# Some Common Fixed Point Theorems in Partially Ordered Sets

Authors: Khadija Bouzkoura; Said Benkaddour
Journal: Journal of Applied Mathematics
 (2020)

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## Abstract

The purpose of this paper is to prove some new fixed point theorem and common fixed point theorems of a commuting family of order-preserving mappings defined on an ordered set, which unify and generalize some relevant fixed point theorems.


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## Body

## 1. Introduction and Preliminaries

The fixed points theory has experienced a great development in recent years, among enormous results which have enriched this theory, Tarski’s theorem on ordered sets is “any application which preserves order on a complete lattice admits a fixed point and the set of its fixed points is a complete lattice.”In our previous work, we have introduced the concept of complete T-lattice to show that a complete lattice is a special case of a complete T-lattice. So, the reader may wonder if we can get a generalization of Tarski’s theorem.In one of our articles [1, 2], we proved that any application that preserves the order on a complete T-lattice has a fixed point. In this work, we give a structure of the set of these fixed points and some other results.So, we begin by describing the relevant notation and terminology. LetE,≤ be a partially ordered set and M⊂E a nonempty subset. Recall that an upper (resp. lower) bound for M is an element p∈E with m≤p (resp. p≤m) for each m∈M; the least upper (resp. greatest-lower) bound of M will be denoted supM (resp. infM).Next, letE,≤ be a partially ordered set with the least element 0 and greatest element 1 and T be a given operator on E,≤ reversing the order such that x≤TxorTx≤x for all x∈E. We consider the following subsets Kr and Kl of E, Kr=x∈E,Tx≤x and Kl=x∈E,x≤Tx, so Kr and Kl are not empty (since T1≤1 and T1≤T21). In what follows, let us say that the set E,≤ a complete T-lattice if every subset A of E admits the greatest lower (resp. least upper) bound as soon as TA≺A (resp. A≺TA), where TA designates the image of A under the map T for A⊂E, and we denote by A≺B if for every a∈A, b∈B, we have a≤b for A and B be two subsets of E.


