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Representability of Interval Orders
Journal of Economic Theory ET2346 journal of economic theory 78, 219227 (1998) article no. ET972346
Representability of Interval Orders Esteban Oloriz Departamento de Matematica e Informatica, Universidad Publica de Navarra, Campus Arrosad@ a s.n., E-31006, Pamplona, Spain
Juan Carlos Candeal Facultad de Ciencias Economicas y Empresariales, Departamento de Analisis Economico, Universidad de Zaragoza, C Doctor Cerrada 1-3, E-50005 Zaragoza, Spain
and Esteban Indurain Departamento de Matematica e Informatica, Universidad Publica de Navarra, Campus Arrosad@ a s.n., E-31006, Pamplona, Spain Received November 3, 1996; revised July 19, 1997
We obtain a full characterization of the representability of interval orders by a pair of real-valued functions. Journal of Economic Literature Classification numbers: 022, 025, 213. 1998 Academic Press
1. INTRODUCTION The concept of interval order appeared in the economics literature to deal with preferences without making the assumption of transitivity of the indifference. It was analyzed in the Journal of Economic Theory trilogy due to Bridges [13]. In those papers, and also in Chap. 6 of the recent book by Bridges and Mehta [4], the question of finding a full characterization of the existence of representability of an interval order appears as an open problem. In this Note we solve that problem.
2. PRELIMINARIES Let X be a nonempty set. An interval order is a reflexive binary relation ``'' defined on X such that ``xz and yt'' O ``xt or yz'', for every 219 0022-053198 25.00 Copyright 1998 by Academic Press All rights of reproduction in any form reserved.
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x, y, z, t # X. The relation ``'' is usually called the weak preference relative to the interval order. An interval order ``'' is said to be representable if there exist two real-valued functions u, v: X R such that xy u(x) v( y) for every x, y # X. A particular case of interval order appears when ``'' is transitive. In such case (X, ) is said to be totally preordered. A totally preordered structure (X, ) is called representable by a utility function if there exists a function u : X R such that xy u(x) u( y)(x, y # X ). (The map ``u'' is called utility). It is perfectly separable if there exists a countable subset D X such that for every x, y # X with x O y there exists d # D such that xdy. Remark 1. (i) Observe that every interval order ``'' is total, because being x, y # X, by reflexivity we have xx and yy, so by definition of interval order, it follows that either xy or yx must hold. (ii) Being ``'' an interval order on X, the binary relation ``O'' defined by xO y c( yx) is said to be the strict interval order (or strict preference) associated to ``''. It is straightforward to see that `` O '' is asymmetric and also ``xO z and y O t'' O ``x O t or y O z''. (iii) Being ``'' an interval order on X, for every x, y, z # X such that xyz it follows from definition that ``xO y or y Oz'' O ``xz''. Moreover it is straightforward to see that the last property is equivalent to the transitivity of `` O ''. This property is called quasitransitivity of the relation ``''. (iv) In general, an interval order ``'' defined on X is not necessarily transitive. In addition, its associated indifference ``t'' given by xty ``xy and yx'' may also fail to be transitive. As an example, consider the interval order ``'' defined on R by xy x< y+1. (v) If u, v is a pair of functions that represent an interval order (X, ), i.e.: xy u(x) v( y)(x, y # X ), then in particular u(x) v(x) for every x # X since ``'' is reflexive. Lemma 1. Let X be a nonempty set, and let F : X_X R be a function. Then the following statements are equivalent: (i) There exist real-valued functions u, v: X R such that F(x, y)= v( y)&u(x), for every x, y # X, (ii) F is a solution of the functional equation F(x, y)+F( y, z)= F(x, z)+F( y, y), for every x, y, z # X. Proof. If in (ii) we fix z=z 0 # X and call u(x)=&F(x, z 0 ), v( y)=F( y, y)+u( y), we obtain (i), and if we call F(x, y)=v( y)&u(x) in (i) we get (ii). K
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Notation. Let X be a nonempty set and ``'' an interval order on X. Given x, y # X, denote: A(x, y)=[s # X : there exists a # X such that xa Osy]. Remark 2. At this point it is necessary to explain why these sets A(x, y) have been introduced. The key idea to achieve the main result (Theorem 1) that provides a characterization of representability of intervalorders by means of the existence of solutions of the functional equation F(x, y)+F( y, z)=F(x, z)+F( y, y) is the translation of such functional equation into some equalities between such subsets of X. The process goes as follows: A suitable countable subset D=(d n ) n # N of points of X will be considered (see the definition of i.o.-separability, that we shall introduce later). Then F(x, y) will be defined as the sum of the series F(x, y)=
:
2 &n
if
xy,
if
y O x,
[n # N : dn # D & A(x, y)]
F(x, y)=&F( y, x)+F(x, x)+F( y, y)
understanding that F(x, y)=0 whenever xy and [n # N : d n # D & A(x, y)]=<. In order to check that the function F so defined satisfies the functional equation, we must prove some introductory results about the presence of the elements of D in the sets A(x, y). Lemma 3 will show that an element in A(x, y) & A(z, t) must also be in A(x, t) & A(z, y) whereas Lemma 4 will prove that A(x, y) _ A( y, z)=A(x, z) _ A( y, y), for every x, y, z # X such that xyz and xz. Observe that all this is necessary in order to guarantee that the terms of the series that may appear in F(x, y)+F( y, z) when F(x, y), F( y, z) 0 are exactly the same that the ones that may appear in F(x, z)+F( y, y) when F(x, z), F( y, y) 0, and, in addition, they appear the same number of times. The underlying idea of solving a functional equation by translating it into one or more equalities on subsets is inspired by several typical constructions through series in order to find suitable functions in utility theory: See, e.g., Theorems 1.4.8, 3.2.6, and 6.1.5 in [4]. Finally, an additional lemma (Lemma 5) will be introduced to prove that the representation xy F(x, y) 0(x, y # X ) works. Lemma 2. Let X be a nonempty set and ``'' an interval order on X. Then the following conditions are equivalent: (i) (ii)
``''is transitive, A(x, x)=< for every x # X.
