Two Factor Additive Conjoint Measurement With One
Solvable Component
Christophe Gonzales∗
LIP6. Université Paris 6
tour 46-0, 2ème étage
4, place Jussieu 75252 Paris Cedex 05, France
e-mail: [email protected]
fax: (33) 1-44-27-70-00
Proofs should be sent to:
Christophe GONZALES
See the address above.
∗
I am greatly indebted to Jean-Yves Jaffray, Peter Fishburn and Peter Wakker for their very useful
comments and suggestions.
1
Abstract
This paper addresses conditions for the existence of additive separable utilities.
It considers mainly two-dimensional Cartesian products in which restricted solvability holds w.r.t. one component, but some results are extended to n-dimensional
spaces. The main result shows that, in general, cancellation axioms of any order
are required to ensure additive representability. More precisely, a generic family of
counterexamples is provided, proving that the (m + 1)st-order cancellation axiom
cannot be derived from the mth order cancellation axiom when m is even. However, a special case is considered, in which the existence of additive representations
can be derived from the independence axiom alone. Unlike the classical representation theorems, these representations are not unique up to strictly positive affine
transformations, but follow Fishburn’s (1981) uniqueness property.
2
1
Introduction
Additive conjoint measurement has been extensively studied since 1960 and Debreu’s
celebrated paper. For instance, Jaffray (1974a, 1974b) give necessary and sufficient
conditions ensuring the existence of additive representations, which are closely related
to cancellation axioms (see Krantz et al. (1971, p.340–344)). Other necessary and
sufficient conditions can be found in Krantz et al. (1971, p.430) and in Fishburn (1992).
However, the lack of testability of high order cancellation axioms led researchers
to look for other axioms that were nonnecessary but easier to test. Thus, Luce and
Tukey (1964), Krantz (1964), Luce (1966), Fishburn (1970, p.58) and Krantz et al.
(1971, p.257) show sufficient conditions of an algebraic nature, as well as uniqueness
properties of the additive representations. Debreu (1960) and Wakker (1989, p.49) give
their counterpart in a topological framework. Fuhrken and Richter (1991) gave necessary and sufficient conditions for the existence of continuous additive representations on
connected topological spaces. All these papers assume that some structural nonnecessary conditions — called unrestricted or restricted solvability in the former approach and
connectedness of the topological spaces in the latter — hold w.r.t. every component of
the Cartesian product. But these assumptions are particularly restrictive because they
simply do not hold in many practical situations. For instance, in economics, the first
component may describe a waiting time and the second one the object waited for (in
this case, using an exponential transform, one gets a multiplicative model). In medical
decision making, the first component may refer to the duration of a treatment, and the
second one to the level of quality of life induced by it. Another example can be found
in Gonzales and Jaffray (1997), in which certainty effects make a restricted solvability
assumption unrealistic.
Therefore, it is of interest to study Cartesian products in which solvability or connectedness does not hold w.r.t. every component. Gonzales (1996a, 1996b) describes
testable conditions ensuring additive representability when unrestricted or restricted
solvability holds w.r.t. at least two components. But, to my knowledge, the only paper
dealing with spaces in which restricted solvability holds w.r.t. exactly one component is
Fishburn (1981), which studies the uniqueness property of additive utilities. In this paper, no conditions are mentioned ensuring the existence of such representations. Those
are addressed in the present paper. The choice of using solvability rather than connectedness was recommended by Wakker (1988).
In section 2, axioms needed in the paper are given. Section 3 presents a generic
family of counterexamples showing that, in general, cancellation axioms of any order
are needed to entail additive representability on two-dimensional Cartesian products
in which unrestricted solvability holds w.r.t. only one component. This can be seen
as a complement to Fishburn (1997a, 1997b, 1997c), which gives the same result for
finite sets, and as an extension of Scott and Suppes (1958). The result is trivially
extended to n-dimensional Cartesian products in which only one component satisfies
unrestricted solvability. Section 3 also studies the relationship between cancellation
axioms of different orders. In particular, it is proved that the second order cancellation
axiom is equivalent to the conjunction of independence and the Thomsen condition; this
result was already well known when restricted solvability holds w.r.t. all the components
(see e.g., Krantz et al. (1971, p.257)) or when unrestricted solvability holds w.r.t. at
3
least two components (see Gonzales (1996a)) or even when a component is finite and
the other one is binary (see Krantz et al. (1971, p.452)).
Section 4 describes how a “weak” structural assumption w.r.t. the second component,
namely equal spacedness, combined with restricted solvability w.r.t. the first component,
independence and an Archimedean axiom, implies additive representability. This is an
extension of Krantz et al. (1971, p.452) and Wakker (1991).
All proofs are given in the appendix.
2
The axiomatic system
Consider a Cartesian product X = X1 × X2 and a weak order % over X, i.e., % is
complete (for all x, y ∈ X, x % y or y % x) and transitive. Let x ∼ y ⇔ [x % y and
y % x], x y ⇔ [x % y and Not(y % x)], and x - y ⇔ y % x. % is representable
by an additive utility function u : X → R if there exist mappings u1 : X1 → R and
u2 : X2 → R such that:
u(x1 , x2 ) = u1 (x1 ) + u2 (x2 ) for all (x1 , x2 ) ∈ X and
(x1 , x2 ) % (y1 , y2 ) ⇔ u(x1 , x2 ) ≥ u(y1 , y2 ) for all (x1 , x2 ), (y1 , y2 ) ∈ X.
The following four axioms are well known to be necessary for additive representability
on any set X:
Axiom 1 (independence) for all x, y ∈ X, (x1 , x2 ) % (x1 , y2 ) ⇔ (y1 , x2 ) % (y1 , y2 )
and (x1 , x2 ) % (y1 , x2 ) ⇔ (x1 , y2 ) % (y1 , y2 ).
By axiom 1, weak orders %1 and %2 , induced by % respectively on X1 and X2 , can
be defined for all x1 , y1 ∈ X1 , x2 , y2 ∈ X2 as: x1 %1 y1 ⇔ [there exists c2 ∈ X2 such that
(x1 , c2 ) % (y1 , c2 )], and x2 %2 y2 ⇔ [there exists c1 ∈ X1 such that (c1 , x2 ) % (c1 , y2 )].
Axiom 2 (Thomsen condition) For every x1 , y1 , z1 ∈ X1 , x2 , y2 , z2 ∈ X2 , if (x1 , z2 )
∼ (z1 , y2 ) and (z1 , x2 ) ∼ (y1 , z2 ), then (y1 , y2 ) ∼ (x1 , x2 ).
The Thomsen condition can be generalized by substituting indifference relations ∼
by weak preference relations %, thus resulting in the following axiom:
Axiom 3 (Generalized Thomsen condition) For every x1 , y1 , z1 ∈ X1 , x2 , y2 , z2 ∈
X2 , if (x1 , z2 ) % (z1 , y2 ) and (z1 , x2 ) % (y1 , z2 ), then (y1 , y2 ) - (x1 , x2 ).
Axiom 4 ((C2 ) : second order cancellation axiom)Consider (xi1 , xi2 ) and (y1i , y2i ),
i ∈ {1, 2, 3}, such that (x1j , x2j , x3j ) is a permutation of (yj1 , yj2 , yj3 ) for j = 1, 2. If
(x11 , x12 ) % (y11 , y21 ) and (x21 , x22 ) % (y12 , y22 ), then (x31 , x32 ) - (y13 , y23 ).
The following two structural conditions are usually assumed w.r.t. every component
of X, but, in what follows, they will only be w.r.t. one component.
Axiom 5 (unrestricted solvability w.r.t. the first component) For every y ∈ X
and every x2 ∈ X2 , there exists x1 ∈ X1 such that y ∼ (x1 , x2 ).
4
Axiom 6 (restricted solvability w.r.t. the first component) For every y ∈ X, if
(a1 , x2 ) - y - (b1 , x2 ), then there exists x1 ∈ X1 such that y ∼ (x1 , x2 ).
Clearly, when unrestricted solvability holds, then restricted solvability holds as well,
so that representation theorems involving restricted solvability are more general than
the ones involving unrestricted solvability. These two axioms are structural conditions
because: i) they are nonnecessary for additive representability; and ii) they strengthen
axioms 1 through 4 to the point that they induce additive representability.
Throughout, when f and g are functions, f ◦ g(·) = f (g(·)); if, moreover, p is a
strictly positive integer, f p = f ◦ f ◦ · · · ◦ f (p times), and f −1 denotes the inverse
function of f , i.e., f ◦ f −1 is the identity function.
3
Counterexamples
3.1
Independence and the Thomsen conditions
Lemma 1 Let % be a weak order on a two-dimensional Cartesian product X. Then
the independence axiom (axiom 1)
the second order
+
⇔
the generalized Thomsen condition (axiom 3)
cancellation axiom (C2 ).
This lemma is rather general since, unlike Krantz et al. (1971), it requires no structural condition such as finiteness of Xi or solvability w.r.t. X1 . However, when the latter
holds, the generalized Thomsen condition can be substituted by the classical Thomsen
condition, hence the following lemma:
Lemma 2 Let % be a weak order on a two-dimensional Cartesian product X satisfying
unrestricted solvability w.r.t. the first component (axiom 5). Then
the independence axiom (axiom 1)
the second order
+
⇔
the Thomsen condition (axiom 2)
cancellation axiom (C2 ).
This lemma is a slight generalization of property P1* ⇒ P1 of theorem 5.3 in
Fishburn (1970) (because, in lemma 2, solvability is not required w.r.t. the second
component). However, the proof I present here is completely different from the one
written by Fishburn.
It is a classical result that in two-dimensional Cartesian products, the second order
cancellation axiom is not implied by independence alone: one must add the Thomsen
condition to get this implication. A well known counterexample (see e.g., Wakker (1989,
p.71)1 ) is the following one: X = R × R and % is defined by the utility function
U (x1 , x2 ) = x1 + x2 + min{x1 , x2 }.
1
in this example, the Thomsen condition is substituted by the hexagon condition, which is somewhat less restrictive. But according to Karni and Safra (1997), under restricted solvability w.r.t. both
components and independence, the Thomsen condition is equivalent to the hexagon condition.
5
3.2
Relations between different cancellation axioms
In this subsection, a new condition, called a symmetric cancellation axiom and denoted
(Sm+1 ), is introduced. This is a particular symmetric case of the classical cancellation
axiom (Cm+1 ) (defined below). As we shall see, it fills the gap between (Cm ) and (Cm+1 )
when unrestricted solvability holds w.r.t. one component of X.
Axiom 7 ((Cm ) : mth order cancellation axiom) Let m ≥ 2. Consider m+1 elements (xi1 , xi2 ) of X, i = 1, . . . , m+1, and m+1 elements (y1i , y2i ) such that (y11 , . . . , y1m+1 )
and (y21 , . . . , y2m+1 ) are permutations of (x11 , . . . , xm+1
) and (x12 , . . . , xm+1
) respectively.
1
2
m+1 m+1
i
i
i
i
If (x1 , x2 ) % (y1 , y2 ) for all i = 1, 2, . . . , m, then (x1 , x2 ) - (y1m+1 , y2m+1 ).
Note that the order of the cancellation axiom follows the notation of Krantz et al.
(1971, p.343) (Fishburn (1970) and Jaffray (1974a) would have called it the (m + 1)st
order cancellation axiom).
Axiom 8 ((Sm+1 ) : (m + 1)st order symmetric cancellation axiom) Suppose
∈ X1 ,
that m is an even number. Let k = (m + 2)/2. Consider elements x11 , . . . , xm+2
1
j
k
1
k
i
1
x2 , . . . , x2 ∈ X2 and y2 , . . . , y2 ∈ X2 such that, for all i ≤ k and j > k, x1 6= x1 and for
all i, j < k, xi2 6= y2j . Then:


