Additively separable representations on non-convex sets

JOURNAL

OF ECONOMIC

56, 89-99

THEORY

Additively

(1992)

Separable Representations on Non-convex UZI SEGAL*

Department 150 St. George

of Economics, Street, Toronto,

University Ontario,

qf Toronto, Canada M5S

IA1

Received May 15, 1990; revised May 6. 1991

This paper proves sufficient conditions under which a completely separable order on non-convex sets can be represented by an additively separable function. The two major requirements are that indifference curves are connected and that intersections of the domain of the order with parallel-to-the-axes hyperplanes are connected. Jourrzal of Economic Literature Classification Numbers: 020, 022. c 1992 Academic ?ress. Inc.

1. INTRODUCTION

It follows from a famous theorem by Debreu [4] that a reflexive, complete, transitive, and continuous order 2 on a product of intervals rrl x ... x xN c RN, N> 3, is completely separable (in the sense that the induced order on the product njzini for any Jo (I, ...) Nj does not depend on the outcome in 711)if and only if it can be represented by an additively separable function of the form V(x,) .... x,) = Cy= 1 ui(xl). There are situations where the domain of the function V is not necessarily a Cartesian product. Some recent generalizations of expected utility theory make use of additively separable functions that are defined on such domains; see, for example, [2].’ The assumption that the domain of preferences is a Cartesian product is common. However, less restrictive assumptions (e.g., convexity) appear in general equ.ilibrium theory. Moreover, considerations of survival suggest that the consumption possibilities set is often not expressible as a Cartesian product, for example, when leisure is one of the goods and greater labor supply requires * I am grateful to Chew Soo Hong and Peter Wakker. and especially to Larry Epstein, for their comments. ‘ As pointed out by Wakker [7], there are mistakes in the proofs of some of the results in this work. I believe that the representation theorems proved below may help in correcting these mistakes. 89 00224X31/92

$3.00

Copyright Q 1992 by Academic Press. Inc. All rights of reproduction in any form reserxkd.

90

UZI

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increased nourishment. (See Blackorby, Primont, and Russell [ 1 ] for a survey of applications of additively separable functions in economics.) Debreti’s theorem was recently extended by Wakker [6] to ordered cones (i.e., sets in [WN,where x1 2 ... 3 xN > 0). In Theorem 1 below I add a monotonicity assumption and prove this theorem for more general nonconvex sets. The crucial conditions concerning the domain of the order 2 are that all its indifference surfaces are connected sets (see [7] for the importance of this condition) and that the intersection of this domain with any parallel-to-the-axes hyperplane is a connected set. Section 2 deals with the case of an open domain, and Section 3 proves a sufficient condition for the existence of an additively separable representation on a closed set given that such a representation exists on its interior. Section 4 concludes with three examples illustrating the importance of some of the assumptions used throughout. Topological definitions and claims are taken from Dugundji [ 5 1. 2.

OPEN

DOMAIN

Let the connected [S, p. 1071 and open subset SC IW”‘, N> 3,2 satisfy the condition that for every i and c, the set S(i, c) = Sn {(xi, .... xN) : xi= c} is a connected subset of RN. Of course, the connectedness of S does not imply, nor is it implied, by the connectedness of the sets S(i, c), as is demonstrated by the sets S=(0,3)3\([1,3)x(0,3)x[1,2]) and S= (0, 1)3 u (2, 3)3. Let k be a reflexive, complete, and transitive order on S. For x, y E S, define x- y if and only if x z y and y 2 x, and x> y if and only if xky but not ykx. The order 2 is called strictly monotonic if [V’~E (1, .... N}yi>,xi and 3i~{l,...,N} such that yi>x,]*y= xN). It is called continuous if for every XE S the (Y 1, .... YN) > x = (Xl, ...> two sets {y~S:y>x} and (y~S:x>y} are open in S. Assume throughout that 2 is strictly monotonic and continuous. For every x E S, let Z(x) = {y E S : y-x} be the indifference surface of 2 through x. Assume further that all indifference surfaces of k are connected subsets of LW? By Debreu [3] there exists a continuous function I’: S --+ [w representing the order 2, that is, x 2 y if and only if V(x) > V(y). 1. The function V: S --) [w is called completely separable if, for every (Xl, .... Xi-l, xi, xi+l, .... XN), (YI, .... Yip1, xi, Yi+l, .... yN), (x 1, +-, xi- 1, Y:, Xi+ 1, -, x,), and (~1, --., Yi- 1, Yi, yi+ 1, .... YN) in S, DEFINITION

V(x 19 *

’ When

v(x13...,

“‘3

xi&l,

xi9

xi+l,

xi-l,

Yi,

xi+19.-*9

N = 2, strict

monotonicity

...? xN)2

xN)3

implies

v(.Yl,

...t

v(Yl,**., complete

Yi-l,

Xi,

,Yj+l,

...T YN)

Yi-12

Yi,

Yi+l,

***> YN).

separability

(see Definition

1).

