Generalizations of Szpilrajn's Theorem in economic and game theories
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Generalizations of Szpilrajn’s Theorem in economicand game theories
Athanasios Andrikopoulos
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Abstract
Szpilrajn’s Lemma entails that each partial order extends to a lin-ear order. Dushnik and Miller use Szpilrajn’s Lemma to show that each partialorder has a relizer. Since then, many authors utilize Szpilrajn’s Theorem andthe Well-ordering principle to prove more general existence type theorems onextending binary relations. Nevertheless, we are often interested not only inthe existence of extensions of a binary relation R satisfying certain axioms oforderability, but in something more: (A) The conditions of the sets of alter-natives and the properties which R satisfies to be inherited when one passesto any member of a subfamily of the family of extensions of R and: (B) Thesize of a family of ordering extensions of R , whose intersection is R , to be thesmallest one. The key to addressing these kinds of problems is the szpilrajn in-herited method. In this paper, we define the notion of Λ ( m )-consistency, where m can reach the first infinite ordinal ω , and we give two general inherited typetheorems on extending binary relations, a Szpilrajn type and a Dushnik-Millertype theorem, which generalize all the well known existence and inherited typeextension theorems in the literature. Keywords
Consistent binary relations, Extension theorems, Intersection ofbinary relations.
JEL Classification
C60, D00, D60, D71.
One of the most fundamental results on extensions of binary relations is thefollowing theorem proved by E. Szpilrajn in 1930 [30].
Athanasios AndrikopoulosUniversity of PatrasDepartment of Computer Engineering & InformaticsTel.: +302610992113E-mail: [email protected] Athanasios Andrikopoulos
Theorem 1
Let ≤ be a partial order on a set X and let x and y be twoincomparable elements of X (neither x ≤ y nor y ≤ x ). Then, there existsa linear order ≤ ∗ on X which contains all pairs of ≤ and all pairs ( κ, λ ) forwhich κ ≤ y and x ≤ λ holds.The original proof of the theorem splits into two steps: In the first step, if x and y are two incomparable elements of X , then it is constructed a partialorder ≤ ′ such that every element of A = { κ ∈ X | κ ≤ y } must lie below, withrespect to ≤ ′ , to every element of B = { λ ∈ X | x ≤ λ } . So we take ≤ ′′ = ≤∪ ≤ ′ = ≤ ∪ ( A × B ), in other words, we include these obvious consequences ofputting y below x and no others. Clearly, R ′ is transitive. In the second step,we see that if ⊑ is a maximal element (under inclusion) in the set of partialorders extending ≤ ′ , then, ⊑ must be a total order. Because otherwise, if x and y are incomparable in ⊑ , then, by the first step, we have an extension ⊑ ′ of ⊑ such that y ⊑ ′ x , a contradiction to the maximality of ⊑ . (We havefirst to show that the union of a chain of posets is a poset. This is a standardZorn’s Lemma argument.)The crucial point in the original proof of Szpilrajn’s Theorem is the rela-tionship of the pair ( x, y ) with the pairs ( κ, λ ) ∈ A × B which concludes thetransitive axiom for the relation ≤ ′ . In fact: ( α ) For every linear extension ⊑ of ≤ , y ⊑ x implies κ ⊑ λ and: ( β ) x < ′ λ implies y < ′ λ and κ < ′ y implies κ < ′ x . In case ( α ), we say that ( y, x ) covers ( κ, λ ) and in case ( β ),we say that ( y, x ) ia an uncovered pair of ≤ ′ (see [23, Lemma 5]). The truemeaning of the Szpilrajn theorem is that, although it is not constructive, itpreserves prescribed properties in the extended relation. In fact, by extendinga binary relation R , it is interesting to see whether the conditions of the un-derlying space X or the properties which R satisfies should be inherited whenone passes to any member of some family of linear extensions of R . More-over, in extending a binary relation relation R , the problem will often be howto incorporate some additional data depending on the binary relation with aminimum of disruption of the existing structure or how to extend the relationso that some desirable new condition is fulfilled. For example, as shows case( β ) above, if we might wish to adjoin the pair ( x, y ) to a transitive relation R that does not already relates x and y , in order to preserve the relation R the axiom of transitivity, we must also adjoin all other pairs of the form ( κ, λ )where ( κ, y ) ∈ R and ( x, λ ) ∈ R . Generally speaking, a natural question in aextension process is to ask, when a given binary relation R defined on a set ofalternatives X will preserve the properties and the characteristics of X , or of R . For instance, if we refer to a property ( P ) of R , the answer is affirmative if(P) is the property that the chains generated by R are well-ordered (see [4]) orif (P) is the property that x ∗ ∈ X is a maximal element of R (see Proposition4 below). Addressing a slightly different question, we might wish to find con-ditions under which the properties which R or X satisfies to be inherited whenone passes to a linear extension of R . For example, Fucks [15, Corollary 13]finds conditions under which homogeneous partial orders can be extended tohomogeneous linear orders. Kontolatou and Stabakis [21] give an analogue of eneralizations of Szpilrajn’s Theorem in economic and game theories 3 the Szpilrajn Lemma for partially ordered abelian groups. On the other hand,many other papers in the literature deal with the characterization of the set ofbinary relations which have an ordering extension that satisfies some additionalconditions. See, among others, Demuynck [8] for the additional conditions ofconvexity, monotonicity and homotheticity and Demuynck and Lauwers [9]for the condition of linearity. If X is endowed with some topology τ one ismainly interested in continuous or semicontinuous linear orders or preordersinstead of only linear orders or preorders. In this case, two natural problemshave to be discussed: (a) Let R be a continuous binary relation defined ona topological space ( X, τ ). Determine necessary and sufficient conditions for R to have a continuous linear order extension; (b) Determine necessary andsufficient conditions for τ on X to have the Szpilrajn property (every contin-uous binary relation R on X has a continuous linear order extension). In thisdirection, some authors utilize the method of Szpilrajn to find the conditionsunder which τ is preserved in the extended relation. For example, Jaffray [20]and Bossert, Sprumont, and Suzumura [5] provide conditions for the existenceof upper semicontinuous extensions of strict (or weak) orders and consistentbinary relations, respectively and Herden and Pallack [18] provide conditionsfor the existence of continuous extensions.In conclusion, there are many types of conditions that one may wish topreserve, or to achieve, in an extension process. These include:(i) Order theoretic conditions (consistency, acyclicity, transitivity, com-pleteness, e.t.c.);(ii) Topological conditions (continuity, openness or closedness of the pref-erence sets);(iii) linear-space conditions (convexity, homogeneity, translation-inva-riance).In the following, we call as inherited type extensions theorems , all thesetheorems that preserve, or achieve in an extension process the properties ofthe original space. The main feature of these theorems, is that they don’t usein their proof the Szpilrajn’s Theorem.On the other hand, many authors give generalizations of the Szpilrajn’sresult by utilizing the original theorem. In what follows, we refer to such resultsas existence type extension theorems . Arrow [1, page 64], Hansson [17] andFishburn [14] prove on the basis of the original Szpilrajn’s extension theoremthat the result remains true if asymmetry is replaced with reflexivity, thatis, any quasi-ordering has an ordering extension. While the property of beinga quasi-ordering is sufficient for the existence of an ordering extension of arelation, this is not necessary. As shown by Suzumura [29], consistency, as itis defined by him, is necessary and sufficient for the existence of an orderingextension. The existence type extension theorems have played an importantrole in the theory of choice. One way of assessing whether a preference relationis rational is to check whether it can be extended to a transitive and complete It is well known that the economic approach to rational behaviour traditionally beginswith a preference relation R and determines the optimal choice function F from R . Revealed Athanasios Andrikopoulos relation (see [7] and [24] ). In addition, the Szpilrajn’s existence type theoremsare applied: (i) By Stehr [27] to characterize the global orientability; (ii) BySholomov [26] to characterize ordinal relations; (iii) By Nehring and Puppe [22]on a unifying structure of abstract choice theory; (iv) By Blackorby, Bossertand Donaldson [3] in pure population problems e.t.c.Dushnik and Miller [13] use the Szpilrajn’s Theorem to prove the followingresult: Theorem 2