## 2. Complete Sub-T-Lattice and Fixed Point Theorems

### 2.1. Complete Sub-T-Lattice

Inspired by the success of the complete lattice subset concept introduced by Birkhoff [3] in 1940, we propose a similar concept in the partially ordered sets. The aim of the concept is to structure the set of fixed points of an increasing application on an ordered set.LetE,≤ be a partially ordered set and T be a given operator on E,≤ reversing the order such that x≤TxorTx≤x for all x∈E.Let us denote by≤A the restriction of the ordered relation of E on A with A a subset of E.Definition 1.
We say thatA,≤A is a complete sub-T-lattice of E if A,≤A admits a least and a greatest element and every party X of AinfAXresp.supAX exists as soon as TX≺X (resp. X≺TX):
(1)
Every finite subsetA of the set ℕ ordered by the usual order relation is a sub-T-lattice with Ta=0A for all a∈A with 0A is the least element of A(2)
We consider Figure1: We takeE=a0,a1,a2,a3,a4,a5;b0,b1,b2,b3and the following order relation ≤ defined by:a5≤ai≤a0for alli∈1,2,3,4,ai≤a1, andai≤a2fori=3,4. Moreover, we have b3≤b2≤b1≤b0 and a3≤b2≤a1≤b0 where Tai=a5 for i=0,2; Tai=b2 for i=1,3; and Tai=a0 for i=4,5. Tbi=b2 for i=0,1,2, Tb3=b0 Let A=a0,a1,a4,a5,a5, so (A,≤A) is a complete sub-T-latticeRecall. Let us note thatinfBX (resp. supBX) is equal to the greatest-lower (resp. least-upper) bound of X in B for every party X of B.Proposition 2.
LetA,≤A be a complete sub-T-lattice with 0A the least element and 1A the greatest element. Then, for every party X⊂Kr∩A (resp. X⊂Kl∩A) supAX (resp. infAX) exists in Kr(resp. in Kl).Proof.
LetX be a nonempty party of Kr∩A. Let us define the setM=y∈A:x≤yforallx∈X. We have M≠∅, since 1A∈M because Kr∩A≠∅ in this case 1A∈Kr, and for every m0,m1∈M, we have Tm0≤Tx≤x≤m0; thus, Tm0≤Tx≤x≤m1 for all x∈X which shows that TM≺M. Consequently, mg=infAM exists since A,≤A is a complete sub-T-lattice. As we have x≤mg for all x∈X, therefore Tmg≤Tx≤x≤mg which implies mg∈Kr. If X was a party of Kl∩A, the proof of existence of infEX in Kl would be symmetric.Figure 1
Example of a complete sub-T-lattice.By this proposition, we can easily prove that each chainC=ci of a complete sub-T-lattice has the least upper bound and the greatest lower bound; indeed, we assume that C=ci is an increasing chain if C∩Kr≠∅; then, the supC∩Kr=supC exists in Kr by Proposition 2, but if C∩Kr=∅, we have ci≤cj≤Tcj for all j≥i which shows that C≺TC and since C=ci is a chain of a complete sub-T-lattice then supC exists. If C is a decreasing sequence, the proof of existence of infC is symmetric.Recall that a chainM of E is maximal if there is an element a of E comparable to every element of M then a∈M.In the following, we define←,m=x∈E,x≤m and the m,→=x∈E,m≤x for any m∈E called intervals.IfE,≤ is a complete T-lattice; then, E,≤ is a complete sub-T-lattice, and we can show that each interval of E is a complete sub-T-lattice, indeed. We have for all x∈m,→, m≤x≤1E which implies that m is the least element and 1E is the greatest element of m,→.LetB⊂m,→ be a nonempty set such that B≺TB. Since E is a complete lattice, so b0=supEB exists in E. We will only prove that supEB exists in m,→. Then, we have for all b∈B, m≤b≤b0 which gives b0∈m,→. Consequently, supEB∈m,→. The proof for the existence of the greatest-lower if TB≺B follows identically.In the same fashion, we can prove thata,b=x∈X:a≤x≤b is a complete sub-T-lattice.Proposition 3.
In a complete sub-T-lattice, any decreasing family of nonempty intervals has a nonempty intersection and it is an interval.Proof.
LetA,≤A be a complete sub-T-lattice and ai,bii∈I a decreasing family of nonempty intervals in E, so we have ai≤ai+1 and bi+1≤bi. Consequently, aii∈I is an increasing sequence, and bii∈I is a decreasing sequence; furthermore, we have ai≤bj for all i,j∈I; indeed, ai≤aj≤bj if j≥i and ai≤bi≤bj if i≥j. Therefore, since A is a complete T-lattice which shows that Supi∈Iai=a and infi∈Ibi=b exist in A.For alli,j∈I, we have ai≤a≤b≤bj. Consequently,⋂i∈Iai,bi=a,b.