Proof. By definition and Remark 1(i), ``'' is reflexive and total. If it is also transitive and there exist x, s, a # X such that xa Osx, it follows
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that xO x, a contradiction. Conversely, if A(x, x)=< for every x # X then for every a, s # X with xa and sx it follows by hypothesis that c(a O s) must hold, so that sa. K Lemma 3. Let X be a nonempty set, and ``'' an interval order on X. Then A(x, y) & A(z, t)=A(x, t) & A(z, y), for every x, y, z, t # X. Proof. If s # A(x, y) & A(z, t), there exist a, b # X such that xa Osy and zb Ost. Thus xa O st and zb Osy. Therefore s # A(x, t) & A(z, y). Similarly A(x, t) & A(z, y) A(x, y) & A(z, t). Lemma 4. Let X be a nonempty set, and ``'' an interval order on X. Then A(x, y) _ A( y, z)=A(x, z) _ A( y, y), for every x, y, z # X such that xyz and xz. Proof. First let us prove that A(x, y) _ A( y, z) A(x, z) _ A( y, y), for every x, y, z # X such that xyz: If s belongs to A(x, y)"(A(x, z) _ A( y, y)) then there exists a # X such that xa O sy and also z O s, since s A(x, z). In addition, for every b # X with bO s it must hold that b O y, because s A( y, y). Remember now that z Os, so taking b=z it follows that z O y, which is a contradiction. If we suppose now that s # A( y, z)"(A(x, z) _ A( y, y)), there exists a # X such that ya O sz and also yO s. Moreover, for all b # X with bO s it must hold that b O x. Thus, we obtain y O x, a new contradiction. To conclude, let us show that A(x, z) _ A( y, y) A(x, y) _ A( y, z) for every x, y, z # X such that xz: If s # A(x, z)"(A(x, y) _ A( y, z)) there exists a # X satisfying xa O sz and y O s. Also, for every b # X with bO s it must hold b O y since otherwise it would follow yb Osz and this implies s # A( y, z), which is a contradiction. Now, being y Os by hypothesis, we obtain, for the particular case b= y, that y O y, which is a new contradiction. Finally, if s # A( y, y)"(A(x, y) _ A( y, z)), there exists a # X satisfying ya O sy and z Os. Moreover, for every b # X with b O s it must hold that b O x. Now we get z Ox, which is again a contradiction. Lemma 5. Let X be a nonempty set, and ``'' an interval order on X. For every x, y # X with x Oy the following properties hold: (i)
A( y, x)=<,
(ii)
A(x, x) _ A( y, y) / A(x, y),
(iii)
A(x, x) & A( y, y)=<.
Proof. (i) Let x, y # X with x O y, and let us suppose that there exists s # A( y, x). Then there is an element a # X such that ya O sx and by
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definition of interval order, it follows that either yx or sa must hold. Contradiction. (ii) If s # A(x, x) _ A( y, y) either there exists a # X such that xa O sx holds or there exists b # X such that yb O sy holds. Consequently, since x O y, either sx O y or xO yb holds, so that by Remark 1(iii) we obtain that either sy or xb holds. In the first case we conclude xa Osy and, in the second case, we obtain xb Osy. Both possibilities imply that s # A(x, y). In addition, x O y O xx O yy since ``'' is reflexive, so y # A(x, y). Moreover y A( y, y) because, by definition of `` O'', there is no a # X such that yaO yy. Finally y A(x, x) because the existence of a # X such that xa O yx implies, in particular, that yx, which contradicts the hypothesis x O y. Therefore y # A(x, y)"(A(x, x) _ A( y, y)). (iii) This follows directly from Lemma 3 and part (i) of this Lemma 5.
