Completeness and Transitivity of Preferences on Mixture Sets
CCompleteness and Transitivity of Preferences on Mixture Sets ∗ Tsogbadral Galaabaatar, † M. Ali Khan ‡ and Metin Uyanık § September 4, 2018
Abstract:
In this paper, we show that the presence of the Archimedean and the mixture-continuity properties of a binary relation, both empirically non-falsifiable in principle, foreclosethe possibility of consistency (transitivity) without decisiveness (completeness), or decisivenesswithout consistency, or in the presence of a weak consistency condition, neither. The basicresult can be sharpened when specialized from the context of a generalized mixture set to thatof a mixture set in the sense of Herstein-Milnor (1953). We relate the results to the antecedentliterature, and view them as part of an investigation into the interplay of the structure of thechoice space and the behavioral assumptions on the binary relation defined on it; the ES researchprogram due to Eilenberg (1941) and Sonnenschein (1965), and one to which Schmeidler (1971)is an especially influential contribution.
Journal of Economic Literature
Classification Numbers: C00, D00, D01
Classification Numbers: 91B55, 37E05.
Key Words:
ES research program, generalized mixture set, completeness, transitivity, convexity
Running Title:
Completeness and Transitivity on Mixture Sets ∗ This version of a preliminary draft circulated in January 28, 2018 was completed during Khan’s visit tothe Department of Economics at the University of Queensland during July 27 to August 13, 2018. The authorsare grateful to Rabee Tourky for his insightful, off-the-cuff remarks, to Priscilla Man, Patrick O’Callaghan andJohn Quiggin for conversation and encouragement, and to H¨ulya Eraslan for her careful reading. Khan shouldalso like to thank the department for its hospitality and to acknowledge inspiration received over the years fromlistening to talks on this broad subject-matter by Edi Karni. † Dept of Economics, Ryerson Univ, Toronto, ON M5B2K3.
E-mail [email protected] ‡ Department of Economics, Johns Hopkins University, Baltimore, MD 21218.
E-mail [email protected] § School of Economics, University of Queensland, Brisbane, QLD 4072.
E-mail [email protected] a r X i v : . [ ec on . T H ] O c t ontents ————————————————————— Khan-Uyanık (2017) highlight what they refer to as the Eilenberg-Sonnenschein (ES) researchprogram, and present a comprehensive treatment of the two-way relationship between the prop-erties of a binary relation and of the set over which the relation is defined. Confining themselvesexclusively to the topological register, they develop a general theory of a complete, transitive,continuous and k -non-trivial ( k any natural number) binary relation defined on a k -connectedtopological space. As is well-known, the transitivity postulate in the presence of completenesswas studied in Eilenberg (1941) and Sonnenschein (1965), and completeness in the presence oftransitivity by Schmeidler (1971), but all only for k = 1 . Khan-Uyanık (2017) also broadenthe theory to include a sufficiency result, based on the work of Sen (1969), that delineates theconsequences of 1-connected (connected) choice sets for transitivity alone of a given continuousbinary relation. All this notwithstanding, the fact remains that the bulk of modern decisiontheory, in of itself and in terms of applications to mathematical economics and mathematicalsocial sciences more generally, does not confine itself solely to the topological register, and it istherefore natural to ask for the added consequences that can be obtained by supplementing itwith the algebraic one. However, this is not the question we ask here.Rather than a supplementation, we inquire into the reformulation of the ES programfor the setting considered in the pioneering work of von Neumann-Morgenstern (1947) andMarschak (1950), and one given a definitive treatment in Herstein-Milnor (1953). There, as iswell-understood, it is not so much a question of supplementing topology by algebra as it is ofworking with a novel mathematical setting of a mixture space, a setting that is endowed witha most minimal topological requirement. In particular, instead of a topology being assumed onthe space of objects, one simply utilizes the standard Euclidean unit interval used to combine the1bjects. This consequence of working with a generalization of a linear to a mixture space leads,as a necessary concomitant, to the fact that the results in Khan-Uyanık (2017), assuming asthey do a topology on the choice set, have no direct application to this richer alternative set-up.New mathematical argumentation along with its associated techniques is required. Even if thechoice set is taken to be the simplex, endowed with both topological and algebraic structures,their results cannot be directly used and applied. Furthermore, given that the “topologicalaction” involves only the unit interval, the question of working with any richer concept ofconnectedness and its natural extensions is excluded by default. And therefore, rather thangoing backwards from behavioral properties of preference relations to deduce properties of thetopology on the choice set, we can only go forwards to deduce consequences for behavior of thecontinuity properties of the given relation, continuity now being formalized by the Archimedeanproperty and by mixture-continuity, both rooted in the unit interval. On moving beyond broad methodological remarks, this paper contributes to a particularstrain within the ES program. This is the work of Dubra (2011), Karni-Safra (2015), andespecially the recent contribution of McCarthy-Mikkola (2018). Its trajectory can be simplylaid out. Dubra, in his specialization of the choice set to the space of lotteries on a set offinite prizes, showed how mixture-continuity and the Archimedean property, together with theindependence axiom, yields Schmeidler’s conclusion that a non-trivial, reflexive and transitiverelation is necessarily complete. Karni-Safra underscore the thrust of Dubra’s contribution byrelaxing independence to the “betweenness” property or cone-monotonicity property, and again,like him, by appealing to Schmeidler’s theorem to obtain completeness of the given relation. Ina move that is not unsurprising, McCarthy-Mikkola ignore the Karni-Safra generalization, andrevert to the original Dubra setting with the independence axiom in full operation, and generalizeDubra’s result to a convex set in a linear space that is not limited to be finite-dimensional.We apply Occam’s razor, and show that these particular results obtain without any linearityassumption on preferences! In particular, all of the results of McCarthy-Mikkola (2018) followas corollaries of the results reported here. But this is perhaps not the primary contribution of the paper. It is rather to showthat once the question is set within the broad outlines of the forward direction of the ES To be sure, there is a literature that imposes a topological structure on the objects themselves, but as wepoint out in Section 4 devoted to applications, this is best discussed in the setting of Khan-Uyanık (2017). This is elaborated below in Section 5 devoted to the proofs of the results. In anticipation, we may point out here that pace
McCarthy-Mikkola (2018), the results concerning mixturespaces are not purely algebraic and topologically-free. As we elaborate in the sequel, they use an algebraiccondition that is equivalent to mixture-continuity. To explicate this a little further, it is not that the backward direction cannot be executed but that infocusing on the topologies on the unit interval, it would involve consequence of little substantive economiccontent. However, we do go backwards in Theorem 4 and in Observation 1, not in the context of topologicalproperties of the choice set, but in terms of its linear structure! Again anticipating somewhat, even though they use an algebraic version of mixture-continuity, their con-dition is equivalent to the usual mixture-continuity notion under the independence assumption. Hence, theirresults can be restated in terms of the usual mixture-continuity notion, and thereby generalized. We also pointout here that unlike Dubra (2011) and Karni-Safra (2015), rather than an appeal to Schmeidler’s theorem, werely on the method of proof of his theorem. On all this, see the first paragraph of Section 4. both transitivity and completeness under a considerably weaker versionof the transitivity postulate. This articulation is then followed up by delineating possibilitiesopened by a further substitution of the weaker transitivity notion by a convexity postulateon preferences. In terms of a more detailed overview of the results, our first theorem showsthat under semi-transitivity and transitivity of the symmetric part of a reflexive and non-trivialbinary relation, mixture-continuity and the Archimedean property yield both completeness andtransitivity of the relation in the context of a generalized mixture space. The two properties arebundled conclusions: under the assumed hypotheses, one property cannot be obtained withoutthe other. Any agent cannot be consistent without being decisive, or decisive without beingconsistent. Next, we specialize the setting to a mixture space, and present a result, and its threecorollaries, that concern only the transitivity (consistency) postulate and its various relaxationsas in Sen (1969). Continuing with the setting of a mixture space, we bring the completenesspostulate into the picture, and show that transitivity of the symmetric part of an Archimedeanor a mixture-continuous binary relation, a convexity assumption on preferences is sufficientfor transitivity. Finally, in what may be the most surprising and anti-climactic finding, weshow (in Theorem 4 below) that if the preference relation is “very nice,” the model essentiallycollapses to a situation where the standard greater-than-or-equal-to relation on a unit interval isbeing investigated. The consequence of this for the Herstein-Milnor representation theorem areunmistakable, and we are thereby led directly to an alternative proof of their result. In sum, it isin this bundling of the results of Eilenberg, Sonnenschein, Sen and Schmeidler, Herstein-Milnorserving as an important subtext, that each individual contribution is mutually illuminated andallows a maturer theory.But mature or otherwise, the question remains as to what precisely these theorems offer interms of the antecedent literature. How can the theory be applied? We have already referred tothe work of Dubra, Karni, Safra, McCarthy and Mikkola (henceforth DKSMM), but it is Section4 below that we attempt a more careful reading and systematic discussion of the literature withthese theorems in pure theory in hand. We do so under the criterial rubric of redundancy and hiddenness on the one hand, and of fragility and flimsiness on the other. We have already men-tioned the taking of Occam’s razor to the theorems and of removing redundancies in them; thecriteria of hidenness is only a little less straightforward – the Malinvaud-Samuelson exposing ofthe independence axiom in von Neumann-Morgenstern (1947) being the archetypical example. As regards the other two criteria, they are robustness criteria inspired by Gerasimou (2013),and perhaps ought to be seen as further elaboration of the incorporated hiddenness criterion.In any case, we turn to their formal explication and discussion below. See Fishburn’s references to von Neumann-Morgenstern (1947), Fishburn (1964), Chipman (1971) andFishburn-Roberts (1978). Khan-Uyanık (2017) refer these relaxations as “Sen’s deconstruction of the transitivity postulate;” also seeFishburn’s (1970) survey. See Corollary 4 below, and its proof in Section 5. See Malinvaud (1952) for the example, and Khan-Uyanık (2017) for a more detailed explication. Notational and Conceptual Preliminaries
Here, and later, lower case Greek letters will always denote real numbers in [0,1] which is endowedwith the usual Euclidean topology. This is in keeping with the inspired usage of Herstein-Milnor(1953).Let X be a set. A subset (cid:60) of X × X denote a binary relation on X. We denote an element( x, y ) ∈ (cid:60) as x (cid:60) y. The asymmetric part (cid:31) of (cid:60) is defined as x (cid:31) y if x (cid:60) y and y (cid:54) (cid:60) x , andits symmetric part ∼ is defined as x ∼ y if x (cid:60) y and y (cid:60) x. We call x (cid:46)(cid:47) y if x (cid:54) (cid:60) y and y (cid:54) (cid:60) x .The inverse of (cid:60) is defined as x (cid:52) y if y (cid:60) x . Its asymmetric ≺ is defined analogously and itssymmetric part is ∼ . We provide the descriptive adjectives pertaining to a relation in a tabularform for the reader’s convenience in the table below. reflexive x (cid:60) x ∀ x ∈ X complete x (cid:60) y or y (cid:60) x ∀ x, y ∈ X non-trivial ∃ x, y ∈ X such that x (cid:31) y transitive x (cid:60) y (cid:60) z ⇒ x (cid:60) z ∀ x, y, z ∈ X negatively transitive x (cid:54) (cid:60) y (cid:54) (cid:60) z ⇒ x (cid:54) (cid:60) z ∀ x, y, z ∈ X semi-transitive x (cid:31) y ∼ z ⇒ x (cid:31) z and x ∼ y (cid:31) z ⇒ x (cid:31) z ∀ x, y, z ∈ X Table 1: Properties of Binary RelationsNext, we provide a definition of the mixture set due to Herstein-Milnor (1953).