### 2.2. Fixed Point Theorems

The theory of fixed points is concerned with the conditions which guarantee that a mapF:E→E of a set E into itself admits one or more fixed points, that there are pointsx∈Efor whichFx=x.Now, letE,≤ be an ordered set and T be a given operator on E,≤ reversing the order such that x≤Tx or Tx≤x for all x∈E.We take a partA of E and a monotone maps f:A→A; we note the set of fixed points of f by δf.Theorem 4.
IfA,≤A is a complete sub-T-lattice with the least element 0A and the greatest element 1A, then, δf is a nonempty complete sub-T-lattice.Indeed, we have0A≤f0A; this shows that Δ=x∈A:x≤fx is a nonempty set. Furthermore, we also have for any x∈Δ,fx∈Δ and for any chain C of Δ, we put s=supAC; for all x∈C, we have x≤s apply the mapping f, x≤fx≤fs that gives s≤fs so s∈Δ. Therefore, Δ satisfies the assumptions of Zorn’s lemma. Hence, for every x∈Δ, there exists a maximal element xm∈Δ such that x≤xm. We claim that if xm is a maximal element of Δ, then xm≤fxm which gives us fxm≤f2xm which implies that fxm∈Δ. According to the maximality of xm, we have xm=fxm which proves that δf≠∅.To show thatδf is a complete sub-T-lattice, let S=x∈A:x≤fx≤x∗forallx∗∈δf. It is clear that S≠∅ since 0A∈S. For any chain xnn of S, supAxnn=c exists in A, and we have xn≤x∗ for all x∗∈δ which implies xn≤c≤x∗. We applyfand we obtainxn≤fxn≤fc≤fx∗=x∗; hence,c≤fc≤x∗that provesc∈S. Consequently, we have the existence of a maximal element of S according to Zorn’s lemma. We claim that if xm is a maximal element of S, then xm≤fxm≤x∗ for all x∗∈δ. In the same fashion, we applyfthat givesxm≤fxm≤f2xm≤x∗for allx∗∈δthat impliesfxm∈S. It is easy from here to see that in fact we have xm=fxm by the maximality of xm and xm=fxm≤x∗ for all x∗∈δ that shows that xm is the least element of δf. The proof of existence of the greatest element of δf is symmetrical.Now, letB⊂δf be a nonempty set such that TB≺B. We want to find infδfB, the greatest-lower bound of B in δf. Let us put e=infAB. For all b∈B, e≤b, as fe≤b=fb. Thus fe≤e. Therefore, the decreasing sequence fnen tends to a fixed point of f which will be the greatest lower bounds (or the greatest-lower bound) of B in δf. The construction of supδfB is the least upper bound of B in δf if B≺TB is symmetric. Thereafter, δf,≤ is a complete sub-T-lattice.


## 2.1. Complete Sub-T-Lattice

Inspired by the success of the complete lattice subset concept introduced by Birkhoff [3] in 1940, we propose a similar concept in the partially ordered sets. The aim of the concept is to structure the set of fixed points of an increasing application on an ordered set.LetE,≤ be a partially ordered set and T be a given operator on E,≤ reversing the order such that x≤TxorTx≤x for all x∈E.Let us denote by≤A the restriction of the ordered relation of E on A with A a subset of E.Definition 1.
We say thatA,≤A is a complete sub-T-lattice of E if A,≤A admits a least and a greatest element and every party X of AinfAXresp.supAX exists as soon as TX≺X (resp. X≺TX):
(1)
Every finite subsetA of the set ℕ ordered by the usual order relation is a sub-T-lattice with Ta=0A for all a∈A with 0A is the least element of A(2)
We consider Figure1: We takeE=a0,a1,a2,a3,a4,a5;b0,b1,b2,b3and the following order relation ≤ defined by:a5≤ai≤a0for alli∈1,2,3,4,ai≤a1, andai≤a2fori=3,4. Moreover, we have b3≤b2≤b1≤b0 and a3≤b2≤a1≤b0 where Tai=a5 for i=0,2; Tai=b2 for i=1,3; and Tai=a0 for i=4,5. Tbi=b2 for i=0,1,2, Tb3=b0 Let A=a0,a1,a4,a5,a5, so (A,≤A) is a complete sub-T-latticeRecall. Let us note thatinfBX (resp. supBX) is equal to the greatest-lower (resp. least-upper) bound of X in B for every party X of B.Proposition 2.
LetA,≤A be a complete sub-T-lattice with 0A the least element and 1A the greatest element. Then, for every party X⊂Kr∩A (resp. X⊂Kl∩A) supAX (resp. infAX) exists in Kr(resp. in Kl).Proof.
LetX be a nonempty party of Kr∩A. Let us define the setM=y∈A:x≤yforallx∈X. We have M≠∅, since 1A∈M because Kr∩A≠∅ in this case 1A∈Kr, and for every m0,m1∈M, we have Tm0≤Tx≤x≤m0; thus, Tm0≤Tx≤x≤m1 for all x∈X which shows that TM≺M. Consequently, mg=infAM exists since A,≤A is a complete sub-T-lattice. As we have x≤mg for all x∈X, therefore Tmg≤Tx≤x≤mg which implies mg∈Kr. If X was a party of Kl∩A, the proof of existence of infEX in Kl would be symmetric.Figure 1
Example of a complete sub-T-lattice.By this proposition, we can easily prove that each chainC=ci of a complete sub-T-lattice has the least upper bound and the greatest lower bound; indeed, we assume that C=ci is an increasing chain if C∩Kr≠∅; then, the supC∩Kr=supC exists in Kr by Proposition 2, but if C∩Kr=∅, we have ci≤cj≤Tcj for all j≥i which shows that C≺TC and since C=ci is a chain of a complete sub-T-lattice then supC exists. If C is a decreasing sequence, the proof of existence of infC is symmetric.Recall that a chainM of E is maximal if there is an element a of E comparable to every element of M then a∈M.In the following, we define←,m=x∈E,x≤m and the m,→=x∈E,m≤x for any m∈E called intervals.IfE,≤ is a complete T-lattice; then, E,≤ is a complete sub-T-lattice, and we can show that each interval of E is a complete sub-T-lattice, indeed. We have for all x∈m,→, m≤x≤1E which implies that m is the least element and 1E is the greatest element of m,→.LetB⊂m,→ be a nonempty set such that B≺TB. Since E is a complete lattice, so b0=supEB exists in E. We will only prove that supEB exists in m,→. Then, we have for all b∈B, m≤b≤b0 which gives b0∈m,→. Consequently, supEB∈m,→. The proof for the existence of the greatest-lower if TB≺B follows identically.In the same fashion, we can prove thata,b=x∈X:a≤x≤b is a complete sub-T-lattice.Proposition 3.
In a complete sub-T-lattice, any decreasing family of nonempty intervals has a nonempty intersection and it is an interval.Proof.
LetA,≤A be a complete sub-T-lattice and ai,bii∈I a decreasing family of nonempty intervals in E, so we have ai≤ai+1 and bi+1≤bi. Consequently, aii∈I is an increasing sequence, and bii∈I is a decreasing sequence; furthermore, we have ai≤bj for all i,j∈I; indeed, ai≤aj≤bj if j≥i and ai≤bi≤bj if i≥j. Therefore, since A is a complete T-lattice which shows that Supi∈Iai=a and infi∈Ibi=b exist in A.For alli,j∈I, we have ai≤a≤b≤bj. Consequently,⋂i∈Iai,bi=a,b.