Let ≤ be a partial order on a set X . Then, there exists a collectionof linear extensions F of R such that: ( α ) The intersection of the membersof F coincides with ≤ and: ( β ) for every pair of elements x, y ∈ X with x incomparable to y , there exists an Q ∈ F with ( x, y ) ∈ Q .A family F of linear extensions of R which satisfies conditions ( α ) and( β ) is called a realizer of R . By the theorem of Szpilrajn, for every pair ( x, y )of incomparable elements of R we choose two linear extensions ≤ xy and ≤ yx for which there holds x ≤ xy y and y ≤ yx x . Then, the intersection of alllinear orderings ≤ xy and ≤ yx , where ( x, y ) runs through the set of all pairs ofincomparable elements of R is the relation ≤ . But, this set of linear extensions,has many more elements than necessary. As a consequence of what we have saidabove, by using the notion of uncovered pair, one can obtain a partial order ≤ with the intersection of a reduced number of its linear order extensions. Theconcept of a realizer F of R leads to the definition of dimension of ≤ . Accordingto Dushnik and Miller, the dimension of a partial order ≤ is defined as theminimum size of a realizer of ≤ . In fact, the Dushnik-Miller’s theorem providesa procedure that represent binary relations as an intersection of a number oflinear order extensions equal to its dimension. In what follows, given a binaryrelation R , a Dushnik-Miller existence type extension theorem means that thereexists a collection of linear extensions F of R whose intersection is R and aDushnik-Miller inherited type extension theorem means that R has a realizer.Much of economic and social behavior observed is either group behavior orthat of an individual acting for a group. Group preferences may be regarded asderived from individual preferences, by means of some process of aggregation.For example, if all voters agree that some alternative x is preferred to anotheralternative y , then the majority rule will return this ranking. In this case, thereis one simple condition that is nearly always assumed called the principle ofunanimity or Pareto principle . This declares that the preference relation fora group of individuals should include the intersection of their individual pref-erences. Another example of the use of intersections is in the description of preference theory provides another axiomatic approach to rational behavior by reversing theabove procedure. In particular Szpilrajn theorem is the main tool for proving a known theorem of Richterthat establishes the equivalence between rational and congruous consumers. Let ( R , R , ..., R n ) be a fixed profile of the individual preference relations. A binaryrelation Q , is called Pareto unanimity relation , if xQy ⇔ xR i y for all i ∈ { , , ..., n } and all x, y ∈ X .If R , R , ..., R n are transitive then Q is quasi-transitive.eneralizations of Szpilrajn’s Theorem in economic and game theories 5 simple games which can be represented as the intersection of weighted ma-jority games [16]. Dushnik-Miller existence type theorems have been given bymany authors. For example, the sufficient part of Suzumuras’s extension re-sult, was subsequently used by Donaldson and Weymark [11] in their proofthat every quasi-ordering is the intersection of a collection of orderings; thisresult extends Dushnik and Miller’s fundamental observation on intersectionsof strict linear orders. Duggan [12] proves a general Dushnik-Miller existencetype theorem from which the above results -and several new ones- can be ob-tained as special cases. On the existence of a social welfare ordering for a fixedprofile in the sense of Bergson and Samuelson, Weymark [31] applies Dushnikand Miller extension theorem in order to prove a generalization of Moulin’sPareto extension theorem.In this paper, we introduce the notion of Λ ( m )-consistency, where m be-longs to the set of all ordinals less than or equal to the first infinite ordinal ω and we characterize: ( a ) The existence of a general inherited type theoremon extending binary relations and: ( b ) The existence of a realizer for a bi-nary relation. The results of the two given general inherited type theoremson extending binary relations, namely, the Szpilrajn type extension theoremand the Dushnik-Miller type extension theorem, generalize all the well knownexistence and inherited type extension theorems in the literature. We also giveexamples in a general context to highlight the importance of the inheritedtype extension theorems and to illustrate its difference from the notion of theexistence type extension theorems. Let X be a non-empty universal set of alternatives, and let R ⊆ X × X be a binary relation on X . We sometimes abbreviate ( x, y ) ∈ R as xRy . The composition of two binary relations R and R is given by R ◦ R where ( x, y ) ∈ R ◦ R if and only if there exists z ∈ X such that ( x, z ) ∈ R and ( z, y ) ∈ R .A binary relation R can always be composed with itself, that is R ◦ R = R .This can be generalized to a relation R m on X where R m = R ◦ R ◦ ... ◦ R ( m -times). Let P ( R ) and I ( R ) denote, respectively, the asymmetric part of R andthe symmetric part of R , which are defined, respectively, by P ( R ) = { ( x, y ) ∈ X × X | ( x, y ) ∈ R and ( y, x ) / ∈ R } and I ( R ) = { ( x, y ) ∈ X × X | ( x, y ) ∈ R and ( y, x ) ∈ R } . Let also ∆ = { ( x, x ) | x ∈ X } denotes the diagonal ox X . Anelement x ∈ X is called maximal if for all y ∈ X , yRx implies xRy . We saythat R on X is (i) reflexive if for each x ∈ X ( x, x ) ∈ R ; (ii) irreflexive ifwe never have ( x, x ) ∈ R ; (iii) transitive if for all x, y, z ∈ X , [( x, z ) ∈ R and( z, y ) ∈ R ] = ⇒ ( x, y ) ∈ R ; (iv) antisymmetric if for each x, y ∈ X , [( x, y ) ∈ R and ( y, x ) ∈ R ] = ⇒ x = y ; (v) total if for each x, y ∈ X , x = y we have xRy or yRx . (vi) complete if for each x, y ∈ X , we have xRy or yRx . Itfollows that R is complete if and only if it is reflexive and total. The transitiveclosure of a relation R is denoted by R , that is for all x, y ∈ X, ( x, y ) ∈ R Athanasios Andrikopoulos if there exist m ∈ N and z , ..., z m ∈ X such that x = z , ( z k , z k +1 ) ∈ R forall k ∈ { , ..., m − } and z m = y . Clearly, R is transitive and, because thecase m = 1 is included, it follows that R ⊆ R . Acyclicity says that there donot exist m and z , z , ..., z m ∈ X such that x = z , ( z k , z k +1 ) ∈ R for all k ∈ { , ..., m − } and z m = x . The relation R is S - consistent (consistentin the sense of Suzumura [29]), if for all x, y ∈ X, for all m ∈ N , and forall z , z , ..., z m ∈ X , if x = z , ( z k , z k +1 ) ∈ R for all k ∈ { , ..., m − } and z m = y , we have that ( y, x ) / ∈ P ( R ). The following combination of propertiesare considered in the next theorems. A binary relation R on X is (i) quasi-ordering if R is reflexive and transitive; (ii) ordering if R is a total quasi-ordering; (iii) partial order if R is an antisymmetric quasi-ordering; (iv) linearorder if R is a total partial order; (v) strict partial order if R is irreflexive andtransitive. (vi) strict linear order if R is a total strict partial order. A binaryrelation Q is an extension of a binary relation R if and only if R ⊆ Q and P ( R ) ⊆ P ( Q ). If an extension Q of R is an ordering, we call it an orderingextension of R , and if Q is an extension of R that is a linear order, we refer toit as a linear order extension or R . In fact, an extension Q of R subsumes allthe pairwise information provided by R , and possibly further information.The following definitions may be seen as natural extensions of classicaldefinitions used in the partial order case. Let inc ( R ) = { ( x, y ) ∈ X × X | ( x, y ) / ∈ R and ( y, x ) / ∈ R } be the set of incomparable pairs of R . The set of all of thelinear extensions of R is denoted by Q . For ( x, y ) and ( κ, λ ) ∈ inc ( R ) we write(( x, y ) , ( κ, λ )) ∈ F - in words ( x, y ) covers ( κ, λ )- if for every linear extension Q of R , ( x, y ) ∈ Q implies ( κ, λ ) ∈ Q . We call a maximal element ( x ∗ , y ∗ ) of( inc ( R ) , F ), i.e., ( x ∗ , y ∗ ) in M ( inc ( R ) , F ), an uncovered pair of R . By F ( x,y ) we denote the set { ( κ, λ ) ∈ inc ( R ) | (( x, y ) , ( κ, λ )) ∈ F } . Any subset F ⊆ Q is a realizer of R if and only if: (¯ α ) The intersection of the members of F coincideswith R and: ( ¯ β ) for every pair ( x, y ) ∈ inc ( R ), x, y ∈ X , there exists an Q ∈ F with ( x, y ) ∈ Q . The dimension of a binary relation R (see [13, Page 601]) isthe smallest number of linear orderings whose intersection is R .Let R be a binary relation defined on a topological space ( X, τ ). We saythat R is continuous , if it is a closed subset of X × X . This is the same thing assaying that for every point x ∈ X , both sets { y ∈ X | xRy } and { y ∈ X | yRx } are closed subsets of X (see [25, Proposition 1]). We say that R is uppersemicontinuous if for all y ∈ X , the set { x ∈ X | ( x, y ) ∈ P ( R ) } is open in X .In general, there is no relationship between a binary relation and a topologyon a space. However, there is one topology that is inherently connected witha total order R , called the order topology , which is generated by the subbaseconsisting of all sets of the form { x ∈ X | xP ( R ) a } and { x ∈ X | bP ( R ) x } , where a and b are points of X . The space ( X, τ ) )is compact if for each collection of τ -open sets which cover X there exists a finite subcollection that also covers X . eneralizations of Szpilrajn’s Theorem in economic and game theories 7 In the context of examining if the individualistic assumptions used in eco-nomics can be used in the aggregation of individual preferences ([1, Definition5, Theorem 2], Arrow proved a key lemma that extends the famous Szpilrajn’sTheorem.