3. CHARACTERIZATION OF INTERVAL ORDERS: THE MAIN RESULT The final open question in [3] asks for natural conditions on an interval order to be representable by a pair of continuous functions. However, even disregarding the continuity, the problem of representability is not yet solved, in spite of elegant partial results (see [2, 3]) and a complete solution for the countable case [1]. So it seems reasonable to believe that the analysis of representability of interval orders should start for furnishing natural conditions of ordinal type, that guarantee such representability. We present such conditions through the main theorem in this section. Definition. Let X be a nonempty set endowed with an interval order ``''. Then the structure (X, ) is said to be i.o.-separable if there exists a countable subset D X such that for every x, y # X with x O y there exists d # D & A(x, y)"(A(x, x) _ A( y, y)). Theorem 1. Let ``'' be an interval order defined on a nonempty set X. Then the following statements are equivalent: (i)
The structure (X, ) is i.o.-separable,
(ii) there exists a function F: X_X R such that xy F(x, y) 0 and F(x, y)+F( y, z)=F(x, z)+F( y, y) for every x, y, z # X, (iii)
(X, ) is representable.
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Proof. (i) O (ii) Let D=[d n : n # N] X be a countable set that makes (X, ) i.o.-separable. Given x, y # X with xy set D(x, y)=[n # N : d n # D & A(x, y)], and define the function F : X_X R as follows: 2 &n
if xy,
F(x, y)=&F( y, x)+F(x, x)+F( y, y)
if y O x,
F(x, y)=
: n # D(x, y)
understanding that F(x, y)=0 whenever xy and D(x, y)=<. Let us prove that F(x, y) 0 xy, or equivalently F(x, y)<0 y O x: By definition, F(x, y)<0 O y O x. For the converse, take x, y # X with x O y. By Lemma 5 (ii) and (iii), we know that A(x, x) & A( y, y)=< and A(x, x) _ A( y, y) / A(x, y). Thus D(x, x) _ D( y, y) / D(x, y) because (X, ) is i.o.-separable. Therefore, by definition of F, F( y, x)<0. Now let us check that F satisfies the functional equation F(x, y)+F( y, z)=F( y, z)+F( y, y)(x, y, z # X ). To do so, we analyze the following cases: 1. Suppose that xy, yz and xz. The equality of the equation follows here by Lemma 3 and Lemma 4, since in this case F(x, y), F(x, z), F( y, z) and F( y, y) are all non-negative. 2. Suppose that xy, yz and zO x. By Remark 1 (iii) it follows that z O x; xy O zy. Also yz holds by hypothesis. Thus we have zyz, as in case 1, so that F(z, y)+F( y, z)=F(z, z)+F( y, y). In the same way, we have zO xy with zy, so that, again as in case 1, F(z, x)+F(x, y)=F(z, y)+F(x, x). Now, by definition of F(x, z) we conclude that F(y, y)+F(x, z)=F(y, y)+F(z, z)+F(x, x)&F(z, x)=F(y, z)+ F(z, y)+F(x, x)&F(z, x)=F( y, z)+F(x, y). 3. Suppose that xy, z O y and xz. Thus we have xz O y with xy. By case 1 and definition of F( y, z) it follows F(x, y)+F( y, z)= F(x, y)+F(z, z)&F(z, y)+F( y, y)=F(x, z)+F( y, y). 4. Suppose that xy, zO y and z Ox. In this case we have z Oxy and z O y. Thus, by case 1 and definition of F (x, z) and F ( y, z) we get F(x, y)+F(y, z)=&F(z, x)+F(x, x)+F(z, y)+F(y, z)=&F(z, x)+F(x, x)+ F(z, z)+F( y, y)=F(x, z)+F( y, y). 5. Suppose that yO x, yz and xz. Then yO xz with yz, so by case 1 and definition of F(x, y) we obtain F( y, z)+F(x, y)=F( y, z)& F( y, x)+F(x, x)+F( y, y)=F(x, z)+F( y, y). 6. Suppose that yO x, yz and zO x. It follows now that yz O x with yO x, so by case 1 and definition of F(x, y) and F(x, z) we have that F(x, y)+F(y, z)=F(x, y)+F( y, x)+F(z, z)&F(z, x)=F( y, y)+ F(x, x)+F(z, z)&F(z, x)=F( y, y)+F(x, z).
