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Additive Utilities on Densely Ordered Sets
Journal of Mathematical Psychology 46, 515–530 (2002) doi:10.1006/jmps.2001.1410
Additive Utilities on Densely Ordered Sets Yutaka Nakamura University of Tsukuba
This paper shows sufficient conditions for the existence of additive utilities without a restricted solvability axiom. Our conditions require that each essential component of the underlying Cartesian product be densely ordered. © 2002 Elsevier Science (USA)
1. INTRODUCTION
Three decades ago, a well-known algebraic approach for the axiomatization of additive conjoint measurement had been established by Krantz, Luce, Suppes, and Tversky (1971, Chapt. 6), hereafter abbreviated KLST. Although their theory provides less restrictive (and more favorable) conditions than the topological approach advocated by Debreu (1960) (see Wakker, 1988), their structural condition, known as restricted solvability, is still unnecessarily strong in applications (see for example Gonzales & Jaffray, 1998). In this respect, Jaffray (1974) is the first to present an axiomatic structure for additive conjoint measurement on ‘‘infinite’’ sets without restricted solvability. Unfortunately, his axioms are too complicated to be tested, since they include the independence condition C as Jaffray (1974) dubbed it, also known as a family of the cancellation axioms of any order. Recent papers by Gonzales (1996, 1999, 2000) have uncovered a subtle relationship between (restricted) solvability and additive representability. Gonzales (1996) showed by counterexamples that solvability cannot be dropped for two or more components. Also in Gonzales (2000), it is shown in the 2-dimensional case that cancellation axioms of any order are required to ensure additive representability when restricted solvability holds with respect to (w.r.t.) one component. Furthermore, he proved that an additional structural condition of equal spacedness w.r.t. the second component is sufficient for the existence of additive utilities. In addition, Gonzales (2000) proved that additive representation exists when restricted solvability holds w.r.t. only two components, but the uniqueness property of a cardinal scale fails to hold. The author is gratefully indebted to Christophe Gonzales for many helpful comments and corrections in the original proofs, and aknowledges the action editor for improving readability of the paper. Address correspondence and reprint requests to Yutaka Nakamura, Institute of Policy and Planning Sciences, University of Tsukuba, 1-1-1 Tennoudai, Tsukuba, Ibaraki 305-8573, Japan.
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0022-2496/02 $35.00 © 2002 Elsevier Science (USA) All rights reserved.
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YUTAKA NAKAMURA
The aim of this paper is to study additive representability without assuming restricted solvability for every component. We show that if every essential component of the underlying Cartesian product is densely ordered, then additive representability obtains without restricted solvability, but retains the uniqueness property of a cardinal scale. Contrary to Jaffray’s complicated axioms, it is shown to be sufficient for additive representability that restricted solvability and the Archimedean axiom in KLST’s axiom system are replaced respectively by a coordinate order denseness condition and a strengthened Archimedean axiom. Since restricted solvability need not be satisfied on all components, a linear order that satisfies our axiom system can have an additive representation. The paper is organized as follows. Section 2 introduces preliminary definitions and results. In Section 3, our structural conditions are defined and the main theorem is stated. Examples in which restricted sovability does not necessarily hold are discussed. Section 4 is devoted to the proof of the main theorem.
2. PRELIMINARIES
Let X=X1 × · · · × Xn be an n-dimensional Cartesian product with elements x=(x1 , ..., xn ). We assume that n \ 2. Given a nonempty proper subset N= {i1 , ..., ir } … {1, ..., n} with nonempty complement (N)={1, ..., n} 0 N, let XN = Xi1 × · · · Xir , and write x ¥ X as (xN , x(N) ) with xN ¥ XN and x(N) ¥ X(N) . By Q we denote a binary preference relation on X with O and ’ defined as usual: for x, y ¥ X, x O y if ¬ (y Q x), and x ’ y if x Q y and y Q x. We say that Q is a weak order if it is transitive and complete and that O is a linear order if Q is antisymmetric weak order. A real valued function U on X is said to be additive if there are real valued functions ui on Xi for i=1, ..., n such that U(x)=; ni=1 ui (xi ) for all x ¥ X. If, for all x, y ¥ X, x Q y Z U(x) [ U(y), then a function U is said to represent Q. When U representing Q is additive, it is called an additive representation for Q. A function U representing Q is a cardinal scale if there are real numbers a > 0 and b such that V=aU+b if and only if V also represents Q. In what follows, we introduce a slightly modified version, advocated by Wakker (1989, 1991), of KLST’s classical theory for additive representability of Q. Definition 2.1. The binary relation Q satisfies coordinate independence if, for all x, y, z, w ¥ X and i=1, ..., n, (xi , z(i) ) Q (xi , w(i) ) whenever (yi , z(i) ) Q (yi , w(i) ). For each i=1, ..., n, the binary relation Q i on Xi is defined as follows. For all xi , yi ¥ Xi , xi Q i yi Z (xi , a(i) ) Q (yi , a(i) )
for some a(i) ¥ X(i) .
Similarly, O i and ’ i can be derived from O and ’, respectively. The property that the definition of Q i does not depend on the choice of a specific a(i) is known as weak separability in the literature, a condition which follows from coordinate independence. When n=2, coordinate independence can be replaced by the following definition.
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Definition 2.2. The binary relation Q satisfies generalized triple cancellation if, for all x, y, z, w, a, b, c, d ¥ X, and i=1, ..., n, (ai , x(i) ) Q (bi , y(i) ), (bi , z(i) ) Q (ai , w(i) ), (ci , y(i) ) Q (di , x(i) ) imply that (ci , z(i) ) Q (di , w(i) ). It is easy to verify that generalized triple cancellation implies coordinate independence. The restricted solvability that we shall drop in our axiomatization is defined as follows. Definition 2.3. The binary relation Q satisfies restricted solvability on component i if, for all x, y, z, w ¥ X, there exists an element ai ¥ Xi such that y ’ (ai , w(i) ) whenever (xi , w(i) ) O y and y O (zi , w(i) ). We say that Q satisfies restricted solvability if it satisfies restricted solvability on all components. With restricted solvability, standard sequences in the following definition are meaningful in the sense that they form equally spaced sequences in utility units. For example, every standard sequence consists of a singleton when O is a linear order for which restricted solvability fails to hold. Definition 2.4. For component i and any set K of consecutive integers, a set {x ki : k ¥ K} is a standard sequence (on component i, w.r.t. Q) if there exist (ci , a(i) ), (ci , b(i) ) ¥ X such that ¬ ((ci , a(i) ) ’ (ci , b(i) )), and (x ki , a(i) ) ’ (x k+1 , b(i) ) i for all k, k+1 ¥ K. A standard sequence may be finite or infinite of any length. A standard sequence (on component i) is bounded if there exist x 0i , x gi ¥ Xi such that x 0i Q i x ki and x ki Q i x gi for all k. The Archimedean axiom requires that every bounded standard sequence be finite. Definition 2.5. Component i is essential if there are a, b, c ¥ X such that (ai , c(i) ) O (bi , c(i) ). Component i is inessential if it is not essential. The following theorem, due to Wakker (1991), gives a characterization of additive representation which does not require separate formulations for a different number of essential components. Theorem 2.1. Suppose that at least two components are essential, and the restricted solvability is satisfied. Then the following statements are equivalent: (1) there exists an additive representation for Q. (2) the binary relation Q is a weak order that satisfies the Archimedean axiom and generalized triple cancellation. If at least three components are essential, then the generalized triple cancellation in (2) can be replaced by the coordinate independence. Moreover, the additive representation in (1) is a cardinal scale.
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3. THE MAIN THEOREM
This section presents the main theorem for additive representability without imposing restricted solvability on our axiomatic structure for Q. The key axiom is the following denseness condition. Definition 3.1. The binary relation Q satisfies coordinate order denseness if, for all x, y, a, b ¥ X and i=1, ..., n, there exists an element ci ¥ Xi such that x O (ci , z(i) ) O y whenever (ai , z(i) ) Q x O y Q (bi , z(i) ). When coordinate independence holds, coordinate order denseness requires that, for essential i, Q i be order dense, i.e., for all xi , yi ¥ Xi , xi O i ai and ai O i yi for some ai ¥ Xi whenever xi O i yi . Since restricted solvability may not hold, there may not exist many standard sequences as in Definition 2.4. As an alternative, we introduce a strengthened notion of the standard sequence which provides a variably spaced utility scale with a minimum length. A similar generalization of the standard sequence is independently introduced by Gonzales (1999, Definition 5), who dubbed it an overstandard sequence. Definition 3.2. For component i and any set K of consecutive integers, a set {x ki : k ¥ K} is a strong standard sequence (on component i, w.r.t. Q ) if there exist (ci , a(i) ), (ci , b(i) ) ¥ X such that (ci , b(i) ) O (ci , a(i) ), and (x ki , a(i) ) Q (x k+1 , b(i) ) for i all integers k, k+1 ¥ K. A strong standard sequence may be finite or infinite of any length even if O is a linear order. A strong standard sequence (on component i) is bounded if there exist x 0i , x gi ¥ Xi such that x 0i Q i x ki and x ki Q x gi for all k. Our strengthening of the Archimedean axiom requires that every bounded strong standard sequence be finite. The main theorem is stated as follows. The proof will be deferred to the next section. Theorem 3.1. Suppose that at least two components are essential and that coordinate order denseness is satisfied. Then the following statements are equivalent: (1) there exists an additive representation for Q; (2) the binary relation Q is a weak order that satisfies the strong Archimedean axiom and generalized triple cancellation. If at least three components are essential, then the generalized triple cancellation in (2) can be replaced by coordinate independence. Moreover, the additive representation in (1) is a cardinal scale. We shall consider two simple examples where restricted solvability does not hold. Let Q be the set of all rational numbers. Example 3.1. Let X1 =R and X2 =Q. Let a weak order Q on X1 × X2 be defined by an additive function U(x)=x1 +x2 . There is no x2 ¥ X2 such that (0, x2 ) ’ (`2, 0), but (0, 0) O (`2, 0) O (0, 2), so restricted solvability on
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component 2 fails to hold. Jaffray (1974) noted that this Q satisfies all conditions in his Theorem 7.2 for additive representability. It is also verified that Q satisfies coordinate order denseness and the conditions (2) in Theorem 3.1 above. Although Q satisfies restricted solvability on component 1, such a condition is not required to verify additive representability. The next example shows that a linear order O has an additive representation. Example 3.2. Let X1 =Q, X2 ={a `2: a ¥ Q}, and X3 ={a `3: a ¥ Q}. Let a weak order Q on X1 × X2 × X3 be represented by the additive function U(x)= x1 +x2 +x3 . It is easy to verify that Q satisfies coordinate order denseness and the conditions (2) in Theorem 3.1 above. A notable feature of this example is that O on X=X1 × X2 × X3 is a linear order, so that O does not satisfy restricted solvability on all components. To see that O is a linear order, it suffices to show that, for all integers m1 , m2 , m3 , m1 +m2 `2+m3 `3=0 S m1 =m2 =m3 =0. We have m1 +m2 `2+m3 `3=0 S m1 +m2 `2=−m3 `3 S m 21 +2m 22 +2m1 m2 `2=3m 23 , whose solutions must be m1 =m2 =m3 =0. We note that there also exist cases in which coordinate order denseness does not hold whereas solvability does. For example, let X=Z × Z, where Z is the set of all integers. Let Q on X be represented by an additive function U(x)=x1 +x2 . Then unrestricted solvability holds but not coordinate order denseness.
4. PROOF OF THE MAIN THEOREM
Theorem 3.1 is proved in this section. Suppose that at least two components are essential and coordinate order denseness is satisfied. It is easy to see that (1) implies (2). Thus we assume that (2) holds to prove (1); i.e., the binary relation Q is a weak order that satisfies the strong Archimedean axiom and generalized triple cancellation. A preference interval Ai (on component i, w.r.t. Q i ) is defined to be a nonempty subset of Xi for which zi ¥ Ai whenever xi , yi ¥ Ai , xi Q i zi , and zi Q i yi . A preference interval Ai on component i is said to be left exhausted if yi ¥ Ai whenever xi ¥ Ai and yi Q i xi , and bounded above if there is a yi ¥ Xi such that xi Q yi for all xi ¥ Ai . Since we are concerned with only preference intervals on each component that are left exhausted and bounded above, let Wi denote the set of all preference intervals on component i that are left exhausted and bounded above, and also let W=W1 × · · · × Wn with elements A=(A1 , ..., An ). We shall identify A ¥ W with a product set A1 × · · · × An and write a ¥ A when ai ¥ Ai for i=1, ..., n. Let
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A 0i ={xi ¥ Ai : xi O i yi for some yi ¥ Ai } and A 0=(B1 , ..., Bn ), where, for i=1, ..., n,
˛ AA
Bi =
0 i
i
if A 0i ] ”, otherwise.