Arrow’s lemma .[1, pp. 64-68]. Let R be a quasi-ordering on X , Y a subsetof X such that, if x = y and x, y ∈ Y , then ( x, y ) / ∈ R , and T an ordering on Y . Then, there exists an ordering extension Q such that Q/Y = T .In fact, the lemma says that, if R is a binary relation defined on a set ofalternatives X , then given any ordering T to any subset Y of incomparableelements of R , there is a way of ordering all the alternatives which will becompatible both with R and with the given ordering T in Y . In this case, itis important that the linear extension of R inherits the relationship we putbetween the incomparable elements of R .Arrow’s generalization of the Szpilrajn’s extension theorem as well as allthe well known generalizations of this theorem, use in their proof the Szpilrajntheorem itself. This procedure lead us in existence type extension theorems.In the following ω denotes the first infinite ordinal which comes after allnatural numbers, that is, the order type of the natural numbers under theirusual linear ordering. By Ω we denote the set { ω, , , , ...... } .A great deal of work in computational economics and Computational socialscience has been done in an attempt to find a fast algorithm to count the exactnumber of linear extensions of a partial order, as well as, to find an efficientalgorithm to compute the dimension of a partial order. In this direction, we givetwo general inherited type extension theorems, by reducing the path length ofthe transitive closure in the definition of S -consistency to a minimum, withoutlosing information. To be more precise, we give the following definition. Definition 1
Let R be a binary relation on a set X , let m ∈ Ω and let x, y ∈ X . We say that: (i) R is m - consistent , if for all x, y ∈ X and for all z , z , ..., z m ∈ X , if x = z , ( z k , z k +1 ) ∈ R for all k ∈ { , ..., m − } and z m = y , we have that ( y, x ) / ∈ P ( R ); (ii) R has the m - rank of symmetry if foreach n ≥ m we have I ( R n ) = I ( R m ); (iii) R is Λ ( m )- consistent if m is thelargest natural satisfying m -consistency and m -rank of symmetry. Remark 1 If R is Λ ( m )-consistent, then it is m ′ -consistent for all 1 ≤ m ′ ≤ m .Therefore, if there exist x, g , g , ..., g m ′ ∈ X such that x = g , ( g k , g k +1 ) ∈ R for all k ∈ { , ..., m ′ − } , and g m ′ = x , then we have that ( g k , g k +1 ) ∈ I ( R ) . The following proposition is evident from Definition 1(i).
Proposition 1
Let X be a non empty set and let m ∈ Ω . A binary relation R on X is m -consistent if and only if P ( R ) ⊆ P ( R m ). Athanasios Andrikopoulos If m = ω , then R ω = ∞ [ k =1 R k = R . Since I ( R ω ) = I ( R ) = I ( R ω ′ ) holdsfor all ordinals ω ′ ≥ ω , Definition 1 and Proposition 1 imply the followingproposition. Proposition 2
A binary relation R is Λ ( ω )-consistent if and only if R is S -consistent.As shows in the following example, an m -consistent binary relation is notan S -consistent one. Example 1
Let X = { x , x , x , x , x } and G = { ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) } .Clearly, G is a 2-consistent binary relation but not an S -consistent one. Theorem 3
Let R be a binary relation on X , m ∈ Ω and Y a subset of X such that, if x = y and x, y ∈ Y , then ( x, y ) / ∈ R , and T an ordering on Y .Then, there exists an ordering extension Q of R such that Q/Y = T if andonly if R is a Λ ( m )-consistent binary relation. Proof
To prove necessity, let R be a Λ ( m )-consistent binary relation on X .Without loss of generality, we can assume that R is reflexive (see [30, Lemma,Page 387]). We put R ∗ = R ∪ { ( κ, λ ) | κRy and xRλ, where x, y ∈ Y and ( y, x ) ∈ T } = R ∪ b R. Since R is reflexive, we have ( y, x ) ∈ b R and x = y for all x, y ∈ Y . It is easyto check that( R ∗ ) m = R m ∪{ ( κ, λ ) | κRy and xRλ, where x, y ∈ Y and ( y, x ) ∈ T } = R m ∪ b R. By the definition of R ∗ , we have κ = λ , because otherwise ( x, y ) ∈ R ◦ R = R ,a contradiction. We first prove that R ∗ R ∗ is Λ ( m )-consistent. Indeed, supposeto the contrary that there are alternatives ν, z , z , z , ..., z m ∈ X such that ν = z P ( R ∪ b R ) z ( R ∪ b R ) z ... ( R ∪ b R ) z m = ν. Since R is Λ ( m )-consistent, there must exists k = 0 , , , ..., m − z k , z k +1 ) ∈ b R and for all i ∈ { , , , ..., m − } with i = k , ( z i , z i +1 ) ∈ R ∗ if and only if ( z i , z i +1 ) ∈ R . It follows that ( x, y ) ∈ R , a contradiction. Itremains to prove that for each n ≥ m there holds I (( R ∗ ) n ) = I (( R ∗ ) m ). Let κ, λ ∈ X and n ∈ N be such that ( κ, λ ) ∈ I (( R ∗ ) n ). Then, we have four casesto consider: Case 1. ( κ, λ ) ∈ R n and ( λ, κ ) ∈ R n . It follows that ( κ, λ ) ∈ I ( R m ) ⊆ I (( R ∗ ) m ). Case 2. ( κ, λ ) ∈ R n , ( λ, κ ) ∈ b R . It follows that ( κ, λ ) ∈ R m , ( x, κ ) ∈ R and( λ, y ) ∈ R . Therefore, ( x, y ) ∈ R which is impossible. eneralizations of Szpilrajn’s Theorem in economic and game theories 9 Case 3.
It is similar to case 2.
Case 4.