Definition 1.
A set S is said to be a mixture set if for any x, y ∈ S and for any µ we canassociate another element, which we write as xµy, which is again in S , and where for all x, y ∈ S and all λ, µ, (S1) x y = x , (S2) xµy = y (1 − µ ) x , (S3) ( xµy ) λy = x ( λµ ) y . Note that the following property of a mixture set is implied by S1-S3 above. (S4) ( xλy ) µ ( xβy ) = x ( µλ + (1 − µ ) β ) y for all x, y ∈ S and all λ, µ, β. The notion of a mixture set can be routinely generalized by replacing equalities between thepairs of mixtures by indifference in the definition above. Definition 2.
Let M be a set and (cid:60) a reflexive binary relation on it with a transitive symmetricpart ∼ . Then, M is said to be a generalized mixture set ( induced by ∼ ) if for any x, y ∈ M In deference to Herstein-Milnor (1953), lower case Greek letters consistently denote real numbers in [0,1]. See Luce-Suppes (1965, p288) or Fishburn (1982, Section 2.4) for a proof. This definition is due to Fishburn (1982, Section 2.3). A complete axiomatization of a generalized mixture setis first provided, to the best of the authors’ knowledge, by Fishburn (1964, p8). A form of (M3) is used in Luce-Raiffa (1957, p26) in the context of the reduction of compound lotteries. Even though von Neumann-Morgenstern(1947, Section 3.6) use equality in their algebra of combining axioms, their interpretation is consistent with theuse of indifference; see also Chipman (1971) and Fishburn-Roberts (1978) for applications and discussion ofgeneralized mixture sets. Even though Chipman seems unaware of the work of Sonnenschein and Schmeidler,his discussion of the Archimedean assumption with Samuelson’s writings as the relevant background, and hismuted claim that continuity does not have behavioral implications, surely merits further engagement. Anotherreference that merits future engagement regarding applications is surely Gudder (1977); see Footnote 13 below. nd for any µ we can associate another element, which we write as xµy, which is again in M , and where for all x, y ∈ M and all λ, µ, β, (M1) x y ∼ x, (M2) xµy ∼ y (1 − µ ) x, (M3)( xµy ) λy ∼ x ( λµ ) y, (M4) ( xλy ) µ ( xβy ) ∼ x ( µλ + (1 − µ ) β ) y . Next, we turn to the various properties of a binary relation on a generalized mixture set,and develop the following notation for subsets of [0 , . For any (cid:60) on M and for any x, y, z ∈ M , let A (cid:60) ( x, y, z ) = { λ | xλy (cid:60) z } and A (cid:52) ( x, y, z ) = { λ | z (cid:60) xλy } . The sets A (cid:31) ( x, y, z ) , A ≺ ( x, y, z ) , A ∼ ( x, y, z ) and A (cid:46)(cid:47) ( x, y, z ) are analogously defined. Definition 3.
We call a binary relation (cid:60) on a generalized mixture set M (i) mixture-continuous if for all x, y, z ∈ M , the sets A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) are closed; (ii) Archimedean if for all x, y, z, w ∈ M with x (cid:31) y , x (cid:46)(cid:47) w and y (cid:46)(cid:47) z there exist λ, δ ∈ (0 , such that λ ∈ A (cid:31) ( x, z, y ) and δ ∈ A ≺ ( y, w, x );(iii) strongly Archimedean if for all x, y, z ∈ M with x (cid:31) y , there exists λ, δ ∈ (0 , such that λ ∈ A (cid:31) ( x, z, y ) and δ ∈ A ≺ ( y, z, x ) . Note that the Archimedean property above is weaker than strong Archimedean property – thelatter is the version that is usually assumed in the literature.
Definition 4.
A binary relation (cid:60) on a generalized mixture set M is (i) linear if for all x, y ∈ M , all z ∈ { x, y } and all λ ∈ (0 , x ∼ y if and only if λ ∈ A ∼ ( x, z, yλz ) , (ii) convex if for all x, y, z ∈ M and all λ, x (cid:60) z and y (cid:60) z implies λ ∈ A (cid:60) ( x, y, z ) , (iii) concave if for all x, y, z ∈ M and all λ, z (cid:60) x and z (cid:60) y implies λ ∈ A (cid:52) ( x, y, z ) , (iv) star-convex if for all distinct x, y ∈ M and all λ ∈ (0 , , if x (cid:60) y, then λ ∈ A (cid:31) ( x, y, y ) , (v) star-concave if for all distinct x, y ∈ M and all λ ∈ (0 , , if y (cid:60) x, then λ ∈ A ≺ ( x, y, y ) . It is well-known that the conventional independence assumption, or the weaker property of betweenness, implies that the preference relation is linear. Under completeness and transitivity,star-convexity implies convexity and star-concavity implies concavity. However, without thecompleteness assumption, there is no inclusion relationship between convexity and star-convexityas well as between concavity and star-concavity. For example, any preference relation with thickindifference curves illustrates that convexity does not imply star-convexity. In order to see thatstar-convexity does not imply convexity, let X = { x ∈ R | (cid:80) i x i = 1 } . Assume (cid:60) is a reflexivebinary relation on X such that (1 , , λ (0 , , (cid:31) (0 , ,
0) and (0 , , λ (0 , , (cid:31) (0 , ,
0) forall λ ∈ (0 , (cid:60) is star-convexand not convex. Analogous arguments illustrate that there is no inclusion relationship betweenconcavity and star-concavity.Our laying-out of the conceptual preliminaries would not be complete without any mentionof results that explore the somewhat more subtle converse to the assertion that every convex set5s a mixture set. The fact that the converse does not hold is hardly esoteric, and deserves to bemore widely known in the mathematical social science literature; see for example Wakker (1989,Section V.II) and Mongin (2001, p61). Stone (1949, Theorem 2) and Hausner (1954, Theorems3.2 and 3.4) provide axiomatic characterization of convex sets by showing that a mixture set S is isomorphic to a convex subset of some linear space if and only if it satisfies the following twoaxioms. (C1) For all x ∈ S and all λ ∈ (0 , , xλy = xλy (cid:48) implies y = y (cid:48) . (C2) For all x, y, z ∈ S and all λ, µ ∈ [0 , with λµ (cid:54) = 1 , ( xλy ) µz = x ( λµ ) (cid:18) y µ (1 − λ )1 − λµ z (cid:19) . In a more recent work, Mongin (2001) introduces the concept of non-degeneracy and showsthat it is equivalent to (C1)-(C2). A function u : S → R is said to be mixture preserving iffor all x, y ∈ S and all λ ∈ [0 , , u ( xλy ) = λu ( x ) + (1 − λ ) u ( y ) . Denote by L ( S ) the set of allmixture preserving functions on S , and define a relation ≈ on S as x ≈ y if and only if for all u ∈L ( S ) , u ( x ) = u ( y ) . It is clear that L ( S ) is a vector space, ≈ is an equivalence relation, andthe quotient space S| ≈ is a mixture set with the mixture operation [ x ] λ [ y ] = [ xλy ] for all[ x ] , [ y ] ∈ S| ≈ and all λ ∈ [0 , Definition 5.
A mixture set S is non-degenerate if all classes of S| ≈ are singletons, i.e., forall x, y ∈ S , x ≈ y implies x = y . The following proposition encapsulates the main results of Stone-Hausner-Mongin and is copiedfrom Mongin (2001, Proposition) for the reader’s convenience.
Proposition 0 (Stone-Hausner-Mongin) . The following are equivalent for any mixture set S . (a) S satisfies (C1) and (C2). (b) S is non-degenerate. (c) S is isomorphic to a convex subset of some linear space. We end this section by presenting the relationship between Archimedean and mixture-continuity properties without the completeness and full transitivity assumptions. We first showthat Archimedean property is equivalent to a topological condition under mixture-continuity.
Proposition 1.
Let (cid:60) be a semi-transitive binary relation on a generalized mixture set M withmixture-continuity. Then, the following are equivalent. (a) (cid:60) is Archimedean, (b) (cid:60) is strongly Archimedean, (c) A (cid:31) ( x, y, z ) and A ≺ ( x, y, z ) are open for all x, y, z ∈ M . See Gudder (1977) for an expository article on abstract convexity and its applications to behavioral, socialand physical sciences. Note that there is no reference to Stone in Hausner’s paper, and given that Mongin isalso silent about the relationship between the two works, we regard them as independent. ⇒ (a) does not require the mixture-continuity assumption, hence condition (c) is stronger than the Archimedean properties. Weelaborate these points in the Appendix by providing examples.The Archimedean property is weaker than mixture-continuity, even under the complete-ness and transitivity assumptions being in force – either a convexity condition, or a furthercontinuity property, needs to be imposed on preferences in order to obtain mixture-continuityof an Archimedean relation. The following result provides sufficient conditions for mixture-continuity under the strong Archimedean property without completeness, full transitivity andconvexity of preferences.
Proposition 2.
Let (cid:60) be a semi-transitive and strongly Archimedean binary relation on ageneralized mixture set M such that for all x, y, z ∈ M , A (cid:46)(cid:47) ( x, y, z ) is open, and A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) have finitely many components. Then, (cid:60) is mixture-continuous. Under the completeness assumption, openness of A (cid:46)(cid:47) ( x, y, z ) trivially holds. The finitenessof the components is implied by concavity and convexity of the preference relation, which aresatisfied under independence hypothesis. We show in the Appendix that each of the assumptionsof this proposition is not redundant.