ADDITIVELYSEPARABLEREPRESENTATIONS

Let xi(S) be the projection ni(S)=

91

of 5’ on the ith axis, (Xi: 3(X,, ...) Xj, ...) XN) E S>.

Since S is an open connected subset of RN, xi(S) is an open interval, i=l , .... N. DEFINITION 2. The function Y: S + R is called additively separable if there exist continuous and strictly increasing functions ui: ~~(5’) --f R, i = 1, ...7N and a strictly increasing function i: Rng(Cy= 1 ui(. )) -+ R! such t

V(x 1, where Rng(Cy=, cf+l 1 %(Xi) 1. THEOREM

I.

ui(.))

...> xN)=i

($:

ui(xil)3

= {x E R : 3(x,, .... xN) E RN

such

that

x =

Let N 3 3 and let (S, 2) satisfv the following conditions:

e The set S is an open and connected subset of IWN c For every i and c, the set S(i, c) is a connected subset of IWN 0 The order 2 on S is continuous and strictly monotonic 9 All indifference surfaces of ,> are connected subsetsof WN. Then a representation function V of 2 is completely separable if and only if it is additively separable.

Obviously, if V is additively separable, then it is completely separable. The proof that the opposite is also true follows from Lemmas 14. In sequel, the term open box means a set of the form ny= 1 Ji, where bj nonempty bounded open interval in zi(S), i = 1, .... N. For each x0 E S, let R(x’) be an open box in S containing x’.~ By Debreu’s f4] theorem, there are strictly increasing and continuous functions uj( .; x0) and i( .; x0) such that on R(xO), vxt >.... x,)=(

5 uj(xi;xo);xo ( i=t

)

.

(1)

LEMMA 1. Let x0 = (xy, .... xt, ...>x”,), y” = (yy, .... x&, .... y”,) E S and let R” and RY be two open boxes in S such that X’E R” and y’~ RY. Let the order 2 on R’ be representedby C;“_ 1uf(xi), z = x, Y.~ Then there are a > 0. b, and E> 0, such that for every xi0 E (X:-E, xi + E), ui’,(xjO)= au;(x,) + b.

Proof

The set S(io, xt) is (isomorphic

3 There are, of course, a lot of open boxes containing of these possible boxes to be R(x”). 4 Such representations exist by Debreu [4].

to) an open set in 18~~” an Y’ in 5’. For every

x0 we choose

one

92

UZI

SEGAL

connected, hence path connected. (A set TE RN is called path connected if for every x, y E T there exists a continuous mappingf: [0, 1] -+ T such that f(O)=x and f(l)= y; see [S, pp. 114-1161.) There is, therefore, a continuous mapping f: [0, 1] + S( zO, . xi) such that f(0) = x0 and f( 1) = y”. Denote its image by L. The curve L c S is compact [S, p. 2241; hence there is a finite set of open boxes R’, .... R” c S covering L. Assume, without loss of generality, that R’ = R”, R” = RY, and Rjn Rjf 1 # @, j= 1, .... m - 1. By the above-mentioned theorem of Debreu, on Rj the order can be represented by Uj(x,, .... xN) = Cy= i u{(xJ, j = 1, .... m, where of = ~7 and u’ z UT, i= 1, .... N. Let 16 j-cm. On the set Rjn R j+l the order can be represented by both Uj and Uj+‘; hence there exists a positive linear transformation such that Ui+l = ajUj + bj (see [4]). In particular, on this intersection, $’ ’ = ajui + bj. Let (X&-E, xi + E), E> 0, be contained in the projection on the ioth axis of the open boxes R’, .... R”. This E satisfies the conditions of the lemma. Q.E.D. Remark. The proof of Lemma 1 is the only place in the proof of Theorem 1 where the assumption concerning the connectedness of the sets S(i, c) is used. LEMMA 2. There are N strictly increasing and continuous functions ui: 7ci(S) -+ R, i= 1, .... N, such that for every x0, OIZ R(x’), V(x,, .... xN) = CCC;“=1 ui(xi); xo).