In this case we have ( κ, y ) ∈ R , ( x, λ ) ∈ R , ( λ, y ) ∈ R and ( x, κ ) ∈ R .It follows that ( x, y ) ∈ R which is impossible. Therefore, I (( R ∗ ) n ) = I ( R n ) = I ( R m ) ⊆ I (( R ∗ ) m ). The last conclusion shows that R ∗ is a Λ ( m )-consistentbinary relation on X satisfying R ∪ T ⊆ R ∗ . We now prove that R ∗ is anextension of R , that is, R ⊆ R ∗ and P ( R ) ⊆ P ( R ∗ ). The first is obvious fromthe definition of R ∗ . To prove the second, let ( κ, λ ) ∈ P ( R ). Then, ( κ, λ ) ∈ P ( R ) ⊆ R ⊆ R ∗ . Suppose to the contrary that ( κ, λ ) / ∈ P ( R ∗ ). It follows that( λ, κ ) ∈ R ∗ . We have two cases to consider: ( α ) ( λ, κ ) ∈ R ; ( β ) λRy and xRκ .In case (i), we have a contradiction to ( κ, λ ) ∈ P ( R ). In case (ii), λRy and xRκ jointly to and ( κ, λ ) ∈ P ( R ) implies that ( x, y ) ∈ R ◦ P ( R ) ◦ R ⊆ R whichis impossible. The last contradiction shows that ( κ, λ ) ∈ P ( R ∗ ) which impliesthat P ( R ) ⊆ P ( R ∗ ).Suppose that e R = { e R i | i ∈ I } denote the set of Λ ( m )-consistent extensionsof R such that R ∪ T ⊆ e R i . Since R ∗ ∈ e R we have that e R 6 = ∅ . Let Q = ( Q i ) i ∈ I be a chain in e R , and let b Q = [ i ∈ I Q i . We prove that b Q ∈ e R . Clearly, R ∪ T ⊆ b Q .To prove that b Q is a Λ ( m )-consistent extension of R , we first show that b Q is m -consistent. Indeed, suppose to the contrary that there are alternatives µ, γ , γ , γ , ..., γ m ∈ X such that µ = γ P ( b Q ) γ b Qγ ... b Qγ m b Qγ m = µ. Consider the largest i for which there exist such µ, γ , γ , γ , ..., γ m +1 . It followsthat Q i is non m -consistent, a contradiction. Therefore, b Q is m -consistent.On the other hand, if n ≥ m , then I ( Q ni ) = I ( Q mi ) for all i ∈ I impliesthat I (( [ i ∈ I Q i ) n ) = I (( [ i ∈ I Q i ) m ). Indeed, let κ, λ ∈ X such that ( κ, λ ) ∈ I (( [ i ∈ I Q i ) n ) and ( κ, λ ) / ∈ I (( [ i ∈ I Q i ) m ). Since ( Q i ) i ∈ I is a chain, there exists i ∗ ∈ I such that ( κ, λ ) ∈ I ( Q n i ∗ ) and ( κ, λ ) / ∈ I ( Q m i ∗ ), a contradiction to I ( Q n i ∗ ) = I ( Q m i ∗ ). The last conclusion shows that b Q is Λ ( m )-consistent. Wenow prove that P ( R ) ⊆ P ( b Q ). Take any ( κ, λ ) ∈ P ( R ) and suppose to thecontrary that ( κ, λ ) / ∈ P ( b Q ). Clearly, κ = λ and for each i ∈ I , ( κ, λ ) ∈ Q i .Since ( κ, λ ) / ∈ P ( [ i ∈ I Q i ) we conclude that ( λ, κ ) ∈ [ i ∈ I Q i . Hence, ( λ, κ ) ∈ Q i ∗ for some i ∗ ∈ I , a contradiction to ( κ, λ ) ∈ P ( R ) ⊆ P ( Q i ∗ ). Therefore, b Q isa Λ ( m )-consistent extension of R such that R ∪ T ⊆ b Q . By Zorn’s lemma e R possesses an element, say Q , that is maximal with respect to set inclusion.We prove that Q is a ordering extension of R satisfying the requirements oftheorem. Since Q is reflexive and transitive, it remains to prove that: ( b α ) Q is total and ( b β ) P ( Q ) ⊆ P ( Q ). To prove ( b α ), take any x, y ∈ X such that( x, y ) / ∈ Q and ( y, x ) / ∈ Q . Then, we have two subcases to consider: ( c α ) x, y ∈ Y ; ( c α ) x / ∈ Y or y / ∈ Y . In subcase ( c α ) we have ( x, y ) / ∈ T . By the completeness of T we conclude that ( y, x ) ∈ T ⊆ Q ⊆ Q , a contradiction. Insubcase ( c α ), if ( x, y ) / ∈ Q and ( y, x ) / ∈ Q , we define Q ∗ = Q ∪ { ( κ, λ ) | κQy and xQλ } .Then, as in the case of R ∗ above, where Y ∗ = { x, y } and T ∗ = { ( y, x ) } playthe role of Y and T respectively, we conclude that Q ∗ is a Λ ( m )-consistentextension of R , a contradiction to the maximality of Q . The last contradic-tion shows that Q is complete (total and reflexive). To prove ( β ), we firstprove that Q = Q m [ ( ∞ [ p = m +1 P ( Q p )). Clearly, Q m [ ( ∞ [ p = m +1 P ( Q p )) ⊆ Q .To prove the converse, suppose to the contrary that ( κ, λ ) ∈ Q and ( κ, λ ) / ∈ Q m [ ( ∞ [ p = m +1 P ( Q p )). Since ( κ, λ ) ∈ Q \ Q m , there exists ρ > m such that( κ, λ ) ∈ Q ρ . On the other hand, ( κ, λ ) / ∈ ∞ [ p = m +1 P ( Q p ) implies that ( κ, λ ) ∈ I ( Q ρ ). By m -rank equivalence we have I ( Q ρ ) = I ( Q m ), a contradiction to( κ, λ ) / ∈ Q m . Therefore, Q = Q m [ ( ∞ [ p = m +1 P ( Q p )). To prove that P ( Q ) ⊆ P ( Q ), suppose to the contrary that ( κ, λ ) ∈ P ( Q ) ⊆ P ( Q m ) and ( κ, λ ) / ∈ P ( Q ). It follows that ( λ, κ ) ∈ Q m [ ( ∞ [ p = m +1 P ( Q p )). Since ( κ, λ ) ∈ P ( Q m ) weconclude that ( λ, κ ) ∈ ∞ [ p = m +1 P ( Q p ). It follows that ( κ, λ ) ∈ I ( Q q ) for some q > m . But then, ( κ, λ ) ∈ I ( Q q ) = I ( Q m ), a contradiction to ( λ, κ ) / ∈ Q m .Therefore, P ( Q ) ⊆ P ( Q ). To complete the sufficiency part we show that Q/Y = T . Evidently, T ⊆ Q/Y . To prove the converse, let ( κ, λ ) ∈ Q/Y .Suppose to the contrary that ( κ, λ ) / ∈ T . Since T is complete ( λ, κ ) ∈ T holdswhich implies ( λ, κ ) ∈ R ∗ . On the other hand, ( κ, λ ) / ∈ T and ( κ, λ ) / ∈ Q ⊇ R ( κ, λ ∈ Y ) imply that ( κ, λ ) / ∈ R ∗ . Since Q is an ordering extension of R ∗ , wehave that ( λ, κ ) ∈ P ( R ∗ ) ⊆ P ( Q ) ⊆ P ( Q ). It follows that ( κ, λ ) / ∈ Q/Y , acontradiction. The last contradiction shows that