## 2.2. Fixed Point Theorems

The theory of fixed points is concerned with the conditions which guarantee that a mapF:E→E of a set E into itself admits one or more fixed points, that there are pointsx∈Efor whichFx=x.Now, letE,≤ be an ordered set and T be a given operator on E,≤ reversing the order such that x≤Tx or Tx≤x for all x∈E.We take a partA of E and a monotone maps f:A→A; we note the set of fixed points of f by δf.Theorem 4.
IfA,≤A is a complete sub-T-lattice with the least element 0A and the greatest element 1A, then, δf is a nonempty complete sub-T-lattice.Indeed, we have0A≤f0A; this shows that Δ=x∈A:x≤fx is a nonempty set. Furthermore, we also have for any x∈Δ,fx∈Δ and for any chain C of Δ, we put s=supAC; for all x∈C, we have x≤s apply the mapping f, x≤fx≤fs that gives s≤fs so s∈Δ. Therefore, Δ satisfies the assumptions of Zorn’s lemma. Hence, for every x∈Δ, there exists a maximal element xm∈Δ such that x≤xm. We claim that if xm is a maximal element of Δ, then xm≤fxm which gives us fxm≤f2xm which implies that fxm∈Δ. According to the maximality of xm, we have xm=fxm which proves that δf≠∅.To show thatδf is a complete sub-T-lattice, let S=x∈A:x≤fx≤x∗forallx∗∈δf. It is clear that S≠∅ since 0A∈S. For any chain xnn of S, supAxnn=c exists in A, and we have xn≤x∗ for all x∗∈δ which implies xn≤c≤x∗. We applyfand we obtainxn≤fxn≤fc≤fx∗=x∗; hence,c≤fc≤x∗that provesc∈S. Consequently, we have the existence of a maximal element of S according to Zorn’s lemma. We claim that if xm is a maximal element of S, then xm≤fxm≤x∗ for all x∗∈δ. In the same fashion, we applyfthat givesxm≤fxm≤f2xm≤x∗for allx∗∈δthat impliesfxm∈S. It is easy from here to see that in fact we have xm=fxm by the maximality of xm and xm=fxm≤x∗ for all x∗∈δ that shows that xm is the least element of δf. The proof of existence of the greatest element of δf is symmetrical.Now, letB⊂δf be a nonempty set such that TB≺B. We want to find infδfB, the greatest-lower bound of B in δf. Let us put e=infAB. For all b∈B, e≤b, as fe≤b=fb. Thus fe≤e. Therefore, the decreasing sequence fnen tends to a fixed point of f which will be the greatest lower bounds (or the greatest-lower bound) of B in δf. The construction of supδfB is the least upper bound of B in δf if B≺TB is symmetric. Thereafter, δf,≤ is a complete sub-T-lattice.