When A is interpreted as a product set, A 0 means B1 × · · · × Bn . Given a nonempty proper subset N={i1 , ..., ir } … {1, ..., n}, we shall let WN =Wi1 × · · · × Wir and write A ¥ W as (AN , A(N) ) with AN ¥ WN and A(N) ¥ W(N) . A certain preference interval associated with a, b ¥ X will be identified as follows: Ii (a, b)={xi ¥ Xi : x Q a and x(i) =b(i) }. We note that Ii (a, b) may not be in Wi if either Ii (a, b)=Xi and Xi ¨ Wi , or Ii (a, b)=”. However, Ii (a, a) always belongs to Wi . Let Ii (a, b) 0={xi ¥ Xi : x O a and x(i) =b(i) } when Ii (a, b) ¥ Wi , where Ii (a, a) 0 may be empty. Let I(a, b)=(I1 (a, b), ..., In (a, b)), I(i) (a, b)=(I1 (a, b), ..., Ii − 1 (a, b), Ii+1 (a, b), ..., In (a, b)). For all A ¥ W, we define NA ={i : A 0i ] ”}. If NA ={1, ..., m}, then I(a, a)=(I1 (a, a), ..., In (a, a)), I(a, a) 0=(I1 (a, a) 0, ..., Im (a, a) 0, Im+1 (a, a), ..., In (a, a)). When x O y for all x ¥ A and all y ¥ B, we shall write A O B. When x O y (respectively, y O x) for all x ¥ A, we shall write A O y (respectively, y O A). We define a binary relation O g on W as follows. For all A, B ¥ W, A O g B Z there is an element b ¥ B 0 such that A 0 O b, with Q g and ’ g also defined as usual. Thus we have A Q g B Z for each x ¥ A 0, x Q b for some b ¥ B 0, A ’ g B Z for each x ¥ A 0 and each y ¥ B 0, x Q b and y Q a for some a ¥ A 0 and some b ¥ B 0. It is easy to verify that A ı B S A Q g B. We define A g=A 0 {a ¥ A : a O b for no b ¥ A}. Note by definition that A 0 ı A g ı A if A g ] ”, and by Lemma 4.1(1) below that A 0=A if A g=”. Lemma 4.1. (1) If ai Q i bi for i=1, ..., n, denoted a [ D b, then a Q b. If, in addition, ak O k bk for some 1 [ k [ n, denoted a < D b, then a O b. (2) If A ¥ W and A 0 O b, then A Q b. If, in addition, A g ] ”, then A g O b. (3) For all A, B ¥ W, A O g B if and only if A O b for some b ¥ B g. (4) If a < D b, then I(a, a) O g I(b, b). (5) If a O b and ¬ (a < D b), then I(a, a) O g I(b, b).
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Proof. (1) Suppose that a [ D b. Then by definition, a1 Q 1 b1 Z (a1 , a(1) ) Q (b1 , a(1) ), a2 Q 2 b2 Z (b1 , a2 , a(12) ) Q (b1 , b2 , a(12) ), x an Q n bn Z (b1 , ..., bn − 1 , an ) Q (b1 , ..., bn − 1 , bn ). By transitivity of Q , a Q b. If ak O k bk for some 1 [ k [ n, (b1 , ..., bk − 1 , ak , a(K) ) O (b1 , ..., bk , a(K) ), where K={1, ..., k}. Thus a O b.
then
(2) Suppose that A ¥ W and A 0 O b. If A g=”, then A=A 0, so that A Q b. In the following we shall assume that A g ] ”. First we show that A g O b. Note that NA ] ”. If A g=A 0, there is nothing to prove. Thus let a ¥ A g 0 A 0. It suffices to show that a O b. Since we have ai ¥ Ai 0 A 0i
for all i ¥ NA S a ¨ A g,
ai ¥ A 0i
for all i ¥ NA S a ¥ A 0,
there are distinct k, a ¥ NA such that ak ¥ Ak 0 A 0k and aa ¥ A 0a . With no loss of generality we assume that {i ¥ NA : ai ¥ Ai 0 A 0i }={1, ..., m} and am+1 ¥ A 0m+1 . Take any ci ¥ A 0i for i=1, ..., m+1. For k=1, ..., m+1, we define a k=(a1 , ..., ak , ck+1 , ..., cm+1 , am+2 , ..., an ). By induction, we show that a k O b for k=1, ..., m. Thus a O b, since a m=a if cm+1 =am+1 . Suppose that b Q a 1. Since c2 ¥ A 02 , coordinate order denseness implies that c2 O 2 c −2 for some c −2 ¥ A 02 . By (1), a 1 O (a1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) ¥ A g 0 A 0. Therefore, (c1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) ¥ A 0 and (c1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) O b O (a1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ). Then, by coordinate order denseness, there is a c −1 ¥ A 01 such that b O (c −1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) ¥ A 0, a contradiction. Hence a 1 O b. We note that this holds for all ci ¥ A 0i (i=2, ..., m+1). Now we assume that a i O b for i=1, ..., k − 1 with 1 < k [ m for all ci ¥ A 0i (i=k, ..., m+1). Suppose that b Q a k. Since ck+1 ¥ A 0k+1 , coordinate order denseness implies that ck+1 O k+1 c −k+1 for some c −k+1 ¥ A 0k+1 . By (1), a k O (a1 , ..., ak , c −k+1 , ck+2 , ..., cm+1 , am+2 , ..., an ) ¥ A g 0 A 0. By induction hypothesis, (a1 , ..., ak − 1 , ck , c −k+1 , ck+2 , ..., cm+1 , am+2 , ..., an ) O b Q a k.
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YUTAKA NAKAMURA
Then by coordinate order denseness, there is a c −k ¥ A 0k such that b O (a1 , ..., ak − 1 , c −k , c −k+1 , ck+2 , ..., cm+1 , am+2 , ..., an ), which contradicts A 0 O b. Hence a k O b. Since ci ¥ A 0i for i=1, ..., m+1 are arbitrary chosen, we conclude that a O b. It remains to show that A Q b. It suffices to show that A 0 A g Q b. Suppose on the contrary that b O a for an a ¥ A 0 A g. For i ¥ NA , take any a −i ¥ A 0i . Since (a −i , a(i) ) ¥ A g, (a −i , a(i) ) O b O a. By coordinate order denseness, b O (a 'i , a(i) ) O a for some a 'i ¥ A 0i . Since (a 'i , a(i) ) ¥ A g, this is a contradiction. Hence A Q b. (3) Suppose that A O g B. Then by definition, A 0 O b for some b ¥ B 0. Note B 0 ı B g. If NA =”, then A=A 0, so A O b. If NA ] ”, then A g ] ”. Therefore, (2) gives that A g O b. Assume that A 0 A g ] ” and b ’ A 0 A g. Since b ¥ B 0, it follows from coordinate order denseness that b O bŒ for some bŒ ¥ B 0. Hence A O bŒ. Suppose next that A O b for some b ¥ B g. If B 0 O b, then, by (2), B g O b, which gives b O b, a contradiction. Hence b Q bŒ for some bŒ ¥ B 0. Since A 0 O bŒ, A 0 O bœ. Hence A O g B. (4) Suppose that a < D b. With no loss of generality, we assume that a1 O 1 b1 and ai Q i bi for i=2, ..., n. Then by coordinate order denseness, we have a1 O 1 b1 S (a1 , b(1) ) O (b1 , b(1) ) S (a1 , b(1) ) O (c1 , b(1) ) O (b1 , b(1) )
for some c1 ¥ X1
S a1 O 1 c1 O 1 b1 . It follows from (1) that a O (c1 , b(1) ) < D b. Thus (c1 , b(1) ) ¥ I(b, b) g. Since I(a, a) Q a, I(a, a) O (c1 , b(1) ). Hence, by (3), I(a, a) O g I(b, b). (5) Suppose that a O b and ¬ (a < D b). If b [ D a, then by (1), b Q a, a contradiction. Thus ¬ (b [ D a), so that ak O k bk for some 1 [ k [ n. By (1), (ak , b(k) ) O b. If (ak , b(k) ) Q a, then by coordinate order denseness, a O (ck , b(k) ) O b for some ck ¥ Xk , so let b −k =ck . If a O (ak , b(k) ), then let b −k =ak . Thus a O (b −k , b(k) ) < D b and (b −k , b(k) ) ¥ I(b, b) g. Hence, by (3), I(a, a) O g I(b, b). L Lemma 4.2. (1) If x O y, Ii (x, z) ¥ Wi , and Ii (y, z) ¥ Wi , then Ii (x, z) … Ii (y, z). (2) If x Q a O b Q (xi , y(i) ), then there is a c(i) ¥ I(i) ((xi , y(i) ), (xi , y(i) )) such that a O (xi , c(i) ) O b. (3) If A ¥ W, x O y, and x, y ¥ A, then x O a O y for some a ¥ A. Proof. (1) Suppose that the hypotheses of (1) hold. By definition of Ii , Ii (x, z) ı Ii (y, z). Since Ii (y, z) is bounded above, Ii (y, z) Q i ai for some ai ¥ Xi . Since Ii (x, z) ] ”, we take any bi ¥ Ii (x, z). Then (bi , z(i) ) Q x O y Q (ai , z(i) ). By coordinate order denseness, x O (ci , z(i) ) O y for some ci ¥ Xi . Therefore, ci ¥ Ii (y, z) and ci ¨ Ii (x, z). Hence Ii (x, z) … Ii (y, z).
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(2) Suppose that x Q a O b Q (xi , y(i) ). With no loss of generality, let i=n. For 1 [ k < n, let
˛ yx
k
y −k =
k
if yk Q k xk , otherwise.
Then y −k Q k xk for all 1 [ k < n. By Lemma 4.1(1), (xn , y −(n) ) Q x. If a O (y −1 , ..., y −k , yk+1 , ..., yn − 1 , xn ) O b for some 1 [ k < n, then the desired result obtains by letting c(n) =(y −1 , ..., y −k , yk+1 , ..., yn − 1 ). Suppose that there is no 1 [ k < n such that a O (y −1 , ..., y −k , yk+1 , ..., yn − 1 , xn ) O b. Then, up to a permutation of indices, there exists 1 [ a < n such that (y −1 , ..., y −a , ya+1 , ..., yn − 1 , xn ) Q a O b Q (y −1 , ..., y −a − 1 , ya , ..., yn − 1 , xn ). By coordinate order denseness, there is a za ¥ Xa such that a O (y −1 , ..., y −a − 1 , za , ya+1 , ..., yn − 1 , xn ) O b and of course za O a ya . So the desired result obtains by letting c(n) = (y −1 , ..., y −a − 1 , za , ya+1 , ..., yn − 1 , xn − 1 ). (3) Suppose that A ¥ W, x O y, and x, y ¥ A. For 1 [ k [ n, let
˛ xy
y −k =
k k
if yk O k xk , otherwise.
