorem 9, there exists a point u∈intJ with an orbit uii=0∞ with respect to F2 which is dense in I.
LetL=c,d be a compact subinterval in I where c,d∈intI and c<d. Since the orbit uii=0∞ with respect to F2 is dense in I, for some positive integers n1<n2, we obtain
(8)un1<minz,minL<maxz,maxL<un2.
By Proposition21, we know that the set of periodic points is dense in I. Since u∈intJ, there exists a periodic point p∈J with period m which is close to point u and the set contains all points of the periodic orbit of p which is contained in intI. Then, we have
(9)un1≈pn1<minz,minL<maxz,maxL<pn2≈un2.
Consequently, there is a positive integers where the interval FsJ contains the fixed point z and the periodic point ps with period m. For the orbit pii=0∞, we can see that the even iterates are distributed on both sides of z. Hence, by Lemma 25, we have FjFsJ⊃p,p1,⋯,pm−1 for all j≥2m−2. Therefore, FnJ⊃minp,p1,⋯,pm−1,maxp,p1,⋯,pm−1⊃L for all n≥s+2m−2 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.
LetI=a,b and F:I⟶2I be a set-valued function. If F is topologically transitive, then the following statements are equivalent:
(1)
F is topologically bitransitive(2)
F is totally transitive(3)
F is topologically weakly mixing(4)
F is topologically mixing(5)
For any nondegenerate subintervalJ⊆I and any pair c,d∈intI such that a<c<d<b, there exists a positive integer M such that c,d⊂FnJ for all n≥MProof 30.
. It follows from Theorem27 that 1⇒5. By Proposition 23, we have 5⇔4 for the case c<d. Finally, by the implication diagram at the end of Section 3, we have 4⇒3⇒2⇒1.


## 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 Section3. 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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