Q/Y = T .In order to prove sufficiency, let us assume that R has an ordering extension Q satisfying the requirements of the theorem. Then, R is S -consistent and thus Λ ( ω )-consistent. Indeed, suppose to the contrary that there are alternatives τ, π , π , π , ..., π σ ∈ X such that τ = π P ( R ) π Rπ R...Rπ σ Rπ = τ. Since Q is an ordering extension of R we have τ P ( Q ) τ which is impossible.Therefore, R is S -consistent. The last conclusion completes the proof. Remark 2
According to Theorem 3, m -consistency ensures the existence of areflexive and complete (tournament) extension of R and it has nothing to do eneralizations of Szpilrajn’s Theorem in economic and game theories 11 with the existence of transitivity. As we can see, the relation G of example 1has as a complete extension, the relation G ∗ = G [ ( ∞ [ p =3 P ( G p )) = G ∪ P ( G ) ∪ P ( G ) = { ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) , ( x , x ) . This means that Theorem 3 guarantees a complete extension G ∗ without G being S -consistent.The following corollary is an immediate consequence of Theorem 3 for Y = { x, y } and T = { ( y, x ) } . Corollary 1
Let R be a Λ ( m )-consistent binary relation on X , m ∈ Ω . Then,for every pair ( x, y ) ∈ inc ( R ), x, y ∈ X , there exists an ordering extension Q xy of R such that ( x, y ) ∈ Q xy .By interchanging the roles of x and y ( inc ( R ) is symmetric), Corollary 1gives an analogous result.Since transitivity implies Λ ( m )-consistency, m ∈ Ω , Arrow’s Lemma is animmediate consequence of the sufficient part of Theorem 3 for m = 1. Definition 2
For each m ∈ N , a Λ ( m )-consistent binary relation R on X is ∆ ( m )- consistent if I ( R m ) = ∆ .A consequence of Proposition 2 and Theorem 3 for m = ω is also theSuzumura’s existence type extension theorem in [28, Page 5]. The followingcorollary shows this fact. Corollary 2
Let R be a binary relation on X , Y a subset of X such that, if x = y and x, y ∈ Y , then ( x, y ) / ∈ R , and T an ordering on Y . Then, thereexists an ordering extension Q of R such that Q/Y = T if and only if R is S -consistent.As a consequence of the proof of Theorem 3, the following result is alsotrue: Corollary 3
Let R be a binary relation on X , m ∈ Ω and Y a subset of X such that, if x = y and x, y ∈ Y , then ( x, y ) / ∈ R , and T a linear order on Y .Then, there exists a linear order extension Q of R such that Q/Y = T if andonly if R is a ∆ ( m )-consistent. Proof
According to Proposition 1 and Theorem 3, there exists an orderingextension Q of R such that P ( R ) ⊆ P ( R m ) ⊆ P ( Q ) and Q/Y = T . Let ≈ bethe equivalence relation defined by x ≈ y if and only if ( x, y ) ∈ I ( Q ). The quotient set by this equivalence relation ≈ will be denoted X ≈ = X ≈ , andits elements (equivalence classes) by [ x ]. There exists on X a linear order Q defined by:( ∀ x, y ∈ X ) ( x Q y ⇔ ∃ x ′ , y ′ ∈ X, x ∈ [ x ′ ] , y ∈ [ y ′ ] , [ x ′ ] Ry ′ )An asymmetric, Λ ( m )-consistent binary relation R ≈ is defined by: ∀ [ x ] , [ y ] ∈ X ≈ ([ x ] R ≈ [ y ] ⇔ ∃ x ′ ∈ [ x ] , ∃ y ′ ∈ [ y ] , x ′ Ry ′ ).According to Corollary 4, there exists a strict linear order extension f Q ≈ of R ≈ . Therefore, as in the proof of Theorem 1 [20, Pages 399-400], by using asubbase of τ , we construct a strict linear order extension R ∗ of R such thatfor each y ∈ X the set { x ∈ X | xR ∗ y } belongs to τ .We prove that ( x, y ) ∈ I ( R m ) = ∆ , which implies that Q is antisymmetricand thus it is a linear order. Suppose to the contrary that ( x, y ) / ∈ I ( R ) = I ( R m ). Then, since ( x, y ) ∈ P ( R m ) ⊆ P ( Q ) and ( y, x ) ∈ P ( R m ) ⊆ P ( Q ) isimpossible, we conclude that ( x, y ) / ∈ R m and ( y, x ) / ∈ R m .But then, x, y ∈ Y and ( x, y ) ∈ T or ( y, x ) ∈ T which implies that xP ( Q ) y or yP ( Q ) x which is impossible. The last contradiction shows that Q is a linearextension of R .Conversely, if there exists a linear order extension Q of R , then by Theorem3, R is Λ ( ω )-consistent. It remains to show that I ( R ) = ∆ . Suppose to thecontrary that I ( R ) = ∆ . This implies that, there exist x, y ∈ X , x = y , suchthat ( x, y ) ∈ I ( R ) ⊆ I ( Q ) and ( x, y ) ∆ which contradicts the anti-symmetry of Q .If R is a partial order, Corolarry 3 implies one of the main results of [19,Theorem 2.2].As a corollary to Theorem 3, we also obtain the following well knowninherited type extension theorem of Szpilrajn [30]. Corollary 4
Every (strict) partial order R possesses a (strict) linear orderextension Q . Moreover, if x and y are any two non-comparable elements of R , then there exists a (strict) linear order extension Q ′ in which xQ ′ y and a(strict) linear order extension Q ′′ in which yQ ′′ x . Proof
It is an immediate result of Theorem 3 for Y = { x, y } , T = { ( x, y ) } , m = 1 and R being (asymmetric and transitive) reflexive, transitive and anti-symmetric.Since transitivity implies Λ ( m )-consistency for al m ∈ Ω , the followingcorollary is an immediate consequence of Theorem 3. Corollary 5 (Hanson [17] and Fishburn [14]). Every quasi-ordering has anordering extension. eneralizations of Szpilrajn’s Theorem in economic and game theories 13
In this paragraph, we give a general Dushnik-Miller inherited type extensiontheorem in which all the well known Dushnik-Miller extension theorems areobtained as special cases.