Then y [ D yŒ and yŒ ¥ A. By Lemma 4.1(1), y Q yŒ, so that x O yŒ. If x O (x1 , y −(1) ) O y, then the desired result obtains, since (x1 , y −(1) ) ¥ A. If (x1 , y −(1) ) ’ x, then by coordinate order denseness, x O (a1 , y −(1) ) O y for some a1 ¥ A1 . Then the desired result follows from (a1 , y −(1) ) ¥ A. If y Q (x1 , y −(1) ), then by (2), x O (x1 , a(1) ) O y for some a(1) ¥ X(1) . Since x and (x1 , y −(1) ) are in A, it follows from the construction of a(1) in (2) that (x1 , a(1) ) ¥ A. L Lemma 4.3. For all a, b ¥ X, a O b if and only if I(a, a) O g I(b, b). Proof. If a O b, then it follows from Lemma 4.1(4) and 4.1(5) that I(a, a) O g I(b, b). Next we assume that I(a, a) O g I(b, b). Then, by definition, I(a, a) 0 O c for some c ¥ I(b, b) 0. If I(a, a) g=”, then I(a, a) O c. If I(a, a) g ] ”, then, by Lemma 4.1(2), I(a, a) Q c. Therefore, a Q c. Since c ¥ I(b, b) 0, c < D b, so, by Lemma 4.1(1), c O b. Hence a O b. L Lemma 4.4. (1) If A ¥ W and i ¥ NA , then (A 0i , A(i) ) ’ gA. If, in addition, a ¥ (A 0i , A(i) ), then (Ii (a, a), A(i) ) O gA. (2) If A, B ¥ W and A O g B, then A O g I(b, b) for some b ¥ B 0. Proof. (1) Suppose that A ¥ W and i ¥ NA . Since (A 0i , A(i) ) ı A, (A 0i , A(i) ) Q g A. Assume that (A 0i , A(i) ) O g A. Then there is a b ¥ A 0 such that (A 0i , A(i) ) 0 O b. However, A 0=(A 0i , A(i) ) 0, which gives b O b, a contradiction. Hence (A 0i , A(i) ) ’ g A. Suppose further that a ¥ (A 0i , A(i) ). Since ai ¥ A 0i , (Ii (a, a), A(i) ) Q gA. By coordinate order denseness, ai O i bi for some bi ¥ A 0i . Assume that NA ={i}. Then for all
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b(i) ¥ A(i) , aj ’ j bj for j ] i. Thus by Lemma 4.1(1), a O b. Since b ¥ A g and (Ii (a, a), A(i) ) Q a, Lemma 4.1(3) implies that (Ii (a, a), A(i) ) O gA. Assume that j ¥ NA for some j ] i. We recursively define a sequence a 1j , a 2j , ... on component j as follows. Fix any a 1j ¥ Aj and for each a kj , take any a k+1 to satisfy j that (bi , a kj , a(ij) ) O (ai , a k+1 , a(ij) ). j Since Aj is bounded above, the set {a kj } is bounded. Therefore, by strong Archimedean axiom, this set consists of a 1j , ..., a mj j for some positive integer mj \ 1. Since (ai , a mj j , a(ij) ) ¥ (Ii (a, a), A(i) ) and (ai , a mj j , a(ij) ) O (bi , a mj j , a(ij) ), we obtain that (Ii (a, a), Aj , a(ij) ) Q (bi , a mj j , a(ij) ). Similarly, for k ¥ NA 0 {i, j}, we can define a sequence a 1k , a 2k , ... on component k such that (bi , a mj j , a ak , a(ijk) ) O (ai , a mj j , a a+1 k , a(ijk) ) for all a. By the strong Archimedean axiom, there is a lowest upper bound on a, say mk . Hence (Ii (a, a), Ajk , a(ijk) ) Q (bi , a mj j , a mk k , a(ijk) ). By induction on every j ¥ NA , m there exist mj ’s such that (Ii (a, a), ANA 0 {i} , a(NA ) ) Q (bi , a {jj¥ NA 0 {i}} , a(NA ) ). Since for all m j ¨ NA , Aj ={aj }, we get (Ii (a, a), A(i) ) Q (bi , a {jj¥ NA 0 {i}} , a(NA ) ). By coordinate order denseness and the fact that i ¥ NA , bi Oi b −i for some b −i ¥ A 0i , so m m that, by Lemma 4.1(1), (Ii (a, a), A(i) ) O (b −i , a {jj¥ NA 0{i}} , a(NA ) ). Since (b −i , a {jj¥ NA 0{i}} , a(NA ) ) ¥ A g, Lemma 4.1(3) gives that (Ii (a, a), A(i) ) O gA. (2) Suppose that A O g B. Then A 0 O bŒ for some bŒ ¥ B 0. Thus by Lemma 4.1 (2), A Q bŒ. If there exists i ¥ NB , then, by definition of B 0i , there exists b −i ¥ Bi such that b O (b −i , b(i) ); hence by Lemma 4.2 (3), there exists b 'i ¥ B 0i such that b O (b 'i , b(i) ) O (b −i , b(i) ) and (b 'i , b(i) ) ¥ B 0. Hence A O g I((b 'i , b(i) ), (b 'i , b(i) )). If, on the contrary, NB =0, then B 0={b}. But since A 0 O b, if a ¥ A 0, then there exists an i such that ai O i bi , which is impossible since i ¨ NB . L We are to show that there is an additive representation for Q g; i.e., there are real valued functions Ui on Wi for i=1, ..., n such that, for all A, B ¥ W, n
n
A Q g B Z C Ui (Ai ) [ C Ui (Bi ). i=1
i=1
Moreover, the additive representation is a cardinal scale. Since, for all a, b ¥ X, I(a, a) and I(b, b) are in W, it follows from Lemma 4.3 that a Q b Z I(a, a) Q g I(b, b) n
n
Z C Ui (Ii (a, a)) [ C Ui (Ii (b, b)). i=1
i=1
Letting ui (ai )=Ui (Ii (a, a)) for all ai ¥ Xi and all i=1, ..., n, we obtain the desired result. By Theorem 2.1, it suffices to show the following lemma. Lemma 4.5. (1) (2)
Q g is a weak order;
Q g satisfies restricted solvability;
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(3)
525
Q g satisfies generalized triple cancellation;
(4) every bounded standard sequence w.r.t. Q g is finite; (5) at least two components are essential w.r.t. Q g. Proof. (1) Transitivity of Q g follows from definition of Q g. To show completeness of Q g, suppose that ¬ (A Q g B) and ¬ (B Q gA) for some A, B ¥ W. Then B O gA and A O g B. By definition, B 0 O a for some a ¥ A 0 and A O b for some b ¥ B 0. Therefore, ¬ (a Q b) from the former, and ¬ (b Q a) from the latter, contradicting completeness of Q. (2) To show restricted solvability of Q g, we assume that (Ai , A(i) ) O g B O g(Ci , A(i) ). We are to show that B ’g
1 0 3 I (x, y), A 2 . i
(i)
x¥B y¥A
Let Ji = 2 x ¥ B 5 y ¥ A Ii (x, y). First we need to verify that Ji ¥ Wi ; i.e., Ji ] ” and Ji is a preference interval which is left exhausted and bounded above. By defnition, A 0 O bŒ for some bŒ ¥ B 0. Thus NB ] ”. Therefore, it follows from coordinate order denseness and Lemma 4.1(2) that A Q bŒ O b for some b ¥ B 0. Similarly, B O (ci , a(i) ) for some (ci , a(i) ) ¥ (Ci , A(i) ) 0. Therefore, for all bŒ ¥ B such that b Q bŒ, and for all (ci , a −(i) ) ¥ (Ci , A(i) ) such that (ci , a(i) ) Q (ci , a −(i) ), A O bŒ O (ci , a −(i) ). Take any ai ¥ Ai . Then Ai ı Ii (bŒ, (ai , a −(i) )) ı Ii (c, c), since Ii (bŒ, (ci , a −(i) ))= Ii (bŒ, (ai , a −(i) )). Therefore, Ii (bŒ, (ai , a −(i) )) ¥ Wi . Given b ¥ B 0, a ¥ A, and ci ¥ C 0i obtained in the preceding paragraph, let Xb ={x ¥ B : b Q x}, Ya ={(ai , y(i) ) ¥ A : (ai , a(i) ) Q (ai , y(i) )}, so that Ai ı Ii (x, y) ı Ii (c, c) for all x ¥ Xb and all y ¥ Ya . Since, for all x, xŒ ¥ B and all y, yŒ ¥ A, x O xŒ S Ii (x, y) ı Ii (xŒ, y), (yi , y(i) ) Q (yi , y −(i) ) S Ii (x, (yi , y(i) )) ` Ii (x, (yi , y −(i) )), we obtain that ” ] Ji = 0 3 Ii (x, y) ı Ii (c, c). x ¥ Xb y ¥ Ya
Hence Ji ¥ Wi . We shall fix Xb and Ya defined in the preceding paragraph throughout the rest of the proof of (2). To show that B ’ g (Ji , A(i) ), we examine two other cases and derive contradictions for those cases. Case 1 (B O g (Ji , A(i) )). By definition, B 0 O (ci , a −(i) ) for some (ci , a −(i) ) ¥ (Ji , A(i) ) 0. By Lemma 4.1(2), B Q (ci , a −(i) ). Let a −i ¥ Ai .
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Suppose first that ci ¥ J 0i . Then, by coordinate order denseness, there is a c −i ¥ J 0i such that ci O i c −i , so, by Lemma 4.1(1), (ci , a −(i) ) O (c −i , a −(i) ). Since Ji ı 1x ¥ B Ii (x, (a −i , a −(i) )), there is a bŒ ¥ B such that c −i ¥ Ii (bŒ, (a −i , a −(i) )). Therefore, (c −i , a −(i) ) Q bŒ, so (ci , a −(i) ) O bŒ, a contradiction. Suppose next that J 0i =”. Then there exists k ¥ NA 0 {i} such that a −k ¥ A 0k , else B O g (Ji , A(i) ) is impossible. By coordinate order denseness, there is a a 'k ¥ A 0k such that a −k O k a 'k . Take any ci ¥ Ji . We obtain that B Q (ci , a −(i) ) O (ci , a 'k , a −(ik) ) ¥ (Ji , A(i) ) 0. Since ci ¥ Ji ı 1x ¥ B Ii (x, (a −i , a 'k , a −(ik) )), there is a bŒ ¥ B such that ci ¥ Ii (bŒ, (a −i , a 'k , a −(ik) )). Therefore, (ci , a 'k , a −(ik) ) Q bŒ, so (ci , a −(i) ) O bŒ, a contradiction. Case 2 ((Ji , A(i) ) O g B). By definition, (Ji , A(i) ) 0 O bŒ for some bŒ ¥ B 0. Since (Ji , A(i) ) O g B and NB ] ”, by coordinate order denseness and Lemma 4.1(2), (Ji , A(i) ) Q bŒ O bœ for some bœ ¥ B 0. Recall that b ¥ B 0 is arbitrarily chosen as long as A O b. Thus we choose b ¥ B 0 to satisfy (Ji , A(i) ) O b. For such a b, it holds that Ji = 0 3 Ii (x, y), x ¥ Xb y ¥ Ya
so that, for all x ¥ Xb , 3 Ii (x, y) ı Ji . y ¥ Ya
Assume that ci ¥ Ji 0 4y ¥ Ya Ii (bŒ, y) for some bŒ ¥ Xb . Then there is a yŒ ¥ Ya such that bŒ O (ci , y(i) ) ¥ (Ji , A(i) ), a contradiction. Therefore, Ji =4y ¥ Ya Ii (x, y) for all x ¥ Xb . Take any c ¥ Xb for which b O c. It follows from Lemma 4.2(3) that there are b 0, b 1, b 2, ... and cŒ, c 0, c 1, c 2, ... in Xb such that b O b i O b i − 1 O cŒ O c i − 1 O c i O c for all i=1, 2, ... . Take any a 1 ¥ Ya , so that Ii (x, a 1) ¥ Wi for all x ¥ Xb . Then we define recursively x ki ¥ Xi and a k+1 (i) ¥ A(i) for k=1, 2, ... to satisfy that (ai , a k+1 (i) ) ¥ Ya , and (i) x ki ¥ Ii (b k − 1, (a 1i , a k(i) )) 0 Ii (b k, (a 1i , a k(i) )), k (ii) c k − 1 O (x ki , a k+1 (i) ) O c .
