 Research
 Open Access
 Published:
On externally complete subsets and common fixed points in partially ordered sets
Fixed Point Theory and Applications volume 2011, Article number: 97 (2011)
Abstract
In this study, we introduce the concept of externally complete ordered sets. We discuss the properties of such sets and characterize them in ordered trees. We also prove some common fixed point results for order preserving mappings. In particular, we introduce for the first time the concept of Banach Operator pairs in partially ordered sets and prove a common fixed point result which generalizes the classical De Marr's common fixed point theorem.
2000 MSC: primary 06F30; 46B20; 47E10.
1. Introduction
This article focuses on the externally complete structure, a new concept that was initially introduced in metric spaces as externally hyperconvex sets by Aronszajn and Panitchpakdi in their fundamental article [1] on hyperconvexity. This idea developed from the original work of Quilliot [2] who introduced the concept of generalized metric structures to show that metric hyperconvexity is in fact similar to the complete lattice structure for ordered sets. In this fashion, Tarski's fixed point theorem [3] becomes Sine and Soardi's fixed point theorems for hyperconvex metric spaces [4, 5]. For more on this, the reader may consult the references [6–8].
We begin by describing the relevant notation and terminology. Let (X, ≺) be a partially ordered set and M ⊂ X a nonempty subset. Recall that an upper (resp. lower) bound for M is an element p ∈ X with m ≺ p (resp. p ≺ m) for each m ∈ M; the leastupper (resp. greatestlower) bound of M will be denoted sup M (resp. inf M). A nonempty subset M of a partially ordered set X will be called Dedekind complete if for any nonempty subset A ⊂ M, sup A (resp inf A) exists in M provided A is bounded above (resp. bounded below) in X. Recall that M ⊂ X is said to be linearly ordered if for every m_{1}, m_{2} ∈ M we have m_{1} ≺ m_{2} or m_{2} ≺ m_{1}. A linearly ordered subset of X is called a chain. For any m ∈ X define
Recall that a connected partially ordered set X is called a tree if X has a lowest point e, and for every m ∈ X, the subset [e, m] is well ordered.
A subset Y of a partially ordered set X is called convex if the segment [x, y] = {z ∈ X; x ≺ z ≺ y} ⊂ Y whenever x, y ∈ Y. A map T: X → X is order preserving (also called monotone, isotone, or increasing) if T(x) ≺ T(y) whenever x ≺ y.
2. Externally complete sets
Inspired by the success of the concept of the externally hyperconvex subsets introduced by Aronszajn and Panitchpakdi [1], we propose a similar concept in partially ordered sets.
Definition 2.1. Let (X, ≺) be a partially ordered set. A subset M of X is called externally complete if and only if for any family of points (x_{ α })_{α∈Г}in X such thatI\left({x}_{\alpha}\right)\cap I\left({x}_{\beta}\right)\ne \varnothingfor any α, β ∈ Γ, andI\left({x}_{\alpha}\right)\cap M\ne \varnothing, we have
Where I(x) = (←, x] or I(x) = [x, →).
The family of all nonempty externally complete subsets of X will be denoted by \mathcal{E}\mathcal{C}\left(X\right).
Proposition 2.1. Let X be a partially ordered set. Then, anyM\in \mathcal{E}\mathcal{C}\left(X\right)is Dedekind complete and convex.
Proof. Let A\subset M\in \mathcal{E}\mathcal{C}\left(X\right) be nonempty and bounded above in X. The set U(A) = {b ∈ X; A ⊂ (←, b]} is not empty since A is bounded above. It is clear that the families (I(a))_{a∈A}, where I(a) = [a, →) and (I(b))_{b∈U(A)}, where I (b) = (←, b], intersect 2by2. Moreover, we have I\left(a\right)\cap M\ne \varnothing and I\left(b\right)\cap M\ne \varnothing, for any (a, b) ∈ A × U(A). Since M is in \mathcal{E}\mathcal{C}\left(X\right), we conclude that
Let m ∈ J. Then, for any a ∈ A, we have a ≺ m. So, m is an upper bound of A. Let b be any upper bound of A, then b ∈ U(A). Hence, m ≺ b which forces m to be the least upper bound of A, i.e. m = sup A. Similarly, one can prove that inf A also exists and belongs to M provided A is bounded below in X. Next, we prove that M is convex. Let x, y ∈ M. Obviously, if x and y are not comparable, then \left[x,y\right]=\varnothing and we have nothing to prove. So, assume x ≺ y. Let a ∈ [x, y]. Obviously, we have (←, a] ⋂ [a, →) = {a}. And, since \left(\leftarrow ,a\right]\cap M\ne \varnothing and \left[a,\to \right)\cap M\ne \varnothing, then
This obviously implies that a ∈ M, i.e. [x, y] ⊂ M, which completes the proof of our proposition.
