Fixed Points and Common Fixed Points Theorems in Pseudo-ordered Sets

Under suitable conditions, we establish the existence of the greatest and the least fixed points of monotone maps defined on nonempty pseudo-ordered sets. Also, we prove that the set of all common fixed points of two categories of finite commutative family F of monotone maps f defined on a nonempty complete trellis is also a nonempty complete trellis.


Introduction
In 1971, H. Skala introduced the notions of pseudo-ordered sets and trellises and gave two fixed points theorems in this setting (see [Theorems 36 and 37,11]).Unfortunately this subject was a long time without serious study because in this setting we have not the essential property used in all proofs for fixed point theory in the case of partially ordered sets namely the property of transitivity (for examples see: [3,5,12,13,15]).Later on, S. Parameshwara Bhatta and all [2,3] studied the fixed point property in pseudo-ordered sets.In the present paper, we first establish the existence of the least and the greatest fixed points of monotone maps defined on nonempty pseudo-ordered sets (see Theorems 3.1 and 3.3).Note that our results ensure the existence of the least and the greatest fixed points under weaker hypothesis than the assumptions of [Theorem 36,11] which was established by H. Skala.We prove also that the set of all common fixed points of two categories of finite commutative family F of monotone maps f defined on a nonempty complete trellis is also a nonempty complete trellis (see Theorems 4.1 and 4.2).Note that our reslts could be used as a tool for solving differential equations in metric pseudo-ordered spaces (for the case of metric ordered spaces see for examples: [1,2,4,6,7,15]).

Preliminaries
Let X be a nonempty set and ≥ be a binary relation defined on its.If the binary relation ≥ is reflexive and antisymmetric, we say that (X, ≥) is a pseudo-ordered set or a psoset.
Let A be a nonempty subset of a psoset (X, ≥).
An element u of X is said to be an upper bound of A (respectively v a lower bound of A) if x ≥ u for every x ∈ A. An element s of X is called a greatest element or the maximum of A and denoted by s = max(A, ≥) if s is an upper bound of A and s ∈ A.
An element v of X is said to be a lower bound of A if v ≥ x for every x ∈ A. An element of X is called a least or the minimum element of A and denoted by = min(A, ≥) if is a lower bound of A and ∈ A.
When the least upper bound (l.u.b.) s of A exists, we shall denoted its by s = sup(A, ≥).Dually if the greatest lower bound (g.l.b.) of A exists, we shall denoted its by = inf(A, ≥).
Note that the greatest lower bound and the least upper bound when they exist they are unique.
A psoset (X, ≥) is said to be a trellis if every pair of elements of (X, ≥) has a greatest lower bound (g.l.b) and a least upper bound (l.u.b).A psoset (X, ≥) is said to be a complete trellis if every nonempty subset of X has a g.l.b and a l.u.b.For more details for these notions can be found in H.L. Skala [10,11].
Let (X, ≥) be a nonempty pseudo-ordered sets and let f : X → X be a map.We shall say that f is monotone if for every x, y ∈ X, with x ≥ y, then we have The set of all fixed points of f is denoted by F ix(f ).

Example.
Let A the set defined by A = {0, a, b, c}.We define a pseudo-order relation on A by setting: (i) for every x ∈ A, we have 0 ≥ x and (ii) a ≥ b ≥ c ≥ a.Then, (A, ≥) is a trellis having the minimum element 0 but (A, ≥) is not complete.
In this paper, we shall need the following notion of inverse relation.Definition 2.1.Let X be a nonempty set and let ≥ be a relation on its.The inverse relation ≤ of ≥ is defined for every x, y ∈ X by: The proofs of the following two lemmas are obvious.Lemma 2.2.Let ≥ be a pseudo-order relation defined on a nonempty set X and let ≤ be the inverse relation of ≥.Then, ≤ is a pseudo-order relation on X. Lemma 2.3.Let ≥ be a pseudo-order relation defined on a nonempty set X, let ≤ be the inverse relation of ≥ and let A be a nonempty subset of X.Then, we have (i) if sup(A, ≥) exists, so inf(A, ≤ ) exists too and sup(A, ≥) = inf(A, ≤ ) ; (ii) if inf(A, ≥) exists, hence also sup(A, ≤ ) exists too and inf(A, ≥) = sup(A, ≤ ) ; (iii) if min(A, ≥) exists, then also max(A, ≤ ) exists and min(A, ≥) = max(A, ≤ ) ; (iv) if max(A, ≥) exists, so also min(A, ≤ ) exists too and max(A, ≥) = min(A, ≤ ) ; (v) if (X, ≥) is a nonempty complete trellis, so (X, ≤ ) is also a nonempty complete trellis ; (vi) if f : X → X is a monotone map for ≥, then f is also a monotone map for ≤ .