Given a k(i) , the existence of x ki in (i) easily follows from Lemma 4.2(1). Given x ki , the existence of a k+1 in (ii) is assured as follows. Suppose that (x ki , y(i) ) Q c k − 1 for all (i) y ¥ Ya . Then x ki ¥ 4y ¥ Ya Ii (c k − 1, y)=Ji . Thus x ki ¥ 4y ¥ Ya Ii (b k, y)=Ji , so that (x ki , a k(i) ) Q b k. But then (x ki , a k(i) ) ¥ Ii (b k, (a 1i , a k(i) )). This is a contradiction. Consequently, there exists (ai , a −(i) ) ¥ Ya such that c k − 1 O (x ki , a −(i) ). Either c k − 1 O (x ki , a −(i) ) − k−1 O c k and the result obtains by letting a k+1 O c k Q (x ki , a −(i) ). But then (i) =a (i) , or c (x ki , a k(i) ) Q b k − 1 O c k − 1 O c k Q (x ki , a −(i) ) and by Lemma 4.2(2), there exists k−1 k a k+1 O (x ki , a k+1 (i) ¥ A(i) such that c (i) ) O c . k k k k k k−1 By (i), b O (x i , a (i) ) Q b for all integers k > 0. By (i), (x k+1 , a k+1 i (i) ) O (x i , a (i) ) 1 2 k k+1 1 for all integers k > 0. By (ii), (x i , a (i) ) O c O (x i , a (i) ) for all integers k > 1. Hence
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(x ki , a k(i) ) O (x 1i , a 1(i) ), (x 1i , a 2(i) ) O (x ki , a k+1 (i) ), k k (x k+1 , a k+1 i (i) ) O (x i , a (i) ),
which implies by generalized triple cancellation that (x k+1 , a 2(i) ) O (x ki , a 1(i) ). Therei k O i x ki for all k. fore, {x i : k=1, 2, ...} is a strong standard sequence with x k+1 i k We know that A O b and a (i) ¥ A(i) . This implies that for all (ai , a k(i) ) ¥ Ya , (ai , a k(i) ) O b. Then since b O (x ki , a k(i) ) for all k, ai O i x ki for all k. Thus ai is a lower bound for the strong standard sequence. Hence the sequence is bounded but infinite, contradicting the strong Archimedean axiom. (3) Suppose that (Ai , A(i) ) Q g (Bi , B(i) ), (Bi , C(i) ) Q g (Ai , D(i) ), (Ci , B(i) ) Q g (Di , A(i) ). Then we are to show that (Ci , C(i) ) Q g (Di , D(i) ); i.e., for each c ¥ C 0, there is a d ¥ D 0 such that c Q d. We have the following two cases to examine: A O g B; A ’ g B. Fix a c ¥ C 0. Case 1 (A O g B). By definition, A 0 O b for some b ¥ B 0. Since (bi , c(i) ) ¥ (Bi , C(i) ) 0 and (ci , b(i) ) ¥ (Ci , B(i) ) 0, it follows from definition of Q g that there are (ai , d(i) ) ¥ (Ai , D(i) ) 0 and (di , a(i) ) ¥ (Di , A(i) ) 0 such that (bi , c(i) ) Q (ai , d(i) ), (ci , b(i) ) Q (di , a(i) ). Since (ai , a(i) ) ¥ A 0, a O b. By generalized triple cancellation, c Q d, where d=(di , d(i) ) ¥ D 0. Case 2 (A ’ g B). If NC =”, then C Q X for all X ¥ W, so that C Q gD. Therefore, we assume that NC ] ”. We have the following three subcases to examine: for some j ] i, Subcase 1: {i, j} ı NC ; Subcase 2: i ¨ NC and j ¥ NC ; Subcase 3: NC ={i}. Subcase 1. Note that C 0i and C 0j are not empty. By Lemma 4.4(1), (Ii (c, c), B(i) ) O g (Ci , B(i) ) Q g (Di , A(i) ), (Bi , I(i) (c, c)) O g (Bi , C(i) ) Q g (Ai , D(i) ). Thus, by definition, there are (di , a(i) ) ¥ (Di , A(i) ) 0 and (ai , d(i) ) ¥ (Ai , D(i) ) 0 such that (Ii (c, c), B(i) ) 0 O (di , a(i) ), (Bi , I(i) (c, c)) 0 O (ai , d(i) ),
528
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so d ¥ D 0. Since (ai , a(i) ) ¥ A 0, the definition of ’ g gives that a Q b for some b ¥ B 0. Since (ci , b(i) ) ¥ (Ii (c, c), B(i) ) and (bi , c(i) ) ¥ (Bi , I(i) (c, c)), it follows from Lemma 4.1(2) that (ci , b(i) ) Q (di , a(i) ), (bi , c(i) ) Q (ai , d(i) ). Hence by generalized triple cancellation, c Q d. Subcase 2. Note that C 0i =” and C 0j ] ”. By Lemma 4.4(1), (Bi , I(i) (c, c)) O g (Bi , C(i) ) Q g (Ai , D(i) ). Thus, by definition, there is an (ai , d(i) ) ¥ (Ai , D(i) ) 0 such that (Bi , I(i) (c, c)) 0 O (ai , d(i) ). Take any a(i) ¥ A(i) for which a ¥ A 0. Then by definition of ’ g, a Q b for some b ¥ B 0. Thus, by Lemma 4.1(2), (Bi , I(i) (c, c)) Q (ai , d(i) ), so that (bi , c(i) ) Q (ai , d(i) ). If k ¨ NB for all k ] i, then (ci , b(i) ) Q X for all X ¥ W, so that (ci , b(i) ) Q (Di , A(i) ) 0. If k ¥ NB for some k ] i, then, by Lemma 4.4(1), (Ci , I(i) (b, b)) O g (Ci , B(i) ) Q g (Di , A(i) ). Thus there is a (di , a(i) ) ¥ (Di , A(i) ) 0 such that (Ci , I(i) (b, b)) 0 O (di , a(i) ), so, by Lemma 4.1(2), (Ci , B(i) ) Q (di , a(i) ). Therefore, in both cases, (ci , b(i) ) Q (di , a(i) ), where (di , d(i) ) ¥ D 0. Hence, by generalized triple cancellation, c Q d. Subcase 3. Note that C 0i ] ” and C 0j =” for all j ] i. By Lemma 4.4(1), (Ii (c, c), B(i) ) O g (Ci , B(i) ) Q g (Di , A(i) ). Thus, by definition, there is a (di , a(i) ) ¥ (Di , A(i) ) 0 such that (Ii (c, c), B(i) ) 0 O (di , a(i) ), so, by Lemma 4.1(2), (Ii (c, c), B(i) ) Q (di , a(i) ). Take any ai ¥ Ai for which a ¥ A 0. Then by definition of ’ g, a Q b for some b ¥ B 0. Assume first that B 0i ] ”. Then bi ¥ B 0i , so by Lemma 4.4(1), (Ii (b, b), C(i) ) O g (Bi , C(i) ) Q g (Ai , D(i) ). Thus, by definition, there is an (ai , d(i) ) ¥ (Ai , D(i) ) 0 such that (Ii (b, b), C(i) ) 0 O (ai , d(i) ),
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529
so by Lemma 4.1(2), (Ii (b, b), C(i) ) Q (ai , d(i) ). Therefore, (bi , c(i) ) Q (ai , d(i) ), (ci , b(i) ) Q (di , a(i) ), where d ¥ D 0. By generalized triple cancellation, c Q d. Assume next that B 0i =”. Then (Bi , C(i) ) Q X for all X. Therefore, (ci , b(i) ) Q (di , a(i) ), (bi , c(i) ) Q (ai , d(i) ) for all d(i) ¥ D(i) for which (ai , d(i) ) ¥ (Ai , D(i) ) 0. Note d ¥ D 0. By generalized triple cancelaltion, c Q d. (4) Suppose that there exists a bounded infinite standard sequence w.r.t. Q g. Let {A ki : k ¥ K} be such a sequence on component i w.r.t. Q g, where K is an infinite set of consecutive integers. Then there are (Ci , A(i) ), (Ci , B(i) ) ¥ W such that , B(i) ) for all k, k+1 ¥ K. Assume ¬ ((Ci , A(i) ) ’ g (Ci , B(i) )), and (A ki , A(i) ) ’ g (A k+1 i that K is the set of all positive integers and (Ci , B(i) ) O g (Ci , A(i) ). The proofs for the other cases are similar. Then by (3), (A ki , B(i) ) O g (A ki , A(i) ) for all k ¥ K, and (A ki , D(i) ) O g (A k+1 , D(i) ) for all k, k+1 ¥ K and all D(i) ¥ W(i) . i First we show that there are ci ¥ Ci and d(i) , e(i) ¥ X(i) such that (Ii (c, c), B(i) ) Q g I((ci , d(i) ), (ci , d(i) )) O g I((ci , e(i) ), (ci , e(i) )) Q g (Ii (c, c), A(i) ). Since (Ci , B(i) ) O g (Ci , A(i) ), it follows from Lemma 4.4 (2) that (Ci , B(i) ) O g I((ci , a(i) ), (ci , a(i) )) for some (ci , a(i) ) ¥ (Ci , A(i) ) 0. By Lemma 4.1 (1) and the definition of O g, (Ii (c, c), B(i) ) Q g (Ci , B(i) ), so by (1), (Ii (c, c), B(i) ) O g I((ci , a(i) ), (ci , a(i) )). By Lemma 4.4 (2), (Ii (c, c), B(i) ) O g I((c −i , a −(i) ), (c −i , a −(i) )) for some (c −i , a −(i) ) ¥ I((ci , a(i) ), (ci , a(i) )) 0. Note that I((c −i , a −(i) ), (c −i , a −(i) )) O g I((ci , a(i) ), (ci , a(i) )). If c −i ’ i ci , then letting ci =c −i , d(i) =a −(i) , and e(i) =a(i) gives the desired result. Assume that c −i ¥ Ii (c, c) 0. Since (ci , b(i) ) O (c −i , a −(i) ) O (ci , a(i) ) for some b(i) ¥ B(i) , Lemma 4.2(2) implies that (c −i , a −(i) ) O (ci , d(i) ) O (ci , a(i) ) for some d(i) ¥ A(i) . Hence letting e(i) =a(i) , we obtain the desired result. −1 Since (A 2k , B(i) ) O g (A 2k i i , B(i) ) for all k ¥ K, it follows from Lemma 4.1(3) that 2k − 1 g for every k ¥ K, there is an (a ki , b k(i) ) ¥ (A 2k , B(i) ) O (a ki , b k(i) ). i , B(i) ) such that (A i k 2k 2k+1 k+1 k+1 g Thus (a i , e(i) ) ¥ (A i , A(i) ) ’ (A i , B(i) ) O (a i , b (i) ), so that (a ki , e(i) ) O (a k+1 , b k+1 i (i) ). By the preceding two paragraphs and Lemma 4.3, (ci , b(i) ) Q (ci , d(i) ) for all b(i) ¥ B(i) . Therefore, in particular for b(i) =b k+1 (i) , generalized triple cancellak+1 k+1 tion implies that (a k+1 , b ) Q (a , d ). Hence (a ki , e(i) ) O (a k+1 , d(i) ) for all (i) i (i) i i k k ¥ K, so that {a i : k ¥ K} is a strong standard sequence on component i w.r.t. Q. Since {A ki : k ¥ K} is bounded, so is {a ki : k ¥ K}. By the strong Archimedean
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axiom, K must be finite, a contradiction. Hence there exist no bounded infinite standard sequence w.r.t. Q g. (5) Suppose that the coordinates 1 and 2 are essential w.r.t. Q. For coordinate 1, there are a, b, c ¥ X such that (a1 , c(1) ) O (b1 , c(1) ). Then it follows form Lemma 4.3 that I((a1 , c(1) ), (a1 , c(1) )) O g I((b1 , c(1) ), (b1 , c(1) )). Since I(1) ((a1 , c(1) ), (a1 , c(1) ))=I(1) ((b1 , c(1) ), (b1 , c(1) )), the coordinate 1 is essential w.r.t. Q g. Essentiality of 2 w.r.t. Q g similarly follows. L
REFERENCES Debreu, G. (1960). Topological methods in cardinal utility theory. In K. J. Arrow, S. Karlin, and P. Suppes (1959, Eds.), Mathematical methods in the social sciences. Stanford: Stanford Univ. Press. Gonzales, C. (1996). Additive utilities when some components are solvable and others are not. Journal of Mathematical Psychology, 40, 141–151. Gonzales, C. (1999). Additive utility without restricted solvability on all components. Mimeograph. Gonzales, C. (2000). Two factor additive conjoint measurement with one solvable component. Journal of Mathematical Psychology, 44, 285–303. Gonzales, C., & Jaffray, J. Y. (1998). Imprecise sampling and direct decision making. Annals of Operations Research, 80, 207–235. Jaffray, J. Y. (1974). On the extension of additive utilities to infinite sets. Journal of Mathematical Psychology, 11, 431–452. Krantz, D. H., Luce, R. D., Suppes, P., & Tvesky, A. (1971). Foundations of measurement (Vol. 1). New York: Academic Press. Wakker, P. P. (1988). The algebraic versus the topological approach to additive representations. Journal of Mathematical Psychology, 32, 421–435. Wakker, P. P. (1989). Additive representation of preferences. Dordrecht: Kluwer Academic Publishers. Wakker, P. P. (1991). Additive representations of preferences, a new foundations of decision analysis: the algebraic approach. In J. P. Doignon and J. C. Falmagne (Eds.), Mathematical psychology: Current developments. Berlin/New York: Springer-Verlag. Received: August 30, 1999; published online: June 13, 2002
Additive Utilities on Densely Ordered Sets Yutaka Nakamura University of Tsukuba
This paper shows sufficient conditions for the existence of additive utilities without a restricted solvability axiom. Our conditions require that each essential component of the underlying Cartesian product be densely ordered. © 2002 Elsevier Science (USA)
1. INTRODUCTION
Three decades ago, a well-known algebraic approach for the axiomatization of additive conjoint measurement had been established by Krantz, Luce, Suppes, and Tversky (1971, Chapt. 6), hereafter abbreviated KLST. Although their theory provides less restrictive (and more favorable) conditions than the topological approach advocated by Debreu (1960) (see Wakker, 1988), their structural condition, known as restricted solvability, is still unnecessarily strong in applications (see for example Gonzales & Jaffray, 1998). In this respect, Jaffray (1974) is the first to present an axiomatic structure for additive conjoint measurement on ‘‘infinite’’ sets without restricted solvability. Unfortunately, his axioms are too complicated to be tested, since they include the independence condition C as Jaffray (1974) dubbed it, also known as a family of the cancellation axioms of any order. Recent papers by Gonzales (1996, 1999, 2000) have uncovered a subtle relationship between (restricted) solvability and additive representability. Gonzales (1996) showed by counterexamples that solvability cannot be dropped for two or more components. Also in Gonzales (2000), it is shown in the 2-dimensional case that cancellation axioms of any order are required to ensure additive representability when restricted solvability holds with respect to (w.r.t.) one component. Furthermore, he proved that an additional structural condition of equal spacedness w.r.t. the second component is sufficient for the existence of additive utilities. In addition, Gonzales (2000) proved that additive representation exists when restricted solvability holds w.r.t. only two components, but the uniqueness property of a cardinal scale fails to hold. The author is gratefully indebted to Christophe Gonzales for many helpful comments and corrections in the original proofs, and aknowledges the action editor for improving readability of the paper. Address correspondence and reprint requests to Yutaka Nakamura, Institute of Policy and Planning Sciences, University of Tsukuba, 1-1-1 Tennoudai, Tsukuba, Ibaraki 305-8573, Japan.
