Arrow's choice axiom

Economics Letters 01651765/94/$07.00

44 (1994) 381-384 0 1994 Elsevier

381 Science

B.V. All rights

reserved

Arrow’s choice axiom Donald

E. Campbell*

Department

of Economics,

Received Accepted

22 July 1993 22 September

The College of William and Mary, PO Box 8795, Williamsburg, VA, 23187-8795,

USA

1993

Abstract If a choice correspondence the domain does not include JEL

satisfies Arrow’s choice axiom then it has a complete and transitive rationalization, any sets with fewer than m members, where m is a given positive integer.

even if

classification: DOO, D70

Arrow’s choice axiom is equivalent to rationalization of the choice correspondence by a complete and transitive binary relation if the domain of eligible feasible sets includes all two-element and all three-element subsets of the action space X; the relation that generates the choice correspondence is unique [Arrow (1959)]. This paper proves that Arrow’s choice axiom implies that a choice correspondence has a complete and transitive rationalization even if there are no sets in its domain with fewer than m members, where m is a given positive integer greater than unity. As long as the domain includes all m-element sets and all sets with exactly m + 1 members, the choice correspondence 4 is generated by some complete and transitive binary relation. If two different binary relations rationalize 4 they must be identical over the range of 4. Let X denote the universal set of actions. 2 represents the family of candidate feasible sets. A choice correspondence 4 for 3 is a correspondence from 23 into X such that 4(Y) is a non-empty subset of Y for all YE 2. We set X+ = l-l,,, $(Y), the range of 4. Of special interest is the domain Z,,, consisting of all m-element subsets of X and all subsets with exactly m + 1 members. At little extra cost we can prove our rationalizability theorem for any domain consisting of the members of _!& and any collection of super-sets of members of 2,,, (assuming m > 1). The paper concludes with an example of a domain 23 that includes .Z4 and a choice correspondence on 2’ that satisfies Arrow’s choice axiom but which cannot be rationalized by an acyclic binary relation. The domain contains some two-element sets; if there were no admissible sets with fewer than four members we would have a counterexample to our theorem. If 2 is a binary relation on X we let z denote its asymmetric factor. If 2 is complete and transitive we say that it is a preorder. If Y is a subset of X, by the restriction of 2 to Y we mean the relation 1 O Y2 on Y. Let 4, denote the choice correspondence generated by 2. That is, 4,(Y) = {x E Y : y > x implies y j? Y}. The domain for 4, must be a subset of 2!(>) = {Y C x:~,(Y)#0}.w e say that choice correspondence 4 with domain 2’ has a transitive rationalization if there exists a preorder 2 on X such that 4 and 4 coincide on 2. 2 * Financial

support

from

edged. SSDI

0165-1765(93)00337-N

the National

Science

Foundation,

grants

no. SES 9007953 and 9209039,

is gratefully

acknowl-

382

D.E.

Arrow’s 4(Z) Arrow sense.

choice

axiom

(ACA).

Campbell

For

I Economics

any

Letters 44 (1994) 381-384

Y,Z E 3,

if Y C Z and

4(Z)

fl Y# 0, then

4(Y)

=

fl Y. (1959)

established

that ACA

is equivalent

to preorder-generated

choice

in the following

If

4 is a choice correspondence with a domain that includes all (non-empty) finite subsets of X, then 4 satisfies ACA if and only if it has a transitive rationalization.

Actually, it is not necessary to assume that the domain contains every finite set. If it contains &, then ACA will hold if and only if 4 has a transitive rationalization. This is easily seen to be a consequence of our theorem below. Hansson (1968) proved that ACA is equivalent to transitive rationalization on a wide family of restricted domains-specifically, domains 3 that are closed under finite union. ’ The closure assumption is unrealistic, however, because it denies the existence of an upper bound on the cardinality of a feasible set. 2 In applications, limits on computational capacity would impose an upper bound on the size of a member of 3. The hypothesis of our theorem is consistent with such an upper bound. The lower bound, m, on the size of a feasible set can also be motivated by computational considerations. It may be too costly to render a decision by partitioning the actual feasible set into tiny subsets, choosing from these small subsets and then progressing in steps towards the global choice, or best alternative. The possibility of a lower bound is part of the treatment of non-binary social choice by Grether and Plott (1982). Now, here is our theorem. Theorem. Let m 2 2 be a given integer, and suppose that 3 contains 3,,, but not any subsets of X with fewer than m members. A choice correspondence 4 on 3 satisfies ACA if and only if there is some preorder 1 on X such that 4, = 4 over 3. Furthermore, if 4,0 and 4 coincide over 3 for preorder )‘, then the restriction of > to X+ equals the restriction of 2’ to X+, with x > y and x >’ y both holding whenever x E X+, and y ,@X,. Proof. It will be convenient to let Z + x denote Z U {x}, to print Z - x instead of Z - {x}, and 4(Z) = x instead of 4(Z) = {x}, etc. (i) Suppose that the choice correspondence 4 is rationalized by 2. Then 4, = 4 on .3. Obviously, the choice correspondence generated by any preorder satisfies ACA on-any domain. Suppose that 2’ also rationalizes 4, but x > y and y >‘x with x E X4. There exists B E 9 such that x E 4(B). Choose some m-element subset A of B containing x. We have x E 4(A) by ACA. Then x E 4,0(A), because 4, = 4 = 4,0. Therefore, XL’ a for all a E A. Then y E 4,o(A + y) > and because wehave y >‘x. Theny E 4,(A) if 4? = 4? o , and this contradicts x > y. Therefore, _ 1’ coincide on X4. Obviously, x E X+, and y$X, implies x > y: choose any m-element subset Z of X such that x E 4(Z). We know that yeX,+, so yjE’4(Z + y), which implies x > y. (ii) Suppose that m is the smallest integer greater than 1 such that 3 contains Tm, and that 4 satisfies ACA. Define the binary relation 2 on X by setting x2 y if and only if (1) there is some Z E 3’ such that x E 4(Z) and y E Z, or