## 3. Common Fixed Points for Commuting Family of Order-Preserving Mappings

### 3.1. Main Results

In this section, we give a generalization of Tarski’s theorem [4] that implies that any finite commuting family of order-preserving mappings has a common fixed point.The existence of a common fixed point for a finite family of order-preserving applications is related to the intersection of a finite decreasing sequence of complete sub-T-lattices.Theorem 5.
LetE,≤ be a complete T-lattice. Then, any finite commuting family of order-preserving mappings (monotone mappings) Tii∈I, Ti:E→E, has a common fixed point. Moreover, if we denote by FixTi the set of the common fixed points, then FixTi is a complete sub-T-lattice of E.Proof.
We takeI=0,2..,n, where n∈ℕ. Let E,≤ be a complete T-lattice so the set δ0=δT0 of fixed points of T0 is a complete sub-T-lattice by Theorem 4. As T1∘T0=T0∘T1, we have T0T1x=T1T0x=T1T0x=T1x that implies T1x∈δT0 for all x∈δT0. Therefore we can restrict the maps T1 to δ0, and since δ0 is a complete sub-T-lattice, the set δ1=FixT1/δ0=x∈δT0,T1x=x of fixed points of T1 in δ0=δT0 is a complete sub-T-lattice. It is easy to see that the family δii∈I such that δi=FixTi/δi−1=x∈δi−1,Tix=x for all i∈I with δ0=δT0 is a decreasing sequence.Now, we consider the map(1)F:δ0×δ1×⋯×δn→δ0×δ1×⋯×δn,x0,x1,..,xn→x1,x2,..,xn,xn.As the mapF is monotone and the set δ0×δ1×⋯×δn is a complete T-lattice, then there exists x0,x1,..,xn∈δ0×δ1×⋯×δn such that Fx0,x1,..,xn=x0,x1,..,xn that means that x0,x1,..,xn=x1,x2,..,xn,xn that gives x0=x1=⋯=xn and x0∈FixTi. Consequently, FixTi≠∅.It remains therefore to show thatFixTi is a complete sub-T-lattice. For this, it is easy to see that FixTi=δn which shows that FixTi is a complete T-lattice.