Example 2.1. Let ℕ = {0, 1,...}. we consider the order 0 ≺ 2 ≺ 4 ≺ ··· and 0 ≺ 1 ≺ 3 ≺ ···, and no even number (different from 0) is comparable to any odd number. Then, (ℕ, ≺) is a tree. The set M = {0, 1, 2} is in\mathcal{E}\mathcal{C}\left(\mathbb{N}\right). Note that M is convex and is not linearly ordered.
In the next result, we characterize the externally complete subsets of trees.
Theorem 2.1. Let X be a tree. A subset M of X is externally complete if and only if M is convex, Dedekind complete, and any chain C ⊂ M has a least upper bound in M.
Proof. Let M\in \mathcal{E}\mathcal{C}\left(X\right). Then, M is convex and Dedekind complete. Let C be a nonempty chain of M. Let c_{1}, c_{2} ∈ C, then we have c_{1} ≺ c_{2} or c_{2} ≺ c_{1}. Hence,
Since M\in \mathcal{E}\mathcal{C}\left(X\right)
Obviously, any c ∈ J is an upper bound of C. Since M is Dedekind complete, sup C exists in M. Assume conversely that M is a convex, and Dedekind complete subset of X such that any chain in M has an upper bound in M. Let x, y ∈ X such that there exist m_{1}, m_{2} ∈ M with x ≺ m_{1} and m_{2} ≺ y. Define P(x) = inf{m ∈ M;x ≺ m}, and P(y) = sup{m ∈ M; m ≺ y}. Both P(x) and P(y) exist and belong to M since M is Dedekind complete. Let (x_{ i })_{i∈I}in X be such that for any i ∈ I there exists m_{ i }∈ M such that x_{ i }≺ m_{ i }. Also, we have \left[{x}_{i},\to \right)\cap \left[{x}_{j},\to \right)\ne \varnothing, for any i, j ∈ I. This condition forces the set {x_{ i }; i ∈ I} to be linearly ordered since X is a tree. Consider the subset M_{ I }= {P(x_{ i }); i ∈ I} of M. It is easy to check that M_{ I }is linearly ordered. Since any linearly ordered subset of M is bounded above, there exists m ∈ M such that P(x_{ i }) ≺ m for any i ∈ I. Since x_{ i }≺ P(x_{ i }) then
Next, let (y_{ j })_{j∈J}in X such that for any j ∈ J there exists m_{ j }∈ M such that m_{ j }≺ y_{ j }. Consider the subset M_{ J }= {P(y_{ j }); j ∈ J} of M. Since X is a tree, the set M_{ J }is bounded below, so m_{0} = inf M_{ J }exists in M. It is obvious that m_{0} ≺ P(y_{ j }) ≺ y_{ j }for any j ∈ J. This implies
Finally, assume that we have (x_{ i })_{i∈I}and (y_{ j })_{j∈J}in X such that the subsets ([x_{ i }, →))_{i∈I}, and ((←, y_{ j }])_{j∈J}intersect 2by2 and \left[{x}_{i},\to \right)\cap M\ne \varnothing and \left(\leftarrow ,{y}_{j}\right]\cap M\ne \varnothing for any (i, j) ∈ I × J. As before, set
For any (i, j) ∈ I × J,we have P(x_{ i }) ≺ P(y_{ j }), which implies m_{ I }≺ m_{ J }. Obviously we have
Hence, M is in \mathcal{E}\mathcal{C}\left(X\right).
The above proof suggests that externally complete subsets are proximinal. In fact in [1], the authors introduced externally hyperconvex subsets as an example of proximinal sets other than the admissible subsets, i.e. intersection of balls. Before we state a similar result, we need the following definitions.