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The aim of this paper is to study additive representability without assuming restricted solvability for every component. We show that if every essential component of the underlying Cartesian product is densely ordered, then additive representability obtains without restricted solvability, but retains the uniqueness property of a cardinal scale. Contrary to Jaffray’s complicated axioms, it is shown to be sufficient for additive representability that restricted solvability and the Archimedean axiom in KLST’s axiom system are replaced respectively by a coordinate order denseness condition and a strengthened Archimedean axiom. Since restricted solvability need not be satisfied on all components, a linear order that satisfies our axiom system can have an additive representation. The paper is organized as follows. Section 2 introduces preliminary definitions and results. In Section 3, our structural conditions are defined and the main theorem is stated. Examples in which restricted sovability does not necessarily hold are discussed. Section 4 is devoted to the proof of the main theorem.
2. PRELIMINARIES
Let X=X1 × · · · × Xn be an n-dimensional Cartesian product with elements x=(x1 , ..., xn ). We assume that n \ 2. Given a nonempty proper subset N= {i1 , ..., ir } … {1, ..., n} with nonempty complement (N)={1, ..., n} 0 N, let XN = Xi1 × · · · Xir , and write x ¥ X as (xN , x(N) ) with xN ¥ XN and x(N) ¥ X(N) . By Q we denote a binary preference relation on X with O and ’ defined as usual: for x, y ¥ X, x O y if ¬ (y Q x), and x ’ y if x Q y and y Q x. We say that Q is a weak order if it is transitive and complete and that O is a linear order if Q is antisymmetric weak order. A real valued function U on X is said to be additive if there are real valued functions ui on Xi for i=1, ..., n such that U(x)=; ni=1 ui (xi ) for all x ¥ X. If, for all x, y ¥ X, x Q y Z U(x) [ U(y), then a function U is said to represent Q. When U representing Q is additive, it is called an additive representation for Q. A function U representing Q is a cardinal scale if there are real numbers a > 0 and b such that V=aU+b if and only if V also represents Q. In what follows, we introduce a slightly modified version, advocated by Wakker (1989, 1991), of KLST’s classical theory for additive representability of Q. Definition 2.1. The binary relation Q satisfies coordinate independence if, for all x, y, z, w ¥ X and i=1, ..., n, (xi , z(i) ) Q (xi , w(i) ) whenever (yi , z(i) ) Q (yi , w(i) ). For each i=1, ..., n, the binary relation Q i on Xi is defined as follows. For all xi , yi ¥ Xi , xi Q i yi Z (xi , a(i) ) Q (yi , a(i) )
for some a(i) ¥ X(i) .
Similarly, O i and ’ i can be derived from O and ’, respectively. The property that the definition of Q i does not depend on the choice of a specific a(i) is known as weak separability in the literature, a condition which follows from coordinate independence. When n=2, coordinate independence can be replaced by the following definition.
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Definition 2.2. The binary relation Q satisfies generalized triple cancellation if, for all x, y, z, w, a, b, c, d ¥ X, and i=1, ..., n, (ai , x(i) ) Q (bi , y(i) ), (bi , z(i) ) Q (ai , w(i) ), (ci , y(i) ) Q (di , x(i) ) imply that (ci , z(i) ) Q (di , w(i) ). It is easy to verify that generalized triple cancellation implies coordinate independence. The restricted solvability that we shall drop in our axiomatization is defined as follows. Definition 2.3. The binary relation Q satisfies restricted solvability on component i if, for all x, y, z, w ¥ X, there exists an element ai ¥ Xi such that y ’ (ai , w(i) ) whenever (xi , w(i) ) O y and y O (zi , w(i) ). We say that Q satisfies restricted solvability if it satisfies restricted solvability on all components. With restricted solvability, standard sequences in the following definition are meaningful in the sense that they form equally spaced sequences in utility units. For example, every standard sequence consists of a singleton when O is a linear order for which restricted solvability fails to hold. Definition 2.4. For component i and any set K of consecutive integers, a set {x ki : k ¥ K} is a standard sequence (on component i, w.r.t. Q) if there exist (ci , a(i) ), (ci , b(i) ) ¥ X such that ¬ ((ci , a(i) ) ’ (ci , b(i) )), and (x ki , a(i) ) ’ (x k+1 , b(i) ) i for all k, k+1 ¥ K. A standard sequence may be finite or infinite of any length. A standard sequence (on component i) is bounded if there exist x 0i , x gi ¥ Xi such that x 0i Q i x ki and x ki Q i x gi for all k. The Archimedean axiom requires that every bounded standard sequence be finite. Definition 2.5. Component i is essential if there are a, b, c ¥ X such that (ai , c(i) ) O (bi , c(i) ). Component i is inessential if it is not essential. The following theorem, due to Wakker (1991), gives a characterization of additive representation which does not require separate formulations for a different number of essential components. Theorem 2.1. Suppose that at least two components are essential, and the restricted solvability is satisfied. Then the following statements are equivalent: (1) there exists an additive representation for Q. (2) the binary relation Q is a weak order that satisfies the Archimedean axiom and generalized triple cancellation. If at least three components are essential, then the generalized triple cancellation in (2) can be replaced by the coordinate independence. Moreover, the additive representation in (1) is a cardinal scale.
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3. THE MAIN THEOREM
This section presents the main theorem for additive representability without imposing restricted solvability on our axiomatic structure for Q. The key axiom is the following denseness condition. Definition 3.1. The binary relation Q satisfies coordinate order denseness if, for all x, y, a, b ¥ X and i=1, ..., n, there exists an element ci ¥ Xi such that x O (ci , z(i) ) O y whenever (ai , z(i) ) Q x O y Q (bi , z(i) ). When coordinate independence holds, coordinate order denseness requires that, for essential i, Q i be order dense, i.e., for all xi , yi ¥ Xi , xi O i ai and ai O i yi for some ai ¥ Xi whenever xi O i yi . Since restricted solvability may not hold, there may not exist many standard sequences as in Definition 2.4. As an alternative, we introduce a strengthened notion of the standard sequence which provides a variably spaced utility scale with a minimum length. A similar generalization of the standard sequence is independently introduced by Gonzales (1999, Definition 5), who dubbed it an overstandard sequence. Definition 3.2. For component i and any set K of consecutive integers, a set {x ki : k ¥ K} is a strong standard sequence (on component i, w.r.t. Q ) if there exist (ci , a(i) ), (ci , b(i) ) ¥ X such that (ci , b(i) ) O (ci , a(i) ), and (x ki , a(i) ) Q (x k+1 , b(i) ) for i all integers k, k+1 ¥ K. A strong standard sequence may be finite or infinite of any length even if O is a linear order. A strong standard sequence (on component i) is bounded if there exist x 0i , x gi ¥ Xi such that x 0i Q i x ki and x ki Q x gi for all k. Our strengthening of the Archimedean axiom requires that every bounded strong standard sequence be finite. The main theorem is stated as follows. The proof will be deferred to the next section. Theorem 3.1. Suppose that at least two components are essential and that coordinate order denseness is satisfied. Then the following statements are equivalent: (1) there exists an additive representation for Q; (2) the binary relation Q is a weak order that satisfies the strong Archimedean axiom and generalized triple cancellation. If at least three components are essential, then the generalized triple cancellation in (2) can be replaced by coordinate independence. Moreover, the additive representation in (1) is a cardinal scale. We shall consider two simple examples where restricted solvability does not hold. Let Q be the set of all rational numbers. Example 3.1. Let X1 =R and X2 =Q. Let a weak order Q on X1 × X2 be defined by an additive function U(x)=x1 +x2 . There is no x2 ¥ X2 such that (0, x2 ) ’ (`2, 0), but (0, 0) O (`2, 0) O (0, 2), so restricted solvability on
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component 2 fails to hold. Jaffray (1974) noted that this Q satisfies all conditions in his Theorem 7.2 for additive representability. It is also verified that Q satisfies coordinate order denseness and the conditions (2) in Theorem 3.1 above. Although Q satisfies restricted solvability on component 1, such a condition is not required to verify additive representability. The next example shows that a linear order O has an additive representation. Example 3.2. Let X1 =Q, X2 ={a `2: a ¥ Q}, and X3 ={a `3: a ¥ Q}. Let a weak order Q on X1 × X2 × X3 be represented by the additive function U(x)= x1 +x2 +x3 . It is easy to verify that Q satisfies coordinate order denseness and the conditions (2) in Theorem 3.1 above. A notable feature of this example is that O on X=X1 × X2 × X3 is a linear order, so that O does not satisfy restricted solvability on all components. To see that O is a linear order, it suffices to show that, for all integers m1 , m2 , m3 , m1 +m2 `2+m3 `3=0 S m1 =m2 =m3 =0. We have m1 +m2 `2+m3 `3=0 S m1 +m2 `2=−m3 `3 S m 21 +2m 22 +2m1 m2 `2=3m 23 , whose solutions must be m1 =m2 =m3 =0. We note that there also exist cases in which coordinate order denseness does not hold whereas solvability does. For example, let X=Z × Z, where Z is the set of all integers. Let Q on X be represented by an additive function U(x)=x1 +x2 . Then unrestricted solvability holds but not coordinate order denseness.
4. PROOF OF THE MAIN THEOREM
Theorem 3.1 is proved in this section. Suppose that at least two components are essential and coordinate order denseness is satisfied. It is easy to see that (1) implies (2). Thus we assume that (2) holds to prove (1); i.e., the binary relation Q is a weak order that satisfies the strong Archimedean axiom and generalized triple cancellation. A preference interval Ai (on component i, w.r.t. Q i ) is defined to be a nonempty subset of Xi for which zi ¥ Ai whenever xi , yi ¥ Ai , xi Q i zi , and zi Q i yi . A preference interval Ai on component i is said to be left exhausted if yi ¥ Ai whenever xi ¥ Ai and yi Q i xi , and bounded above if there is a yi ¥ Xi such that xi Q yi for all xi ¥ Ai . Since we are concerned with only preference intervals on each component that are left exhausted and bounded above, let Wi denote the set of all preference intervals on component i that are left exhausted and bounded above, and also let W=W1 × · · · × Wn with elements A=(A1 , ..., An ). We shall identify A ¥ W with a product set A1 × · · · × An and write a ¥ A when ai ¥ Ai for i=1, ..., n. Let
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YUTAKA NAKAMURA
A 0i ={xi ¥ Ai : xi O i yi for some yi ¥ Ai } and A 0=(B1 , ..., Bn ), where, for i=1, ..., n,
˛ AA
Bi =
0 i
i
if A 0i ] ”, otherwise.