Fixed points in pseudo-ordered sets
In this section we shall establish the existence of the greatest and the least fixed points of monotone maps defined on a nonempty pseudo-ordered sets.Note that our results (Theorem 3.1 and 3.3) ensure the existence of the least and the greatest fixed points under weaker hypothesis than the assumptions of [Theorem 36,11] which was established by H. Skala.First, we shall give our key result in this paper.
Theorem 3.1.Let (X, ≥) be a nonempty pseudo-ordered set with a least element .Assume that every nonempty subset of X has a supremum in (X, ≥).Then, the set of all fixed points of every monotone map f : (X, ≥) → (X, ≥) is nonempty and has a least element.
Proof.Let (X, ≥) be a nonempty pseudo-ordered set with a least element and f : (X, ≥) → (X, ≥) be a monotone map.
First step.We have: F ix(f ) 6 = ∅.Indeed, let A the family of all subsets A of X satisfying the following three conditions: Thus, we get f (C) ⊂ C. Now, let D ⊂ C such that D 6 = ∅.Then, D ⊂ A for every A ∈ A. So, sup(D, ≥) ∈ A for every A ∈ A. Hence, we obtain sup(D, ≥) ∈ C. Therefore, C is the least nonempty element of A. Then, we set s = sup(C, ≥).
Claim 2. We have: s ∈ F ix(f ).Indeed, since s ∈ C and f (C) ⊂ C, then f (s) ∈ C. As s = sup(C, ≥), so we get f (s) ≥ s. (2.1) Now, we consider the following subset D of C defined by As ∈ D, so D 6 = ∅.We claim that D ∈ A. As D ⊂ C, then we get x ≥ s for every x ∈ D. By the monotonicity of f we obtain f (x) ≥ f (s), for every x ∈ D.Then, we get f (D) ⊂ D.
Now, if E ⊂ D and E 6 = ∅, then we set m = sup(E, ≥).As E ⊂ D, so E ⊂ C and m ∈ C. By our definition of D, we deduce that f (s) is an upper bound of (E, ≥).Then, we get m ≥ f (s).Thus, we have m ∈ D. Therefore, we obtain D ∈ A. As D ⊂ C and C is the least nonempty element of A, so we get C = D. On the other hand, we know that s = sup(C, ≥) ∈ C, then we obtain By combining (2.1) and (2.2) and using the antisymmetry of ≥, we obtain that f (s) = s.Thus, we have s ∈ F ix(f ).Therefore, F ix(f ) 6 = ∅.
Second step.We have: (F ix(f ), ≥) has a least element.Indeed, from the first step above, we know that F ix(f ) 6 = ∅.Next, we shall show that (F ix(f ), ≥) has a least element.Now we consider the following subset S of X defined by As = min(X, ≥), then ∈ S. Hence, = min(S, ≤ ).Now let L be a nonempty subset of S. Hence, by our hypothesis L has a supremum in (X, ≤ ).Let t = sup(L, ≥).As L ⊂ S, so every element z of F ix(f ) is an upper bound of (L, ≥).Since t = sup(L, ≥), hence we get t ≥ z for every z ∈ F ix(f ).Then, we have t ∈ S. So, sup(L, ≥) ∈ S. Hence, sup(L, ≥) is the supremum of L in (S, ≥).Now, let x ∈ S.So by our definition of S, we have x ≥ z for every z ∈ F ix(f ).Hence, by the monotonicity of f we get f (x) ≥ z for every z ∈ F ix(f ).Then, f (x) ∈ S for every x ∈ S. Thus, we have f (S) ⊂ S.
Therefore, all hypothesis of the first step above are satisfied for the map f / S : (S, ≥) → (S, ≥) defined by f / S (x) = f (x) for every x ∈ S.Then, we get F ix(f / S ) 6 = ∅.So, there exists a ∈ S such that f (a) = a and a ≥ z for every z ∈ F ix(f ).Thus, a ∈ F ix(f ) and a is a lower bound of (F ix(f ), ≥).Thus, a is the minimum of (F ix(f ), ≥).
As a corollary of Theorem 3.1, we get the following result.
Corollary 3.2.Let (X, ≤) be a nonempty partially ordered set with a least element .Assume that every nonempty subset of X has a supremum in (X, ≤).Then, the set of all fixed points of every monotone map f : (X, ≤ ) → (X, ≤) is nonempty and has a least element.
Next, by using Theorem 3.1 and Lemmas 2.2 and 2.3, we obtain the following result.Theorem 3.3.Let (X, ≥) be a nonempty pseudo-ordered set with a greatest element g.Assume that every nonempty subset of X has an infimum in (X, ≥).Then, the set of all fixed points of every monotone map f : (X, ≥) → (X, ≥) is nonempty and has a greatest element.
Proof.Let (X, ≥) be a nonempty pseudo-ordered set with a greatest element g, let f : (X, ≥) → (X, ≥) be a monotone map and let ≤ be the inverse relation of ≥.From Lemma 2.2, we know that ≤ is a pseudo-order relation on X.On the other hand, by Lemma 2.3, min(X, ≤ ) exists and we have min(X, ≤ ) = g.As by our hypothesis f : (X, ≥) → (X, ≥) is a monotone map, so from Lemma 2.3 the map f : (X, ≤ ) → (X, ≤ ) is also monotone.Thus, all hypothesis of Theorem 3.1 are satisfied.Therefore, The set F ix(f ) of all fixed points of the map f is nonempty and has a least element in (X, ≤ ), m, say.Then from Lemma 2.3, we get m = min(F ix(f ), ≤ ) = max(F ix(f ), ≥).
As a consequence of Theorem 3.3, we get the following result.
Corollary 3.4.Let (X, ≤) be a nonempty partially ordered set with a greatest element g.Assume that every nonempty subset of X has an infimum in (X, ≤).Then, the set of all fixed point of every monotone map f : (X, ≤) → (X, ≤) is nonempty and has a greatest element.