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7. Suppose that yO x, z O y and xz. By Remark 1 (iii), z O y and yO x implies that z O x, which contradicts the hypothesis xz. Therefore, this case 7 is impossible. 8. Suppose that y O x, z O y and z O x. Then z O y O x and zO x, so by case 1 again, and definition of F(x, y), F( y, z), F(x, z) it follows that F(x, y)+F( y, z)=F(x, y)+F( y, y)&F(z, y)+F(z, z)=F(x, y)+ F( y, x)&F(z, x)+F(z, z)=F(y, y)+F(x, x)&F(z, x)+F(z, z)=F( y, y)+ F(x, z). (ii) O (iii) This follows directly from Lemma 1 and the definition of representability of interval orders. (iii) O (i) For every pair of rational numbers ( p, q) # Q_Q consider, when it exists, an element d p, q # X such that p u(d p, q ) q. In this way we get a countable subset B=[d p, q : p, q # Q] X. Consider now the subset of the real line u(X )=[u(x) : x # X] R. Define a gap in u(X ) as a maximal nondegenerate interval disjoint from u(X) and with a lower bound and an upper bound in u(X ). Observe then that the set of gaps of u(X ) of the types (a, b) or [a, b) with a, b # R is countable. Let [G n : n # N] be that set of gaps. Thus G n =[a n , b n ) or (a n , b n ) for suitable a n , b n # R. For every n # N choose an element d n # X such that u(d n )=b n . Let C=[d n # X : n # N]. Let us prove now that the countable subset D=B _ C X provides the i.o.-separability of the structure (X, ), that is, for every x, y # X with xO y we must prove the existence of an element d # D & (A(x, y)" (A(x, x) _ A( y, y))). Since A(x, y) is nonempty because xx O yy O y # A(x, y), for every s # A(x, y) we have u(x) v(a s )
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s lies in the open interval (x, y+1) of the real line. Now observe that every x # R satisfies x Ox+1. Therefore, if we assume that (R, ) is separable, there will exist a countable subset D and an element d # D & (A(x, x+1)" (A(x, x) _ A(x+1, x+1))). But A(x, x+1)"(A(x, x) _ A(x+1, x+1)))= [x+1]. This is a contradiction because R is uncountable. Some classical results follow now as direct corollaries of the main theorem just introduced. Corollary 1. Let X be a nonempty set endowed with a total preorder ``''. Then the following statements are equivalent: (i)
(X, ) is representable by a utility function,
(ii) there exists a function F : X_X R such that xy O F(x, y) 0 and F(x, y)+F( y, z)=F(x, z) for every x, y, z # X, (iii)
(X, ) is perfectly separable.
Proof. The equivalence (i) (iii) is a classical result in utility theory. (See, e.g, Theorem 1.4.8 in [4, p. 14]). In the particular case in which an interval order ``'' is a total preorder, conditions (i) and (ii) in Theorem 1 are equivalent to perfect separability because, by Lemma 2, A(x, x)=<, (x # X). Moreover, the function F provided by Theorem 1 (iii) satisfies F(x, x)=0 for every x # X, so that the functions u, v of a representation of the interval order must coincide, and then u=v is a utility function representing (X, ). K Remark 3. Despite in [4] the proof of the equivalence (ii) (iii) in Corollary 1 is not included, and nothing is said about the functional equation F(x, y)+F( y, z)=F(x, z), the proof of (i) (iii) given there (see [4], Theorem 1.4.8 on p. 14) uses the idea, as pointed out in Remark 2, of choosing suitable subsets of X and then, working with some convergent series, turn the functional equation into some equalities between such subsets. Corollary 2. Every interval order ``'' defined on a countable set X is representable. Proof. Take D=X and observe that (X, ) is i.o.-separable by Lemma 5 (ii). (This Corollary 2 is the main result in [1].) K Corollary 3. Let `` O '' be an strict interval order defined on a nonempty set X and suppose that there exists a countable subset B X with the following property: If xO y then there exists d # B and z # X such that x O dz O y. Then ``'' is representable.
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Proof. This Corollary 3 is essentially Lemma 6.1.4 in [4, p. 89]. Given x, y # X with x Oy, denote B(x, y)=[s # X : there exists z # X such that xO sz O y]. Let s # B(x, y). It follows that s # A(x, y) since xx Osy. Plainly s A(x, x) because x O s. Finally, if s # A( y, y) there exists a # X such that ya Osy. Now observe that sz and ya. Hence, by definition of interval order, either sa or yz must hold. This is again a contradiction. So we have proved that B(x, y) A(x, y)"(A(x, x) _ A( y, y)). Therefore, the condition in the statement of Corollary 3 implies the i.o.separability of ``'', so that by Theorem 1 the structure (X, ) is representable. Remark 4. We recall that a totally ordered set (X, ) is Cantorseparable if there exists a countable subset D X such that for every x, y # X with x O y one can find an element d # D such that x O d O y. Keeping this concept in mind, another idea that has inspired the main result in the present paper is searching for an strengthening of the classical extension of Cantor's results about representability of Cantor-separable total orders, to the concept of perfect separability, which is a necessary and sufficient condition for representability of totally ordered sets, whereas Cantor-separability is sufficient but not necessary. (For further details see [4, pp. 16 and ff.]) Observe that, in the extended frame of interval orders, the condition that appears in the statement of Corollary 3 looks like the condition of Cantor-separability for total orders. Actually, it coincides with Cantor-separability when restricted to total orders. Thus, in Theorem 1 we have found a condition of ordinal type (namely, the i.o.-separability) that reminds the perfect separability and is characteristic of the representability of interval orders through a pair of real valued functions.
REFERENCES 1. D. S. Bridges, Numerical representation of intransitive preferences on a countable set, J. Econ. Theory 30 (1983), 213217. 2. D. S. Bridges, Representing interval orders by a single real-valued function, J. Econ. Theory 36 (1985), 149155. 3. D. S. Bridges, Numerical representation of interval orders on a topological space, J. Econ. Theory 38 (1986), 160166. 4. D. S. Bridges and G. B. Mehta, ``Representations of preference orderings,'' SpringerVerlag, Berlin, 1995.