In this section, we present four theorems and with the help of four observations draw out fourcorollaries from them. We begin with our first result that pertains to a generalized mixture set.
Theorem 1.
Any non-trivial, reflexive, semi-transitive, mixture-continuous and Archimedeanbinary relation (cid:60) with a transitive symmetric part ∼ on a generalized mixture set induced by ∼ ,is complete and transitive. For the case of an anti-symmetric relation, we can re-state the above result without any referenceto any form of transitivity.
Observation 1.
Any anti-symmetric, non-trivial, reflexive, mixture-continuous and Archimedeanbinary relation (cid:60) on a generalized mixture set induced by ∼ , is complete and transitive. A simple elaboration shows that any anti-symmetric relation satisfies semi-transitivity and itssymmetric part is transitive. The reader who does not want to worry about different transitivity Moreover, in mixture sets, this direction of the equivalence result does not require semi-transitivity; indeed,assertion (a) ⇒ (c) does not require semi-transitivity even in generalized mixture sets. See for example Dubra (2011) and Karni-Safra (2015) for the former, and Karni (2007) for the latter. As in Proposition 1, semi-transitivity is not needed for this proposition in mixture sets. can refer to this simpler version of the theorem. Moreover, this observation hasimportant implications on the structure of the mixture set to which we return at the end ofthis section. The following corollary shows that, under completeness assumption, a weak formof transitivity implies full transitivity. Corollary 1.
Any complete, mixture-continuous and Archimedean binary relation (cid:60) on a gen-eralized mixture set M induced by ∼ , is transitive if any or both of the following holds: (a) x (cid:31) y ∼ z implies x (cid:31) z for all x, y, z ∈ M , (b) x ∼ y (cid:31) z implies x (cid:31) z for all x, y, z ∈ M . Our second result shows that the transitivity of the asymmetric part of a complete,strongly Archimedean and star-convex (or star-concave) binary relation is sufficient for its tran-sitivity.
Theorem 2.
Any complete and strongly Archimedean binary relation (cid:60) with a transitive asym-metric part ∼ on a generalized mixture set induced by ∼ , is transitive if any or both of thefollowing holds: (a) (cid:60) is star-convex, (b) (cid:60) is star-concave. Our third result shows that, in a mixture set, certain convexity properties are sufficientfor semi-transitivity.
Theorem 3.
The following are true for a reflexive, mixture-continuous and Archimedean binaryrelation (cid:60) on a mixture set S whose symmetric part is transitive. (a) If (cid:60) is linear, then it is semi-transitive. (b) If (cid:60) is convex, then x ∼ y (cid:31) z implies x (cid:31) z for all x, y, z ∈ S . (c) If (cid:60) is concave, then x (cid:31) y ∼ z implies x (cid:31) z for all x, y, z ∈ S . (d) If (cid:60) is complete, then its convexity or its concavity implies its semi-transitivity. We now rely on the following observation to bring out the fuller implications of the result.
Observation 2.
If a relation is convex and concave, then it is linear. The converse is trueunder mixture continuity and the Archimedean axiom. Figure 1 below illustrates examples of linear, convex, concave and semi-transitive preferences.It is easy to see that, when the choice set is a convex subset of a linear space, linearity impliesthat the indifference sets are convex (thick indifference curves are allowed), convexity that theupper section of the weak preference relation is convex and concavity that the lower section ofthe weak preference relation is convex.The following two corollaries illustrate that, in a mixture set, semi-transitivity can besubstituted by a convexity property in the hypothesis of Theorem 1. First, we show that wecan replace semi-transitivity with linearity. Khan-Uyanık (2017, Section 3) discuss in detail the relationship between different transitivity concepts. See Lemma 1 in the Appendix below for a proof. a) (b) (d)(c) X = { x ∈ R | (cid:80) x i = 1 } is the two dimensional unit simplex. The curves illustrate the indifference curves andthe arrow indicates the direction of preferences. Panel (a) illustrates a linear preference relation, (b) convex butneither linear, nor concave (c) concave but neither linear, nor convex and (d) semi-transitive that is not linear,convex and concave. Figure 1: Linear, Convex, Concave and Semi-transitive Preferences
Corollary 2.
Any non-trivial, reflexive, mixture-continuous and Archimedean binary relation (cid:60) with a transitive symmetric part on a mixture set, is complete and transitive if any or bothof the following holds: (a) (cid:60) is linear, (b) (cid:60) is semi-transitive.
Next, we show that linearity hypothesis in the corollary above can be replaced with one ofconvexity or concavity in the presence of completeness.
Corollary 3.
Any complete, mixture-continuous and Archimedean binary relation (cid:60) with atransitive symmetric part on a mixture set, is transitive if any or both of the following holds: (a) (cid:60) is convex, (b) (cid:60) is concave.
We end this section by presenting a joint implication of Theorem 1 and Observation 1.
Theorem 4.
If there exists a non-trivial, complete, transitive, anti-symmetric and mixture con-tinuous binary relation (cid:60) on a generalized mixture set M induced by ∼ , then M is isomorphicto an interval in R and (cid:60) is equivalent to the usual “greater-than-or-equal-to” or “less-than-or-equal-to” relation. Theorem 4 implies that ( R, ≥ ( ≤ )) is the only linear space with binary relation(s) that satisfiesall of the above properties, hence it provides a characterization result for ( R, ≥ ( ≤ )). It followsfrom Theorem 1 that we can equivalently state the theorem above as follows. Observation 3.
If there exists a non-trivial, reflexive, anti-symmetric, mixture continuousand Archimedean binary relation (cid:60) on a generalized mixture set M induced by ∼ , then M isisomorphic to an interval in R and (cid:60) is equivalent to the usual “greater-than-or-equal-to” or“less-than-or-equal-to” relation. It follows from the definition of an anti-symmetric relation that if a generalized mixture set M is inducedby an anti-symmetric relation, then M is a mixture set. Hence, we can state Theorem 4 with a mixture setwithout loss of generality. Moreover, the non-triviality assumption in Theorem 4 is not restrictive. If a binaryrelation satisfying other assumptions of the theorem is trivial, then the space contains at most one element, andis hence isomorphic to a (possibly empty) interval in R . R . Moreover, the preference relation satisfies the inde-pendence axiom, hence it satisfies all assumptions of Herstein-Milnor. As already emphasizedin the introduction, this is an important result, one that provides a litmus-test for evaluatingresults that are set up in what appears to be a generalized setting. xyz A B xyz ( ] ] A B ] Figure 2: A Non-convex Mixture Set with a Relation satisfying HM Axioms
Example 1.
Let A = { ( x , x ) ∈ R | x = 0 , ≤ x ≤ } , B = { ( x , x ) ∈ R | < x ≤ , x =0 } and S = A ∪ B . Set x = (1 , , y = (0 , , z = (0 , a, a (cid:48) ∈ A or a, a (cid:48) ∈ B , then aλa (cid:48) = λa + (1 − λ ) a (cid:48) for all λ ∈ [0 , a ∈ A and b ∈ B , then aλb = λy + (1 − λ ) b for all λ ∈ [0 , a b = a and bλa = a (1 − λ ) b for all λ ∈ [0 , S is a mixture set. Define a binary relation (cid:60) on S as follows: any pair in A are indifferent, anypoint in A is strictly worse than any point in B and xλy (cid:60) xδy if and only if λ ≥ δ . It is easyto see that (cid:60) is non-trivial, complete, transitive, mixture continuous and satisfies independenceaxiom It follows from xλz = xλy for all λ ∈ (0 ,
1) and z (cid:54) = y that axiom (C1) does not hold.Therefore, Proposition 0 implies S is not isomorphic to a convex set.We have not yet formally defined the well-known notion of an independent relation: forall x, y, z ∈ S and all λ ∈ (0 , x ∼ y if and only if xλz ∼ yλz . We can now derive the classicexpected utility representation theorem of Herstein-Milnor on a generalized mixture set as aconsequence of Theorem 4. Corollary 4.
Let (cid:60) be a complete, transitive and mixture continuous binary relation with theindependence axiom on a generalized mixture set M induced by ∼ . Then, there exists a function u : M → R such that for all x, y ∈ M , x (cid:60) y ⇐⇒ u ( x ) ≥ u ( y ) where u ( xλy ) = λu ( x ) + (1 − λ ) u ( y ) for all λ ∈ [0 , . The alternative proof is relegated to Section 5, and here we limit ourselves to the observationthat is responsible for the basic idea underlying the alternative proof.
Observation 4.