(x11 , x12 )
%
(xk1 , y21 )

(x21 , x22 )
%
(x11 , y22 ) 


3
3

(x1 , x2 )
%
(x21 , y23 ) 


..
..
..




.
.
.


k−1 k 
k , xk )

%
(x
,
y
)
(x
1
2
2
1
, xk2 ).
, y2k ) - (xm+1
 ⇒ (xm+2

1
1


 (xk+1 , y 1 )
, x12 ) 
% (xm+2
2
1
1


k+1 2 
 (xk+2 , y 2 )
,
x
)
%
(x

2 
2
1
1


..
..
..


.
.
.
k−1
m+1 k−1
m
(x1 , y2 ) % (x1 , x2 )
Note the symmetry between the left hand side and the right hand side of preference
relations %, and between the elements above the horizontal line and those under the
line. Note also that when the %’s are substituted by ∼’s, the third order symmetric
cancellation axiom corresponds to the hexagon condition (see Wakker (1989, p.47–48)).
Theorem 1 Let % be a weak order on X = X1 × X2 . Assume that unrestricted solvability
w.r.t. the first component (axiom 5) holds. Then, for any m ≥ 2,
if m is odd, then (Cm ) ⇔ (Cm+1 )
if m is even, then (Cm ) + (Sm+1 ) ⇔ (Cm+1 ).
3.3
Second and third order cancellation axioms
In this subsection, an example is given which shows that, when unrestricted solvability
holds w.r.t. only one component, (C2 ) may hold without (C3 ) holding. Suppose that %
is represented on X = R × {0, 2, 4, 6} by the following utility function:
x1 + x2
if x2 ≤ 4,
U (x1 , x2 ) =
(1)
0.5(x1 mod 2)2 + 2bx1 /2c + 6.5 if x2 = 6.
6
Remark that for a fixed value x2 of X2 , U (x, x2 ) — henceforth denoted Ux2 (x) —
is a continuous function on R. Hence, working in X1 × R, where R represents the set of
values that can be taken by U (x1 , x2 ) over X, is very attractive because in this space
properties of % can be easily visualized (see figure 1).
Now, the problem is to translate the definition of independence and the Thomsen
condition into this new space. (x1 , x2 ) % (y1 , x2 ) ⇔ (x1 , y2 ) % (y1 , y2 ) is equivalent to
Ux2 (x1 ) ≥ Ux2 (y1 ) ⇔ Uy2 (x1 ) ≥ Uy2 (y1 ). This condition clearly holds since functions
Ux (·) are strictly increasing for every x ∈ X2 . (x1 , x2 ) % (x1 , y2 ) ⇔ (y1 , x2 ) % (y1 , y2 )
is equivalent to Ux2 (x1 ) ≥ Uy2 (x1 ) ⇔ Ux2 (y1 ) ≥ Uy2 (y1 ). So, since we consider only
continuous functions, it comes to the non intersection of the graph of Ux2 by that of
Uy2 . This condition also holds because 0.5(x mod 2)2 + 2bx/2c + 6.5 > x + 4 for every
x ∈ R. Therefore %, as represented by (1), satisfies the independence axiom.
Let us now translate the Thomsen condition. Figure 1 helps in understanding the
process: [(x11 , x22 ) ∼ (x21 , x12 ) and (x11 , x32 ) ∼ (x31 , x12 )] ⇒ (x21 , x32 ) ∼ (x31 , x22 ). So, in terms
X2
x32
Ux32 (x1 )
R
C
E
E
Ux22 (x1 )
F
Ux12 (x1 )
C
x22
A
F
B
x12
x11
D
D
x21
x31
B
A
x21
x11
X1
x31
X1
Figure 1: Translation of the Thomsen condition
of utility functions, Ux22 (x11 ) = Ux12 (x21 ), Ux32 (x11 ) = Ux12 (x31 ) and Ux32 (x21 ) = Ux22 (x31 ). By
−1
1
1
3
the first equality, x21 = Ux−1
1 ◦ Ux2 (x1 ), and by the second equality, x1 = Ux1 ◦ Ux3 (x1 ).
2
2
2
2
−1
1
1
Hence, the third equality can be written as Ux32 ◦ Ux−1
1 ◦ Ux2 (x1 ) = Ux2 ◦ Ux1 ◦ Ux3 (x1 ).
2
2
2
2
2
But this must be true for all x11 . Hence the Thomsen condition can be written as:
−1
for all x12 , x22 , x32 ∈ X2 and all x ∈ R, Ux32 ◦ Ux−1
1 ◦ Ux2 (x) = Ux2 ◦ Ux1 ◦ Ux3 (x).
2
2
2
2
2
Now, to show that %, as represented by (1), satisfies the Thomsen condition, let me
introduce the following lemma:
Lemma 3 Let % be a weak order on a X = X1 × X2 . Assume unrestricted solvability
w.r.t. the first component (axiom 5) and independence (axiom 1). Suppose that
for
x12 -2 x22 -2 x32 and every x11 , x21 ,x31 ∈ X1 ,
1every
(x1 , x22 ) ∼ (x21 , x12 ) and (x11 , x32 ) ∼ (x31 , x12 ) ⇒ (x21 , x32 ) ∼ (x31 , x22 ).
Then % satisfies the Thomsen condition (axiom 2).
7
(2)
Thus, to prove that % satisfies the Thomsen condition, it is sufficient to prove that:
−1
for all x12 -2 x22 -2 x32 and all x ∈ X1 , Ux32 ◦ Ux−1
1 ◦ Ux2 (x) = Ux2 ◦ Ux1 ◦ Ux3 (x). (3)
2
2
2
2
2
Suppose that x32 6= 6. Then (3) is obviously equivalent to ((x + x22 ) − x12 ) + x32 =
((x + x32 ) − x12 ) + x22 ; this relation is clearly satisfied by U , as defined by (1). Now
suppose that x32 = 6; then if x22 = 6, (3) trivially holds, otherwise (3) is equivalent
to: U6 (x + x22 − x12 ) = U6 (x) + x22 − x12 . Since x22 − x12 are multiples of 2, this can be
summarized as: U6 (x + 2) = U6 (x) + 2. By (1), U obviously satisfies this condition.
Hence the Thomsen condition holds for % represented by U . By subsection 3.1, we can
conclude that the second order cancellation axiom holds.
However, % cannot be represented by an additive utility function because the third
order cancellation axiom does not hold. Indeed, U (1, 6) = U (3, 4) = 7, U (3, 0) =
U (1, 2) = 3, U (1.5, 2) = U (3.5, 0) = 3.5 and U (3.5, 4) = 7.5 < U (1.5, 6) = 7.625.
3.4
Cancellation axioms of order m + 1 and m + 2
In this subsection we shall see an example proving that the (m + 1)st order cancellation axiom does not imply the (m + 2)nd order cancellation axiom for m odd when
unrestricted solvability holds w.r.t. only one component. Consequently, no cancellation axiom is sufficient to ensure additive representability. Let % be a weak order on
X = R×{0, 2, 4, . . . , 2m}, with m ≥ 3 odd, represented by the following utility function:
x1 + x2 − m2 − 2m + 2.5 if x2 < 2m,
(4)
U (x1 , x2 ) =
0.5(x1 mod 2)2 + 2bx1 /2c if x2 = 2m.
Then the following lemma holds:
Lemma 4 (Ci ) holds for all i ∈ {0, 1, . . . , m + 1}.
However, (Cm+2 ) cannot hold because:

(2(m − 1) + 1, 0)

(4(m − 1) + 1, 0)


..

.

 ((m + 1)(m − 1) + 1, 0)


(1, 2m)


(0,
2(m − 1))


(2(m − 1), 2(m − 1))


..

.

 ((m − 1)(m − 1), 2(m − 1))
∼
∼
..
.
∼
∼
∼
∼
..
.
∼
((m + 1)(m − 1), 2(m − 1)) ≺
3.5
(1, 2(m − 1))
(2(m − 1) + 1, 2(m − 1))
..
.






((m − 1)(m − 1) + 1, 2(m − 1)) 

((m + 1)(m − 1) + 1, 2(m − 1)) 
.

(2(m − 1), 0)


(4(m − 1), 0)


..

.


((m + 1)(m − 1), 0)
(0, 2m)
An extension to n-dimensional spaces
The preceding results can be extended rather straightforwardly to n-dimensional Cartesian products. Thus, the mth-order cancellation axiom, the (m + 1)st-order symmetric
cancellation axiom and unrestricted solvability w.r.t. the first component can be extended respectively as follows:
8
Axiom 9 ((Cm ) for n-dimensional spaces) Let m ≥ 2. Let (xi1 , . . . , xin ) and
(y1i , . . . , yni ), i = 1, . . . , m + 1, be some elements of X such that (yj1 , . . . , yjm+1 ) are
permutations of (x1j , . . . , xm+1
) for all j ∈ {1, . . . , n}. If (xi1 , . . . , xin ) % (y1i , . . . , yni ) for
j
all i = 1, 2, . . . , m, then (xm+1
, . . . , xm+1
) - (y1m+1 , . . . , ynm+1 ).
n
1
Axiom 10 ((Sm+1 ) for n-dimensional spaces) Let m be an even number and let
k = (m + 2)/2. Let x11 , . . . , xm+2
∈ X1 , and x1i , . . . , xm+2
, yi1 , . . . , yim+2 ∈ Xi , i =
1
i
2, . . . , n, be such that:
• for all j ≤ k and all p > k, xj1 6= xp1 ;
• for all i ∈ {2, . . . , n}, there exists an index hi ∈ {0, . . . , k} such that
p=j
if j ≤ hi or if k < j ≤ k + hi ;
j
p
xi = yi ⇔
p = j + k if hi < j ≤ k;
• at least one index hi is equal to 0.
Then:

(x11 , x12 , . . . , x1n )
(x21 , x22 , . . . , x2n )
(x31 , x32 , . . . , x3n )
..
.
%
%
%
..
.
(xk1 , y21 , . . . , yn1 )
(x11 , y22 , . . . , yn2 )
(x21 , y23 , . . . , yn3 )
..
.








k−1 k
k

%
(x1 , y2 , . . . , yn )


m+2 k+1
k+1
% (x1 , y2 , . . . , yn ) 

k+2
, . . . , ynk+2 ) 
% (xk+1

1 , y2

..
..

.
.
m+1
, . . . , ynm+1 )
, . . . , xm+1
) % (xm
, xm+1
(xm+1
n
1 , y2
2
1








(xk1 , xk2 , . . . , xkn )


 (xk+1 , xk+1 , . . . , xk+1 )
n
2
1

 (xk+2 , xk+2 , . . . , xk+2 )

n
2
1

..