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Representability of Interval Orders Esteban Oloriz Departamento de Matematica e Informatica, Universidad Publica de Navarra, Campus Arrosad@ a s.n., E-31006, Pamplona, Spain
Juan Carlos Candeal Facultad de Ciencias Economicas y Empresariales, Departamento de Analisis Economico, Universidad de Zaragoza, C Doctor Cerrada 1-3, E-50005 Zaragoza, Spain
and Esteban Indurain Departamento de Matematica e Informatica, Universidad Publica de Navarra, Campus Arrosad@ a s.n., E-31006, Pamplona, Spain Received November 3, 1996; revised July 19, 1997
We obtain a full characterization of the representability of interval orders by a pair of real-valued functions. Journal of Economic Literature Classification numbers: 022, 025, 213. 1998 Academic Press
1. INTRODUCTION The concept of interval order appeared in the economics literature to deal with preferences without making the assumption of transitivity of the indifference. It was analyzed in the Journal of Economic Theory trilogy due to Bridges [13]. In those papers, and also in Chap. 6 of the recent book by Bridges and Mehta [4], the question of finding a full characterization of the existence of representability of an interval order appears as an open problem. In this Note we solve that problem.
2. PRELIMINARIES Let X be a nonempty set. An interval order is a reflexive binary relation ``'' defined on X such that ``xz and yt'' O ``xt or yz'', for every 219 0022-053198 25.00 Copyright 1998 by Academic Press All rights of reproduction in any form reserved.
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x, y, z, t # X. The relation ``'' is usually called the weak preference relative to the interval order. An interval order ``'' is said to be representable if there exist two real-valued functions u, v: X R such that xy u(x) v( y) for every x, y # X. A particular case of interval order appears when ``'' is transitive. In such case (X, ) is said to be totally preordered. A totally preordered structure (X, ) is called representable by a utility function if there exists a function u : X R such that xy u(x) u( y)(x, y # X ). (The map ``u'' is called utility). It is perfectly separable if there exists a countable subset D X such that for every x, y # X with x O y there exists d # D such that xdy. Remark 1. (i) Observe that every interval order ``'' is total, because being x, y # X, by reflexivity we have xx and yy, so by definition of interval order, it follows that either xy or yx must hold. (ii) Being ``'' an interval order on X, the binary relation ``O'' defined by xO y c( yx) is said to be the strict interval order (or strict preference) associated to ``''. It is straightforward to see that `` O '' is asymmetric and also ``xO z and y O t'' O ``x O t or y O z''. (iii) Being ``'' an interval order on X, for every x, y, z # X such that xyz it follows from definition that ``xO y or y Oz'' O ``xz''. Moreover it is straightforward to see that the last property is equivalent to the transitivity of `` O ''. This property is called quasitransitivity of the relation ``''. (iv) In general, an interval order ``'' defined on X is not necessarily transitive. In addition, its associated indifference ``t'' given by xty ``xy and yx'' may also fail to be transitive. As an example, consider the interval order ``'' defined on R by xy x< y+1. (v) If u, v is a pair of functions that represent an interval order (X, ), i.e.: xy u(x) v( y)(x, y # X ), then in particular u(x) v(x) for every x # X since ``'' is reflexive. Lemma 1. Let X be a nonempty set, and let F : X_X R be a function. Then the following statements are equivalent: (i) There exist real-valued functions u, v: X R such that F(x, y)= v( y)&u(x), for every x, y # X, (ii) F is a solution of the functional equation F(x, y)+F( y, z)= F(x, z)+F( y, y), for every x, y, z # X. Proof. If in (ii) we fix z=z 0 # X and call u(x)=&F(x, z 0 ), v( y)=F( y, y)+u( y), we obtain (i), and if we call F(x, y)=v( y)&u(x) in (i) we get (ii). K
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Notation. Let X be a nonempty set and ``'' an interval order on X. Given x, y # X, denote: A(x, y)=[s # X : there exists a # X such that xa Osy]. Remark 2. At this point it is necessary to explain why these sets A(x, y) have been introduced. The key idea to achieve the main result (Theorem 1) that provides a characterization of representability of intervalorders by means of the existence of solutions of the functional equation F(x, y)+F( y, z)=F(x, z)+F( y, y) is the translation of such functional equation into some equalities between such subsets of X. The process goes as follows: A suitable countable subset D=(d n ) n # N of points of X will be considered (see the definition of i.o.-separability, that we shall introduce later). Then F(x, y) will be defined as the sum of the series F(x, y)=
:
2 &n
if
xy,
if
y O x,
[n # N : dn # D & A(x, y)]
F(x, y)=&F( y, x)+F(x, x)+F( y, y)
understanding that F(x, y)=0 whenever xy and [n # N : d n # D & A(x, y)]=<. In order to check that the function F so defined satisfies the functional equation, we must prove some introductory results about the presence of the elements of D in the sets A(x, y). Lemma 3 will show that an element in A(x, y) & A(z, t) must also be in A(x, t) & A(z, y) whereas Lemma 4 will prove that A(x, y) _ A( y, z)=A(x, z) _ A( y, y), for every x, y, z # X such that xyz and xz. Observe that all this is necessary in order to guarantee that the terms of the series that may appear in F(x, y)+F( y, z) when F(x, y), F( y, z) 0 are exactly the same that the ones that may appear in F(x, z)+F( y, y) when F(x, z), F( y, y) 0, and, in addition, they appear the same number of times. The underlying idea of solving a functional equation by translating it into one or more equalities on subsets is inspired by several typical constructions through series in order to find suitable functions in utility theory: See, e.g., Theorems 1.4.8, 3.2.6, and 6.1.5 in [4]. Finally, an additional lemma (Lemma 5) will be introduced to prove that the representation xy F(x, y) 0(x, y # X ) works. Lemma 2. Let X be a nonempty set and ``'' an interval order on X. Then the following conditions are equivalent: (i) (ii)
``''is transitive, A(x, x)=< for every x # X.