In the context of the objects in Corollary 4, it is easy to show that the quotientspace
M| ∼ is a mixture set with the mixture operation defined as [ x ] λ [ y ] = [ xλy ] for all [ x ] , [ y ] ∈ | ∼ and all λ ∈ [0 , . Moreover, the derived relation ˆ (cid:60) on the quotient set, defined as [ x ] ˆ (cid:60) [ y ] if x (cid:48) (cid:60) y (cid:48) for all x (cid:48) ∼ x, y (cid:48) ∼ y , is anti-symmetric, complete, transitive and mixture continuous. It is worth noting that our alternative proof is not straightforward, an is directly based on theembedding theorem of Stone-Hausner-Mongin presented as Proposition 0. This being said, theproof itself is not difficult and bears comparison with the proofs presented in Sections 2.4 and2.5 in Fishburn (1982). We begin this section with the results of Dubra, Karni-Safra and McCarthy-Mikkola (DKSMM),already referred to in the introduction. Theorem 1 considerably generalizes these results bydropping any form of convexity assumption on preferences, by weakening transitivity and byallowing the choice space to be a generalized mixture set. Moreover, Corollary 2 shows that whenthe choice space is a mixture set, the semitransitivity of the preference relation can be substitutedby its linearity, which is implied both by the independence and betweenness properties assumedin the papers above. To elaborate a little more, Dubra uses the independence assumptionto show that preferences satisfy Schmeidler’s continuity assumption, and thereby deduces hisresult as a corollary of Schmeidler’s theorem; KS show that independence can be replaced by betweenness or cone-monotonicity, with the same method of proof. Dubra’s argumentation isbased on Rockafellar (1970, Theorem 6.1), a result Rockafellar refers to as “fundamental”. MMgeneralize Dubra’s theorem to convex subsets of arbitrary real linear spaces. They use equivalentalgebraic versions of the continuity assumptions and use an algebraic proof technique, and theydo not use Schmeidler’s theorem for their argument.Next, we provide some application of our results to the antecedent literature which high-lights the hiddenness of completeness and transitivity. von Neumann-Morgenstern (1947),Herstein-Milnor (1953) and their followers show that the following four conditions are nec-essary and sufficient for representation of preferences with an expected utility function: com-pleteness, transitivity, independence, mixture-continuity and Archimedean; see Fishburn (1970,1982), Kreps (1988) and Gilboa (2009) for a survey. Our results show that a weak form of thetransitivity postulate, along with the two continuity properties, implies both completeness and transitivity; hence both of them are hidden assumptions. It is also worth pointing out in thisconnection that even though each paper in this line of literature assumes either one of these twocontinuity assumptions, it is by now well known that they are equivalent under the completenessand independence hypotheses. Furthermore, there is one other delicate point worth stressing: Fishburn (1982, p. 20) writes, “The proof of Theorem 2 [representation in generalized mixture sets] is similarto the proof given above for the construction of linear, order-preserving utilities on the basis of M1-M3 and J1-J5. Our main concern in modifying the preceding proof is to make sure that the uses of = from M1-M3 can bereplaced by ∼ on the basis of the axioms in Theorem 2. [T]he construction of the desired u then parallels theconstruction given above with a few changes from = to ∼ , and the uniqueness proof is likewise straightforward.”In our lterantive proof, we bypass the construction, however natural, and all the checking that it requires. Alsosee Footnote 12 in the context of this textual exegesis. I personally believe the archimidean ( sic ) principle to be very compelling, not withstandingsome of the counter-intuitive examples that have been offered in the literature. [T]here maycertainly be situations in which the lexicographic order or something similar constitutesthe most convenient model, so it is desirable to have a theory that covers it. There is surely no problem with this given the rich and considerable analysis of incompletepreferences that is now available in the literature. However, the issue from the point of viewbeing emphasized in this paper is slightly deeper than this. It is not a matter of the literaturelacking one of the two assumptions but also of violating, by necessity, one of the two equallyplausible alternatives they represent. If not, our results show that the preferences are necessarilycomplete! Thus, we tend to see the results that we present above in a somewhat “negative”vein in the modeling of incomplete and/or non-transitive preferences: full continuity does notallow incompleteness and/or non-transitivity of the preferences. In any case, the seeminglyrather innocuous continuity assumptions have both behavioral and empirical implications. Itis presumably for this reason of strong continuity assumptions precluding the study of suchquestions that models with incomplete preferences in decision theory do not assume full con-tinuity, but only one of the two continuity axioms. This sacrifice of continuity imposes anundesirable property on preferences regarding which we attempt a conceptual extraction. Westudy the robustness of the structure of such preferences by introducing the concepts of fragility and flimsiness : we identify those that violate the Archimedean property as fragile , and thosewhich violates mixture-continuity as flimsy , and take each in turn.
Definition 6.
A binary relation (cid:60) on a generalized mixture set M is fragile if there exist x, y, z ∈ M and λ ∈ A (cid:31) ( x, y, z ) ∪ A ≺ ( x, y, z ) such that every open neighborhood of λ contains anon-empty open set V such that V ⊂ A (cid:46)(cid:47) ( x, y, z ) . See Aumann (1962, Footnote 25). We invite the reader to use Propositions 1, 2 and Footnote 5 above to seethe relationship between the continuity assumptions of this paper and those in Aumann (1962, (4.1) and (4.2)). See Karni (2014) for a recent survey and Hara et al. (2015) for the state-of-the-art results in this line ofliterature.
Example 2.
Assume a decision maker chooses between two alternatives, a and b . Let X = [0 , { a, b } where x ∈ X denote the probability of a and (1 − x ) is the probability of b . Assume the agent has a reflexive preference relation (cid:52) on X such that 0 ≺
1, i.e., she strictly prefers a for sure to b for sure. No other alternatives arecomparable. Note that every neighborhood of 1 contains an open interval of alternatives whichare incomparable to 0 and vice-a-versa. Therefore, even though she prefers a to b , she cannotcompare a with any lottery that assigns a slightly positive weight on a . In this example, it isclear that all assumptions of Theorem 1 hold except the Archimedean property. However, if weadd Archimedean, then we know that the preference relation has to be complete.The following proposition shows that dropping Archimedean assumption from Theorem 1 yieldsa fragile preference relation. Proposition 3.
Any incomplete, non-trivial, reflexive, transitive and mixture-continuous binaryrelation on a generalized mixture set M , is fragile. Remark 1.
We leave it to the reader to check that Shapley-Baucells (1998) drop theArchimedean property and assume mixture continuity, and hence, that the preferences in theirresults are fragile. Indeed, there is an extensive literature which drops the Archimedean pos-tulate and assumes a continuity assumption that is stronger than mixture-continuity; see forexample Ghirardato et al. (2003), Dubra et al. (2004), Evren-Ok (2011), Ok et al. (2012). Inall these papers, the choice set is endowed with a topological structure and the sections of theweak preference relation is closed. For the models which drop mixture-continuity but keep strong Archimedean property, thefollowing concept is useful, which is originally due to Khan-Uyanık (2017).
Definition 7.
A binary relation (cid:60) on a generalized mixture set M is flimsy if there exist x, y, z ∈ M and λ ∈ A (cid:46)(cid:47) ( x, y, z ) such that every open neighborhood V λ of λ contains λ (cid:48) in A (cid:60) ( x, y, z ) ∪ A (cid:52) ( x, y, z ) . Flimsiness implies that limit of some sequence of comparable alternatives is non-comparable.The following simple example illustrates a flimsy preference relation. For a comprehensive discussion of the structure of incomplete preferences on a topological space withoutany algebraic structure; see for example Khan-Uyanık (2017). xample 3. Let X = [0 ,
3] and the agent has a reflexive preference relation on X whichsatisfy the following: any pair in [0 ,
1) is indifferent to each other; similarly any pair in (2 , Proposition 4.
Any incomplete, non-trivial, reflexive, transitive, strongly Archimedean binaryrelation (cid:60) on a generalized mixture set induced by ∼ such that A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) havefinitely many components for all x, y, z , is flimsy. Remark 2.
Note that papers that drop mixture-continuity and assume the Archimedean prop-erty typically assume the strict preference relation to be the relevant primitive, and hence,strictly speaking, our flimsiness result does not apply; see for example Bewley (2002), Manzini-Mariotti (2008), Galaabaatar-Karni (2012; 2013) and Evren (2014). The earlier works ofAumann (1962) and Kannai (1963) do assume the weak preference relation as primitive, buttheir continuity assumption is weaker than the Archimedean and mixture-continuity versionswe assume. Nevertheless, it is worth emphasizing that the conceptual notions of fragility andflimsiness can be reworked and re-calibrated to apply to them.
We provide the proofs in the same order in which the results presented above in Sections 3 and4 above.
Proof of Proposition 1. ( a ) ⇒ ( c ) Assume (cid:60) is mixture-continuous and Archimedean. Pick x, y, z ∈ M . If A (cid:31) ( x, y, z ) is empty, then it is open. Otherwise, pick λ ∈ A (cid:31) ( x, y, z ) . It follows from mixture-continuity and λ / ∈ A (cid:52) ( x, y, z ) that there exists t > N t ( λ ) = { β | | β − λ | < t } is contained in the complement of A (cid:52) ( x, y, z ) . Assume there exists β ∈ N t ( λ ) ∩ A (cid:46)(cid:47) ( x, y, z ) . It follows from mixture-continuity that A (cid:46)(cid:47) ( x, y, z ) is open. Therefore, as an open set in [0,1], A (cid:46)(cid:47) ( x, y, z ) is union of at most countablymany mutually disjoint open intervals such that the intervals of type [0 , α ) or ( α,
1] are allowedfor any α ∈ (0 , I such that β ∈ I. If β > λ, then set δ = inf I, otherwise δ = sup I. Then δ ∈ A (cid:31) ( x, y, z ) . It follows from M4 that( xδy ) γ ( xβy ) ∼ x ( γδ + (1 − γ ) β ) y for all γ. Since ( δ, β ) ⊂ I, therefore γδ + (1 − γ ) β ∈ A (cid:46)(cid:47) ( x, y, z )for all γ ∈ (0 , . This furnishes us a contradiction with (cid:60) is Archimedean. Therefore, A (cid:31) ( x, y, z )is open. An analogous argument implies A ≺ ( x, y, z ) is open.14 c ) ⇒ ( b ) Assume (cid:60) is mixture-continuous and A (cid:31) ( x, z, y ) and A ≺ ( x, z, y ) are open for all x, y, z. Pick x, y, z ∈ M such that x (cid:31) y. Then, semi-transitivity and M1 imply 1 ∈ A (cid:31) ( x, z, y ) . It follows from A (cid:31) ( x, z, y ) is open that there exists λ < λ, ⊂ A (cid:31) ( x, z, y ) . Similarly, 1 ∈ A ≺ ( y, z, x ) . Since A ≺ ( y, z, x ) is open, therefore there exists δ < δ, ⊂ A ≺ ( y, z, x ) . Hence, (cid:60) is strongly Archimedean.The assertion ( b ) ⇒ ( a ) is immediate from the definition of strong Archimedean property.Therefore, the proof of Proposition 1 is complete. Proof of Proposition 2.
Pick x, y, z ∈ M . If A (cid:60) ( x, y, z ) is empty, then it is closed. Then, assume A (cid:60) ( x, y, z ) (cid:54) = ∅ . Since A (cid:60) ( x, y, z ) has finitely many components, it is the union of a finitely manynon-empty, disjoint, convex sets { C i } ki =1 which are closed in the subspace A (cid:60) ( x, y, z ). Pick i ≤ k. Assume λ = inf C i is not contained in C i . Then, C i is a non-degenerate interval. It follows from C i is a component of A (cid:60) ( x, y, z ) that xλy (cid:54) (cid:60) z . Since A (cid:46)(cid:47) ( x, y, z ) is open and disjoint from A (cid:60) ( x, y, z ) , therefore λ ∈ A ≺ ( x, y, z ) . Pick δ ∈ C i . Then, δ > λ.
It follows from (cid:60) is stronglyArchimedean that there exists β ∈ (0 ,
1) such that z (cid:31) ( xλy ) β ( xδy ) . Since (cid:60) is semi-transitiveand ( xλy ) β ( xδy ) ∼ x ( βλ + (1 − β ) δ ) y, therefore z (cid:31) x ( βλ + (1 − β ) δ ) y. It follows from δ > λ that λ < βλ + (1 − β ) δ < δ. Hence, βλ + (1 − β ) δ ∈ C i . This furnishes us a contradiction.Therefore, λ ∈ C i . An analogous argument shows that C i contains its supremum. Hence, C i isclosed. As a union of finitely many closed sets, A (cid:60) ( x, y, z ) is closed.An analogous argument shows that A (cid:52) ( x, y, z ) is closed. (Note that, for mixture sets, wedo not need any transitivity property in order to prove this proposition.) Proof of Theorem 1.