Proof. I show first how to construct ur. Let x1 = inf n,(S) and X, = sup rcr(S). If S is unbounded, x1 may be ---co and Xi may be co. Let # +x1 and X; --f X, such that for every n, $+I < C&C 2x1~ 2;’ ‘. Let Y” = [&, Xy]. For every t E Y” there is a point x(t) = (t, x2, .... xN) E S with the open box R(x( t)) c S. Let the open interval X(t) be the projection of R(x(t)) on the xi(S). Since Y” is compact, there is a finite set of these open intervals covering Y”. Denote them by X1, .... X”, and assume, without loss of generality, that ~7 E X’, X; E X”, X’n Xj+ ’ # a, j = 1, .... m - 1, and XjnXj+‘=@, j= 1, .... m - 2. Otherwise, if for some j, Xj n Xj+ 2 f 0, then {Xl, .... Xj, Xj+‘, ..,, Xm} is also a cover of Y”. Note that this implies Xi) = @. For every 1 < j< m there is that for k
ADDITIVELY

SEPARABLE

REPRESENTATIONS

93

follows that there is an open interval T around I in z,(s) on which u;+yxJ = &‘;.” (x1) + bk with uk > 0. Assume therefore, without loss of generality, that on R(x(t”)) the order is represented by ak xy= I am* + bk rather than by Cr’ 1 uF~(xJ, and redefine u:*, .~.,7~:” accordingly. Hence on T, ~7” = uf+‘ln = u’;. Suppose now that there exists xl* E XknXk+’ such that z@(x~) # u:’ ““(x:). Assume first that XT > Z Let t* = infix, > i : u?“(x~) + u~“,~(x~)>. Since on T, u F” = u/;+I,~, it follows that t* > Z y Lemma 1 there is an open interval T* around t* on which ~4,~ is a positive linear transformation of U: + ‘.n. Since these two functions coincide on [i, f*]* they must also coincide on T*, a contradiction. Similar proof deals with the case XT < 7. It thus follows that on Xk n XC”+ ‘, ~7” = M’;+ ‘.n = u;. Extend now the function UT such that on Xk, u; = u:“. For every k < nz - 1, Xj) = 0, hence this extension of ~4 does not violate its Xkn (i&k+2 already defined values. In a finite number of steps we obtain u; defined for every x1 E Y”. Since for every k, u F” is strictly increasing and continuous, so is u;. Suppose now that another choice of points a boxes yields the function v:~ By Lemma 1 there is a finite open cover o n such that on each of its members, UT is a positive linear transformation of UT. Since these open sets have nonempty open intersections, it must be the same transformation on each neighbouring set, hence on all of these sets. The function u; is thus unique up to positive linear transformations. That is, there are a > 0 and h such that on Y”, v; = au; + b. Consider now the functions u; and u;’ ‘. By the above arguments, we may assume, without loss of generality, that on Y”, uinfl_ - ti;. x1 E [xi, %:I, let u~(x,)=u~(x,) and, for xi E [x;+‘, 3;)~ (27, x’;+“], let ui(xi) = u’;’ ‘(xi). The function U, is thus defined on (xi, Xi), and since for every n, u; is strictly increasing and continuous, so is ui. Let x’ES. By (l), the order k on R(x’) can be represented by C;“= 1 u,(x,; x0). I sh ow next that on its domain, ul( .; x0) is an increasing linear transformation of ui. Let Z be the projection of R(x”) on z,(S). Let 01,/3 E 2, and let n such that [a, /?I c ($,5;). For each z E [a, /J] there is, by Lemma 1, an open interval around z in Z n ($, 2;) on which ui( .; 2) is an increasing linear transformation of UT, and hence also of U! By the same arguments used in the construction of u; it is easy to verify that ul(.; x0) is an increasing linear transformation of u1 on [a, /?]. Let [P, /3”] 3 Z to get the desired result. Repeat the above procedure for each coordinate. arguments we have now obtained the condition for V on vx

15 ...2 XN)

= [

; c i=l

[ai

Ui(Xi)]

+ b(XO);

X0

, 1

(2)