.
, . . . , xm+2
) - (xm+1
, y2m+2 , . . . , ynm+2 ).
, xm+2
⇒ (xm+2
n
2
1
1
Axiom 11 (unrestricted solvability w.r.t. the first component) For all y ∈ X
and all xj ∈ Xj , j = 2, . . . , n, there exists x1 ∈ X1 such that y ∼ (x1 , x2 , . . . , xn ).
Theorem 1 is extended by the following theorem:
Q
Theorem 2 Let % be a weak order on X = ni=1 Xi and assume unrestricted solvability
w.r.t.
the first component (axiom 11). Then, for any m ≥ 2,
if m is odd, then (Cm ) ⇔ (Cm+1 )
if m is even, then (Cm ) + (Sm+1 ) ⇔ (Cm+1 ).
The generic family of counterexamples of subsection 3.4 can also be extended to
n-dimensional Cartesian products, thus showing that, in general, no cancellation axiom
is sufficient to ensure additive representability:
9
Let m be any odd number greater than or equal to 3. Let % be a weak order on
X = R × {0, 2, 4, . . . , 2m} × {0, 2}n−2 represented by the following utility function:

n
X



xi if x2 < 2m,

 H + x1 + x2 +
i=3
U (x1 , . . . , xn ) =
(5)
n
X


 V (x1 ) +
xi
if x2 = 2m.


i=3
where
H = −m2 − mn − 3n + 8.5,
V (x1 ) = .5(x1 mod 2)2 + 2bx1 /2c.
Then, the following lemma holds:
Lemma 5 (Ci ) holds for all i ∈ {0, 1, . . . , m + 1}.
However, (Cm+2 ) cannot hold because:

(1 + 2(m + n − 3), 0, 0, . . . , 0)

(1
+ 4(m + n − 3), 0, 0, . . . , 0)


..

.

 (1 + (m + 1)(m + n − 3), 0, 0, . . . , 0)


(1, 2m, 0, . . . , 0)


(0,
2(m − 1), 2, . . . , 2)

 (2(m + n − 3), 2(m − 1), 2, . . . , 2)


..

.

((m − 1)(m + n − 3), 2(m − 1), 2, . . . , 2)
∼
∼
..
.
∼
∼
∼
∼
..
.
∼
((m + 1)(m + n − 3), 2(m − 1), 2, . . . , 2) ≺
4
(1, 2(m − 1), 2, . . . , 2)
(1 + 2(m + n − 3), 2(m − 1), 2, . . . , 2)
..
.





(1 + (m − 1)(m + n − 3), 2(m − 1), 2, . . . , 2)

(1 + (m + 1)(m + n − 3), 2(m − 1), 2, . . . , 2)
.

(2(m + n − 3), 0, 0, . . . , 0)


(4(m + n − 3), 0, 0, . . . , 0)


..

.


((m + 1)(m + n − 3), 0, 0, . . . , 0)
(0, 2m, 0, . . . , 0)
Representation theorem
The preceding section provided arguments showing that additive representability required cancellation axioms of every order when the only structural assumption made is
solvability w.r.t. one component. This section describes how this can be circumvented
by adding a “weak” assumption w.r.t. the second component, namely equal spacedness.
To provide results as general as possible, restricted solvability is assumed w.r.t. the first
component of X (instead of unrestricted solvability, as in the preceding section). Moreover, X2 is assumed to be a sequence, (xi2 )i∈N , where N is a set of consecutive integers
(positive or negative, finite or infinite) such that Card(N ) ≥ 3. For convenience, assume
that xi+1
2 xi2 , for all i, i + 1 ∈ N .
2
Cancellation axioms, though necessary, are not generally sufficient2 to ensure additive representability: the latter still requires an Archimedean property. Classically, in
additive conjoint measurement, this is stated in terms of standard sequences:
2
When X is a finite set, the set of all cancellation axioms is sufficient to ensure additive representability; see Scott (1964) or Adams (1965).
10