Assume (cid:60) is a non-trivial, reflexive, semi-transitive, mixture-continuousand Archimedean binary relation with a transitive symmetric part ∼ on a generalized mixtureset M . Recall that (cid:31) denote the asymmetric part of (cid:60) . Fist consider the following claim.
Claim . (cid:31) is negatively transitive. It is easy to see that this claim implies (cid:31) is transitive. Then, it follows from the transitivityof ∼ and semi-transitivity of (cid:60) that (cid:60) is transitive. The following claim implies (cid:60) is completeand transitive.
Claim . (cid:60) is complete. It remains to prove Claims 1 and 2 in order to complete the proof.
Proof of Claim 1.
Note that (cid:31) is negatively transitive if and only if x (cid:31) y implies x (cid:31) z or z (cid:31) y for all x, y, z ∈ M . Assume (cid:31) is not negatively transitive, i.e. there exists x, y, z ∈ M such that x (cid:31) y and neither x (cid:31) z nor z (cid:31) y. See Khan-Uyanık (2017, Proposition 2) for a proof and a detailed discussion on the interdependence betweendifferent transitivity conditions.
15t follows from semi-transitivity of (cid:60) , transitivity of ∼ and M1 that 0 / ∈ A ∼ ( x, z, x ) ∪ A ∼ ( y, z, y ) . Since (cid:60) is reflexive, therefore M1 implies 1 ∈ A ∼ ( x, z, x ) ∪ A ∼ ( y, z, y ) . Moreover,mixture-continuity implies A ∼ ( x, z, x ) and A ∼ ( y, z, y ) are closed subsets of [0,1], hence compact.Define λ x = min A ∼ ( x, z, x ) and λ y = min A ∼ ( y, z, y ) . It is clear that λ x , λ y ∈ (0 , . Define¯ x = xλ x z and ¯ y = yλ y z. By construction, ¯ x ∼ x and ¯ y ∼ y, and it follows from semi-transitivitythat ¯ x (cid:31) ¯ y. It follows from Proposition 1 that A ≺ ( z, ¯ y, ¯ x ) is open and from mixture-continuity that A (cid:52) ( z, ¯ y, ¯ x ) is closed. Since ¯ x (cid:31) ¯ y, therefore semi-transitivity of (cid:60) , transitivity of ∼ and M1imply 0 ∈ A ≺ ( z, ¯ y, ¯ x ) . It follows from semi-transitivity and M1 that 1 / ∈ A ≺ ( z, ¯ y, ¯ x ) . Since A ≺ ( z, ¯ y, ¯ x ) is a non-empty strict subset of a connected set [0,1], therefore it cannot be bothopen and closed, hence A ≺ ( z, ¯ y, ¯ x ) is not closed. Therefore, A ≺ ( z, ¯ y, ¯ x ) (cid:54) = A (cid:52) ( z, ¯ y, ¯ x ) . Thisimplies, there exists λ ¯ x ∈ A (cid:52) ( z, ¯ y, ¯ x ) \ A ≺ ( z, ¯ y, ¯ x ) , i.e. zλ ¯ x ¯ y ∼ ¯ x. It is clear that λ ¯ x ∈ (0 , . Define ¯ x = ¯ y (1 − λ ¯ x ) z. It follows from M2 that ¯ x ∼ zλ ¯ x ¯ y. An analogous argument impliesthere exists λ ¯ y ∈ A (cid:52) ( z, ¯ x, ¯ y ) \ A ≺ ( z, ¯ x, ¯ y ) , i.e. zλ ¯ y ¯ x ∼ ¯ y. It is clear that λ ¯ y ∈ (0 , . Define¯ y = ¯ x (1 − λ ¯ y ) z. It follows from M2 that ¯ y ∼ zλ ¯ y ¯ x. The transitivity of ∼ implies ¯ x ∼ x and¯ y ∼ y. It follows from semi-transitivity that ¯ x (cid:31) ¯ y . Repeating the construction in the preceding paragraph one more time implies there exists λ ¯ x ∈ (0 ,
1] such that zλ ¯ x ¯ y ∼ ¯ x . Since ∼ is transitive, therefore M2 implies ¯ y (1 − λ ¯ x ) z ∼ x. It follows from transitivity of ∼ and M3 that ¯ y (1 − λ ¯ x ) z ∼ x [(1 − λ ¯ x )(1 − λ ¯ y ) λ x ] z, hence(1 − λ ¯ x )(1 − λ ¯ y ) λ x ∈ A ∼ ( x, z, x ) . Then (1 − λ ¯ x )(1 − λ ¯ y ) < λ x (cid:54) = min A ∼ ( x, z, x ) whichfurnishes us a contradiction. Therefore, (cid:31) is negatively transitive. Proof of Claim 2.
Assume there exists u, v ∈ M such that u (cid:46)(cid:47) v.
It follows from non-trivialitythat x (cid:31) y for some x, y ∈ M . Then, Claim 1 implies x (cid:31) u or u (cid:31) y .Let x (cid:31) u. Then, Claim 1 implies x (cid:31) v or v (cid:31) u . Since u (cid:46)(cid:47) v, therefore x (cid:31) v. Hence, x (cid:31) u and x (cid:31) v. Next, we show that A (cid:31) ( x, u, v ) ∩ A (cid:31) ( x, u, u ) = A (cid:60) ( x, u, v ) ∩ A (cid:60) ( x, u, u ) . One of the inclusion relationship is trivial. In order to prove the other direction, pick λ ∈ A (cid:60) ( x, u, v ) ∩ A (cid:60) ( x, u, u ) . If xλu ∼ v, then it follows from transitivity of ∼ , semi-transitivity of (cid:60) and xλu (cid:60) u that v (cid:60) u. This furnishes us a contradiction with u (cid:46)(cid:47) v.
Hence, xλu (cid:31) v, i.e. λ ∈ A (cid:31) ( x, u, v ) . Similarly, if xλu ∼ u, then u (cid:60) v which contradicts u (cid:46)(cid:47) v. Hence, λ ∈ A (cid:31) ( x, u, u ) . It follows from M1, semi-transitivity and u (cid:46)(cid:47) v that 1 ∈ A (cid:31) ( x, u, v ) ∩ A (cid:31) ( x, u, u ) and0 / ∈ A (cid:31) ( x, u, v ) ∩ A (cid:31) ( x, u, u ) . Mixture-continuity imply A (cid:60) ( x, u, v ) ∩ A (cid:60) ( x, u, u ) is closed andProposition 1 imply A (cid:31) ( x, u, v ) ∩ A (cid:31) ( x, u, u ) is open. Therefore, we obtain a non-empty propersubset of [0,1] which is both open and closed. This furnishes us a contradiction with connect-16dness of [0,1].The proof is analogous for u (cid:31) y. Therefore, (cid:60) is complete. The proof of Theorem 1 is complete.
Proof of Corollary 1. If (cid:60) is semi-transitive and ∼ is transitive, then applying Theorem 1 fin-ishes the proof. Therefore, it remains to prove (cid:60) is semi-transitive and ∼ is transitive. First,assume ( a ), i.e., x ∼ y (cid:31) z implies x (cid:31) z for all x, y, z ∈ M . In order to show that (cid:60) issemi-transitive pick x, y, z ∈ M such that x (cid:31) y ∼ z. Assume x (cid:7) z. Then, completeness of (cid:60) implies either z (cid:31) x or z ∼ x. If z (cid:31) x, then it follows from y ∼ z that y (cid:31) x, which furnishesus a contradiction. If z ∼ x, then x (cid:31) y implies z (cid:31) y which contradicts z ∼ y. Hence, x (cid:31) z .Therefore, (cid:60) is semi-transitive. In order to show that ∼ is transitive pick x, y, z ∈ M such that x ∼ y ∼ z. Assume x (cid:28) z. Then completeness implies either x (cid:31) z or z (cid:31) x. Then, it followsfrom semi-transitivity that either x (cid:31) y or z (cid:31) y, which contradict x ∼ y and y ∼ z. Therefore, ∼ is transitive. An analogous argument shows that assertion ( b ) implies semi-transitivity of (cid:60) and transitivity of ∼ . Proof of Theorem 2.
Assume there exists x, y, z ∈ M such that x (cid:60) y (cid:60) z and x (cid:54) (cid:60) z . Then,completeness of (cid:60) implies z (cid:31) x. It follows from (cid:60) is strongly Archimedean that there exists λ ∈ (0 ,
1) such that z (cid:31) xλy. Since (cid:60) is star-convex and λ ∈ (0 , , therefore xλy (cid:31) y. Then,transitivity of (cid:31) implies z (cid:31) y which contradicts y (cid:60) z. Therefore (cid:60) is transitive. An analogousargument proves the sufficiency of star-concavity for transitivity of (cid:60) .The following result shows that convexity properties of preferences are characterized bythe convexity of the certain subsets of [0,1].
Lemma 1.