Theorem 4
Let R be a binary relation on X . Then, R has as realizer theset of ordering extensions of R if and only if R is Λ ( m )-consistent for some m ∈ Ω . Proof
To prove necessity, let R be a Λ ( m )-consistent binary relation on X forsome m ∈ Ω and let Q be the set of all order extensions of R . By Theorem3, Q is non-empty. We show that R = \ Q ∈Q Q . Indeed, since R ⊆ \ Q ∈Q Q , wehave to show that \ Q ∈Q Q ⊆ R . Suppose to the contrary that there exists an( x, y ) ∈ \ Q ∈Q Q with ( x, y ) / ∈ R . We first prove that ( y, x ) / ∈ R . Suppose tothe contrary that ( y, x ) ∈ R . Since ( y, x ) ∈ P ( R ) ⊆ P ( Q ) contradicts the factthat ( x, y ) ∈ Q , we conclude that ( y, x ) / ∈ R . Define R ′ = R ∪ { ( y, x ) } .Clearly, R ⊂ R ′ ⊆ R . We also have P ( R ) ⊂ P ( R ′ ) ⊆ P ( R ). To prove thesecond inclusion suppose to the contrary that ( κ, λ ) ∈ P ( R ′ ) and ( κ, λ ) / ∈ P ( R ). It follows that ( λ, κ ) ∈ R . If ( κ, λ ) = ( y, x ), then ( x, y ) ∈ R which isimpossible. If ( κ, λ ) ∈ P ( R ′ ) \ { ( y, x ) } = P ( R ), then ( κ, λ ) ∈ I ( R ) = I ( R m ) acontradiction to ( κ, λ ) ∈ P ( R ) and ( λ, κ ) ∈ R m . Therefore, P ( R ) ⊂ P ( R ′ ) ⊆ P ( R ). If R is complete then R ∈ Q . But then, ( y, x ) ∈ P ( R ) implies that( x, y ) / ∈ \ Q ∈Q Q , a contradiction). Thus, R is incomplete. Let T be the set oftransitive extensions of R . Since R ∈ T , this set is non-empty. Then, as in theproof of Theorem 1,there exists a maximal element b T of T . We prove that b T is complete. Suppose to the contrary that ( x ∗ , y ∗ ) / ∈ b T and ( y ∗ , x ∗ ) / ∈ b T forsome x ∗ , y ∗ ∈ X . Then, it is easy to check that the relation e T = b T ∪ { ( κ, λ ) | κ b T y ∗ and x ∗ b T λ } is transitive, a contradiction to the maximal character of b T . Therefore, b T ∈ Q .Since ( y, x ) ∈ P ( R ) ⊆ P ( b T ) we have ( x, y ) / ∈ b T , again a contradiction to( x, y ) ∈ \ Q ∈Q Q . Therefore, in any case we have ( y, x ) / ∈ R . We now provethat ( x, y ) / ∈ R jointly to ( y, x ) / ∈ R leads again to a contradiction, and thus, \ Q ∈Q Q ⊆ R . Indeed, let S = R ∪ { ( κ, λ ) | κRy and xRλ } . Then, since R is Λ ( m )-consistent, as in the proof of Theorem 1, there exists anordering extension b S of R such that P ( R ) ⊆ P ( S ) ⊆ P ( b S ). Since ( y, x ) ∈ P ( S )(( x, x ) ∈ R , ( y, y ) ∈ R , ( x, y ) / ∈ R and ( y, x ) R ) we have that ( x, y ) / ∈ b S , acontradiction to ( x, y ) ∈ \ Q ∈Q Q . This contradiction confirms that \ Q ∈Q Q ⊆ R .To finish the proof of necesity, it remains to show that Q is a realizer. But,this is an immediate consequence of the Corollary 1.To prove sufficiency, suppose that R has as realizer the set of all orderextensions of R , let Q . We prove that R is Λ ( ω )-consistent. Indeed, since \ Q ∈Q Q = R we have P ( R ) ⊆ \ Q ∈Q P ( Q ) ⊆ P ( \ Q ∈Q Q ) = P ( R ) = P ( R ω ).Therefore, R is Λ ( ω )-consistent. The last conclusion completes the proof.Theorems 3 and 6 and remark 2 imply the following corollary. Corollary 6
Let R be a binary relation on X . Then, R m has as realizer theset of reflexive and complete (tournament) extensions of R if and only if R is m -consistent for some m ∈ Ω .The following result is an immediate consequence of Corollary 3. Corollary 7
Let R be a binary relation on X and let m ∈ Ω . Then, R hasas realizer the set of linear order extensions of R if and only if R is ∆ ( m )-consistent.The following corollary is an irreflexive variant of Corollary 7. Corollary 8
Let R be a binary relation on X and let m ∈ Ω . Then, R hasas realizer the set of strict linear order extensions of R if and only if R isasymmetric and Λ ( m )-consistent. Proof
Suppose that R is an asymmetric and Λ ( m )-consistent binary relationfor some m ∈ Ω . Then, R ∪ ∆ is ∆ ( m )-consistent. Let Q be the class of linearorder extensions of R . Then, R ∪ ∆ = \ Q ∈Q Q . It follows that R = \ Q ∈Q Q \ ∆ ,where Q \ ∆ is a strict linear order extension of R . Conversely, suppose that R is the intersection of all strict linear order extensions of R . Then, R is Λ ( ω )-consistent and asymmetric since I ( R ) ⊆ I ( R ) ⊆ I ( Q ) = ∅ .Next is a result due to Dushnik and Miller [13, Theorem 2.32]. Corollary 9 If R is any (strict) partial order on a set X , then there exists acollection Q of (strict) linear orders on X which realize R . Proof
This follows immediately from Theorem 6, by letting R to be (strict)partial order.The next result, proved by Donaldson and Weymark [11], strengthens Fish-burn’s Lemma 15.4 in [14] and Suzumura’s Theorem A(4) in [29]. eneralizations of Szpilrajn’s Theorem in economic and game theories 15 Corollary 10
Every quasi-ordering is the intersection of a collection of or-derings.
Proof
It is an immediate consequence of the sufficiency part of Theorem 6, byletting m = 1. Definition 3 [12, Definition 6]. Given relations R and R ′ , R ′ is a compatibleextension of R if R ⊆ R ′ and P ( R ) ⊆ P ( R ′ ).In what follows, R denotes the class of binary relations which are compatibleextensions of R .We recall the following definitions from [12]. Definition 4
The class R is closed upward if, for all chains C in R , S { R ′ | R ′ ∈ C} ∈ R . Definition 5
The class R is arc-receptive if, for all distinct s and t and forall transitive R ′ ∈ R , ( t, s ) / ∈ R ′ implies R ∪ { ( s, t ) } ∈ R . Proposition 3
Assume R is closed upward and arc-receptive. If R is Λ ( m )-consistent for some m ∈ Ω and R ∈ R , then R = T { R ′ ∈ R| R ′ is a complete, transitive extension of R } . Proof
To prove the corollary, let m ∈ N and R be a Λ ( m )-consistent binaryrelation such that R ∈ R . It follows from Theorem 6 that, R = T { R ′ | R ′ is a complete, transitive extension of R } .It remains to prove that R ′ ∈ R . Because R ⊆ R ′ by transitivity of R ′ , weobtain R ⊆ R ′ . If R ′ = R , then R ′ ∈ R . Otherwise, suppose that R ⊂ R ′ . Wefirst show that there exists a transitive extension of R , let Q , such that Q ∈ R and R ⊂ Q ⊆ R ′ . Indeed, assume that s, t ∈ X are such that ( s, t ) ∈ R ′ \ R .There are two cases to consider: (i) ( t, s ) ∈ R ′ ; (ii) ( t, s ) / ∈ R ′ .Case (i). ( t, s ) ∈ R ′ . In this case, since R is arc-receptive, R ∈ R and ( s, t ) / ∈ R we conclude that Q = R ∪ { ( t, s ) } ∈ R . We now prove that Q is a transitiveextension of R . Since Q is transitive, it suffices to show that Q is an extensionof R . Clearly, R ⊂ Q . To verify that P ( R ) ⊂ P ( Q ), take any ( p, q ) ∈ P ( R )and suppose ( p, q ) / ∈ P ( Q ).Since ( p, q ) ∈ R ⊂ R ∪ { ( t, s ) } , this means that ( q, p ) ∈ R ∪ { ( t, s ) } . Hence,there exists z , z , z , ..., z m ∈ X such that q = z { R ∪ { ( t, s ) }} z { R ∪ { ( t, s ) }} z ... { R ∪ { ( t, s ) }} z m = p .Thus, there exists at least one k ∈ { , , ..., m − } such that ( z k , z k +1 ) = ( t, s ),for otherwise ( q, p ) ∈ R , a contradiction. Let z λ be the first occurrence of t and let z µ the last occurrence of s . Then, since ( p, q ) ∈ P ( R ) ⊆ R , s = z µ Rz µ +1 ...Rz m = pRqRz ...Rz λ = t .Hence, ( s, t ) ∈ R , a contradiction. Since R ′ is transitive, R ⊂ Q ⊆ R ′ .Case (ii). ( t, s ) / ∈ R ′ . In this case, we must have ( t, s ) / ∈ R , since otherwise, wemust have ( t, s ) ∈ R ′ , a contradiction.Let Q = R ∪ { ( s, t ) } . Then, as in the case (i), we obtain Q ∈ R and R ⊂ Q ⊆ R ′ . Let b Q = ( c Q i ) i ∈ I be the set of transitive extensions of R such that R ⊂ c Q i ⊆ R ′ and c Q i ∈ R . Let C be a chain in b Q , and b C = S C . Clearly, R ⊂ b C ⊆ R ′ . Since R is closed upward, b C ∈ R . Therefore, by Zorn’s lemma, b Q has an element, say e Q , that is maximal with respect to set inclusion. Then, R ′ = e Q ∈ R . Otherwise, there exists ( s, t ) ∈ R ′ \ e Q such that Q ′ = e Q ∪ { ( s, t ) } or Q ′ = e Q ∪ { ( t, s ) } is a transitive extension of R satisfying R m ⊂ e Q ⊂ Q ′ ⊆ R ′ , which is impossible by maximality of e Q . This completes the proof.Since S -consistency is equivalent to Λ ( ω )-consistency, the following resultis an immediate corollary of the previous proposition. Corollary 11 (Duggan’s General Extension Theorem [12]). Assume R is closedupward and arc-receptive. If R is S -consistent and R ∈ R , then R = T { R ′ ∈ R| R ′ is a complete, transitive extension of R } .Clearly, Theorem 6 concludes all the extension theorems referred to Duggan[12, pp. 13-14]. Actually, it is well known that the notion of maximal element has interestingapplications to the study of economic and game theories. In fact, it plays acentral role in many economic models, including global maximum of a utilityfunction and Nash equilibrium of a noncooperative game or equilibrium ofan economy (Debreu [10]). We prove the following propositions as a generalapplication of the notion of inherited type extension theorems.