1 See Chapter 2 of Suzumura (1983) for related results, Hansson’s theorem. *For example, we should be able to stipulate that there elementary particles in the accessible universe.

and Theorem are no feasible

A(8)

on page

sets that

have

52 of that more

book

members

for a proof than

there

of are

D.E.

Campbell

I Economics

383

Letters 44 (1994) 381-384

(2) ra,.

First we prove that 2 is complete. If xj?!X+ or y $X+,, then either y )x or x > y holds by part (2) of the definition. Suppose that x and y belong to X+. We have x E d(Z) for some Z E 58’. Choose some m-element subset A of Z containing x. Then x E 4(A) by ACA. And +(A + y) = y or +(A + y) = +(A) or +(A + y) = +(A) + y by ACA. Then xe 4(A + y) implies y E +(A + y). Therefore, either x or y belongs to +(A + y). Then either x 2 y or y )x, and hence 2 is complete. To show that 2 is transitive, assume that we have x2 y )z. If z~X,, then x,z by definition. If z E X,, then we must have y E X+ and hence x E X4. Then by ACA there exist m-element subsets B and C of x such that x E 4(B), y E B n 4(C), and z E C. Suppose B = {x,

Y, a,, a29 . . . , a,, b,, b,,

. . . > b,J

and C= {z,y,a,,a,,

. . . ,a,,c,,c,,

. . . ?C,),

with {b,, b,, . . . , bk} n {cl, c2, . . . , ck} = 0. If z E B we have x,z Suppose z$B. Set B, = B and

by part (1) of the definition.

for f > 1, a subset of X with exactly m members. We know that x E 4(B,). following: zfx~+(B,)

or

4(B,)c{c,,c,,

. . . ,c~_~>, then xE~@,+,)

Now we prove the

or d@t+l)C{c~~~2~~-~

,CP).

If x E +(B,) and x$4(B, + c,), then +(Be + cc) = cp by ACA. If 4(B,) C {cl, c2,. . . , c~-~), then ACA implies that +(B, + ce) equals +(B,) or {c,} or +(B,) + ct. Therefore, either x belongs + ce) or else +(B,+, + ce) C {cl, c2,. . . , ct}. And B,,, = B, + c, -b,, so either x to do,+, belongs to $(B,+,) or +(B,+,) C (5, c2,. . . , ce} by ACA. By induction we have either x E C,, z}. The latter contradicts ACA and the fact that ~(B,+z) or 4(B,+z)C{c,,c,,..., y E 4(C). Therefore, x E +(Bk + z) and hence xlz. We have proved that 2 is a preorder. Now we show that 4, = 4 on 9. It is easy to prove that 4(Y) C 6,(Y) for all Y in 3’. If x E 4(Y), then by definition of 1 we haver x 1 y for all y E Y, and thusx E &_(Y). Finally, we prove 4, (Y) C 4(Y) for all Y in 3. Suppose x E 4, (Y) and Y E _5?.Then x E X,+,as we now show. If z$Xi for arbitrary Z E x, then a > z for all a Ejr, by part (2) of the definition. If X - X+, contains more than m - 1 alternatives, then it has a subset belonging to 3m C 3, a contradiction. Therefore, if Y contains m or more alternatives, including z, there must be some a E Y such that a > z. Therefore, x E &(Y) and YE 3 implies x E X+. Let Y be an arbitrary subset of X with exactly m members. Choose x E 4,(Y) and y E 4(Y). We have xl y and x E Y+ so x E 4(Z) for some Z E .3 containing y. ACA allows us to assume that Z has exactly m members, including y. Let Y = {yr, y,, . . . , y,} and Z = {zi, z2, . . . , zm} withy,=x=z, andy,=y=z,. Set Y,{z,, 22,.

f. > ze-1,

Y,, Ye+1,.

Then Y, = Y and Y,, 1 = Z. We {Ye, Ye+17. . . 7 y,}. Then either by an argument identical to the {Y3>Y4,. . . , y,} contradicts y E

. . 9 Y,)

.

have x E +(Y,+r). Suppose @(Y,) contains x or else +(Ye) C +(Y,_,) contains x or else +(I’_,) C {y,_,, ye, Y(+~, . . . , y,> one that we used to show that 2 is transitive. But +(Y3) C +(Y,). Therefore, x E 4(Y,) = 4(Y).

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