Let (cid:60) be a reflexive binary relation with a transitive symmetric part ∼ on a mixtureset S . Then the following are valid. (a) (cid:60) is linear ⇔ A ∼ ( x, y, z ) is convex for all x, y, z ∈ S . (b) (cid:60) is convex ⇔ A (cid:60) ( x, y, z ) is convex for all x, y, z ∈ S . (c) (cid:60) is concave ⇔ A (cid:52) ( x, y, z ) is convex for all x, y, z ∈ S . (d) Under mixture-continuity and Archimedean, (cid:60) is linear ⇔ it is convex and concave. It is possible to apply the method of proof of Claim 1 in order to prove this claim. The proof we providehere is simpler. This part of the proof does not use the properties of a generalized mixture set. Different versions areprovided in Sonnenschein (1965, Theorems 3 and 3 (cid:48) ), Lorimer (1967, Theorem 1) and Sen (1969, Theorem I).For completeness, we also provide a proof here. roof of Lemma 1. ( a ) Assume (cid:60) is linear. Pick λ, δ ∈ A ∼ ( x, y, z ) and β ∈ [0 , . Define w λ = xλy and w δ = xδy. By construction, w λ ∼ z and w δ ∼ z. It follows from transitivityof ∼ that w λ ∼ w δ . It follows from (cid:60) is linear that w λ ∼ w λ βw δ . A simple algebra andS4 imply w λ βw δ = x ( βλ + (1 − β ) δ ) y. It follows from transitivity of ∼ and w λ ∼ z that x ( βλ + (1 − β ) δ ) y ∼ z. Therefore, βλ + (1 − β ) δ ∈ A ∼ ( x, y, z ) . Hence, A ∼ ( x, y, z ) is convex.Now assume A ∼ ( x, y, z ) is convex for all x, y, z ∈ S . Pick x, y ∈ S and λ ∈ [0 ,
1] such that x ∼ y. It follows from reflexivity that x ∼ x. Therefore, S1 implies 0 , ∈ A ∼ ( x, y, x ) . Then theconvexity assumption implies A ∼ ( x, y, x ) = [0 , . Therefore, λ, (1 − λ ) ∈ A ∼ ( x, y, x ) . It followsfrom the transitivity of ∼ and x ∼ y that xλy ∼ y and x (1 − λ ) y ∼ x. Since x (1 − λ ) y = yλx, therefore (cid:60) is linear.( b ) Assume (cid:60) is convex. Pick λ, δ ∈ A (cid:60) ( x, y, z ) and β ∈ [0 , . Define w λ = xλy and w δ = xδy. By construction, w λ (cid:60) z and w δ (cid:60) z. It follows from (cid:60) is convex that w λ βw δ (cid:60) z. A simplealgebra and S4 imply w λ βw δ = x ( βλ + (1 − β ) δ ) y. Therefore, βλ + (1 − β ) δ ∈ A (cid:60) ( x, y, z ) . Hence, A (cid:60) ( x, y, z ) is convex.Now assume A (cid:60) ( x, y, z ) is convex for all x, y, z ∈ S . Pick x, y, z ∈ S and λ ∈ [0 ,
1] suchthat x (cid:60) z and y (cid:60) z. It follows from S1 that 0 , ∈ A (cid:60) ( x, y, z ) . Since A (cid:60) ( x, y, z ) is convex,therefore A (cid:60) ( x, y, z ) = [0 , . Hence, xλy (cid:60) z. Therefore, (cid:60) is convex.( c ) The proof is analogous to the proof of assertion (b) above.( d ) Assume (cid:60) is linear. Assertion (a) above implies A ∼ ( x, y, z ) is convex for all x, y, z ∈ S , hencea connected subset of [0 , . Note that [0 , \ A ∼ ( x, y, z ) = A (cid:31) ( x, y, z ) ∪ A ≺ ( x, y, z ) ∪ A (cid:46)(cid:47) ( x, y, z ) . It is clear that the three sets are mutually disjoint. It follows from Proposition 1 and mixture-continuity that they are open. By definition A (cid:60) ( x, y, z ) = A (cid:31) ( x, y, z ) ∪ A ∼ ( x, y, z ) . It followsfrom assertions (b) and (c) above that convexity of (cid:60) is equivalent to the convexity of A (cid:60) ( x, y, z )and concavity of (cid:60) is equivalent to the convexity of A (cid:52) ( x, y, z ) for all x, y, z ∈ S . Therefore, thefollowing result due to Wilder (1949, Theorem 9.9, p. 20) completes the proof of this part.
Claim . If C is a connected subset of a connected topological space X such that X \ C is theunion of n ( n > ) non-empty, pairwise disjoint sets A i which are open in X \ C, then C ∪ A i isconnected for all i. Now assume A (cid:60) ( x, y, z ) is convex and A (cid:52) ( x, y, z ) is concave for all x, y, z ∈ S . By definition A ∼ ( x, y, z ) = A (cid:60) ( x, y, z ) ∩ A (cid:52) ( x, y, z ) for all x, y, z ∈ S . Since intersection of two convex sets isconvex, therefore A ∼ ( x, y, z ) is convex for all x, y, z ∈ S . Then, assertion (a) above implies (cid:60) islinear. (Note that we do not use the continuity assumptions in order to prove this direction ofthe assertion.) Note that semi-transitivity is not needed in Proposition 1 in a mixture set.
Proof of Theorem 3. ( a ) Assume (cid:60) is linear. Pick x, y, z ∈ S such that x ∼ y and y (cid:31) z. Assume x (cid:7) z. If x ∼ z, then the transitivity of ∼ implies y ∼ z which contradicts y (cid:31) z. Therefore, either z (cid:31) x or x (cid:46)(cid:47) z. Proposition 1 implies A (cid:31) ( x, y, z ) and A ≺ ( x, y, z ) are open. It follows from mixture-continuity that A (cid:46)(cid:47) ( x, y, z ) is open. It is easy to see that the sets A (cid:31) ( x, y, z ) , A ≺ ( x, y, z ) and A (cid:46)(cid:47) ( x, y, z ) are pairwise disjoint. It follows from ∼ is transitive, x ∼ y and x (cid:28) z that A ∼ ( x, y, x )and A ∼ ( x, y, z ) are disjoint. Therefore, A ∼ ( x, y, x ) = [ A (cid:31) ( x, y, z ) ∩ A ∼ ( x, y, x )] ∪ [( A ≺ ( x, y, z ) ∪ A (cid:46)(cid:47) ( x, y, z )) ∩ A ∼ ( x, y, x )] (1)It is clear that the two sets in square brackets in Equation 1 are pairwise disjoint and openin A ∼ ( x, y, x ) . Since (cid:60) is reflexive and x ∼ y, therefore S1 implies 0 , ∈ A ∼ ( x, y, x ) . It followsfrom y (cid:31) z ans S1 that 0 ∈ A (cid:31) ( x, y, z ) . Therefore, 0 ∈ A (cid:31) ( x, y, z ) ∩ A ∼ ( x, y, x ) . It follows fromeither z (cid:31) x or x (cid:46)(cid:47) z , and S1 that either 1 ∈ ( A ≺ ( x, y, z ) ∪ A (cid:46)(cid:47) ( x, y, z )) ∩ A ∼ ( x, y, x ) . Therefore, A ∼ ( x, y, x ) is the union of two non-empty, disjoint and open sets which contradicts Lemma 1(a). Therefore, x (cid:31) z .An analogous argument shows that x (cid:31) y and y ∼ z implies x (cid:31) z for all x, y, z ∈ S .Therefore, (cid:60) is semi-transitive.( b ) Assume (cid:60) is convex. Pick x, y, z ∈ S such that x ∼ y and y (cid:31) z. Assume x (cid:7) z. If x ∼ z, then the transitivity of ∼ implies y ∼ z which contradicts y (cid:31) z. Therefore, either z (cid:31) x or x (cid:46)(cid:47) z. By definition[0 , \ A ∼ ( x, y, z ) = A (cid:31) ( x, y, z ) ∪ A ≺ ( x, y, z ) ∪ A (cid:46)(cid:47) ( x, y, z ) . It follows from Proposition 1 and mixture-continuity that A (cid:31) ( x, y, z ) , A ≺ ( x, y, z ) and A (cid:46)(cid:47) ( x, y, z )are open. It is clear that these three sets are pairwise disjoint. Moreover, it follows from y (cid:31) z and S1 along with either z (cid:31) x or x (cid:46)(cid:47) z that 0 ∈ A (cid:31) ( x, y, z ) and 1 ∈ ( A ≺ ( x, y, z ) ∪ A (cid:46)(cid:47) ( x, y, z )).Therefore, there exists λ (cid:48) ∈ A ∼ ( x, y, z ) , otherwise this yields a contradiction with the connect-edness of [0 , . It is clear that λ (cid:48) ∈ (0 , . It follows from S1, x ∼ y and reflexivity of (cid:60) that 0 , ∈ A (cid:60) ( x, y, y ) . Hence, Lemma 1(b)implies A (cid:60) ( x, y, y ) = [0 , . Therefore, λ (cid:48) ∈ A (cid:60) ( x, y, y ) . It follows from y (cid:31) z and ∼ is transitivethat A ∼ ( x, y, z ) ∩ A ∼ ( x, y, y ) = ∅ . Hence, λ (cid:48) ∈ A (cid:31) ( x, y, y ) . Define z (cid:48) = xλ (cid:48) y. Then[0 , \ A ∼ ( z (cid:48) , z, y ) = A (cid:31) ( z (cid:48) , z, y ) ∪ A ≺ ( z (cid:48) , z, y ) ∪ A (cid:46)(cid:47) ( z (cid:48) , z, y ) . It follows from S1, z (cid:48) (cid:31) y and y (cid:31) z that 1 ∈ A (cid:31) ( z (cid:48) , z, y ) and 0 ∈ A ≺ ( z (cid:48) , z, y ) . Analogous tothe above argument, connectedness of [0 ,
1] implies there exists δ (cid:48) ∈ A ∼ ( z (cid:48) , z, y ) . It is clear that19 (cid:48) ∈ (0 , . It follows from λ (cid:48) ∈ A ∼ ( x, y, z ) that z (cid:48) ∼ z. Then, S1 and reflexivity of (cid:60) implies 0 , ∈ A (cid:60) ( z (cid:48) , z, z (cid:48) ) . Hence, Lemma 1(b) implies A (cid:60) ( z (cid:48) , z, z (cid:48) ) = [0 , . Therefore, δ (cid:48) ∈ A (cid:60) ( z (cid:48) , z, z (cid:48) ) . Itfollows from y (cid:31) z and ∼ is transitive that A ∼ ( z (cid:48) , z, y ) ∩ A ∼ ( z (cid:48) , z, z (cid:48) ) = ∅ , therefore δ (cid:48) ∈ A (cid:31) ( z (cid:48) , z, z (cid:48) ) . Define y (cid:48) = z (cid:48) δ (cid:48) z. It follows from δ (cid:48) ∈ A ∼ ( z (cid:48) , z, y ) and transitivity of ∼ that y ∼ y (cid:48) and x ∼ y (cid:48) . Then, it follows from Lemma 1(b) that A (cid:60) ( x, y, y (cid:48) ) = [0 , . Since z (cid:48) = xλ (cid:48) y, therefore z (cid:48) (cid:60) y (cid:48) . This furnishes us a contradiction with δ (cid:48) ∈ A (cid:31) ( z (cid:48) , z, z (cid:48) ) , i.e. y (cid:48) (cid:31) z (cid:48) . ( c ) The proof is analogous to the proof of (b) above.( d ) Let (cid:60) be complete and convex. Then, it follows from assertion (b) above that x ∼ y and y (cid:31) z implies x (cid:31) z for all x, y, z ∈ S . In order to show (cid:60) is semi-transitive, pick x, y, z ∈ S suchthat x (cid:31) y and y ∼ z. Assume x (cid:7) z. Then either z (cid:31) x or z ∼ x. If z (cid:31) x, the it follows from y ∼ z and convexity that y (cid:31) x, which furnishes us a contradiction. If z ∼ x, then convexityimplies z (cid:31) y which contradicts z ∼ y. Hence, x (cid:31) z . Therefore, (cid:60) is semi-transitive. The proofof the sufficiency of concavity for semi-transitivity is analogous.The proof of Theorem 3 is complete.The proofs of Corollaries 2 and 3 are immediate from Theorems 1 and 3. Proof of Theorem 4.