Proof. By definition and Remark 1(i), ``'' is reflexive and total. If it is also transitive and there exist x, s, a # X such that xa Osx, it follows
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that xO x, a contradiction. Conversely, if A(x, x)=< for every x # X then for every a, s # X with xa and sx it follows by hypothesis that c(a O s) must hold, so that sa. K Lemma 3. Let X be a nonempty set, and ``'' an interval order on X. Then A(x, y) & A(z, t)=A(x, t) & A(z, y), for every x, y, z, t # X. Proof. If s # A(x, y) & A(z, t), there exist a, b # X such that xa Osy and zb Ost. Thus xa O st and zb Osy. Therefore s # A(x, t) & A(z, y). Similarly A(x, t) & A(z, y) A(x, y) & A(z, t). Lemma 4. Let X be a nonempty set, and ``'' an interval order on X. Then A(x, y) _ A( y, z)=A(x, z) _ A( y, y), for every x, y, z # X such that xyz and xz. Proof. First let us prove that A(x, y) _ A( y, z) A(x, z) _ A( y, y), for every x, y, z # X such that xyz: If s belongs to A(x, y)"(A(x, z) _ A( y, y)) then there exists a # X such that xa O sy and also z O s, since s A(x, z). In addition, for every b # X with bO s it must hold that b O y, because s A( y, y). Remember now that z Os, so taking b=z it follows that z O y, which is a contradiction. If we suppose now that s # A( y, z)"(A(x, z) _ A( y, y)), there exists a # X such that ya O sz and also yO s. Moreover, for all b # X with bO s it must hold that b O x. Thus, we obtain y O x, a new contradiction. To conclude, let us show that A(x, z) _ A( y, y) A(x, y) _ A( y, z) for every x, y, z # X such that xz: If s # A(x, z)"(A(x, y) _ A( y, z)) there exists a # X satisfying xa O sz and y O s. Also, for every b # X with bO s it must hold b O y since otherwise it would follow yb Osz and this implies s # A( y, z), which is a contradiction. Now, being y Os by hypothesis, we obtain, for the particular case b= y, that y O y, which is a new contradiction. Finally, if s # A( y, y)"(A(x, y) _ A( y, z)), there exists a # X satisfying ya O sy and z Os. Moreover, for every b # X with b O s it must hold that b O x. Now we get z Ox, which is again a contradiction. Lemma 5. Let X be a nonempty set, and ``'' an interval order on X. For every x, y # X with x Oy the following properties hold: (i)
A( y, x)=<,
(ii)
A(x, x) _ A( y, y) / A(x, y),
(iii)
A(x, x) & A( y, y)=<.
Proof. (i) Let x, y # X with x O y, and let us suppose that there exists s # A( y, x). Then there is an element a # X such that ya O sx and by
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definition of interval order, it follows that either yx or sa must hold. Contradiction. (ii) If s # A(x, x) _ A( y, y) either there exists a # X such that xa O sx holds or there exists b # X such that yb O sy holds. Consequently, since x O y, either sx O y or xO yb holds, so that by Remark 1(iii) we obtain that either sy or xb holds. In the first case we conclude xa Osy and, in the second case, we obtain xb Osy. Both possibilities imply that s # A(x, y). In addition, x O y O xx O yy since ``'' is reflexive, so y # A(x, y). Moreover y A( y, y) because, by definition of `` O'', there is no a # X such that yaO yy. Finally y A(x, x) because the existence of a # X such that xa O yx implies, in particular, that yx, which contradicts the hypothesis x O y. Therefore y # A(x, y)"(A(x, x) _ A( y, y)). (iii) This follows directly from Lemma 3 and part (i) of this Lemma 5.