Definition 2.2. Let X be a partially ordered set. Let M be a nonempty subset of X. Define the lower and upper cones by
and
The cone generated by M will be defined by\mathcal{C}\left(M\right)={\mathcal{C}}_{l}\left(M\right)\cup {\mathcal{C}}_{u}\left(M\right).
Theorem 2.2. Let X be a partially ordered set and M a nonempty externally complete subset of X. Then, there exists an order preserving retractP:\mathcal{C}\left(M\right)\to Msuch that

(1)
for any x\in {\mathcal{C}}_{l}\left(M\right) we have x ≺ P(x), and

(2)
for any x\in {\mathcal{C}}_{u}\left(M\right) we have P(x) ≺ x.
Proof. First set P(m) = m for any m ∈ M. Next, let x\in {\mathcal{C}}_{l}\left(M\right). We have
where m ∈ M. Using the external completeness of M, we get
It is easy to check that this intersection is reduced to one point. Set
Similarly, let x\in {\mathcal{C}}_{u}\left(M\right). We have
where m ∈ M. Using the external completeness of M, we get
It is easy to check that this intersection is reduced to one point. Set
In particular, this prove (1) and (2). In order to finish the proof of the theorem, let us show that P is order preserving. Indeed, let x,y\in \mathcal{C}\left(M\right) with x ≺ y. If y\in {\mathcal{C}}_{l}\left(M\right), then x\in {\mathcal{C}}_{l}\left(M\right). Since y ≺ P(y) then x ≺ P(y) which implies P(x) ≺ P(y). Similarly, if x\in {\mathcal{C}}_{u}\left(M\right), then y\in {\mathcal{C}}_{u}\left(M\right) and again it is easy to show P(x) ≺ P(y). Assume x\in {\mathcal{C}}_{l}\left(M\right) and y\in {\mathcal{C}}_{u}\left(M\right). Since
and M is externally complete, we have
where m ∈ M. But \left[x,\to \right)\cap \left(\underset{x\prec m}{\cap}\left(\leftarrow ,m\right]\right)=\left\{P\left(x\right)\right\}, this forces P(x) ∈ (←, y]. By definition of P(y), we get P(x) ≺ P(y). In fact, we proved that x ≺ P(x) ≺ P(y) ≺ y. This completes the proof of the theorem.
We have the following result.
Theorem 2.3. Let X be a partially ordered set and M a nonempty externally complete subset of X. Assume that X has a supremum or an infimum, then there exists an order preserving retract P: X → M which extends the retract of\mathcal{C}\left(M\right)into M.
Proof. Let P:\mathcal{C}\left(M\right)\to M be the retract defined in Theorem 2.2. Assume first that X has a supremum e. Then, X={\mathcal{C}}_{l}\left(\mathcal{C}\left(M\right)\right). Indeed, for any x ∈ X, we have x ≺ e and e\in {\mathcal{C}}_{u}\left(M\right). Because x\in \underset{x\prec z}{\cap}\left(\leftarrow ,z\right], where z\in {\mathcal{C}}_{u}\left(M\right), and M is externally complete, we get M\cap \left(\underset{x\prec z}{\cap}\left(\leftarrow ,z\right]\right)\ne \varnothing, where z\in {\mathcal{C}}_{u}\left(M\right). Hence, there exists m ∈ M such that m ≺ z, for any z\in {\mathcal{C}}_{u}\left(M\right) such that x ≺ z. Using the properties of P, we get m ≺ P(z), for any z\in {\mathcal{C}}_{u}\left(M\right) such that x ≺ z. Since M is Dedekind complete, inf{P(z); z\in {\mathcal{C}}_{u}\left(M\right) such that x ≺ z} exists. Set
First note that if x\in \mathcal{C}\left(M\right), then for any z\in {\mathcal{C}}_{u}\left(M\right) such that x ≺ z we have P(x) ≺ P(z). This will imply P\left(x\right)\prec \stackrel{\u0303}{P}\left(x\right). If x\in {\mathcal{C}}_{u}\left(M\right), then by definition of \stackrel{\u0303}{P}, we have \stackrel{\u0303}{P}\left(x\right)\prec P\left(x\right). Hence, P\left(x\right)\prec \stackrel{\u0303}{P}\left(x\right). If x\in {\mathcal{C}}_{l}\left(M\right), then by definition of \stackrel{\u0303}{P}, we have \stackrel{\u0303}{P}\left(x\right)\prec P\left(P\left(x\right)\right) since x ≺ P(x). Hence, \stackrel{\u0303}{P}\left(x\right)\prec P\left(x\right) which implies again P\left(x\right)\prec \stackrel{\u0303}{P}\left(x\right). So, \stackrel{\u0303}{P} extends P. Let us show that \stackrel{\u0303}{P} is order preserving. Indeed, let x, y ∈ X such that x ≺ y. Since
we have
or \stackrel{\u0303}{P}\left(x\right)\prec \stackrel{\u0303}{P}\left(y\right). In order to finish the proof of Theorem 2.3, consider the case when X has an infimum, say e. Then, X={\mathcal{C}}_{l}\left(\mathcal{C}\left(M\right)\right). As for the previous case, define
It is easy to show that \stackrel{\u0303}{P}\left(x\right) exists. In a similar proof, one can show that \stackrel{\u0303}{P} extends P and is order preserving.