Assume (cid:60) is a non-trivial, anti-symmetric, complete, transitive and mix-ture continuous binary relation on a generalized mixture set M induced by ∼ . Since M isinduced by an anti-symmetric relation, it is a mixture set. For the convenience of reader, define S = M . First, consider the following claim. Claim . S is isomorphic to a convex subset of some linear space. Assume without loss of generality that S is a convex subset of some linear space and xλy = λx + (1 − λ ) y for all x, y ∈ S and all λ ∈ [0 , x, ¯ y ∈ S such that ¯ y (cid:31) ¯ x . We nextshow that for any z ∈ S , one and only one of the following is true.(i) there exists λ ∈ (0 ,
1) such that ¯ x = zλ ¯ y ,(ii) there exists λ ∈ [0 ,
1] such that z = ¯ xλ ¯ y ,(iii) there exists λ ∈ (0 ,
1) such that ¯ y = ¯ xλz .For any z ∈ S , it follows from completeness and transitivity of (cid:60) that one and only one ofthe following is true: (i (cid:48) ) z ≺ ¯ x , (ii (cid:48) ) ¯ x (cid:52) z (cid:52) ¯ y , (iii (cid:48) ) ¯ y ≺ z . If (i (cid:48) ) holds, then z ≺ ¯ x ≺ ¯ y .By mixture continuity and connectedness of [0 , λ ∈ (0 ,
1) such that zλ ¯ y ∼ ¯ x .Since (cid:60) is anti-symmetric, ¯ x = zλ ¯ y . Similarly, (ii (cid:48) ) implies that there exists λ ∈ [0 ,
1] such that Since (cid:60) is non-trivial, there exist such pair. = ¯ xλ ¯ y , and (iii (cid:48) ) implies that there exists λ ∈ (0 ,
1) such that ¯ y = ¯ xλz . Therefore, for any z ∈ S , at least one of (i), (iii), and (iii) holds. In order to show that only one of them holds,pick z ∈ S . Assume (i) holds and z (cid:54)≺ ¯ x , i.e., there exists λ ∈ (0 ,
1) such that ¯ x = zλ ¯ y and z (cid:60) ¯ x . Then, either ¯ x (cid:52) z (cid:52) ¯ y or ¯ x ≺ ¯ y ≺ z , i.e., either (ii (cid:48) ) or (iii (cid:48) ) holds. Assume (ii (cid:48) ) holds.Then, it follows from (i) that there exists λ ∈ (0 ,
1) such that ¯ x = λz + (1 − λ )¯ y . Then ¯ x (cid:54) = ¯ y implies ¯ x (cid:54) = z (cid:54) = ¯ y . Since (ii (cid:48) ) implies (ii), therefore it follows from ¯ x (cid:54) = z (cid:54) = ¯ y that there exists δ ∈ (0 ,
1) such that z = δ ¯ x + (1 − δ )¯ y . Then, ¯ x = λδ ¯ x + (1 − λδ )¯ y and 0 < λδ < x (cid:54) = ¯ y . Analogously, (iii (cid:48) ) yields a contradiction. Therefore, z ≺ ¯ x , hence (i)and (i (cid:48) ) are equivalent. Similarly, (ii) implies ¯ x (cid:52) z (cid:52) ¯ y , and (iii) implies ¯ y ≺ z . Therefore, (ii)and (ii (cid:48) ), and (iii) and (iii (cid:48) ) are equivalent.Therefore, the linear space containing S is one dimensional. Without loss of generality,assume S is an interval in R . Since the usual orders ≥ and ≤ on R are complete and anti-symmetric, therefore either ¯ x < ¯ y or ¯ x > ¯ y . Assume ¯ x < ¯ y . Pick a pair x, y ∈ S of distinctpoints. Assume without loss of generality that x ≺ y . Then, one and only one of the followingis true: (1) y ≺ ¯ x , (2) y = ¯ x , (3) ¯ x ≺ y ≺ ¯ y , (4) y = ¯ y , (5) ¯ y ≺ y . If (1) holds, then it followsfrom y ≺ ¯ x ≺ ¯ y and (i) above that there exists λ ∈ (0 ,
1) that ¯ x = λy + (1 − λ )¯ y . Since ¯ x < ¯ y ,therefore y < ¯ x . Analogously, x ≺ y ≺ ¯ x and (i) imply x < y . If (2) holds, then it follows from x ≺ y = ¯ x ≺ ¯ y and (i) that x < ¯ x = y . If (3) holds, then it follows from ¯ x ≺ y ≺ ¯ y and (i)that y < ¯ y . Then, x ≺ y ≺ ¯ y and (i) imply x < y . If (4) holds and x ≺ ¯ x , then it follows from x ≺ ¯ x ≺ ¯ y = y and (i) that x < y . If (4) holds and ¯ x ≺ x , then it follows from ¯ x ≺ x ≺ ¯ y = y and (i) that x < y . The case where (4) holds and x = ¯ x trivially implies x < y . Lastly, assume(5) holds. It follows from ¯ x ≺ ¯ y ≺ y and (i) that ¯ x < ¯ y < y . If x ≺ ¯ y , then it follows from x ≺ ¯ y ≺ y , (i) and ¯ y < y that x < y . Similarly, if ¯ y ≺ x , then it follows from ¯ y ≺ x ≺ y , (i)and ¯ y < y that x < y . The case where x = ¯ y trivially implies x < y . Since x, y are arbitrary, x (cid:48) ≺ y (cid:48) implies x (cid:48) < y (cid:48) for all x (cid:48) , y (cid:48) ∈ S . Conversely, if x < y and x ⊀ y for some x, y ∈ S ,then anti-symmetry and completeness of (cid:52) imply y ≺ x . The argument above implies y < x which furnishes us a contradiction. Then, ≺ and < are identical. Since (cid:52) is anti-symmetric,therefore (cid:52) and ≤ are identical. If ¯ x > ¯ y , then an analogous argument implies that (cid:52) and ≥ are identical. It remains to prove Claim 4. Proof of Claim 4.
Proposition 0 shows that if a mixture set S satisfies conditions (C1) and (C2),then S is isomorphic to a convex subset of some linear space. Therefore showing both conditionshold completes the proof.In order to show (C2), pick x, y, z ∈ S . If x = y or x = z or y = z , then S4 implies(C2). Thus, assume x, y, z are distinct. Anti-symmetry of (cid:60) implies that one and only one ofthe following cases holds: x (cid:31) y (cid:31) z, x (cid:31) z (cid:31) y, y (cid:31) x (cid:31) z, y (cid:31) z (cid:31) x, z (cid:31) x (cid:31) y, z (cid:31) y (cid:31) x .We only show (C2) holds in the first case since other cases can be proved similarly. In follows21rom x (cid:31) y (cid:31) z and (i) above that there exists α ∈ (0 ,
1) such that y = xαz . Then, S4 impliesthat for all λ, µ ∈ [0 ,
1] with λµ (cid:54) = 1,( xλy ) µz = (cid:0) xλ ( xαz ) (cid:1) µz = (cid:0) x ( λ + (1 − λ ) α ) z (cid:1) µz = x (cid:0) µ ( λ + (1 − λ ) α ) (cid:1) z,x ( λµ ) (cid:0) y µ (1 − λ )1 − λµ z (cid:1) = x ( λµ ) (cid:0) ( xαz ) µ (1 − λ )1 − λµ z (cid:1) = x ( λµ ) (cid:0) x αµ (1 − λ )1 − λµ z (cid:1) = x (cid:0) µ ( λ + (1 − λ ) α ) (cid:1) z. Thus, (C2) holds.The following claim is useful in the remaining part of the proof.
Claim . If x (cid:54) = y , then µ ∈ (0 , implies x (cid:54) = xµy .Proof of Claim 5. Assume x (cid:54) = y and x = xµy for some µ ∈ (0 , x = xµy = ( xµy ) µy = xµ y . Repeating this argument n times implies y = yµ n x for any naturalnumber n . Since x (cid:54) = y , we have x (cid:31) y or y (cid:31) x . If x (cid:31) y , then mixture continuity implies { λ ∈ [0 ,
1] : xλy ≺ x } is an open set which contains 0. Thus, there exists δ > λ ∈ [0 , δ ), xλy ≺ x . Then, for big enough n , µ n < δ . Hence, x = xµ n y ≺ x furnishes us acontradiction. Analogously, y (cid:31) x yields a contradiction.In order to show (C1), we first show the following condition of Mongin (2001, (A0 (cid:48) )) holds.(C1 (cid:48) ) If x (cid:54) = y and λ , λ ∈ [0 , , then xλ y = xλ y implies λ = λ . Pick x, y ∈ S and λ , λ ∈ [0 ,
1] such that x (cid:54) = y and λ (cid:54) = λ . Assume, without loss ofgenerality, that λ > λ . Assume xλ y = xλ y . First, let λ = 0. Then, S1, S2 and x (cid:54) = y imply y = xλ y and λ ∈ (0 , µ = 1 − λ . Then, S2 implies y = yµx , hence µ ∈ (0 ,
1) yields acontradiction with Claim 5. Therefore, (C1 (cid:48) ) holds when λ = 0. Now, let λ >
0. If λ = 1,then S1, S2 and x (cid:54) = y imply x = xλ y and λ ∈ (0 , > λ > λ >
0. Since (cid:16) − λ − λ − λ (cid:17) (1 − λ ) = 1 − λ , it follows fromS2 and S3 that xλ y = x λ − λ − λ ( xλ y ). Let µ = λ − λ − λ and z = xλ y . Since λ ∈ (0 , z (cid:54) = x . It follows from xλ y = xλ y = z that z = z (1 − µ ) x . Then, µ ∈ (0 ,
1) yileds acontradiction with Claim 5. Therefore, proof of (C1 (cid:48) ) is complete.Next, we will show that (C1) holds. Pick x, y, y (cid:48) ∈ S such that y (cid:54) = y (cid:48) . Assume xλy = xλy (cid:48) for some x ∈ S and λ ∈ (0 , x = y , then x = xλx = xλy = xλy (cid:48) furnishes us acontradiction with Claim 5. Similarly, x = y (cid:48) yields a contradiction. Hence, assume y (cid:54) = x (cid:54) = y (cid:48) .Then, one and only one of the following cases holds: x ≺ y ≺ y (cid:48) , x ≺ y (cid:48) ≺ y (cid:48) , y ≺ x ≺ y (cid:48) , y ≺ y (cid:48) ≺ x, y (cid:48) ≺ x ≺ y, y (cid:48) ≺ y ≺ x . We provide proof of the first case, since other cases canbe shown similarly. Since x ≺ y ≺ y (cid:48) , it follows from (i) above that there exists µ ∈ (0 , The fact that x = xλx follows from x = x x = x x = ( x x ) λx = ( x x ) λx = xλx ; see Fishburn (1982,Chapter 2) for a detailed discussion on further implications of mixture set axioms. y = xµy (cid:48) . Thus, xλy (cid:48) = xλy = xλ ( xµy (cid:48) ) = x (cid:0) λ + (1 − λ ) µ (cid:1) y (cid:48) . Then, (C1 (cid:48) ) implies λ = λ + (1 − λ ) µ . It follows from λ ∈ (0 ,
1) that µ = 0. This furnishes us a contradiction.Therefore, (C1) holds.The proof of Theorem 4 is complete. Proof of Corollary 4.