### 3.2. Common Fixed Points

In the following, we investigate the existence of a common fixed point of a commuting family of order-preserving mappings defined on a complete sub-T-lattice. The proof of our result follows the ideas of Baillon [5] developed in hyperconvex metric spaces and Abu-Sbeih and Khamsi [6] on the partially ordered sets. It is astonishing that we will develop their ideas in the case of the complete sub-T-trellis despite the difficulty of the demonstration which comes from the fact that in complete sub-T-trellis, we do not always have the existence of sup and inf of its parts.We have the following result in partially ordered sets.LetE,≤ be a partially ordered set and T be a given operator on E,≤ reversing the order such that x≤TxorTx≤x for all x∈E. Then,Theorem 6.
For any decreasing family of nonempty complete sub-T-lattice subsetsXββ∈Γ of E, where Γ is a directed index set, we have ⋂β∈ΓXβ is not empty and it is a complete sub-T-lattice.Proof.
Consider the family(2)P=Aββ∈Γ;AβisanonemptyintervalinXβandAββ∈Γisasequencedecreasing.
The setP is not empty since Xββ∈Γ∈P. If we order P by the inclusion relation, every chain of P has a lower bound since in a complete sub-T-lattice, any decreasing family of nonempty intervals has a nonempty intersection and it is an interval by Proposition 3. Therefore, P satisfies the assumptions of Zorn’s lemma. Hence, for every D∈P, there exists a minimal element A∈P such that A⊂D. We claim that if Aββ∈Γ is minimal, then each Aβ is a singleton. Indeed, let us fix β0∈Γ. We know that Aβ0=mβ0,Mβ0. We consider the new family
(3)Bβ=Aβifβ≥β0orβnotcomparabletoβ0,x∈Xβ,mβ0≤x≤Mβ0ifβ0≥β.
Our assumptions onXβ and Aβ imply that Bβ∈P. Moreover, we have Bβ⊂Aβ for any β∈Γ. Since Aββ∈Γ is minimal, we get Bβ=Aβ for any β∈Γ. In particular, we have
(4)Aβ=x∈Xβ;mβ0≤x≤Mβ0forβ≤β0.
IfAβ=mβ,Mβ, then we must have mβ=mβ0 and Mβ=Mβ0. Therefore, we proved the existence of m,M∈E such that Aβ=x∈Xβ;m≤x≤Mforanyβ∈Γ.It is easy from here to show that in fact we havem=M by the minimality of Aββ∈Γ, which proves our claim. Clearly, we have m∈Aβ for any β∈Γ which implies ϖ=⋂β∈ΓXβ is not empty.We will prove thatϖ is a complete sub-T-lattice. First, we start by proving the existence of the least and the greatest element of ϖ for that we consider the set:
(5)K=Aββ∈Γ;ϖ⊂AβisaintervalinXβandAβisdecreasing.For the same reason above, we have:(1)
K is not empty since Xββ∈Γ∈K(2)
K satisfies the assumptions of Zorn’s lemma. Hence, for every D∈K, there exists a minimal element A∈K such that A⊂D(3)
IfAββ∈Γ is a minimal element of K, then Aβ=Aβ′ for all β,β′ according to the minimality of Aββ∈Γ which gives us Aβ=m0,M0=ϖ for all β. Hence the result we are searchingSecondly, letA⊂ϖ be nonempty such that A≺TA. We will only prove that the sup A exists in ϖ. The proof for the existence of the infimum follows identically if TA≺A. For any β∈Γ, we have A⊂Xβ. Since Xβ is a complete sub-T-lattice, then aβ=supXβA exists in Xβ and the family aββ∈Γ is an increasing chain.Now, we consider the setXβ∗=aβ,→∩Xβ for β∈Γ. Then, Xβ∗ is a nonempty complete sub-T-lattice of Xβ. It is easy to see that the family Xβ∗β is a decreasing chain of complete sub-T-lattice. Hence, ⋂β∈ΓXβå=I is a not empty interval and I⊂ϖ. Obviously, we have supϖA=infI which completes the proof of Theorem 6.As a consequence of this theorem, we obtain the following common fixed point result.Theorem 7.
IfE,≤ is a complete T-lattice, then any commuting family of order-preserving mappings Tii∈I, Ti:E→E, has a common fixed point. Moreover, if we noted by FixTi the set of the common point fixed points, then FixTi is a complete sub-T-lattice of E.Proof.
First, note that the fixed point theorem (Theorem2) implies that any finite commuting family of order-preserving mappingsT1,T2,..,Tn,Ti:E→E, has a common fixed point. Moreover, if we denote byFixTithe set of the common fixed points, i.e.,FixTi=x∈M;Tix=xi=1,..,n,it is a complete sub-T-lattice. Let Γ=β;βisafinitenonemptysubsetofI. Clearly, Γ is downward directed (where the order on Γ is the set inclusion). For any β∈Γ, the set Fβ of common fixed point set of the mappings Ti, i∈β, is a nonempty complete sub-T-lattice. Clearly, the family Fββ∈Γ is decreasing. The theorem above implies that ⋂β∈ΓFβ is nonempty and it is a complete sub-T-lattice.The commutativity assumption may be relaxed using a new concept discovered in [6] (see also [7, 8]). Of course, this new concept was initially defined in the metric setting; therefore, we will extend it to the case of partially ordered sets in the next work where we will arrive at a result similar to De Marr’s result [9] without compactness assumption of the domain.