Since a tree has an infimum, we get the following result.
Corollary 2.1. Let X be a tree and M a nonempty externally complete subset of X. Then, there exists an order preserving retract P: X → M.
A similar result for externally hyperconvex subsets of metric trees maybe found in [9].
3. Common fixed point
In this section, we investigate the existence of a common fixed point of a commuting family of order preserving mappings defined on a complete lattice. Here the proof follows the ideas of Baillon [10] developed in hyperconvex metric spaces. It is amazing that these ideas extend nicely to the case of partially ordered sets. The ideas in question are not the conclusions which maybe known but the proofs as developed in the metric setting. Maybe one of the most beautiful results known in the hyperconvex metric spaces is the intersection property discovered by Baillon [10]. The boundedness assumption in Baillon's result is equivalent to the complete lattice structure in our setting. Indeed, any nonempty subset of a complete lattice has an infimum and a supremum. We have the following result in partially ordered sets.
Theorem 3.1. Let X be a partially ordered set. Let (X_{ β })_{β∈Γ}be a decreasing family of nonempty complete lattice subsets of X, where Γ is a directed index set. Then, ⋂_{β∈Γ}X_{ β }is not empty and is a complete lattice.
Proof. Consider the family
\mathcal{F} is not empty since {\prod}_{\beta \in \Gamma}{X}_{\beta}\in \mathcal{F}. In a complete lattice, any decreasing family of nonempty intervals has a nonempty intersection and it is an interval. Therefore, \mathcal{F} satisfies the assumptions of Zorn's lemma. Hence, for every D\in \mathcal{F}, there exists a minimal element A\in \mathcal{F} such that A ⊂ D. We claim that if ∏_{β∈Γ}A_{ β }is minimal, then each A_{ β }is a singleton. Indeed, let us fix β_{0} ∈ Γ. We know that A_{β 0}= [m_{β 0}, M_{β 0}]. Consider the new family
Our assumptions on (X_{ β }) and (A_{ β }) imply that \left({B}_{\beta}\right)\in \mathcal{F}. Moreover, we have B_{ β }⊂ A_{ β }for any β ∈ Γ. Since ∏_{β∈Γ}A_{ β }is minimal, we get B_{ β }= A_{ β }for any β ∈ Γ. In particular, we have
for β ≺ β_{0}. If A_{ β }= [m_{ β }, M_{ β }], then we must have m_{ β }= m_{β 0}and M_{ β }= M_{β 0}. Therefore, we proved the existence of m, M ∈ X such that A_{ β }= {x ∈ X_{ β }; m ≺ x ≺ M}, for any β ∈ Γ. It is easy from here to show that in fact we have m = M by the minimality of ∏_{β∈Γ}A_{ β }, which proves our claim. Clearly, we have m ∈ A_{ β }for any β ∈ Γ which implies K = ⋂_{β∈Γ}X_{ β }is not empty. Next, we will prove that K is a complete lattice. Let A ⊂ K be nonempty. We will only prove that sup A exists in K. The proof for the existence of the infimum follows identically. For any β ∈ Γ, we have A ⊂ X_{ β }. Since X_{ β }is a complete lattice, then m_{ β }= sup A exists in X_{ β }. The interval [m_{ β }, →) is a complete lattice. Clearly, the family ([m_{ β }, →))_{β∈Γ}is decreasing. From the above result, we know that ⋂_{β∈Γ}[m_{ β }, →) is not empty. Therefore, there exists m ∈ K such that a ≺ m for any a ∈ A. Set B = {m ∈ K; a ≺ m for any a ∈ A}. For any β ∈ Γ, define M_{ β }= inf B in X_{ β }. Set
Then, {X}_{\beta}^{*} is a nonempty complete sublattice of X_{ β }. It is easy to see that the family \left({X}_{\beta}^{*}\right) is decreasing. Hence, {\cap}_{\beta \in \Gamma}{X}_{\beta}^{*} is not empty Obviously, we have
in ⋂_{β∈Γ}X_{ β }. The proof of Theorem 3.1 is therefore complete.