We first show that the mixture operation [ x ] λ [ y ] = [ xλy ] is well definedon the quotient set M| ∼ . Since M is a generalized mixture set, any element in the quotient setis assigned to a point in the set. For all [ x (cid:48) ] = [ x ] and [ y (cid:48) ] = [ y ], it follows from x ∼ x (cid:48) , y ∼ y (cid:48) andthe independence axiom that xλy ∼ x (cid:48) λy (cid:48) , hence [ x ] λ [ y ] = [ xλy ] = [ x (cid:48) λy (cid:48) ] = [ x (cid:48) ] λ [ y (cid:48) ]. Hence,the mixture operation is a well-defined function. It follows from (i) [ x ]1[ y ] = [ x y ] = [ x ], (ii)[ x ] λ [ y ] = [ xλy ] = [ y (1 − λx )] = [ y ](1 − λ )[ x ] and (iii) ([ x ] λ [ y ]) δ [ y ] = [ xλy ] δ [ y ] = [( xλy ) δy ] =[ x ( λδ ) y ] = [ x ]( λδ )[ y ] that the mixture operation defined above satisfies S1–S3, hence M| ∼ is amixture set.Completeness and transitivity of the derived relation ˆ (cid:60) on the quotient set
M| ∼ followfrom those of (cid:60) . In order to show that ˆ (cid:60) satisfies mixture continuity, pick x, y, z ∈ M . We willshow that A (cid:60) ( x, y, z ) = A ˆ (cid:60) ([ x ] , [ y ] , [ z ]) . In order to show the forward direction, pick λ such that xλy (cid:60) z . It follows from transitivity of (cid:60) that for all w ∼ xλy and all z (cid:48) ∼ z , w (cid:60) z (cid:48) . Hence, [ x ] λ [ y ] = [ xλy ] ˆ (cid:60) [ z ]. In order to provethe backward direction, pick λ such that [ x ] λ [ y ] ˆ (cid:60) [ z ]. Then, [ x ] λ [ y ] = [ xλy ] and the definitionof ˆ (cid:60) imply xλy (cid:60) z . Therefore, (cid:60) is mixture continuous if and only if ˆ (cid:60) is mixture continuous.Therefore, ˆ (cid:60) is complete, transitive, anti-symmetric and mixture continuous. Then,following Theorem 4, assume without loss of generality that the mixture set M| ∼ is aninterval in the real line and ˆ (cid:60) is either ≥ or ≤ . Let ˆ (cid:60) = ≥ . Define a utility functionˆ u : M| ∼ → R as ˆ u ([ x ]) = [ x ]. Note that ˆ u ([ x ] λ [ y ]) = [ x ] λ [ y ] = λ ˆ u ([ x ]) + (1 − λ )ˆ u ([ y ]). Define u : M → R as u ( x ) = ˆ u ([ x ]) . Since u ( xλy ) = ˆ u ([ xλy ]) = ˆ u ([ x ] λ [ y ]) = λ ˆ u ([ x ]) + (1 − λ )ˆ u ([ y ]) = λu ( x ) + (1 − λ ) u ( y ), therefore u has the desired properties for an expected utility representationof (cid:60) . If ˆ (cid:60) = ≤ , then the argument above with ˆ u ([ x ]) = − [ x ] completes the proof. Proof of Proposition 3.
It follows from Theorem 1 that (cid:60) is not Archimedean. Then Proposition1 implies there exists x, y, z ∈ M such that A (cid:31) ( x, y, z ) or A ≺ ( x, y, z ) is not open. Assumewithout loss of generality that A (cid:31) ( x, y, z ) is not open. Then there exists λ ∈ A (cid:31) ( x, y, z ) and asequence { λ t } t ∈ N such that λ t → λ and xλ t y (cid:7) z for t large enough.Assume z (cid:60) xλ t y for t large enough. Then, it follows from mixture-continuity that A (cid:52) ( x, y, z ) is closed, hence λ ∈ A (cid:52) ( x, y, z ) . This furnishes us a contradiction with xλy (cid:31) z. Therefore, { λ t } has a subsequence { δ t } such that δ t ∈ A (cid:46)(cid:47) ( x, y, z ) for t large enough. It follows23rom mixture-continuity that A (cid:46)(cid:47) ( x, y, z ) is open. Therefore for all t large enough, A (cid:46)(cid:47) ( x, y, z )contains an open neighborhood V δ t of δ t . Pick an open neighborhood V λ of λ. It follows from δ t → λ that δ t is contained in V λ for t large enough. Therefore, the non-empty open set V λ ∩ V δ t is contained in A (cid:46)(cid:47) ( x, y, z ) for t large enough. Proof of Proposition 4.
Assume (cid:60) satisfies the hypotheses of the proposition and it is not flimsy.Then, for all x, y, z ∈ M and all λ ∈ A (cid:46)(cid:47) ( x, y, z ), there exists an open neighborhood V λ of λ such that V λ ⊂ A (cid:46)(cid:47) ( x, y, z ). Then, A (cid:46)(cid:47) ( x, y, z ) is open, and hence it has open sections. Then,it follows from Proposition 2 that (cid:60) is mixture-continuous. Then, Theorem 1 implies (cid:60) iscomplete. This furnishes us a contradiction. In this section, we provide some examples in mixture sets illustrating how Propositions 1 and 2fail if any of its assumptions is dropped. Let [0 ,
1] be the choice space endowed with the usualstructures. It is clear that [0 ,
1] is a mixture set with the usual linear space operations. Thefollowing example concerns Proposition 1.
Example 4.
Assume (cid:60) is a reflexive relation on [0,1] such that 1 (cid:31) x for all x < A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) either contains at most twoelements or equals [0,1] for all x, y, z ∈ [0 , A (cid:31) (1 , ,
0) = { } is not open. Moreover, 1 (cid:31)
0, but for all x ∈ (0 ,
1) and all λ ∈ (0 , λx (cid:7)
0. Therefore, (cid:60) ismixture-continuous, but violates any of the three equivalent conditions of Proposition 1.We next provide examples concerning Proposition 2.
Example 5.
First, we illustrate A (cid:46)(cid:47) ( x, y, z ) is not open for some x, y, z. Define a binary relation (cid:60) as follows: x ∼ y for all x, y ∈ [0 , .
5) and x (cid:46)(cid:47) y for all other points in [0,1]. Then A (cid:46)(cid:47) (1 , ,
0) = [0 . ,
1] which is not open. Since A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) are equal to one of ∅ andan interval in [0,1] for all x, y, z ∈ [0 , , therefore they are convex. Strong Archimedean propertyis trivially satisfied since there is no x, y such that x (cid:31) y. It follows from A (cid:60) (0 , ,
0) = [0 , . (cid:60) is not mixture-continuous. Example 6.
Second, we illustrate (cid:60) is not strongly Archimedean. Define a binary relation (cid:60) asfollows: x ∼ y for all x, y ∈ [0 , .
5) and all x, y ∈ [0 . , x (cid:31) y for all x ∈ [0 . , , y ∈ [0 , . . Then, (cid:60) is complete. It follows from 0 . (cid:31) . λ ∼ λ ∈ (0 ,
1) that (cid:60) is notstrongly Archimedean. Since A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) are equal to an interval in [0,1] for all x, y, z ∈ [0 , , therefore they are convex. Observing that A (cid:52) (0 , ,
0) = [0 , .
5) is not closedshows that (cid:60) is not mixture-continuous. 24 xample 7.
Third, we illustrate A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) do not have finitely many compo-nents. Define a binary relation (cid:60) as follows. Assume x ∼ y if x, y rational or x, y irrational.Assume x (cid:31) y for all x rational and y irrational. For any rational z , A (cid:60) (1 , , z ) = Q ∩ [0 ,
1] andfor any irrational z (cid:48) , A (cid:52) (1 , , z (cid:48) ) = ( R \ Q ) ∩ [0 , A (cid:46)(cid:47) ( x, y, z ) is empty for all x, y, z, hence open. It follows from the “Archimedean”property in mathematics that (cid:60) is strongly Archimedean. In order to see this, pick x, y, z suchthat x (cid:31) y. Then, it is clear that x is rational and y is irrational. If z is rational, then it followsfrom xλz is rational for all rational λ ∈ [0 ,
1] that xλz (cid:31) y. Similarly, since yλz is irrational forall rational λ ∈ (0 , , therefore x (cid:31) yλz. If z is irrational, then either x < z or z < x. In bothcases the open interval determined by x and z contains a rational number x (cid:48) such that x (cid:48) = xλz for some λ ∈ [0 , . Hence xλz (cid:31) y. Similarly, if z is irrational and z = y, then x (cid:31) yλz for all λ ∈ [0 , z (cid:54) = y. Then the open interval determined by y and z contains an irrationalnumber y (cid:48) such that y (cid:48) = yλz. Hence x (cid:31) yλz. Therefore, (cid:60) is strongly Archimedean. It followsfrom the set of rationals is not closed and that A (cid:60) (1 , , z ) is the set of rationals in [0,1] for anyrational z that (cid:60) is not mixture-continuous.In the last example, both A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) do not have finitely many components.We do not have an example where only one of these fail. Hence it is possible to weaken thisassumption, but we do not know such weaker assumption at present. It is easy to show thatunder independence assumption, both A (cid:60) ( x, y, z ) and A (cid:52) ( x, y, z ) are convex, hence have finitelymany components. We show in Lemma 1 in Section 5 that the convexity of these sets followsfrom linearity under mixture-continuity and Archimedean properties. References
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