3. CHARACTERIZATION OF INTERVAL ORDERS: THE MAIN RESULT The final open question in [3] asks for natural conditions on an interval order to be representable by a pair of continuous functions. However, even disregarding the continuity, the problem of representability is not yet solved, in spite of elegant partial results (see [2, 3]) and a complete solution for the countable case [1]. So it seems reasonable to believe that the analysis of representability of interval orders should start for furnishing natural conditions of ordinal type, that guarantee such representability. We present such conditions through the main theorem in this section. Definition. Let X be a nonempty set endowed with an interval order ``''. Then the structure (X, ) is said to be i.o.-separable if there exists a countable subset D X such that for every x, y # X with x O y there exists d # D & A(x, y)"(A(x, x) _ A( y, y)). Theorem 1. Let ``'' be an interval order defined on a nonempty set X. Then the following statements are equivalent: (i)
The structure (X, ) is i.o.-separable,
(ii) there exists a function F: X_X R such that xy F(x, y) 0 and F(x, y)+F( y, z)=F(x, z)+F( y, y) for every x, y, z # X, (iii)
(X, ) is representable.
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Proof. (i) O (ii) Let D=[d n : n # N] X be a countable set that makes (X, ) i.o.-separable. Given x, y # X with xy set D(x, y)=[n # N : d n # D & A(x, y)], and define the function F : X_X R as follows: 2 &n
if xy,
F(x, y)=&F( y, x)+F(x, x)+F( y, y)
if y O x,
F(x, y)=
: n # D(x, y)
understanding that F(x, y)=0 whenever xy and D(x, y)=<. Let us prove that F(x, y) 0 xy, or equivalently F(x, y)<0 y O x: By definition, F(x, y)<0 O y O x. For the converse, take x, y # X with x O y. By Lemma 5 (ii) and (iii), we know that A(x, x) & A( y, y)=< and A(x, x) _ A( y, y) / A(x, y). Thus D(x, x) _ D( y, y) / D(x, y) because (X, ) is i.o.-separable. Therefore, by definition of F, F( y, x)<0. Now let us check that F satisfies the functional equation F(x, y)+F( y, z)=F( y, z)+F( y, y)(x, y, z # X ). To do so, we analyze the following cases: 1. Suppose that xy, yz and xz. The equality of the equation follows here by Lemma 3 and Lemma 4, since in this case F(x, y), F(x, z), F( y, z) and F( y, y) are all non-negative. 2. Suppose that xy, yz and zO x. By Remark 1 (iii) it follows that z O x; xy O zy. Also yz holds by hypothesis. Thus we have zyz, as in case 1, so that F(z, y)+F( y, z)=F(z, z)+F( y, y). In the same way, we have zO xy with zy, so that, again as in case 1, F(z, x)+F(x, y)=F(z, y)+F(x, x). Now, by definition of F(x, z) we conclude that F(y, y)+F(x, z)=F(y, y)+F(z, z)+F(x, x)&F(z, x)=F(y, z)+ F(z, y)+F(x, x)&F(z, x)=F( y, z)+F(x, y). 3. Suppose that xy, z O y and xz. Thus we have xz O y with xy. By case 1 and definition of F( y, z) it follows F(x, y)+F( y, z)= F(x, y)+F(z, z)&F(z, y)+F( y, y)=F(x, z)+F( y, y). 4. Suppose that xy, zO y and z Ox. In this case we have z Oxy and z O y. Thus, by case 1 and definition of F (x, z) and F ( y, z) we get F(x, y)+F(y, z)=&F(z, x)+F(x, x)+F(z, y)+F(y, z)=&F(z, x)+F(x, x)+ F(z, z)+F( y, y)=F(x, z)+F( y, y). 5. Suppose that yO x, yz and xz. Then yO xz with yz, so by case 1 and definition of F(x, y) we obtain F( y, z)+F(x, y)=F( y, z)& F( y, x)+F(x, x)+F( y, y)=F(x, z)+F( y, y). 6. Suppose that yO x, yz and zO x. It follows now that yz O x with yO x, so by case 1 and definition of F(x, y) and F(x, z) we have that F(x, y)+F(y, z)=F(x, y)+F( y, x)+F(z, z)&F(z, x)=F( y, y)+ F(x, x)+F(z, z)&F(z, x)=F( y, y)+F(x, z).