When A is interpreted as a product set, A 0 means B1 × · · · × Bn . Given a nonempty proper subset N={i1 , ..., ir } … {1, ..., n}, we shall let WN =Wi1 × · · · × Wir and write A ¥ W as (AN , A(N) ) with AN ¥ WN and A(N) ¥ W(N) . A certain preference interval associated with a, b ¥ X will be identified as follows: Ii (a, b)={xi ¥ Xi : x Q a and x(i) =b(i) }. We note that Ii (a, b) may not be in Wi if either Ii (a, b)=Xi and Xi ¨ Wi , or Ii (a, b)=”. However, Ii (a, a) always belongs to Wi . Let Ii (a, b) 0={xi ¥ Xi : x O a and x(i) =b(i) } when Ii (a, b) ¥ Wi , where Ii (a, a) 0 may be empty. Let I(a, b)=(I1 (a, b), ..., In (a, b)), I(i) (a, b)=(I1 (a, b), ..., Ii − 1 (a, b), Ii+1 (a, b), ..., In (a, b)). For all A ¥ W, we define NA ={i : A 0i ] ”}. If NA ={1, ..., m}, then I(a, a)=(I1 (a, a), ..., In (a, a)), I(a, a) 0=(I1 (a, a) 0, ..., Im (a, a) 0, Im+1 (a, a), ..., In (a, a)). When x O y for all x ¥ A and all y ¥ B, we shall write A O B. When x O y (respectively, y O x) for all x ¥ A, we shall write A O y (respectively, y O A). We define a binary relation O g on W as follows. For all A, B ¥ W, A O g B Z there is an element b ¥ B 0 such that A 0 O b, with Q g and ’ g also defined as usual. Thus we have A Q g B Z for each x ¥ A 0, x Q b for some b ¥ B 0, A ’ g B Z for each x ¥ A 0 and each y ¥ B 0, x Q b and y Q a for some a ¥ A 0 and some b ¥ B 0. It is easy to verify that A ı B S A Q g B. We define A g=A 0 {a ¥ A : a O b for no b ¥ A}. Note by definition that A 0 ı A g ı A if A g ] ”, and by Lemma 4.1(1) below that A 0=A if A g=”. Lemma 4.1. (1) If ai Q i bi for i=1, ..., n, denoted a [ D b, then a Q b. If, in addition, ak O k bk for some 1 [ k [ n, denoted a < D b, then a O b. (2) If A ¥ W and A 0 O b, then A Q b. If, in addition, A g ] ”, then A g O b. (3) For all A, B ¥ W, A O g B if and only if A O b for some b ¥ B g. (4) If a < D b, then I(a, a) O g I(b, b). (5) If a O b and ¬ (a < D b), then I(a, a) O g I(b, b).
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Proof. (1) Suppose that a [ D b. Then by definition, a1 Q 1 b1 Z (a1 , a(1) ) Q (b1 , a(1) ), a2 Q 2 b2 Z (b1 , a2 , a(12) ) Q (b1 , b2 , a(12) ), x an Q n bn Z (b1 , ..., bn − 1 , an ) Q (b1 , ..., bn − 1 , bn ). By transitivity of Q , a Q b. If ak O k bk for some 1 [ k [ n, (b1 , ..., bk − 1 , ak , a(K) ) O (b1 , ..., bk , a(K) ), where K={1, ..., k}. Thus a O b.
then
(2) Suppose that A ¥ W and A 0 O b. If A g=”, then A=A 0, so that A Q b. In the following we shall assume that A g ] ”. First we show that A g O b. Note that NA ] ”. If A g=A 0, there is nothing to prove. Thus let a ¥ A g 0 A 0. It suffices to show that a O b. Since we have ai ¥ Ai 0 A 0i
for all i ¥ NA S a ¨ A g,
ai ¥ A 0i
for all i ¥ NA S a ¥ A 0,
there are distinct k, a ¥ NA such that ak ¥ Ak 0 A 0k and aa ¥ A 0a . With no loss of generality we assume that {i ¥ NA : ai ¥ Ai 0 A 0i }={1, ..., m} and am+1 ¥ A 0m+1 . Take any ci ¥ A 0i for i=1, ..., m+1. For k=1, ..., m+1, we define a k=(a1 , ..., ak , ck+1 , ..., cm+1 , am+2 , ..., an ). By induction, we show that a k O b for k=1, ..., m. Thus a O b, since a m=a if cm+1 =am+1 . Suppose that b Q a 1. Since c2 ¥ A 02 , coordinate order denseness implies that c2 O 2 c −2 for some c −2 ¥ A 02 . By (1), a 1 O (a1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) ¥ A g 0 A 0. Therefore, (c1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) ¥ A 0 and (c1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) O b O (a1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ). Then, by coordinate order denseness, there is a c −1 ¥ A 01 such that b O (c −1 , c −2 , c3 , ..., cm+1 , am+2 , ..., an ) ¥ A 0, a contradiction. Hence a 1 O b. We note that this holds for all ci ¥ A 0i (i=2, ..., m+1). Now we assume that a i O b for i=1, ..., k − 1 with 1 < k [ m for all ci ¥ A 0i (i=k, ..., m+1). Suppose that b Q a k. Since ck+1 ¥ A 0k+1 , coordinate order denseness implies that ck+1 O k+1 c −k+1 for some c −k+1 ¥ A 0k+1 . By (1), a k O (a1 , ..., ak , c −k+1 , ck+2 , ..., cm+1 , am+2 , ..., an ) ¥ A g 0 A 0. By induction hypothesis, (a1 , ..., ak − 1 , ck , c −k+1 , ck+2 , ..., cm+1 , am+2 , ..., an ) O b Q a k.
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YUTAKA NAKAMURA
Then by coordinate order denseness, there is a c −k ¥ A 0k such that b O (a1 , ..., ak − 1 , c −k , c −k+1 , ck+2 , ..., cm+1 , am+2 , ..., an ), which contradicts A 0 O b. Hence a k O b. Since ci ¥ A 0i for i=1, ..., m+1 are arbitrary chosen, we conclude that a O b. It remains to show that A Q b. It suffices to show that A 0 A g Q b. Suppose on the contrary that b O a for an a ¥ A 0 A g. For i ¥ NA , take any a −i ¥ A 0i . Since (a −i , a(i) ) ¥ A g, (a −i , a(i) ) O b O a. By coordinate order denseness, b O (a 'i , a(i) ) O a for some a 'i ¥ A 0i . Since (a 'i , a(i) ) ¥ A g, this is a contradiction. Hence A Q b. (3) Suppose that A O g B. Then by definition, A 0 O b for some b ¥ B 0. Note B 0 ı B g. If NA =”, then A=A 0, so A O b. If NA ] ”, then A g ] ”. Therefore, (2) gives that A g O b. Assume that A 0 A g ] ” and b ’ A 0 A g. Since b ¥ B 0, it follows from coordinate order denseness that b O bŒ for some bŒ ¥ B 0. Hence A O bŒ. Suppose next that A O b for some b ¥ B g. If B 0 O b, then, by (2), B g O b, which gives b O b, a contradiction. Hence b Q bŒ for some bŒ ¥ B 0. Since A 0 O bŒ, A 0 O bœ. Hence A O g B. (4) Suppose that a < D b. With no loss of generality, we assume that a1 O 1 b1 and ai Q i bi for i=2, ..., n. Then by coordinate order denseness, we have a1 O 1 b1 S (a1 , b(1) ) O (b1 , b(1) ) S (a1 , b(1) ) O (c1 , b(1) ) O (b1 , b(1) )
for some c1 ¥ X1
S a1 O 1 c1 O 1 b1 . It follows from (1) that a O (c1 , b(1) ) < D b. Thus (c1 , b(1) ) ¥ I(b, b) g. Since I(a, a) Q a, I(a, a) O (c1 , b(1) ). Hence, by (3), I(a, a) O g I(b, b). (5) Suppose that a O b and ¬ (a < D b). If b [ D a, then by (1), b Q a, a contradiction. Thus ¬ (b [ D a), so that ak O k bk for some 1 [ k [ n. By (1), (ak , b(k) ) O b. If (ak , b(k) ) Q a, then by coordinate order denseness, a O (ck , b(k) ) O b for some ck ¥ Xk , so let b −k =ck . If a O (ak , b(k) ), then let b −k =ak . Thus a O (b −k , b(k) ) < D b and (b −k , b(k) ) ¥ I(b, b) g. Hence, by (3), I(a, a) O g I(b, b). L Lemma 4.2. (1) If x O y, Ii (x, z) ¥ Wi , and Ii (y, z) ¥ Wi , then Ii (x, z) … Ii (y, z). (2) If x Q a O b Q (xi , y(i) ), then there is a c(i) ¥ I(i) ((xi , y(i) ), (xi , y(i) )) such that a O (xi , c(i) ) O b. (3) If A ¥ W, x O y, and x, y ¥ A, then x O a O y for some a ¥ A. Proof. (1) Suppose that the hypotheses of (1) hold. By definition of Ii , Ii (x, z) ı Ii (y, z). Since Ii (y, z) is bounded above, Ii (y, z) Q i ai for some ai ¥ Xi . Since Ii (x, z) ] ”, we take any bi ¥ Ii (x, z). Then (bi , z(i) ) Q x O y Q (ai , z(i) ). By coordinate order denseness, x O (ci , z(i) ) O y for some ci ¥ Xi . Therefore, ci ¥ Ii (y, z) and ci ¨ Ii (x, z). Hence Ii (x, z) … Ii (y, z).
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(2) Suppose that x Q a O b Q (xi , y(i) ). With no loss of generality, let i=n. For 1 [ k < n, let
˛ yx
k
y −k =
k
if yk Q k xk , otherwise.
Then y −k Q k xk for all 1 [ k < n. By Lemma 4.1(1), (xn , y −(n) ) Q x. If a O (y −1 , ..., y −k , yk+1 , ..., yn − 1 , xn ) O b for some 1 [ k < n, then the desired result obtains by letting c(n) =(y −1 , ..., y −k , yk+1 , ..., yn − 1 ). Suppose that there is no 1 [ k < n such that a O (y −1 , ..., y −k , yk+1 , ..., yn − 1 , xn ) O b. Then, up to a permutation of indices, there exists 1 [ a < n such that (y −1 , ..., y −a , ya+1 , ..., yn − 1 , xn ) Q a O b Q (y −1 , ..., y −a − 1 , ya , ..., yn − 1 , xn ). By coordinate order denseness, there is a za ¥ Xa such that a O (y −1 , ..., y −a − 1 , za , ya+1 , ..., yn − 1 , xn ) O b and of course za O a ya . So the desired result obtains by letting c(n) = (y −1 , ..., y −a − 1 , za , ya+1 , ..., yn − 1 , xn − 1 ). (3) Suppose that A ¥ W, x O y, and x, y ¥ A. For 1 [ k [ n, let
˛ xy
y −k =
k k
if yk O k xk , otherwise.