Definition 1 (standard sequence w.r.t. the first component) For any set M of
consecutive integers, a set {xk1 ∈ X1 , k ∈ M } is a standard sequence w.r.t. the first
0
component if there exist x02 , x12 ∈ X2 such that Not((x01 , x02 ) ∼ (x01 , x12 )) and (xk+1
1 , x2 ) ∼
(xk1 , x12 ) for all k, k + 1 ∈ M . {x02 , x12 } is called the mesh of the standard sequence.
Axiom 12 (Archimedean axiom w.r.t. the first component) Any bounded standard sequence w.r.t. the first component is finite, i.e., if {xk1 ∈ X1 , k ∈ M } is a standard
sequence with mesh {x02 , x12 } and if there exist y, z ∈ X such that y - (xk1 , x02 ) - z for
all k ∈ M , then M is finite.
Obviously independence and the Archimedean axiom are required. However they
are still not sufficient. Indeed, assume that X = C×{0, 1}, where C is the set of complex
numbers. Let % on X be such that
• for all x1 , y1 ∈ C, (x1 , 0) ≺ (y1 , 1),
• for all x1 + iy1 , z1 + it1 ∈ C and all x2 ∈ {0, 1}, (x1 + iy1 , x2 ) % (z1 + it1 , x2 ) ⇔
x1 ≥ z1 or (x1 = z1 and y1 ≥ t1 ).
Then % is not even representable by a utility function, it is not a fortiori representable
by an additive utility function. Hence, we will also require that % be representable
by a utility function on X. This set of axioms is not yet sufficient to ensure additive
representability because the structure of the second component is too poor. Hence we
also require that X2 be equally spaced and the lemma below follows.
Definition 2 X2 is equally spaced if, for all x1 , y1 ∈ X1 and all i, i + 1, i + 2 ∈ N :
i+2
i+1
(x1 , xi2 ) ∼ (y1 , xi+1
2 ) ⇔ (x1 , x2 ) ∼ (y1 , x2 ).
(6)
Lemma 6 Let % be a weak order on X = X1 × X2 , where X2 = {xi2 , i ∈ N }. Assume
restricted solvability w.r.t. X1 (axiom 6), independence (axiom 1) and the Archimedean
2 xi2
axiom (axiom 12). Suppose that % is representable by a utility function, that xi+1
2
for all i, i + 1 ∈ N , and that X2 is equally spaced (definition 2). Then % is representable
by an additive utility function3 .
Lemma 6 can be viewed as a slight generalization of Wakker (1991): in addition
to equal spacedness4 w.r.t. every component of X, the latter indeed requires restricted
solvability w.r.t. every component.
Of course, the existence of a utility function representing % is not a very appealing
assumption. However, according to a well known theorem (see Debreu (1954, 1964) or
Krantz et al. (1971, p.40)), if there exists a set A ⊂ X, at most denumerable, such that
x ≺ y ⇒ there exist a, b ∈ A such that x - a ≺ b - y, then % is representable by a
3
This lemma is important because it can serve as a basis to prove the classical representation theorem
for two-dimensional Cartesian products — see Fishburn (1970, theorem 5.2, p58) — using neither
topology nor algebra. But this is beyond the scope of this paper and is explained in full detail in
Gonzales (1996b).
4
In fact, the definition of equal spacedness stated in Wakker (1991) is different from definition 2, but
together with restricted solvability and weak separability (which are assumed by Wakker (1991)) both
definitions are equivalent.
11
utility function. In the present framework, the assumption that there exists a utility
function representing % can be substituted by the following condition:
there exists a denumerable set A1 ⊂ X1 such that
x1 ≺1 y1 ⇒ there exist a1 , b1 ∈ A1 such that x1 -1 a1 ≺1 b1 -1 y1 .
As was mentioned in Wakker (1991), although equal spacedness “may have been
considered complicated because it is of a combinatorial nature”, representation theorems
are in fact simpler than in the classical frameworks (in Krantz et al. (1971) and Fishburn
(1970), the Thomsen condition must be assumed to derive the existence of additive
utilities in two-dimensional Cartesian products).
5
Appendix: Proofs
Proof of lemma 1: It is straightforward to see that (C2 ) implies independence and
the generalized Thomsen condition. Let us prove the converse. (C2 ) can be written as:
(x1 , x2 ) % (x01 , x02 )
⇒ (z1 , z2 ) - (z10 , z20 ),
(7)
(y1 , y2 ) % (y10 , y20 )
where (x0i , yi0 , zi0 ) are permutations of (xi , yi , zi ) for i ∈ {1, 2}. By permutation of the
above left hand side (LHS) relations, it is not restrictive to assume that either x02 or y20
is equal to x2 . Assume that y20 = x2 . If y1 = x01 , then the right hand side (RHS) relation
is deduced from the generalized Thomsen condition. If y1 = y10 , then substituting y1
and y10 by x01 and applying the generalized Thomsen condition and then independence
gives the RHS relation. If y1 = z10 , then either x1 = x01 and, substituting x1 and x01 by y1
and applying the generalized Thomsen condition and independence, one gets the RHS
relation, or x1 = y10 and, by transitivity, one gets the RHS relation. A similar reasoning
holds when x02 = x2 . Hence, the conjunction of independence and the generalized
Thomsen condition implies the second order cancellation axiom.
Proof of lemma 2: By lemma 1, to prove lemma 2 it is sufficient to show that independence (axiom 1), unrestricted solvability w.r.t. the first component (axiom 5) and
the Thomsen condition (axiom 2) imply the generalized Thomsen condition (axiom 3).
Suppose that the generalized Thomsen condition does not hold. Then, there exist
x1 , y1 , z1 ∈ X1 and x2 , y2 , z2 ∈ X2 such that (x1 , z2 ) % (z1 , y2 ), (z1 , x2 ) % (y1 , z2 ) and
(y1 , y2 ) % (x1 , x2 ), with at least one strict preference relation. By unrestricted solvability w.r.t. the first component, there exists t1 ∈ X1 such that (x1 , z2 ) ∼ (t1 , y2 ), and by
transitivity, (t1 , y2 ) % (z1 , y2 ). Therefore, by independence, (t1 , x2 ) % (z1 , x2 ) % (y1 , z2 ).
Hence (x1 , z2 ) ∼ (t1 , y2 ), (t1 , x2 ) % (y1 , z2 ) and (y1 , y2 ) % (x1 , x2 ), with at least one strict
preference relation. Again, by unrestricted solvability w.r.t. the first component, there
exists s1 ∈ X1 such that (t1 , x2 ) ∼ (s1 , z2 ) and
(x1 , z2 ) ∼ (t1 , y2 ),
(t1 , x2 ) ∼ (s1 , z2 )
and
(s1 , y2 ) (x1 , x2 ).
But the above implication is impossible because it contradicts the Thomsen condition
(axiom 2). Hence, by contradiction, it entails lemma 2.
12
Proof of lemma 3: Assume that (2) holds. Let x11 , x21 , x31 ∈ X1 and x12 , x22 , x32 ∈ X2
be arbitrary elements such that (x11 , x22 ) ∼ (x21 , x12 ) and (x11 , x32 ) ∼ (x31 , x12 ). By symmetry,
it is not restrictive to assume that x22 -2 x32 .
Let us show that (x21 , x32 ) ∼ (x31 , x22 ). If x12 -2 x22 -2 x32 , then, by (2), (x21 , x32 ) ∼
3
(x1 , x22 ). Now assume that either x22 -2 x32 -2 x12 or x22 -2 x12 -2 x32 , and that
Not((x21 , x32 ) ∼ (x31 , x22 )).
(8)
By unrestricted solvability w.r.t. the first component, there exists g 1 ∈ X1 such that
∼ (x21 , x32 ) and, by (8),
(g 1 , x22 )
Not((g 1 , x22 ) ∼ (x31 , x22 )).
(9)
But, then, (x21 , x32 ) ∼ (g 1 , x22 ), (x21 , x12 ) ∼ (x11 , x22 ) and either x22 -2 x32 -2 x12 or x22 -2
x12 -2 x32 . Hence, by (2), (g 1 , x12 ) ∼ (x11 , x32 ), and so (g 1 , x12 ) ∼ (x31 , x12 ). But this is
impossible according to (9) and independence. Hence (8) cannot hold.
Proof of theorems 1 and 2: Theorem 1 is proved when fixing n = 2. (Cm+1 ) ⇒
(Cm ) is a classical result and since (Sm+1 ) is a subset of (Cm+1 ), the implication from
right to left is obvious. The proof of the converse is adapted from a proof given in
Jaffray (1974a) and Jaffray (1992). We first note that (Cm ) can be stated as follows:
Condition (Cm ): Suppose that, for all j ∈ {1, . . . , n}, (x1j , . . . , xm+1
) is a permutaj
m+1
1
i
i
i
i
tion of (yj , . . . , yj ). Then [(x1 , . . . , xn ) % (y1 , . . . , yn ) for all i ∈ {1, . . . , m + 1}] ⇒
[(xi1 , . . . , xin ) ∼ (y1i , . . . , yni ) for all i ∈ {1, . . . , m + 1}].
Hereafter we will use this definition instead of axiom 9 in subsection 3.5. Suppose that
(Cm ) holds—as well as (Sm+1 ) if m is even—and consider the following sequence of
preferences:
[(xi1 , . . . , xin ) % (y1i , . . . , yni ) for all i ∈ {1, . . . , m + 2}].
(10)
First case: there exists i0 and j0 such that xi0 = y j0 :
Substitute xi0 % y i0 and xj0 % y j0 by xi0 % y j0 and xj0 % y i0 . Since xi0 = y j0 , by
removing the line xi0 % y j0 , the conditions of application of (Cm ) obtain, and xj0 ∼ y i0
and xi ∼ y i for all i 6= i0 , j0 . But xj0 % y j0 = xi0 % y i0 , hence xi0 ∼ y i0 and xj0 ∼ y j0 .
Therefore (Cm+1 ) holds.
If the previous case does not obtain, change the order of the preference relations so
as to get cycles w.r.t. the first component, i.e., a sequence of the form:
(x11 , x12 , . . . , x1n ) % (xk1 , y21 , . . . , yn1 ),
i
i
(xi1 , xi2 , . . . , xin ) % (xi−1
1 , y2 , . . . , yn )
for all i ∈ {2, . . . , k}.
(11)
Second case: k = m + 2:
We know that there exists i2 such that y2i2 = x12 . If i2 6= 1 then, by unrestricted
solvability w.r.t. X1 , there exists g i2 −1 (2) such that (g i2 −1 (2), y2i2 −1 , y3i2 , . . . , yni2 ) ∼
(xi12 −1 , y2i2 , . . . , yni2 ). But then, by (C2 ), the following holds:
13
(g i2 −1 (2), y2i2 −1 , y3i2 , . . . , yni2 ) ∼ (xi12 −1 , y2i2 , . . . , yni2 )
(xi12 −1 , xi22 −1 , xi32 −1 , . . . , xin2 −1 ) % (xi12 −2 , y2i2 −1 , . . . , yni2 −1 )
|
{z
}
⇓
(xi12 −2 , y2i2 , y3i2 −1 , . . . , yni2 −1 ) - (g i2 −1 (2), xi22 −1 , . . . , xin2 −1 ).
(12)
So, the lines corresponding to indices i2 − 1 and i2 in (11) can be substituted by
(g i2 −1 (2), xi22 −1 , xi32 −1 , . . . , xin2 −1 ) % (xi12 −2 , y2i2 , y3i2 −1 , . . . , yni2 −1 )
(xi12 , xi22 , xi32 , . . . , xin2 ) % (g i2 −1 (2), y2i2 −1 , y3i2 , . . . , yni2 ).
(13)
Repeat the process with indices i2 − 1, i2 − 2, . . . , 1, so that y2i2 moves upward till it
reaches the first preference relation. Thus we get a cycle of the following form:
, x12 , y31 , . . . , yn1 ),
(g 1 (2), x12 , . . . , x1n ) % (xm+2
1
σ (i)
(g i (2), xi2 . . . , xin ) % (g i−1 (2), y2 2 , y3i , . . . , yni )
for all i ∈ {2, . . . , m + 2},
where σ2 is the permutation such that σ2 (i) = i − 1 for all i ≤ i2 and σ2 (i) = i for
all i > i2 . Similarly, there exists i3 such that y3i3 = x13 . Repeat with the above cycle
the same process, i.e., find elements g i (3) to move y3i3 upward to the first preference
relation. By induction, repeat the process for all j ≥ 4, hence resulting in the cycle:
, x12 , x13 , . . . , x1n ),
(g 1 (n), x12 , . . . , x1n ) % (xm+2
1
(14)
σ (i) σ (i)
σ (i)
(g i (n), xi2 . . . , xin ) % (g i−1 (n), y2 2 , y3 3 , . . . , ynn ) for all i ∈ {2, . . . , m + 2},
where σj (i) = i − 1 for all i ≤ ij and σj (i) = i for all i > ij . The formulas for modifying