UZI SEGAL

94

where b(x’) = C b,(x’). The functions ui are unique up to positive linear transformations. Assume therefore, without loss of generality, that at a certain point x* ES, aI = ... = a,.,,(~*) = u(x*). I now want to show that for every x0 E S, ai = . . . = aN(xo). Let x0 E S, and let the curve L c S connect x* and x0 (recall that as S is open and connected, it is also path connected). Each point x on the curve defines an open box R(x), and, since L is compact, it can be covered by the finite set of open boxes R(x’), ...) R(x”), where x1 =x*, xm =x0, and R(x’) n R(xj+‘) # 0, j= m- 1. On the open set R(x*)nR(x2), we obtain the two representa:;oz Cr=“=, [U(X*) Ui(Xi)] + b(x*) and Cfi 1 [a,(~‘) Us] + b(x2). By [4] they are cardinally equivalent; that is, Cfi 1 [a;(~‘) ui(xi)] + b(x2) is a positive linear transformation of Cyz I [a(~*) ui(xi)] + b(x*). Hence = a,,,(~~). By finite induction it thus follows that ur(x”) = . . . q(x’) = . . . = UN(XO). We now obtain from (2) that on each R(x’), T/(x,,

. ..) xff)

=

5:

u(x”) (

Define ((s; x0) = [(u(x’)s

g i=l

Ui(Xi)

+

b(xO);

x0

. )

+ b(x’); x0) to obtain the desired representation. Q.E.D.

Next I prove that the above representation can be used as a global one independent of x0. For this, only the connectedness of the indifference surfaces is used. I start with the following technical lemma. LEMMA 3. All indifference

surfaces

of the order k are path connected.

Proof: Let I be an indifference surface of 2, and define f: Z 4 RN- ’ by f(x1, x2, *.*>XN) = (x2, a’.,xN). This function is clearly continuous, and, since the order 2 is strictly monotonic, f is one to one. Let G be the range of f, G= ((x2, .... xN) : 3x, such that (x1, x2, .... xN) E Z}. The inverse f-l: G-+Ioffis th ere f ore well defined. The function f-' is also continuous. Indeed, let {(xz, .... x>)}:=~ c G such that (xi, .... XL) + (xi, .... z&,), but suppose that (x7, x;, .... xk) =fr(xi, .... x;) does not converge to ...) x”,). Assume, without loss of generality, that by, x;, .... x;, =f-‘(x;, E >O. It thus follows by strict monotonicity that Vn,x;>xy+z, 0 0 x”,, - (x7, x;, ...) x%) > (x7 + E, xi, ...) x”,) -+ (x? + E, xi, ...) x”,) > i;p: z;-:y, x”,), a contradiction. Next I show that the set G, the range off, is open in RN-‘. That is, w 2, .... xN) E G there is 6 > 0 such that d((y2, .... yN), (x,, .... xN)) < 6 + Let (Y 2, -*-,yN) E G, where d( ., .) is the Euclidean metric on RN-‘. and let x=(x1,x2 ,..., xN) =f-‘(x2, .... xN). Since S is (x 2, .... x,)EG,

ADDITIVELY

SEPARABLE

REPRESENTATIONS

95

open, there exists E> 0 such that B(x, e) c S, where B(x, E) is the ball of radius E around X. By the strict monotonicity of 2 it follows that

x1=cx,+-;, x2)...) x,)tx>x’=(I,-~,x2....xN) Since k is continuous there is 6 > 0 such that B(x’, 6) u B(x’, ii) c B(x, e)+ and for every y’ E B(x’, 6) and y2 E B(x2, 6), yi > x > y2. Eet (y2, .... yN) E RN-’ such that d((y,, ..‘, yN), (x2, .... xN)) < 6. It follows from the definition of 6 that (x1 +&/2, y,, .... yN), (x1 -e/2, y,, ...) y,)eS and that (Xl + 45 Y2> ‘..> YN) > x > (Xl - $2, y,, .... yN). Hence by the continuity of 2 and the convexity of B(x, E), there exists z1 such that (zr , yz, .... yN) E S and (zr, y,, ..i, yN)-x. Therefore, (y2, .... yN)eG. The range of a continuous function on a connected set is connected p. 108 ], and since G is open, it is also path connected. The function f ~ is continuous, I= f ~ ‘(G); hence the indifference surface I is path co [5, p. 1151. LEMMA 4. In the notations of Lemma 2, V(x,, . ... xN) 3 Vfy, , .... yN) G fL 1 %i%) 2cfi 1 U,(Y,).