As a consequence of this theorem, we obtain the following common fixed point result.
Theorem 3.2. Let X be a complete lattice. Then, any commuting family of order preserving mappings (T_{ i })_{i∈I}, T_{ i }: X → X, has a common fixed point. Moreover, if we denote by Fix((T_{ i })) the set of the common fixed points, then Fix((T_{ i })) is a complete sublattice of X.
Proof. First note that Tarski fixed point theorem [3] implies that any finite commuting family of order preserving mappings T_{1}, T_{2}, .., T_{ n }, T_{ i }: X → X, has a common fixed point. Moreover, if we denote by Fix((T_{ i })) the set of the common fixed points, i.e. Fix((T_{ i })) = {x ∈ M; T_{ i }(x) = x i = 1,..,n}, is a complete sublattice of X. Let Γ = {β; β is a finite nonempty subset of I}. Clearly, Γ is downward directed (where the order on Γ is the set inclusion). For any β ∈ Γ, the set F_{ β }of common fixed point set of the mappings T_{ i }, i ∈ β, is a nonempty complete sublattice of X. Clearly, the family (F_{ β })_{β∈Γ}is decreasing. Theorem 3.1 implies that ⋂_{β∈Γ}F_{ β }is nonempty and is a complete sublattice of X. The proof of Theorem 3.2 is therefore complete.
The commutativity assumption maybe relaxed using a new concept discovered in [11] (see also [12–15]. Of course, this new concept was initially defined in the metric setting, therefore we need first to extend it to the case of partially ordered sets.
Definition 3.1. Let X be a partially ordered set. The ordered pair (S, T) of two selfmaps of the set X is called a Banach operator pair, if the set Fix(T) is Sinvariant, namely S(Fix(T)) ⊆ Fix(T).
We have the following result whose proof is easy.
Theorem 3.3. Let X be a complete lattice. Let T: X → X be an order preserving mapping. Let S: X → X be an order preserving mapping such that (S, T) is a Banach operator pair. Then, Fix(S, T) = Fix(T) ⋂ Fix(S) is a nonempty complete lattice.
In order to extend this conclusion to a family of mappings, we will need the following definition.
Definition 3.2. Let T and S be two selfmaps of a partially ordered set X. The pair (S, T) is called symmetric Banach operator pair if both (S, T) and (T, S) are Banach operator pairs, i.e., T(Fix(S)) ⊆ Fix(S) and S(Fix(T)) ⊆ Fix(T).
We have the following result which can be seen as an analogue to De Marr's result [16] without compactness assumption of the domain.
Theorem 3.4. Let X be a partially ordered set. Let\mathcal{T}be a family of order preserving mappings defined on X. Assume any two mappings from\mathcal{T}form a symmetric Banach operator pair. Then, the family\mathcal{T}has a common fixed point provided one map from\mathcal{T}has a fixed point set which is a complete lattice. Moreover, the common fixed point setFix\left(\mathcal{T}\right)is a complete lattice.