the (r − 1)th and the rth lines of the pth cycle constructed are as follows: (i) before the
modification, the cycle is
14
(g 1 (p − 1), x12 , . . . , x1n ) % (xm+2
, x12 , . . . , x1p−1 , yp1 , . . . , yn1 ),
1
σ
σ (i)
(i)
p−1
, ypi , . . . , yni ) for all i ≤ r − 1,
(g i (p − 1), xi2 , . . . , xin ) % (g i−1 (p − 1), y2 2 , . . . , yp−1
σ (r)
(g r (p), xr2 , . . . , xrn ) % (g r−1 (p − 1), y2 2
σ
p−1
, . . . , yp−1
(r)
i
r , . . . , y r ),
, ypp , yp+1
n
σ (i)
σ (i)
i
(g i (p), xi2 , . . . , xin ) % (g i−1 (p), y2 2 , . . . , yp p , yp+1
, . . . , yni ) for all i > r.
σ (r)
(ii) g r−1 (p) is defined by (g r−1 (p), y2 2
σ (r)
y2 2
σ
p−1
, . . . , yp−1
σ (r)
(g r−1 (p), y2 2
(r)
σ
p−1
, . . . , yp−1
(r)
r , . . . , y r ) ∼ (g r−1 (p − 1),
, ypr−1 , yp+1
n
i
r , . . . , y r ); (iii) formula (13) is generalized by:
, ypp , yp+1
n
σ
p−1
, . . . , yp−1
(r)
σ (r)
r , . . . , y r ) ∼ (g r−1 (p − 1), y 2
, ypr−1 , yp+1
n
2
σ
p−1
, . . . , yp−1
σ
σ (r−1)
(r)
i
r , . . . , yr )
, ypp , yp+1
n
(r−1)
p−1
r−1
r−2 (p − 1), y 2
, ypr−1 , . . . , ynr−1 )
(g r−1 (p − 1), xr−1
, . . . , yp−1
2 , . . . , xn ) % (g
2
{z
⇓
σp−1 (r−1) ip r−1
σ2 (r−1)
r−1
r−2
(g (p − 1), y2
, . . . , yp−1
, yp , yp+1 , . . . , ynr−1 ) - (g r−1 (p), xr−1
2 , . . . , xn ).
|
}
and (iv) the (r − 1)th and rth lines of the cycle are substituted by:
σ (r−1)
r−2 (p − 1), y 2
r−1
(g r−1 (p), xr−1
2
2 , . . . , xn ) % (g
σ (r)
(g r (p), xr2 , . . . , xrn ) % (g r−1 (p), y2 2
σ
p−1
, . . . , yp−1
σ
p−1
, . . . , yp−1
(r)
(r−1)
i
r−1
, . . . , ynr−1 ),
, ypp , yp+1
r , . . . , y r ).
, ypr−1 , yp+1
n
,...,
Now, by independence, the first line of (14) can be substituted by (g 1 (n), xm+2
2
m+2 m+2
m+2
m+2
xn ) % (x1 , x2 , . . . , xn ). Then the first case of this proof can be applied and
%’s can be replaced by ∼’s in (14). Now to propagate indifference relations back to the
initial cycle (11), consider the converse of (12), namely, for the (r − 1)th and the rth
lines of the cycle generated by the g (·) (p), the following implications:
σ (r−1)
r−2 (p − 1), y 2
(g r−1 (p), x2r−1 , . . . , xr−1
n ) ∼ (g
2
σ (r)
(g r−1 (p − 1), y2 2
|
σ
(r)
σ
(r)
σ
p−1
, . . . , yp−1
(r−1)
i
σ (r)
i
σ (r)
i
r−1
, ypp , yp+1
, . . . , ynr−1 )
σ
(r)
σ
(r)
r , . . . , y r ) ∼ (g r−1 (p), y 2 , . . . , y p−1 , y r−1 , y r
r
, ypp , yp+1
n
p
p+1 . . . , yn )
2
p−1
{z
}
⇓
σp−1 (r−1) r−1
σ (r−1)
r−1
(g r−2 (p − 1), y2 2
, . . . , yp−1
, yp , . . . , ynr−1 ) ∼ (g r−1 (p − 1), xr−1
2 , . . . , xn ).
p−1
, . . . , yp−1
and
σ (r)
(g r−1 (p − 1), y2 2
p−1
, . . . , yp−1
r , . . . , y r ) ∼ (g r−1 (p), y 2
, ypp , yp+1
n
2
p−1
, . . . , yp−1
r , . . . , yr )
, ypr−1 , yp+1
n
σp−1 (r) r−1 r
σ (r)
(g r−1 (p), y2 2 , . . . , yp−1
, yp , yp+1 , . . . , ynr )
|
∼ (g r (p), xr2 , . . . , xrn )
{z
⇓
σp−1 (r) ip r
σ (r)
r
r
r
r−1
(g (p), x2 , . . . , xn ) ∼ (g (p − 1), y2 2 , . . . , yp−1
, yp , yp+1 , . . . , ynr ).
But, then the (r − 1)th and the rth lines can be substituted back to:
σ (r−1)
r−2 (p − 1), y 2
(g r−1 (p − 1), x2r−1 , . . . , xr−1
n ) ∼ (g
2
σ (r)
(g r (p), xr2 , . . . , xrn ) ∼ (g r−1 (p − 1), y2 2
σ
p−1
, . . . , yp−1
σ
p−1
, . . . , yp−1
(r)
(r−1)
, ypr−1 , . . . , ynr−1 ),
i
r , . . . , y r ).
, ypp , yp+1
n
By induction this leads to replacing all the %’s in (11) by ∼’s. Hence (Cm+1 ) holds.
Third case: there exist exactly 2 cycles of length k = (m + 2)/2:
15
}
If there exist i, j such that i ≤ m+2
< j and xi1 = xj1 , then the second case can
2
be applied since both cycles can be aggregated to form one cycle of length m + 2. Of
course, xj1 appears in several lines, but the process described in the second case can still
be applied, hence ensuring that (Cm+1 ) holds.
Assume now that there exist no i, j such that i ≤ m+2
< j and xi1 = xj1 . Using
2
i
transformation (12), each element yj , i ≤ k, such that there exists p ≤ k such that
xpj = yji , can be moved in the list (11) such that xpj = yjp . A transformation similar
to (12) would enable to move xpj along the list instead of yji .
If for all j ∈ {2, . . . , n}, there exist i, p ≤ k such that xpj = yji , then transformations
described above enable to modify the initial list (11) so that x1j = yj2 for all j. Thus,
the first case of this proof can be applied and %’s can be substituted by ∼’s. Inverse
transformations entail that this result also holds for the initial list, so that (Cm+1 ) holds.
If there exists an index j ∈ {2, . . . , n} such that, for all i ≤ k, there exists no p ≤ k
such that xpj = yji , then we are in the conditions of the (m + 1)st order symmetric
cancellation axiom, and so (Cm+1 ) holds.
Fourth case: there exists a cycle of length k <
m+2
:
2
k+1 k+1
Suppose that y2k+1 = x12 and that the corresponding relation is (xk+1
1 , x2 , x3 , . . . ,
p 1 k+1
k+1
% (x1 , x2 , y3 , . . . , yn ). Then, by unrestricted solvability w.r.t. the first component, there exists g 1 ∈ X1 such that (g 1 , x12 , x13 , . . . , x1n ) ∼ (xp1 , x12 , y3k+1 , . . . , ynk+1 ).
Again, by unrestricted solvability w.r.t. the first component, there exist g 2 , . . . , g k such
i+1
i+1
that, for all i ∈ {1, . . . , k − 1}, (g i , y2i , y3i , . . . , yni ) ∼ (g i+1 , xi+1
2 , x3 , . . . , xn ) and
(C2k−1 ) implies that (g k , y2k , y3k , . . . , ynk ) - (g 1 , x12 , x13 , . . . , x1n ) ∼ (xp1 , x12 , y3k+1 , . . . , ynk+1 ) k+1 k+1
k+1
(xk+1
1 , x2 , x3 , . . . , xn ).
So one can substitute in the initial list the lines (g 1 , x12 , x13 , . . . , x1n ) % (g k , y2k , y3k , . . . , ynk )
p 1 k+1
k+1 k+1
k+1 k+1
k+1
k+1
, . . . , ynk+1 ) by (xk+1
and (xk+1
1 , x2 , x3 , . . . , xn )
1 , x2 , x3 , . . . , xn ) % (x1 , x2 , y3
p 1 k+1
k+1
k
k
k
k
1
1
1
1
% (g , y2 , y3 , . . . , yn ) and (g , x2 , x3 , . . . , xn ) ∼ (x1 , x2 , y3 , . . . , yn ).
Hence cycle (1 → k) — where (1 → k) denotes the list of relations located on rows
1 through k — and cycle (k + 1 → p) now form only one cycle 1 → p. One can go on
with the same process until either the initial list (11) is transformed into a list with a
cycle of length m + 2 (with some xj1 possibly in several lines), or is transformed into
a list containing two cycles of length m+2
(with some xj1 possibly in several relations).
2
The former corresponds to case 2 and so all the %’s can be substituted by ∼’s and one
can trace the ∼’s back to the initial list. The latter case can be split into two cases:
either no cycle contains several lines with the same xj1 , and we are in the conditions of
the third case, or at least one cycle contains several identical xj1 ’s.
So now we have two cycles of length m+2
2 , i.e., cycles (1 → k) and (k + 1 → m + 2),
of which at least one contains several identical xj1 ’s. Consider that the latter is cycle
(1 → k). If the identical xj1 ’s are not on a same line, i.e., the cycle (1 → k) can be split
into several cycles of length strictly shorter than m+2
2 , then the process described above
can be used to aggregate one of these small cycles with the second cycle of length m+2
2 ,
hence creating a new cycle of overall length m + 2; then using case 2, we can deduce
that (Cm+1 ) holds. If all the identical xj1 ’s are on the same lines, i.e., on some lines like
xk+1
n )
(g i , xp2 , xp3 , . . . , xpn ) % (g i , y2h , y3h , . . . , ynh ),
16
(15)
then, removing these lines from the cycle 1 → k, we get a cycle of length < m+2
2 , that
we can aggregate with cycle k + 1 → m + 2. Then, using independence, lines (15)
can be substituted by (xm+2
, xp2 , xp3 , . . . , xpn ) % (xm+2
, y2h , y3h , . . . , ynh ). But then we have
1
1
created a cycle of length m + 2 and the second case ensures that (Cm+1 ) holds.
Proof of lemmas 4 and 5: Lemma 4 is proved by assigning n = 2 in the rest of the
proof. The proof is organized in three steps: first, it is shown that independence holds;
then the second order cancellation axiom is shown to hold, and, finally, using theorem 2,
the other cancellation axioms are studied.
First step: independence holds
Clearly, by definition of U , independence holds w.r.t. components 3 to n. Independence also holds w.r.t. the second component: suppose that (x1 , 2m, x3 , . . . , xn ) %
(y1 , 2m, y3 , . . . , yn ). Then, by unrestricted solvability w.r.t. the first component, there
exists g1 such that (g1 , 2m, x3 , . . . , xn ) ∼ (y1 , 2m, y3 , . . . , yn ) or, equivalently:
0.5 (g1 mod 2)2 + 2bg1 /2c +
n
X
x3 = 0.5 (y1 mod 2)2 + 2by1 /2c +
i=3
n
X
y3
i=3
and so 0.5 (g1 mod 2)2 − 0.5 (y1 mod 2)2 = −2bg1 /2c + 2by1 /2c +
n
X
(y3 − x3 ).
i=3
The right hand side being a multiple of 2, it is clear that there exists an integer k
(positive or negative) such that y1 = g1 + 2k. But then
−2bg1 /2c + 2by1 /2c +
n
X
(y3 − x3 ) = 0 = −g1 + y1 +
i=3
n
X
(y3 − x3 ) ,
i=3
P
or, equivalently, for all x2 ∈ {0, . . . , 2(m − 1)}, g1 + x2 + ni=3 (x3 ) + H = y1 + x2 +
P
n
i=3 (y3 )+H. Therefore, since U represents %, (g1 , x2 , x3 , . . . , xn ) ∼ (y1 , x2 , y3 , . . . , yn ),
and since 0.5 (x1 mod 2)2 + 2bx1 /2c ≥ 0.5 (g1 mod 2)2 + 2bg1 /2c, (x1 , x2 , x3 , . . . , xn ) %
(y1 , x2 , y3 , . . . , yn ).
Conversely, if (g1 , x2 , x3 , . . . , xn ) ∼ (y1 , x2 , y3 , . . . , yn ), x2 6= 2m, then, since all
the xi ’s and all the yi ’s, i ≥ 2, are multiples of 2, there exists an integer P
k (positive
or negative) such that g1 =Py1 + 2k. Then 0.5 (g1 mod 2)2 + 2bg1 /2c + ni=3 x3 =
0.5 (y1 mod 2)2 + 2by1 /2c + ni=3 y3 and so (g1 , 2m, x3 , . . . , xn ) ∼ (y1 , 2m, y3 , . . . , yn ).
Independence axiom also holds w.r.t. X1 . Indeed, assume that (x1 , x2 , x3 , . . . , xn ) %
(x1 , y2 , y3 , . . . , yn ). If x2 = y2 or if x2 6= 2m and y2 6= 2m, then, clearly, by definition of
U , (y1 , x2 , x3 , . . . , xn ) % (y1 , y2 , y3 , . . . , yn ) for any P
y1 ∈ R. Now, y2 = 2m and x2 6= 2m
is impossible because 0.5 (x1 mod 2)2 + 2bx1 /2c + ni=3 y3 ≥ x1 − 0.5 and
n
X
xi + H ≤ x1 + 2(m − 1) + 2(n − 2) + H ≤ x1 − (m − 1)2 − n(m + 1) + 3.5
i=1
Pn
and so, since m ≥ 3 and n ≥ 2,
i=1 xi + H ≤ x1 − 8.5. Therefore, assume that
2
(x1 , 2m, x3 ,P
. . . , xn ) % (x
,
y
,
y
,
.
.
.
,
y
1
2
3
n ). Then, clearly, for any y1 ∈ R, 0.5 (y1 mod 2) +
Pn
n
2by1 /2c + i=3 y3 >
i=1 yi + H, and so (y1 , 2m, x3 , . . . , xn ) % (y1 , y2 , y3 , . . . , yn ).
Hence independence holds.
17
Second step: (C2 ) holds
Assume that (C2 ) does not hold, i.e., that there exist xji , yij , i ∈ {1, . . . , n}, j ∈ {1, 2, 3},
such that (yi1 , yi2 , yi3 ) are permutations of (x1i , x2i , x3i ) and such that
 1 1 1