Proof: Let V(xl, .... x,)= V(yl, .... y,); that is, the two points x= (X 1, ..., xN) and y = (y,, .... yN) are on the same indifference surface I of 2. Let the curve L ~1 connect these two points, and let the open boxes R(x) = R”, ...) R” = R(y) be a finite open cover of L in S such that for every j,LnRjnR.‘+‘f@ ( recall that L is a connected set). Let zi E L n Rj+l ) j = 1, ..,) m - 1. On each of the open boxes R’, ~.~,R” the order 2 can be represented by W(W) = Cf’_ 1 ui(wi); hence

Suppose now that there are x = (x,, .... xN) and y = (y,, .... y,,,) SW V(x)> V(y) (and x>y), but Ci”=l~i(xi)~Cfil~i(yi). Let the curve L c S connect these two points; that is, there is a con function f: [O, 1 ] -+ L such that f(0) = x and f( 1) = y. may assume that for every TV (0, l), x>f(t) > y. Otherwise, let S=max(tE [0, 1) :x-f(t)} and r=min{tE(s, I] : y-f(t)), and replace x by f(s) and y by f(r). By the first part of this proof, the value of the function Cy! I ui is the same at x and f(s), and at y and f(r). Define g, h: [0, l] + 58 by g = (Cj?! 1 ui) of and h = Vo$ Since the functions S, ‘V3 and ul, .... uN are continuous, so are g and h. Moreover, g(O) < g( 1) and h(O)>h(l). Let the open boxes R(x) = R’, .... R” = R(y) be a finite open cover of L 642/56/l-7

96

UZI SEGAL

in S as in the first part of this proof. On R(y) both V and Cr= i ui represent the order 2. By construction, for every z E L n (R(y)\{ y } ), z > y; hence, for such z, C;“= I ui(zi) > Cy! i u,(y,). There is, therefore, t E (0, 1) such that g(t) > g( 1) > g(0); hence the continuous function g reaches a maximum A on (0, 1). Let t*=min(tE [0, l] : g(t)=A}. On R(f(t*)) both V and xi”= i ui represent the order 2. There is E> 0 such that for t E (t* -F, t*), f(t) E R(f(t*)), and g(t) < g(t*), hence for t E (t* -8, t*), f(t*)>f(t), and h(t)h(t*), there exists SE (0, t*) such that h(s) = h(t*). By the first part of this proof it follows Q.E.D. that g(s) = g(t*), in contradiction with the definition of t*. Theorem

1 now follows by Lemmas l&4.

3. CLOSED DOMAIN The results of the last section do not hold for closed sets (see [7] for counterexamples). Following Wakker’s examples it is possible to show that Theorem 1 does not hold even when the set S is bounded and equals the closure of its interior. For example, let S= Conv{ (0, 0, 0), (10, l,l), (1, 10, 11, (4 1, W), and let V(X, , x2, x3) = x1x2x3. On the interior of S, V(xl, x2, x3) = .pxl+-+‘nx3, but this representation cannot be extended to S. In other words, V is completely separable, but not additively separable. Let SC RN be compact, and define, for i = 1, .... N, Si={(X1,...,Xj--l,Xi,Xi+l,...,XN)ES:

(Y 1,

..*9

Yi-19

Yi3 Yi+l,

sj= ((x,, ...) Xi-l,Xi,Xi+l (Y 19 THEOREM

***2

Yi-13

Yi,

Yi+l>

...T

YN)ES*Xi~Yi}

)...) x,)ES: .“>

YNlESJXi6

Yi}-

2. Let N > 3 and let (S, 2, V) satisfy the following conditions:

l

The set SE RN is compact and equals the closure of its interior

l

The interior of S, denoted by T, is a connected subset of RN

The order 2 on S is continuous and strictly monotonic The continuous function V: S + R represents the order 2. function is completely separable on S and additively separable on T l

l

This

For every i, each of the sets S’ and Si includes y = (yI, .... yN), z = (Zl) ...) zn), and w = (wl, .... wN) such that for every j# i, yj> zj> wi. l

Then the function V is additively separable on S.