## 3.1. Main Results

In this section, we give a generalization of Tarski’s theorem [4] that implies that any finite commuting family of order-preserving mappings has a common fixed point.The existence of a common fixed point for a finite family of order-preserving applications is related to the intersection of a finite decreasing sequence of complete sub-T-lattices.Theorem 5.
LetE,≤ be a complete T-lattice. Then, any finite commuting family of order-preserving mappings (monotone mappings) Tii∈I, Ti:E→E, has a common fixed point. Moreover, if we denote by FixTi the set of the common fixed points, then FixTi is a complete sub-T-lattice of E.Proof.
We takeI=0,2..,n, where n∈ℕ. Let E,≤ be a complete T-lattice so the set δ0=δT0 of fixed points of T0 is a complete sub-T-lattice by Theorem 4. As T1∘T0=T0∘T1, we have T0T1x=T1T0x=T1T0x=T1x that implies T1x∈δT0 for all x∈δT0. Therefore we can restrict the maps T1 to δ0, and since δ0 is a complete sub-T-lattice, the set δ1=FixT1/δ0=x∈δT0,T1x=x of fixed points of T1 in δ0=δT0 is a complete sub-T-lattice. It is easy to see that the family δii∈I such that δi=FixTi/δi−1=x∈δi−1,Tix=x for all i∈I with δ0=δT0 is a decreasing sequence.Now, we consider the map(1)F:δ0×δ1×⋯×δn→δ0×δ1×⋯×δn,x0,x1,..,xn→x1,x2,..,xn,xn.As the mapF is monotone and the set δ0×δ1×⋯×δn is a complete T-lattice, then there exists x0,x1,..,xn∈δ0×δ1×⋯×δn such that Fx0,x1,..,xn=x0,x1,..,xn that means that x0,x1,..,xn=x1,x2,..,xn,xn that gives x0=x1=⋯=xn and x0∈FixTi. Consequently, FixTi≠∅.It remains therefore to show thatFixTi is a complete sub-T-lattice. For this, it is easy to see that FixTi=δn which shows that FixTi is a complete T-lattice.