(x1 , x2 , x3 , . . . , x1n ) % (y11 , y21 , y31 , . . . , yn1 )
(x21 , x22 , x23 , . . . , x2n ) % (y12 , y22 , y32 , . . . , yn2 )
(16)
(x31 , x32 , x33 , . . . , x3n ) % (y13 , y23 , y33 , . . . , yn3 )
with at least one strict preference . Let us prove that (16) cannot hold. If xj2 = 2m
for all j ∈ {1, 2, 3}, then, by independence, one can substitute all the xj2 , y2j by 0;
translating these preference relations
, and P
summing the three
Pin terms
P of utility U P
inequalities thus obtained, one gets 3j=1 ( ni=1 xji + H) > 3j=1 ( ni=1 yij + H). But
the yij ’s are permutations of the xji ’s, so that the last inequality cannot hold. Similarly,
if xj2 6= 2m for all j ∈ {1, 2, 3}, then (16) cannot hold.
If there exist i, j, i 6= j, such that xi2 = xj2 = 2m, then at least one preference relation
has a 2m term on its right and left hand sides. By independence, one can substitute
the 2m’s by 0’s. So, one need only study the following case: there exists a unique index
i such that xi2 = 2m, and, moreover, y2i 6= 2m. Without loss of generality, by reordering
the preference relations, x12 = 2m and y22 = 2m. Hence the following relations:

 1
(x1 , 2m, x13 , . . . , x1n ) % (y11 , y21 , y31 , . . . , yn1 )
 (x21 , x22 , x23 , . . . , x2n ) % (y12 , 2m, y32 , . . . , yn2 )
(x31 , x32 , x33 , . . . , x3n ) % (y13 , y23 , y33 , . . . , yn3 )
with at least one strict preference .
If y12 = x11 , then, summing the utilities over the three preference relations, one gets:
P
2
P
j
n
x
0.5 x11 mod 2 + 2bx11 /2c + x21 + x31 − x12 + 3j=1
i=2 i + 2H >
P
P
2
0.5 x11 mod 2 + 2bx11 /2c + x21 + x31 − x12 + 3j=1 ( ni=2 ) xji + 2H.
Hence y12 = x11 is impossible. If y12 = x21 , then,Ptranslating the secondP
preference relation
n
2
2
2
2
into its utility counterpart, one gets: x1 +x2 + i=3 xi +H ≥ V (x1 )+ ni=3P
yi2 . But this is
2
2
2
2
impossible because V (x1 ) ≥ x1 −0.5 and, since m ≥ 3 and n ≥ 2, x1 +x2 + ni=3 x2i +H ≤
x21 − 8.5. Hence y12 = x31 . Consequently, if (16) holds, then it can be written as:
 1