Then y [ D yŒ and yŒ ¥ A. By Lemma 4.1(1), y Q yŒ, so that x O yŒ. If x O (x1 , y −(1) ) O y, then the desired result obtains, since (x1 , y −(1) ) ¥ A. If (x1 , y −(1) ) ’ x, then by coordinate order denseness, x O (a1 , y −(1) ) O y for some a1 ¥ A1 . Then the desired result follows from (a1 , y −(1) ) ¥ A. If y Q (x1 , y −(1) ), then by (2), x O (x1 , a(1) ) O y for some a(1) ¥ X(1) . Since x and (x1 , y −(1) ) are in A, it follows from the construction of a(1) in (2) that (x1 , a(1) ) ¥ A. L Lemma 4.3. For all a, b ¥ X, a O b if and only if I(a, a) O g I(b, b). Proof. If a O b, then it follows from Lemma 4.1(4) and 4.1(5) that I(a, a) O g I(b, b). Next we assume that I(a, a) O g I(b, b). Then, by definition, I(a, a) 0 O c for some c ¥ I(b, b) 0. If I(a, a) g=”, then I(a, a) O c. If I(a, a) g ] ”, then, by Lemma 4.1(2), I(a, a) Q c. Therefore, a Q c. Since c ¥ I(b, b) 0, c < D b, so, by Lemma 4.1(1), c O b. Hence a O b. L Lemma 4.4. (1) If A ¥ W and i ¥ NA , then (A 0i , A(i) ) ’ gA. If, in addition, a ¥ (A 0i , A(i) ), then (Ii (a, a), A(i) ) O gA. (2) If A, B ¥ W and A O g B, then A O g I(b, b) for some b ¥ B 0. Proof. (1) Suppose that A ¥ W and i ¥ NA . Since (A 0i , A(i) ) ı A, (A 0i , A(i) ) Q g A. Assume that (A 0i , A(i) ) O g A. Then there is a b ¥ A 0 such that (A 0i , A(i) ) 0 O b. However, A 0=(A 0i , A(i) ) 0, which gives b O b, a contradiction. Hence (A 0i , A(i) ) ’ g A. Suppose further that a ¥ (A 0i , A(i) ). Since ai ¥ A 0i , (Ii (a, a), A(i) ) Q gA. By coordinate order denseness, ai O i bi for some bi ¥ A 0i . Assume that NA ={i}. Then for all
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YUTAKA NAKAMURA
b(i) ¥ A(i) , aj ’ j bj for j ] i. Thus by Lemma 4.1(1), a O b. Since b ¥ A g and (Ii (a, a), A(i) ) Q a, Lemma 4.1(3) implies that (Ii (a, a), A(i) ) O gA. Assume that j ¥ NA for some j ] i. We recursively define a sequence a 1j , a 2j , ... on component j as follows. Fix any a 1j ¥ Aj and for each a kj , take any a k+1 to satisfy j that (bi , a kj , a(ij) ) O (ai , a k+1 , a(ij) ). j Since Aj is bounded above, the set {a kj } is bounded. Therefore, by strong Archimedean axiom, this set consists of a 1j , ..., a mj j for some positive integer mj \ 1. Since (ai , a mj j , a(ij) ) ¥ (Ii (a, a), A(i) ) and (ai , a mj j , a(ij) ) O (bi , a mj j , a(ij) ), we obtain that (Ii (a, a), Aj , a(ij) ) Q (bi , a mj j , a(ij) ). Similarly, for k ¥ NA 0 {i, j}, we can define a sequence a 1k , a 2k , ... on component k such that (bi , a mj j , a ak , a(ijk) ) O (ai , a mj j , a a+1 k , a(ijk) ) for all a. By the strong Archimedean axiom, there is a lowest upper bound on a, say mk . Hence (Ii (a, a), Ajk , a(ijk) ) Q (bi , a mj j , a mk k , a(ijk) ). By induction on every j ¥ NA , m there exist mj ’s such that (Ii (a, a), ANA 0 {i} , a(NA ) ) Q (bi , a {jj¥ NA 0 {i}} , a(NA ) ). Since for all m j ¨ NA , Aj ={aj }, we get (Ii (a, a), A(i) ) Q (bi , a {jj¥ NA 0 {i}} , a(NA ) ). By coordinate order denseness and the fact that i ¥ NA , bi Oi b −i for some b −i ¥ A 0i , so m m that, by Lemma 4.1(1), (Ii (a, a), A(i) ) O (b −i , a {jj¥ NA 0{i}} , a(NA ) ). Since (b −i , a {jj¥ NA 0{i}} , a(NA ) ) ¥ A g, Lemma 4.1(3) gives that (Ii (a, a), A(i) ) O gA. (2) Suppose that A O g B. Then A 0 O bŒ for some bŒ ¥ B 0. Thus by Lemma 4.1 (2), A Q bŒ. If there exists i ¥ NB , then, by definition of B 0i , there exists b −i ¥ Bi such that b O (b −i , b(i) ); hence by Lemma 4.2 (3), there exists b 'i ¥ B 0i such that b O (b 'i , b(i) ) O (b −i , b(i) ) and (b 'i , b(i) ) ¥ B 0. Hence A O g I((b 'i , b(i) ), (b 'i , b(i) )). If, on the contrary, NB =0, then B 0={b}. But since A 0 O b, if a ¥ A 0, then there exists an i such that ai O i bi , which is impossible since i ¨ NB . L We are to show that there is an additive representation for Q g; i.e., there are real valued functions Ui on Wi for i=1, ..., n such that, for all A, B ¥ W, n
n
A Q g B Z C Ui (Ai ) [ C Ui (Bi ). i=1
i=1
Moreover, the additive representation is a cardinal scale. Since, for all a, b ¥ X, I(a, a) and I(b, b) are in W, it follows from Lemma 4.3 that a Q b Z I(a, a) Q g I(b, b) n
n
Z C Ui (Ii (a, a)) [ C Ui (Ii (b, b)). i=1
i=1
Letting ui (ai )=Ui (Ii (a, a)) for all ai ¥ Xi and all i=1, ..., n, we obtain the desired result. By Theorem 2.1, it suffices to show the following lemma. Lemma 4.5. (1) (2)
Q g is a weak order;
Q g satisfies restricted solvability;
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(3)
525
Q g satisfies generalized triple cancellation;
(4) every bounded standard sequence w.r.t. Q g is finite; (5) at least two components are essential w.r.t. Q g. Proof. (1) Transitivity of Q g follows from definition of Q g. To show completeness of Q g, suppose that ¬ (A Q g B) and ¬ (B Q gA) for some A, B ¥ W. Then B O gA and A O g B. By definition, B 0 O a for some a ¥ A 0 and A O b for some b ¥ B 0. Therefore, ¬ (a Q b) from the former, and ¬ (b Q a) from the latter, contradicting completeness of Q. (2) To show restricted solvability of Q g, we assume that (Ai , A(i) ) O g B O g(Ci , A(i) ). We are to show that B ’g
1 0 3 I (x, y), A 2 . i
(i)
x¥B y¥A
Let Ji = 2 x ¥ B 5 y ¥ A Ii (x, y). First we need to verify that Ji ¥ Wi ; i.e., Ji ] ” and Ji is a preference interval which is left exhausted and bounded above. By defnition, A 0 O bŒ for some bŒ ¥ B 0. Thus NB ] ”. Therefore, it follows from coordinate order denseness and Lemma 4.1(2) that A Q bŒ O b for some b ¥ B 0. Similarly, B O (ci , a(i) ) for some (ci , a(i) ) ¥ (Ci , A(i) ) 0. Therefore, for all bŒ ¥ B such that b Q bŒ, and for all (ci , a −(i) ) ¥ (Ci , A(i) ) such that (ci , a(i) ) Q (ci , a −(i) ), A O bŒ O (ci , a −(i) ). Take any ai ¥ Ai . Then Ai ı Ii (bŒ, (ai , a −(i) )) ı Ii (c, c), since Ii (bŒ, (ci , a −(i) ))= Ii (bŒ, (ai , a −(i) )). Therefore, Ii (bŒ, (ai , a −(i) )) ¥ Wi . Given b ¥ B 0, a ¥ A, and ci ¥ C 0i obtained in the preceding paragraph, let Xb ={x ¥ B : b Q x}, Ya ={(ai , y(i) ) ¥ A : (ai , a(i) ) Q (ai , y(i) )}, so that Ai ı Ii (x, y) ı Ii (c, c) for all x ¥ Xb and all y ¥ Ya . Since, for all x, xŒ ¥ B and all y, yŒ ¥ A, x O xŒ S Ii (x, y) ı Ii (xŒ, y), (yi , y(i) ) Q (yi , y −(i) ) S Ii (x, (yi , y(i) )) ` Ii (x, (yi , y −(i) )), we obtain that ” ] Ji = 0 3 Ii (x, y) ı Ii (c, c). x ¥ Xb y ¥ Ya
Hence Ji ¥ Wi . We shall fix Xb and Ya defined in the preceding paragraph throughout the rest of the proof of (2). To show that B ’ g (Ji , A(i) ), we examine two other cases and derive contradictions for those cases. Case 1 (B O g (Ji , A(i) )). By definition, B 0 O (ci , a −(i) ) for some (ci , a −(i) ) ¥ (Ji , A(i) ) 0. By Lemma 4.1(2), B Q (ci , a −(i) ). Let a −i ¥ Ai .
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Suppose first that ci ¥ J 0i . Then, by coordinate order denseness, there is a c −i ¥ J 0i such that ci O i c −i , so, by Lemma 4.1(1), (ci , a −(i) ) O (c −i , a −(i) ). Since Ji ı 1x ¥ B Ii (x, (a −i , a −(i) )), there is a bŒ ¥ B such that c −i ¥ Ii (bŒ, (a −i , a −(i) )). Therefore, (c −i , a −(i) ) Q bŒ, so (ci , a −(i) ) O bŒ, a contradiction. Suppose next that J 0i =”. Then there exists k ¥ NA 0 {i} such that a −k ¥ A 0k , else B O g (Ji , A(i) ) is impossible. By coordinate order denseness, there is a a 'k ¥ A 0k such that a −k O k a 'k . Take any ci ¥ Ji . We obtain that B Q (ci , a −(i) ) O (ci , a 'k , a −(ik) ) ¥ (Ji , A(i) ) 0. Since ci ¥ Ji ı 1x ¥ B Ii (x, (a −i , a 'k , a −(ik) )), there is a bŒ ¥ B such that ci ¥ Ii (bŒ, (a −i , a 'k , a −(ik) )). Therefore, (ci , a 'k , a −(ik) ) Q bŒ, so (ci , a −(i) ) O bŒ, a contradiction. Case 2 ((Ji , A(i) ) O g B). By definition, (Ji , A(i) ) 0 O bŒ for some bŒ ¥ B 0. Since (Ji , A(i) ) O g B and NB ] ”, by coordinate order denseness and Lemma 4.1(2), (Ji , A(i) ) Q bŒ O bœ for some bœ ¥ B 0. Recall that b ¥ B 0 is arbitrarily chosen as long as A O b. Thus we choose b ¥ B 0 to satisfy (Ji , A(i) ) O b. For such a b, it holds that Ji = 0 3 Ii (x, y), x ¥ Xb y ¥ Ya
so that, for all x ¥ Xb , 3 Ii (x, y) ı Ji . y ¥ Ya
Assume that ci ¥ Ji 0 4y ¥ Ya Ii (bŒ, y) for some bŒ ¥ Xb . Then there is a yŒ ¥ Ya such that bŒ O (ci , y(i) ) ¥ (Ji , A(i) ), a contradiction. Therefore, Ji =4y ¥ Ya Ii (x, y) for all x ¥ Xb . Take any c ¥ Xb for which b O c. It follows from Lemma 4.2(3) that there are b 0, b 1, b 2, ... and cŒ, c 0, c 1, c 2, ... in Xb such that b O b i O b i − 1 O cŒ O c i − 1 O c i O c for all i=1, 2, ... . Take any a 1 ¥ Ya , so that Ii (x, a 1) ¥ Wi for all x ¥ Xb . Then we define recursively x ki ¥ Xi and a k+1 (i) ¥ A(i) for k=1, 2, ... to satisfy that (ai , a k+1 (i) ) ¥ Ya , and (i) x ki ¥ Ii (b k − 1, (a 1i , a k(i) )) 0 Ii (b k, (a 1i , a k(i) )), k (ii) c k − 1 O (x ki , a k+1 (i) ) O c .