Proposition 4
Let R be a ∆ ( m )-consistent binary relation on some non-empty set X , m ∈ N , and let x ∗ be a maximal element of R in X . Then, thereexists a linear order extension Q of R such that x ∗ is a maximal element of Q in X . Proof
We first show that x ∗ is a maximal element of R . Indeed, suppose tothe contrary that ( y, x ∗ ) ∈ P ( R ) for some y ∈ X . It then follows that thereexists l ∈ N and alternatives t , t , ..., t l such that yRt ...t m − Rt l Rx ∗ . Since( t l , x ∗ ) / ∈ P ( R ), we conclude that ( t l , x ∗ ) ∈ I ( R ) ⊆ I ( R m ). Hence, because of ∆ ( m )-consistency, we conclude that t m = x ∗ . Similarly, ( t m − , x ∗ ) ∈ R , and eneralizations of Szpilrajn’s Theorem in economic and game theories 17 an induction argument based on this logic yields y = x ∗ , a contradiction to( y, x ∗ ) ∈ P ( R ). Hence, x ∗ is a maximal element of R . If R is complete, then itis a linear order extension of R which has x ∗ as maximal element. Otherwise,there are x, y ∈ X such that ( x, y ) / ∈ R and ( y, x ) / ∈ R . Clearly, one of x and y is different from x ∗ . Let x = x ∗ . We define R ∗ = R ∪ { ( κ, λ ) | κRy, xRλ, ( x, y ) ∈ inc ( R ) x, y ∈ X, and x = x ∗ } . Then, as in Theorem 3, we conclude that R ∗ is a Λ ( m )-consistent extensionof R . Since I (( R ∗ ) m ) = I ( R m ) = ∆ , we conclude that R ∗ is ∆ ( m )-consistent.To show that x ∗ is a maximal element of R ∗ , suppose to the contrary that( κ, x ∗ ) ∈ P ( R ∗ ) for some κ ∈ X . Since x ∗ is a maximal element of R , weconclude that κRy and xRx ∗ . It follows that ( x, x ∗ ) ∈ I ( R ) = I ( R m ) = ∆ ,a contradiction to x = x ∗ . Hence, x ∗ is a maximal element of R ∗ . Supposethat e R = { e R i | i ∈ I } denote the set of ∆ ( m )-consistent extensions of R whichhas x ∗ as maximal element. Since R ∗ ∈ e R we have that e R 6 = ∅ . Let Q =( Q i ) i ∈ I be a chain in e R , and let b Q = [ i ∈ I Q i . We show that b Q ∈ e R . As in theproof of Theorem 3, we conclude that b Q is a ∆ ( m )-consistent extension of R .To verify that x ∗ is a maximal element of b Q , take any y ∈ X and suppose( y, x ∗ ) ∈ P ( b Q ) = P ( [ i ∈ I Q i ). Clearly, y = x ∗ and ( y, x ∗ ) ∈ Q i ∗ for some i ∗ ∈ I .Since ( x ∗ , y ) / ∈ [ i ∈ I Q i we conclude that ( x ∗ , y ) / ∈ Q i for each i ∈ I . Hence,( y, x ∗ ) ∈ P ( Q i ∗ ), a contradiction to Q i ∗ ∈ e R . Therefore, b Q ∈ e R . By Zorn’slemma e R possesses an element, say Q , that is maximal with respect to setinclusion. Therefore, as above we can prove that Q is an extension of R whichhas x ∗ as maximal element. We prove that Q is complete. Indeed, take any x, y ∈ X such that ( x, y ) / ∈ Q and ( y, x ) / ∈ Q . We define Q ∗ = Q ∪ { ( κ, λ ) | κQy, xQλ, ( x, y ) / ∈ Q, ( y, x ) / ∈ Q and x = x ∗ } . Then, as in case of R ∗ above, we have that Q ∗ is a ∆ ( m )-consistent binaryrelation which has x ∗ as maximal element, a contradiction to the maximalityof Q . The last contradiction implies that Q is complete.As a corollary of the previous result we have a generalization of SophieBade’s result in [2, Theorem 1](she uses transitive binary relations) whichshows that the set of Nash equilibria of any game with incomplete prefer-ences can be characterized in terms of certain derived games with completepreferences. More general, it is shown a similarity between the theory of gameswith incomplete preferences and the existing theory of games with completepreferences. I put in mind the following definition: In this case, G = { ( A i , R i ) | i ∈ I } is an arbitrary (normal-form) game. Where I is a setof players, player i ’s nonempty action space is denoted by A i and R i is player i ’s preferencerelation on the outcome space A = Y i ∈ I A i .8 Athanasios Andrikopoulos Definition . We say that a game G ′ = { ( A i , R ′ i ) | i ∈ I } is a completion ofa game G = { ( A i , R i ) | i ∈ I } if R ′ i is a complete extension of R i for each i . In what follows, we denote the set of all Nash equilibria of a game G by N ( G ). In the following theorem, each preference relation R i is assumed to be ∆ ( m )-consistent for some m ∈ N . Corollary 12
Let G = { ( A i , R i ) | i ∈ I } be any game. Then N ( G ) = S { N ( G ′ ) | G ′ is a completion of G } .Proof Clearly, S { N ( G ′ ) | G ′ is a completion of G } ⊆ N ( G ). Conversely, let a ∗ ∈ N ( G ), that is, a ∗ is a Nash equilibrium of G . Let us define B i = { ( a i , a ∗− i ) | a i ∈ A i } for all players i . Then, for any player i , a ∗ is a maximal element of R i in B i . By Proposition 4, there exists a completion R ′ i of R i for each player i suchthat a ∗ is maximal point of R ′ i in B i . Consequently a ∗ is a Nash equilibriumof the completion G ′ = { ( A i , R ′ i ) | i ∈ I } . Hence, N ( G ) ⊆ S { N ( G ′ ) | G ′ is acompletion of G } .As we have pointed out in the introduction, we are often interested inparticular binary relations which have an ordering extension that satisfies someadditional conditions. The following proposition, which generalizes the mainresult in [20], is an application to this specific case. Proposition 5
Let (
X, τ ) be a topological space and m ∈ Ω . If R is anasymmetric, Λ ( m )-consistent and upper semicontinuous binary relation on X ,then R has an upper semicontinuous strict linear order extension. Proof
To begin with, we associate to R the equivalence relation ≈ defined by x ≈ y if and only if ( ∀ z ∈ X )[( zRx ⇔ zRy ) and ( xRz ⇔ yRz )],that is, x ≈ y if and only if x covers y and y covers x . The quotient set by thisequivalence relation ≈ will be denoted X ≈ = X ≈ , and its elements (equivalenceclasses) by [ x ].An asymmetric, Λ ( m )-consistent binary relation R ≈ is defined by: ∀ [ x ] , [ y ] ∈ X ≈ ([ x ] R ≈ [ y ] ⇔ ∃ x ′ ∈ [ x ] , ∃ y ′ ∈ [ y ] , x ′ Ry ′ ).According to Corollary 4, there exists a strict linear order extension f Q ≈ of R ≈ . Therefore, as in the proof of Theorem 1 [20, Pages 399-400], by using asubbase of τ , we construct a strict linear order extension R ∗ of R such thatfor each y ∈ X the set { x ∈ X | xR ∗ y } belongs to τ .In direction of the inherited type Szpilrajn extension theorems, Demuynck[8] give results for complete extensions satisfying various additional propertiessuch as convexity, homotheticity and monotonicity. Since Demuynck’s paper An action profile a = ( a , ..., a | I | ) is a Nash equilibrium if for no player i there exists anaction a ′ i ∈ A i such that ( a ′ i , a − i ) R i ( a i , a − i ).eneralizations of Szpilrajn’s Theorem in economic and game theories 19 generalizes S -consistency by replacing the transitive closure R of R with amore general function F , I conjecture that these results can be extended tothe case of Λ ( m )-consistent binary relation for all m ∈ Ω .The following proposition is given as a general application of the Dushnik-Miller’s inherited type extension theorem.For each x ∈ X , we define (see [25, Page 20]): i ( x ) = { y ∈ X | xRy } , d ( x ) = { y ∈ X | yRx } and I x = i ( x ) ∪ d ( x ). For any x ∈ X , x m and x M denotethe minimum and the maximum of I x . Definition 6