ADDITIVELY

97

SEPARABLEREPRESENTATIONS

Proof: The function V is continuous on the compact domain 5’; hence it is bounded. By Theorem 1 there exist strictly increasing and continuous functions ur , .... uN such that on T, the interior of S,

We want to show that the functions ui can be extended to xi= min(.ui: 3(x,, .... xi, .... xN) E S} and Xi= max(xi: 3(x1, .... .‘ci, ...) x,,,) E S), such that condition (3) is still satisfied. Suppose that as xi -+ xi, u(xi) -+ -co. Let y, z, WE Si be as in the fift condition of the theorem. Since T is a connected subset of RN, it follows that for every j# i, zj is in the domain of uj, and uj(zj) < co. Since Sic s\T, there are sequences (z”}~= I and (w”>z= I in T converging to z and w, respectively. Moreover, by the fifth condition of the theorem, we may assume that for every II and j E { 1, .... N}, 2; > WT.The continuous functions uI are strictly increasing; hence lim f u~(w,“)< lim z uI(z/n)= -co. n-m is1 n-c-2 i=l By (3) and the continuity of V it follows that V(w) < V(z). Since the functions uj are continuous on T and limXi,, u(xi) = -XJ, there are two sequences in T, z” -+ z and w” + w, such that for every n, ~,!=, u,(zJ!)
The importance of the assumption concerning the connectedness of the indifference curves is demonstrated in [73. Wakker also shows why it is not sufficient to assume that the set S is a closed set. As demonstrated the following example, the condition that for every i and C, the set S(i, c) is a connected subset of RN, cannot be easily withdrawn. EXAMPLE

1. Let

e ~=((%x,,x,)+L+

:x,>9,2x,+x,<20,

0 B=((xl,x,,x,)~R~+:18<2x1+x~<20,x~<1)

l
98

UZI

l

C={(X1,X2,XJ)ER~+:

l

D=((x1,x2,x3)~R?++:

SEGAL

x,>18,2x,+x,<20, x,+x,+x,<15,

l
x,>l}. Let S=A

uBu

vx,,

x2,

CUD

and define a function V: S+ R by

(X,,X,,X,)EAUBUD

x,+x,+x, x3)= i xi +x,+2x,-

1

(Xl, x2, X3)E c.

Define an order 2 on S by (xi, x1,x3)2 (vi, y2, y3) if and only if V(x,, x2, x3) > V((yi, y2, y3). Obviously, all the indifference surfaces of 2 are connected subsets of R3. It is also easy to verify that V is completely separable, but it is not additively separable. Note that the set S(3,2) = {(xi, x2, x3) E S: x3 = 2) is not a connected subset of R3. Next I show the importance connected subset of UP’. EXAMPLE 2.

of the assumption

Let S= (0, 1)3 u (1, 2)3, and let V: S-+ R be given by

V~Xl~x27x3)= ;::ix;)(x2-1)(x3i V is completely

that the set S is a

l)+ 1

separable, but not additively

(Xl, x2, x3) E(0, II3 (x1, x2, x3) E (1, 213.

separable.

The aim of the next example is to demonstrate the importance of the existence of three points y, z, and w as in the fifth condition of Theorem 2. EXAMPLE 3. l

Let

A = Conv{(

-7d2,71/2,7d2),

(7112,-n/2,

n/2), (7-t/2, n/2, +2),

(~14,

7d4)>

7d4, l

B

71/4>0)>

=

Convi

(

- 742,-742, -7O),

(0,71./4, n/4),

(7~/4,0,7~/4),

(7~14,

Define S = A u B. Let V( - 7~12,-n/2, -n/2) = -n/2, v( - $2,71j2, q2) = v(n(71/2,-n/2, n/2) = v(71/2, n/2, -r/2) = ~12, and at all the other points of S, let V(x,, x2, x3) = arc tan(tan xi + tan x2 + tan x3).

On T, the interior of S, the function V is completely separable, additively separable, and continuous. On S, the function V is completely separable and continuous, but not additively separable. The only non-trivial property of V is its continuity.

ADDITIVELY

SEPARABLE REPRESENTATIONS

99

Let (x;, x;, x’;) -+ ( --z/2,71/2, n/2) such that for every n, (x;, x;‘, xl) f ( - 42, n/2, 742). F or a sufficiently large n, x; -tx; 4-x; 3 n/2. Let E* = x1+ 42, and it follows that either +2-xX;
hence V is continuous

on S.

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