Given a k(i) , the existence of x ki in (i) easily follows from Lemma 4.2(1). Given x ki , the existence of a k+1 in (ii) is assured as follows. Suppose that (x ki , y(i) ) Q c k − 1 for all (i) y ¥ Ya . Then x ki ¥ 4y ¥ Ya Ii (c k − 1, y)=Ji . Thus x ki ¥ 4y ¥ Ya Ii (b k, y)=Ji , so that (x ki , a k(i) ) Q b k. But then (x ki , a k(i) ) ¥ Ii (b k, (a 1i , a k(i) )). This is a contradiction. Consequently, there exists (ai , a −(i) ) ¥ Ya such that c k − 1 O (x ki , a −(i) ). Either c k − 1 O (x ki , a −(i) ) − k−1 O c k and the result obtains by letting a k+1 O c k Q (x ki , a −(i) ). But then (i) =a (i) , or c (x ki , a k(i) ) Q b k − 1 O c k − 1 O c k Q (x ki , a −(i) ) and by Lemma 4.2(2), there exists k−1 k a k+1 O (x ki , a k+1 (i) ¥ A(i) such that c (i) ) O c . k k k k k k−1 By (i), b O (x i , a (i) ) Q b for all integers k > 0. By (i), (x k+1 , a k+1 i (i) ) O (x i , a (i) ) 1 2 k k+1 1 for all integers k > 0. By (ii), (x i , a (i) ) O c O (x i , a (i) ) for all integers k > 1. Hence
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(x ki , a k(i) ) O (x 1i , a 1(i) ), (x 1i , a 2(i) ) O (x ki , a k+1 (i) ), k k (x k+1 , a k+1 i (i) ) O (x i , a (i) ),
which implies by generalized triple cancellation that (x k+1 , a 2(i) ) O (x ki , a 1(i) ). Therei k O i x ki for all k. fore, {x i : k=1, 2, ...} is a strong standard sequence with x k+1 i k We know that A O b and a (i) ¥ A(i) . This implies that for all (ai , a k(i) ) ¥ Ya , (ai , a k(i) ) O b. Then since b O (x ki , a k(i) ) for all k, ai O i x ki for all k. Thus ai is a lower bound for the strong standard sequence. Hence the sequence is bounded but infinite, contradicting the strong Archimedean axiom. (3) Suppose that (Ai , A(i) ) Q g (Bi , B(i) ), (Bi , C(i) ) Q g (Ai , D(i) ), (Ci , B(i) ) Q g (Di , A(i) ). Then we are to show that (Ci , C(i) ) Q g (Di , D(i) ); i.e., for each c ¥ C 0, there is a d ¥ D 0 such that c Q d. We have the following two cases to examine: A O g B; A ’ g B. Fix a c ¥ C 0. Case 1 (A O g B). By definition, A 0 O b for some b ¥ B 0. Since (bi , c(i) ) ¥ (Bi , C(i) ) 0 and (ci , b(i) ) ¥ (Ci , B(i) ) 0, it follows from definition of Q g that there are (ai , d(i) ) ¥ (Ai , D(i) ) 0 and (di , a(i) ) ¥ (Di , A(i) ) 0 such that (bi , c(i) ) Q (ai , d(i) ), (ci , b(i) ) Q (di , a(i) ). Since (ai , a(i) ) ¥ A 0, a O b. By generalized triple cancellation, c Q d, where d=(di , d(i) ) ¥ D 0. Case 2 (A ’ g B). If NC =”, then C Q X for all X ¥ W, so that C Q gD. Therefore, we assume that NC ] ”. We have the following three subcases to examine: for some j ] i, Subcase 1: {i, j} ı NC ; Subcase 2: i ¨ NC and j ¥ NC ; Subcase 3: NC ={i}. Subcase 1. Note that C 0i and C 0j are not empty. By Lemma 4.4(1), (Ii (c, c), B(i) ) O g (Ci , B(i) ) Q g (Di , A(i) ), (Bi , I(i) (c, c)) O g (Bi , C(i) ) Q g (Ai , D(i) ). Thus, by definition, there are (di , a(i) ) ¥ (Di , A(i) ) 0 and (ai , d(i) ) ¥ (Ai , D(i) ) 0 such that (Ii (c, c), B(i) ) 0 O (di , a(i) ), (Bi , I(i) (c, c)) 0 O (ai , d(i) ),
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so d ¥ D 0. Since (ai , a(i) ) ¥ A 0, the definition of ’ g gives that a Q b for some b ¥ B 0. Since (ci , b(i) ) ¥ (Ii (c, c), B(i) ) and (bi , c(i) ) ¥ (Bi , I(i) (c, c)), it follows from Lemma 4.1(2) that (ci , b(i) ) Q (di , a(i) ), (bi , c(i) ) Q (ai , d(i) ). Hence by generalized triple cancellation, c Q d. Subcase 2. Note that C 0i =” and C 0j ] ”. By Lemma 4.4(1), (Bi , I(i) (c, c)) O g (Bi , C(i) ) Q g (Ai , D(i) ). Thus, by definition, there is an (ai , d(i) ) ¥ (Ai , D(i) ) 0 such that (Bi , I(i) (c, c)) 0 O (ai , d(i) ). Take any a(i) ¥ A(i) for which a ¥ A 0. Then by definition of ’ g, a Q b for some b ¥ B 0. Thus, by Lemma 4.1(2), (Bi , I(i) (c, c)) Q (ai , d(i) ), so that (bi , c(i) ) Q (ai , d(i) ). If k ¨ NB for all k ] i, then (ci , b(i) ) Q X for all X ¥ W, so that (ci , b(i) ) Q (Di , A(i) ) 0. If k ¥ NB for some k ] i, then, by Lemma 4.4(1), (Ci , I(i) (b, b)) O g (Ci , B(i) ) Q g (Di , A(i) ). Thus there is a (di , a(i) ) ¥ (Di , A(i) ) 0 such that (Ci , I(i) (b, b)) 0 O (di , a(i) ), so, by Lemma 4.1(2), (Ci , B(i) ) Q (di , a(i) ). Therefore, in both cases, (ci , b(i) ) Q (di , a(i) ), where (di , d(i) ) ¥ D 0. Hence, by generalized triple cancellation, c Q d. Subcase 3. Note that C 0i ] ” and C 0j =” for all j ] i. By Lemma 4.4(1), (Ii (c, c), B(i) ) O g (Ci , B(i) ) Q g (Di , A(i) ). Thus, by definition, there is a (di , a(i) ) ¥ (Di , A(i) ) 0 such that (Ii (c, c), B(i) ) 0 O (di , a(i) ), so, by Lemma 4.1(2), (Ii (c, c), B(i) ) Q (di , a(i) ). Take any ai ¥ Ai for which a ¥ A 0. Then by definition of ’ g, a Q b for some b ¥ B 0. Assume first that B 0i ] ”. Then bi ¥ B 0i , so by Lemma 4.4(1), (Ii (b, b), C(i) ) O g (Bi , C(i) ) Q g (Ai , D(i) ). Thus, by definition, there is an (ai , d(i) ) ¥ (Ai , D(i) ) 0 such that (Ii (b, b), C(i) ) 0 O (ai , d(i) ),
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so by Lemma 4.1(2), (Ii (b, b), C(i) ) Q (ai , d(i) ). Therefore, (bi , c(i) ) Q (ai , d(i) ), (ci , b(i) ) Q (di , a(i) ), where d ¥ D 0. By generalized triple cancellation, c Q d. Assume next that B 0i =”. Then (Bi , C(i) ) Q X for all X. Therefore, (ci , b(i) ) Q (di , a(i) ), (bi , c(i) ) Q (ai , d(i) ) for all d(i) ¥ D(i) for which (ai , d(i) ) ¥ (Ai , D(i) ) 0. Note d ¥ D 0. By generalized triple cancelaltion, c Q d. (4) Suppose that there exists a bounded infinite standard sequence w.r.t. Q g. Let {A ki : k ¥ K} be such a sequence on component i w.r.t. Q g, where K is an infinite set of consecutive integers. Then there are (Ci , A(i) ), (Ci , B(i) ) ¥ W such that , B(i) ) for all k, k+1 ¥ K. Assume ¬ ((Ci , A(i) ) ’ g (Ci , B(i) )), and (A ki , A(i) ) ’ g (A k+1 i that K is the set of all positive integers and (Ci , B(i) ) O g (Ci , A(i) ). The proofs for the other cases are similar. Then by (3), (A ki , B(i) ) O g (A ki , A(i) ) for all k ¥ K, and (A ki , D(i) ) O g (A k+1 , D(i) ) for all k, k+1 ¥ K and all D(i) ¥ W(i) . i First we show that there are ci ¥ Ci and d(i) , e(i) ¥ X(i) such that (Ii (c, c), B(i) ) Q g I((ci , d(i) ), (ci , d(i) )) O g I((ci , e(i) ), (ci , e(i) )) Q g (Ii (c, c), A(i) ). Since (Ci , B(i) ) O g (Ci , A(i) ), it follows from Lemma 4.4 (2) that (Ci , B(i) ) O g I((ci , a(i) ), (ci , a(i) )) for some (ci , a(i) ) ¥ (Ci , A(i) ) 0. By Lemma 4.1 (1) and the definition of O g, (Ii (c, c), B(i) ) Q g (Ci , B(i) ), so by (1), (Ii (c, c), B(i) ) O g I((ci , a(i) ), (ci , a(i) )). By Lemma 4.4 (2), (Ii (c, c), B(i) ) O g I((c −i , a −(i) ), (c −i , a −(i) )) for some (c −i , a −(i) ) ¥ I((ci , a(i) ), (ci , a(i) )) 0. Note that I((c −i , a −(i) ), (c −i , a −(i) )) O g I((ci , a(i) ), (ci , a(i) )). If c −i ’ i ci , then letting ci =c −i , d(i) =a −(i) , and e(i) =a(i) gives the desired result. Assume that c −i ¥ Ii (c, c) 0. Since (ci , b(i) ) O (c −i , a −(i) ) O (ci , a(i) ) for some b(i) ¥ B(i) , Lemma 4.2(2) implies that (c −i , a −(i) ) O (ci , d(i) ) O (ci , a(i) ) for some d(i) ¥ A(i) . Hence letting e(i) =a(i) , we obtain the desired result. −1 Since (A 2k , B(i) ) O g (A 2k i i , B(i) ) for all k ¥ K, it follows from Lemma 4.1(3) that 2k − 1 g for every k ¥ K, there is an (a ki , b k(i) ) ¥ (A 2k , B(i) ) O (a ki , b k(i) ). i , B(i) ) such that (A i k 2k 2k+1 k+1 k+1 g Thus (a i , e(i) ) ¥ (A i , A(i) ) ’ (A i , B(i) ) O (a i , b (i) ), so that (a ki , e(i) ) O (a k+1 , b k+1 i (i) ). By the preceding two paragraphs and Lemma 4.3, (ci , b(i) ) Q (ci , d(i) ) for all b(i) ¥ B(i) . Therefore, in particular for b(i) =b k+1 (i) , generalized triple cancellak+1 k+1 tion implies that (a k+1 , b ) Q (a , d ). Hence (a ki , e(i) ) O (a k+1 , d(i) ) for all (i) i (i) i i k k ¥ K, so that {a i : k ¥ K} is a strong standard sequence on component i w.r.t. Q. Since {A ki : k ¥ K} is bounded, so is {a ki : k ¥ K}. By the strong Archimedean
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axiom, K must be finite, a contradiction. Hence there exist no bounded infinite standard sequence w.r.t. Q g. (5) Suppose that the coordinates 1 and 2 are essential w.r.t. Q. For coordinate 1, there are a, b, c ¥ X such that (a1 , c(1) ) O (b1 , c(1) ). Then it follows form Lemma 4.3 that I((a1 , c(1) ), (a1 , c(1) )) O g I((b1 , c(1) ), (b1 , c(1) )). Since I(1) ((a1 , c(1) ), (a1 , c(1) ))=I(1) ((b1 , c(1) ), (b1 , c(1) )), the coordinate 1 is essential w.r.t. Q g. Essentiality of 2 w.r.t. Q g similarly follows. L
REFERENCES Debreu, G. (1960). Topological methods in cardinal utility theory. In K. J. Arrow, S. Karlin, and P. Suppes (1959, Eds.), Mathematical methods in the social sciences. Stanford: Stanford Univ. Press. Gonzales, C. (1996). Additive utilities when some components are solvable and others are not. Journal of Mathematical Psychology, 40, 141–151. Gonzales, C. (1999). Additive utility without restricted solvability on all components. Mimeograph. Gonzales, C. (2000). Two factor additive conjoint measurement with one solvable component. Journal of Mathematical Psychology, 44, 285–303. Gonzales, C., & Jaffray, J. Y. (1998). Imprecise sampling and direct decision making. Annals of Operations Research, 80, 207–235. Jaffray, J. Y. (1974). On the extension of additive utilities to infinite sets. Journal of Mathematical Psychology, 11, 431–452. Krantz, D. H., Luce, R. D., Suppes, P., & Tvesky, A. (1971). Foundations of measurement (Vol. 1). New York: Academic Press. Wakker, P. P. (1988). The algebraic versus the topological approach to additive representations. Journal of Mathematical Psychology, 32, 421–435. Wakker, P. P. (1989). Additive representation of preferences. Dordrecht: Kluwer Academic Publishers. Wakker, P. P. (1991). Additive representations of preferences, a new foundations of decision analysis: the algebraic approach. In J. P. Doignon and J. C. Falmagne (Eds.), Mathematical psychology: Current developments. Berlin/New York: Springer-Verlag. Received: August 30, 1999; published online: June 13, 2002