Proof. Let {T}_{0}\in \mathcal{T} be the map for which Fix(T_{0}) = X_{0} is a nonempty complete lattice. Since any two mappings from \mathcal{T} form a symmetric Banach operator pair, then for any T\in \mathcal{T}, we have T(X_{0}) ⊂ X_{0}. Since X_{0} is a complete lattice, T has a fixed point in X_{0}. The fixed point set of T in X_{0} is Fix(T) ⋂ Fix(T_{0}) and is a complete sublattice of X_{0}. Let S\in \mathcal{T}. Then, S (Fix(T) ⋂ Fix(T_{0})) ⊂ Fix(T) ⋂ Fix(T_{0}). Since Fix(T) ⋂ Fix(T_{0}) is a complete lattice, then S has a fixed point in Fix(T) ⋂ Fix(T_{0}). The fixed point set of S in Fix(T) ⋂ Fix(T_{0}) is Fix(S) ⋂ Fix(T) ⋂ Fix(T_{0}) which is a complete sublattice of X_{0}. By induction, one will prove that any finite subfamily T_{1}, ..., T_{ n }of \mathcal{T} has a nonempty common fixed point set Fix(T_{1}) ⋂ ··· ⋂ Fix(T_{ n }) ⋂ X_{0} which is a complete sublattice of X_{0}. Theorem 3.1 will then imply that {\bigcap}_{{T}_{\in \mathcal{T}}}Fix\left(T\right)\cap {X}_{0} is not empty and is a complete lattice. Since
we conclude that Fix\left(\mathcal{T}\right) is a nonempty complete lattice.
References
Aronszajn N, Panitchpakdi P: Extensions of uniformly continuous transformations and hyperconvex metric spaces. Pacific J Math 1956, 6: 405–439.
Quilliot A: Homomorphismes, points fixes, rtractions et jeux de poursuite dans les graphes, les ensembles ordonns et les espaces mtriques, Thèse de Doctorat d'Etat. Paris; 1983.
Tarski A: A lattice theoretical fixpoint theorem and its applications. Pacific J Math 1955, 5: 285–309.
Sine R: On nonlinear contraction semigroups in sup norm spaces. Nonlinear Anal Theory Methods Appl 1979, 3: 885–890. 10.1016/0362546X(79)900555
Soardi P: Existence of fixed points of nonexpansive mappings in certain Banach lattices. Proc Amer Math Soc 1979, 73: 25–29. 10.1090/S00029939197905120516
Jawhari E, Misane D, Pouzet M: Retracts: graphs and ordered sets from the metric point of view. Contemp Math 1986, 57: 175–226.
Khamsi MA, Misane D: Fixed point theorems in logic programming. Ann Math Artif Intell 1997, 21: 231–243. 10.1023/A:1018969519807
Khamsi MA, Kirk WA, Martinez Yanez C: Fixed point and selection theorems in hyperconvex spaces. Proc Amer Math Soc 2000, 128: 3275–3283. 10.1090/S0002993900057774
Aksoy AG, Khamsi MA: A selection theorem in metric trees. Proc Amer Math Soc 2006, 134: 2957–2966. 10.1090/S0002993906085558
Baillon JB: Nonexpansive mappings and hyperconvex spaces. Contemp Math 1988, 72: 11–19.
Chen J, Li Z: Common fixed points for Banach operator pairs in best approximation. J Math Anal Appl 2007, 336: 1466–1475. 10.1016/j.jmaa.2007.01.064
Espinola R, Hussain N: Common fixed points for multimaps in metric spaces. Fixed Point Theory Appl 2010., 14: 2010, Article ID 204981
Hussain N: Common fixed points in best approximation for Banach operator pairs with Ciric type Icontractions. J Math Anal Appl 2008, 338: 1351–1362. 10.1016/j.jmaa.2007.06.008
Pathak HK, Hussain N: Common fixed points for Banach operator pairs with applications. Nonlinear Anal 2008, 69: 2788–2802. 10.1016/j.na.2007.08.051
Hussain N, Khamsi MA, Latif A: Banach operator pairs and common fixed points in hyperconvex metric spaces. Nonlinear Anal TMA 2011,74(17):5956–5961. 10.1016/j.na.2011.05.072
De Marr R: Common fixed points for commuting contraction mappings. Pacific J Math 1963, 13: 1139–1141.
Acknowledgements
The authors were grateful to King Fahd University of Petroleum & Minerals for supporting research project IN 101008.
Author information
Authors and Affiliations
Corresponding author
Additional information
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
All authors participated in the design of this work and performed equally. All authors read and approved the final manuscript.
Rights and permissions
Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
About this article
Cite this article
AbuSbeih, M.Z., Khamsi, M.A. On externally complete subsets and common fixed points in partially ordered sets. Fixed Point Theory Appl 2011, 97 (2011). https://doi.org/10.1186/16871812201197
Received:
Accepted:
Published:
DOI: https://doi.org/10.1186/16871812201197
Keywords
 partially ordered sets
 order preserving mappings
 order trees
 hyperconvex metric spaces
 fixed point