## 3.2. Common Fixed Points

In the following, we investigate the existence of a common fixed point of a commuting family of order-preserving mappings defined on a complete sub-T-lattice. The proof of our result follows the ideas of Baillon [5] developed in hyperconvex metric spaces and Abu-Sbeih and Khamsi [6] on the partially ordered sets. It is astonishing that we will develop their ideas in the case of the complete sub-T-trellis despite the difficulty of the demonstration which comes from the fact that in complete sub-T-trellis, we do not always have the existence of sup and inf of its parts.We have the following result in partially ordered sets.LetE,≤ be a partially ordered set and T be a given operator on E,≤ reversing the order such that x≤TxorTx≤x for all x∈E. Then,Theorem 6.
For any decreasing family of nonempty complete sub-T-lattice subsetsXββ∈Γ of E, where Γ is a directed index set, we have ⋂β∈ΓXβ is not empty and it is a complete sub-T-lattice.Proof.
Consider the family(2)P=Aββ∈Γ;AβisanonemptyintervalinXβandAββ∈Γisasequencedecreasing.
The setP is not empty since Xββ∈Γ∈P. If we order P by the inclusion relation, every chain of P has a lower bound since in a complete sub-T-lattice, any decreasing family of nonempty intervals has a nonempty intersection and it is an interval by Proposition 3. Therefore, P satisfies the assumptions of Zorn’s lemma. Hence, for every D∈P, there exists a minimal element A∈P such that A⊂D. We claim that if Aββ∈Γ is minimal, then each Aβ is a singleton. Indeed, let us fix β0∈Γ. We know that Aβ0=mβ0,Mβ0. We consider the new family
(3)Bβ=Aβifβ≥β0orβnotcomparabletoβ0,x∈Xβ,mβ0≤x≤Mβ0ifβ0≥β.
Our assumptions onXβ and Aβ imply that Bβ∈P. Moreover, we have Bβ⊂Aβ for any β∈Γ. Since Aββ∈Γ is minimal, we get Bβ=Aβ for any β∈Γ. In particular, we have
(4)Aβ=x∈Xβ;mβ0≤x≤Mβ0forβ≤β0.
IfAβ=mβ,Mβ, then we must have mβ=mβ0 and Mβ=Mβ0. Therefore, we proved the existence of m,M∈E such that Aβ=x∈Xβ;m≤x≤Mforanyβ∈Γ.It is easy from here to show that in fact we havem=M by the minimality of Aββ∈Γ, which proves our claim. Clearly, we have m∈Aβ for any β∈Γ which implies ϖ=⋂β∈ΓXβ is not empty.We will prove thatϖ is a complete sub-T-lattice. First, we start by proving the existence of the least and the greatest element of ϖ for that we consider the set:
(5)K=Aββ∈Γ;ϖ⊂AβisaintervalinXβandAβisdecreasing.For the same reason above, we have:(1)
K is not empty since Xββ∈Γ∈K(2)
K satisfies the assumptions of Zorn’s lemma. Hence, for every D∈K, there exists a minimal element A∈K such that A⊂D(3)
IfAββ∈Γ is a minimal element of K, then Aβ=Aβ′ for all β,β′ according to the minimality of Aββ∈Γ which gives us Aβ=m0,M0=ϖ for all β. Hence the result we are searchingSecondly, letA⊂ϖ be nonempty such that A≺TA. We will only prove that the sup A exists in ϖ. The proof for the existence of the infimum follows identically if TA≺A. For any β∈Γ, we have A⊂Xβ. Since Xβ is a complete sub-T-lattice, then aβ=supXβA exists in Xβ and the family aββ∈Γ is an increasing chain.Now, we consider the setXβ∗=aβ,→∩Xβ for β∈Γ. Then, Xβ∗ is a nonempty complete sub-T-lattice of Xβ. It is easy to see that the family Xβ∗β is a decreasing chain of complete sub-T-lattice. Hence, ⋂β∈ΓXβå=I is a not empty interval and I⊂ϖ. Obviously, we have supϖA=infI which completes the proof of Theorem 6.As a consequence of this theorem, we obtain the following common fixed point result.Theorem 7.
IfE,≤ is a complete T-lattice, then any commuting family of order-preserving mappings Tii∈I, Ti:E→E, has a common fixed point. Moreover, if we noted by FixTi the set of the common point fixed points, then FixTi is a complete sub-T-lattice of E.Proof.
First, note that the fixed point theorem (Theorem2) implies that any finite commuting family of order-preserving mappingsT1,T2,..,Tn,Ti:E→E, has a common fixed point. Moreover, if we denote byFixTithe set of the common fixed points, i.e.,FixTi=x∈M;Tix=xi=1,..,n,it is a complete sub-T-lattice. Let Γ=β;βisafinitenonemptysubsetofI. Clearly, Γ is downward directed (where the order on Γ is the set inclusion). For any β∈Γ, the set Fβ of common fixed point set of the mappings Ti, i∈β, is a nonempty complete sub-T-lattice. Clearly, the family Fββ∈Γ is decreasing. The theorem above implies that ⋂β∈ΓFβ is nonempty and it is a complete sub-T-lattice.The commutativity assumption may be relaxed using a new concept discovered in [6] (see also [7, 8]). Of course, this new concept was initially defined in the metric setting; therefore, we will extend it to the case of partially ordered sets in the next work where we will arrive at a result similar to De Marr’s result [9] without compactness assumption of the domain.


## 4. Conclusion

In this paper, we have extended some Tarski’s theorems of the fixed point into ordered sets by new fixed point theorems. The original proof of fixed point for complete T-lattice [2] is beautiful and elegant but nonconstructive and somewhat uninformative. In [3], we have given a constructive proof that generalizes the Tarski’s version results. In this paper, we have given a structure to the set of fixed points of an increasing application on an ordered set and we have investigated the existence of a common fixed point of a commuting family of order-preserving mappings defined on a complete sub-T-lattice. Our next work concerns the development of some fixed point theorems in hyperconvex metric spaces.



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