A binary relation R on X has finite decomposition incompa-rability , if there exists n ∈ N and ( x µ , y µ ) ∈ inc ( R ), µ ∈ { , , ..., n } , suchthat:(1) ( x mµ , y Mµ ) / ∈ R and ( y mµ , x Mµ ) / ∈ R , and(2) inc ( R ) = { ( κ, λ ) ∈ I xµ × I yµ | ( κ, λ ) ∈ inc ( R ) , ≤ µ ≤ n } . Proposition 6
Let m ∈ Ω and let R be a continuous Λ ( m )-consistent binaryrelation on a topological space ( X, τ ) having finite decomposition incompara-bility. Then, the dimension of R is finite. Proof
Without loss of generality, assume that R is reflexive. We first showthat for each x ∈ X the sets i ( x ) and d ( x ) are closed. To prove the case of i ( x ), let z / ∈ i ( x ). Then, ( x, z ) / ∈ R ⊇ R . Then, by [25, Proposition 1], thereexists an open R -increasing neighbourhood O x of x and an open R -decreasingneighbourhood O z of z such that O x ∩ O z = ∅ . Since x ∈ O x and O x is R -increasing we conclude that i ( x ) ⊆ O x . It follows that z ∈ O z ⊆ X \ i ( x ).Therefore, i ( x )is closed. Similarly, we prove that d ( x ) is closed which impliesthat I x is closed as well. Hence, I x is compact. Then, I x has a maximumelement x M and a minimum element x m . To see this, note that if I x has nolargest element, then { I x \ d ( z ) | z ∈ I x } is an open cover of I x in subspacetopology with no finite subcover, and if I x has no smallest element, then { I x \ i ( z ) | z ∈ I x } is an open cover of I x in subspace topology with no finite subcover.Since R has finite decomposition incomparability, there exists n ∈ N and( x µ , y µ ) ∈ inc ( R ), µ ∈ { , , ..., n } , such that ( x mµ , y Mµ ) / ∈ R and ( y mµ , x Mµ ) / ∈ R ,and inc ( R ) = { ( κ, λ ) ∈ I xµ × I yµ | ≤ µ ≤ n } . On the other hand, by thecontinuity of R we have ( y Mµ , x mµ ) / ∈ R and ( x Mµ , y mµ ) / ∈ R . It follows that( x mµ , y Mµ ) , ( y mµ , x Mµ ) ∈ inc ( R ). We prove that dim ( R ) ≤ n . We define R µ = R ∪ { ( κ, λ ) ∈ X × X | κRy Mµ and x mµ Rλ } and R Dµ = R ∪ { ( λ, κ ) ∈ X × X | λRx Mµ and y mµ Rκ } .By Theorem 3, there exist linear order extensions Q µ and Q Dµ of R such that inc ( R ) ∩ ( I yµ × I xµ ) ⊆ Q µ and inc ( R ) ∩ ( I xµ × I yµ ) ⊆ Q Dµ . We prove that R = n \ m =1 ( Q m ∩ Q Dm ). Clearly, R ⊆ n \ m =1 ( Q m ∩ Q Dm ). To provethe converse, let ( α, β ) ∈ n \ m =1 ( Q m ∩ Q Dm ) and ( α, β ) / ∈ R . The proof proceedsin a similar way to Theorem 6, as follows: We first prove that ( β, α ) / ∈ R andby the finite decomposition incomparability property of R , there exists µ ∗ ∈{ , , ..., n } such that ( α, β ) ∈ I xµ ∗ × I yµ ∗ . Then, we prove that ( α, β ) / ∈ Q µ ∗ ,a contradiction to ( α, β ) ∈ n \ m =1 ( Q m ∩ Q Dm ). The last conclusion completes theproof.Another example is the following: In the games that are compositions of m individualist games ( N, u i ) ( i = 1 , ..., m ) via unanimity, the usual descriptionof the game, by means of minimal winning coalitions, requires n · ... · n m coalitions (with n i = | N i | ) and if each one of them has m players, then, m · n · ... · n m digits are needed to describe the game. Using [16, Theorem3.1], ( n + 1) · ( m − p ) ( p < m ) digits are required to describe the game.This latter number is generally much smaller than the former, and so, thedescription of the game is much shorter.Many other interesting applications of the dimension of a binary relationare obtained in Economics. For example, Ok [ ? , Proposition 1] shows that if( X, ≻ ) is a preordered set with X countable and dim ( X, ≻ ) < ∞ , then ≻ isrepresentable by means of a real function u in such a way that x ≻ y if andonly if u ( x ) > u ( y ). From the multicriteria point of view, the classical crisp dimension refers to a minimal representation of crisp partial orders as theintersection of linear orders, in the sense that each of one of these linear ordersis a possible underlying criterion. Brightwell and Scheinerman [6], on the basisof Dushnik-Miller’s original theorem, prove that the fractional dimension of If a game with player set N = { , ..., n } admits a partition N , ..., N m in such a waythat W = { S ⊆ N : | S ∩ N i | ≥
1, for all i = 1 , ..., m } we shall say that this game is a composition of m individualist games via unanimity. Let ( N, W ) be a composition of m individualist games ( N i , u i ) ( i = 1 , ..., m ) with1 ≤ n ≤ ... ≤ n m via unanimity and let p < m such that either n p = 1, n p +1 > p = 0if n >
1. Then the dimension of ( N, W ) is m − p . Given a finite set of alternatives X = { x , x , ..., x n } , a crisp partial order set R ⊆ X × X is characterized by a mapping µ : X × X −→ { , } being(i) irreflexive: µ ( x i , x i ) = 0 ∀ x i ∈ X, (ii) antisymmetric: µ ( x i , x j ) = 1 ⇒ µ ( x j , x i ) = 0,(iii) transitive: µ ( x i , x j ) = µ ( x j , x k ) = 1 ⇒ µ ( x i , x k ) = 1. It is therefore assumed that µ ( x i , x j ) = 1 means that alternative x i is strictly better than x j ( µ ( x i , x j ) = 0 otherwise). Brightwell and Scheinerman [6] introduce the notion of fractional dimension of a poset( X, ≻ ). Let F = { L , L , ..., L t } be a nonempty multiset of linear extensions of ( X, ≻ ). Theeneralizations of Szpilrajn’s Theorem in economic and game theories 21 a partially ordered set ( X, ≻ ) arises naturally when considering a particulartwo-person game on ( X, ≻ ), e.t.c. References
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