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7. Suppose that yO x, z O y and xz. By Remark 1 (iii), z O y and yO x implies that z O x, which contradicts the hypothesis xz. Therefore, this case 7 is impossible. 8. Suppose that y O x, z O y and z O x. Then z O y O x and zO x, so by case 1 again, and definition of F(x, y), F( y, z), F(x, z) it follows that F(x, y)+F( y, z)=F(x, y)+F( y, y)&F(z, y)+F(z, z)=F(x, y)+ F( y, x)&F(z, x)+F(z, z)=F(y, y)+F(x, x)&F(z, x)+F(z, z)=F( y, y)+ F(x, z). (ii) O (iii) This follows directly from Lemma 1 and the definition of representability of interval orders. (iii) O (i) For every pair of rational numbers ( p, q) # Q_Q consider, when it exists, an element d p, q # X such that p u(d p, q ) q. In this way we get a countable subset B=[d p, q : p, q # Q] X. Consider now the subset of the real line u(X )=[u(x) : x # X] R. Define a gap in u(X ) as a maximal nondegenerate interval disjoint from u(X) and with a lower bound and an upper bound in u(X ). Observe then that the set of gaps of u(X ) of the types (a, b) or [a, b) with a, b # R is countable. Let [G n : n # N] be that set of gaps. Thus G n =[a n , b n ) or (a n , b n ) for suitable a n , b n # R. For every n # N choose an element d n # X such that u(d n )=b n . Let C=[d n # X : n # N]. Let us prove now that the countable subset D=B _ C X provides the i.o.-separability of the structure (X, ), that is, for every x, y # X with xO y we must prove the existence of an element d # D & (A(x, y)" (A(x, x) _ A( y, y))). Since A(x, y) is nonempty because xx O yy O y # A(x, y), for every s # A(x, y) we have u(x) v(a s )
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s lies in the open interval (x, y+1) of the real line. Now observe that every x # R satisfies x Ox+1. Therefore, if we assume that (R, ) is separable, there will exist a countable subset D and an element d # D & (A(x, x+1)" (A(x, x) _ A(x+1, x+1))). But A(x, x+1)"(A(x, x) _ A(x+1, x+1)))= [x+1]. This is a contradiction because R is uncountable. Some classical results follow now as direct corollaries of the main theorem just introduced. Corollary 1. Let X be a nonempty set endowed with a total preorder ``''. Then the following statements are equivalent: (i)
(X, ) is representable by a utility function,
(ii) there exists a function F : X_X R such that xy O F(x, y) 0 and F(x, y)+F( y, z)=F(x, z) for every x, y, z # X, (iii)
(X, ) is perfectly separable.
Proof. The equivalence (i) (iii) is a classical result in utility theory. (See, e.g, Theorem 1.4.8 in [4, p. 14]). In the particular case in which an interval order ``'' is a total preorder, conditions (i) and (ii) in Theorem 1 are equivalent to perfect separability because, by Lemma 2, A(x, x)=<, (x # X). Moreover, the function F provided by Theorem 1 (iii) satisfies F(x, x)=0 for every x # X, so that the functions u, v of a representation of the interval order must coincide, and then u=v is a utility function representing (X, ). K Remark 3. Despite in [4] the proof of the equivalence (ii) (iii) in Corollary 1 is not included, and nothing is said about the functional equation F(x, y)+F( y, z)=F(x, z), the proof of (i) (iii) given there (see [4], Theorem 1.4.8 on p. 14) uses the idea, as pointed out in Remark 2, of choosing suitable subsets of X and then, working with some convergent series, turn the functional equation into some equalities between such subsets. Corollary 2. Every interval order ``'' defined on a countable set X is representable. Proof. Take D=X and observe that (X, ) is i.o.-separable by Lemma 5 (ii). (This Corollary 2 is the main result in [1].) K Corollary 3. Let `` O '' be an strict interval order defined on a nonempty set X and suppose that there exists a countable subset B X with the following property: If xO y then there exists d # B and z # X such that x O dz O y. Then ``'' is representable.
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Proof. This Corollary 3 is essentially Lemma 6.1.4 in [4, p. 89]. Given x, y # X with x Oy, denote B(x, y)=[s # X : there exists z # X such that xO sz O y]. Let s # B(x, y). It follows that s # A(x, y) since xx Osy. Plainly s A(x, x) because x O s. Finally, if s # A( y, y) there exists a # X such that ya Osy. Now observe that sz and ya. Hence, by definition of interval order, either sa or yz must hold. This is again a contradiction. So we have proved that B(x, y) A(x, y)"(A(x, x) _ A( y, y)). Therefore, the condition in the statement of Corollary 3 implies the i.o.separability of ``'', so that by Theorem 1 the structure (X, ) is representable. Remark 4. We recall that a totally ordered set (X, ) is Cantorseparable if there exists a countable subset D X such that for every x, y # X with x O y one can find an element d # D such that x O d O y. Keeping this concept in mind, another idea that has inspired the main result in the present paper is searching for an strengthening of the classical extension of Cantor's results about representability of Cantor-separable total orders, to the concept of perfect separability, which is a necessary and sufficient condition for representability of totally ordered sets, whereas Cantor-separability is sufficient but not necessary. (For further details see [4, pp. 16 and ff.]) Observe that, in the extended frame of interval orders, the condition that appears in the statement of Corollary 3 looks like the condition of Cantor-separability for total orders. Actually, it coincides with Cantor-separability when restricted to total orders. Thus, in Theorem 1 we have found a condition of ordinal type (namely, the i.o.-separability) that reminds the perfect separability and is characteristic of the representability of interval orders through a pair of real valued functions.
REFERENCES 1. D. S. Bridges, Numerical representation of intransitive preferences on a countable set, J. Econ. Theory 30 (1983), 213217. 2. D. S. Bridges, Representing interval orders by a single real-valued function, J. Econ. Theory 36 (1985), 149155. 3. D. S. Bridges, Numerical representation of interval orders on a topological space, J. Econ. Theory 38 (1986), 160166. 4. D. S. Bridges and G. B. Mehta, ``Representations of preference orderings,'' SpringerVerlag, Berlin, 1995.
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