(x1 , 2m, x13 , . . . , x1n ) % (y11 , y21 , y31 , . . . , yn1 )
 (x21 , x22 , x23 , . . . , x2n ) % (x31 , 2m, y32 , . . . , yn2 )
(x31 , x32 , x33 , . . . , x3n ) % (y13 , y23 , y33 , . . . , yn3 )
with at least one strict preference .
If y13 = x21 , then,
of utility U , the last two preference
relations are equivalent
Pnin terms
Pn
2
3
2
2
2
3
2
to x1 ≥ V (x1 ) + i=3 (yi − xi ) − x2 − H and x1 ≥ x1 + i=2 (yi3 − x3i ). Therefore
x31 ≥ V (x31 ) +
n
n
X
X
(yi3 − x3i ) + y23 − x32 − x22 − H.
(yi2 − x2i ) +
i=3
i=3
18
But
the yij ’s are permutations of the xji ’s,
Pn since
1
1
i=3 (xi − yi ). And since m ≥ 3 and n ≥ 2,
Pn
2
i=3 (yi
− x2i ) +
Pn
3
i=3 (yi
− x3i ) =
−H = [(m − 2)2 + n(m + 1) − 4.5] + 2(n − 2) + 4(m − 1)
n
X
≥
(yi1 − x1i ) + x32 + x22 + [(m − 2)2 + n(m + 1) − 4.5]
≥
i=3
n
X
(yi1 − x1i ) + x32 + x22 + 4.5,
i=3
one gets: x31 ≥ V (x31 ) + 4.5, which is impossible since V (x31 ) ≥ x31 − 0.5. Therefore,
y13 = x11 and (16) can be written as:

 1
(x1 , 2m, x13 , . . . , x1n ) % (x21 , y21 , y31 , . . . , yn1 )
 (x21 , x22 , x23 , . . . , x2n ) % (x31 , 2m, y32 , . . . , yn2 )
(x31 , x32 , x33 , . . . , x3n ) % (x11 , y23 , y33 , . . . , yn3 )
with at least one strict preference .
Now, by unrestricted solvability w.r.t. the first component, there exists g 2 ∈ R
such that (g 2 , x22 , x23 , . . . , x2n ) ∼ (x31 , 2m, y32 , . . . , yn2 ) and there exists g 1 ∈ R such that
(g 1 , 2m, x13 , . . . , x1n ) ∼ (g 2 , y21 , y31 , . . . , yn1 ). But then,

 1
(g , 2m, x13 , . . . , x1n ) ∼ (g 2 , y21 , y31 , . . . , yn1 )
 (g 2 , x22 , x23 , . . . , x2n ) ∼ (x31 , 2m, y32 , . . . , yn2 ) .
(x31 , x32 , x33 , . . . , x3n ) (g 1 , y23 , y33 , . . . , yn3 )
Therefore, in terms of utility function U ,
V (g 1 ) +
n
X
x1i = g 2 + y21 +
i=3
n
X
yi1 + H
and g 2 + x22 +
i=3
n
X
x2i + H = V (x31 ) +
n
X
i=3
yi2 .
i=3
Replacing
in the first
g 2 by its value P
given by the second equality, V (g 1 ) +
Pn
Pnequality
n
1
1
3
2
2
2
1
i=3 xi = V (x1 ) +
i=3 (yi − xi ) − x2 + y2 +
i=3 yi or, equivalently,
1
V (g ) = V
(x31 )
+
n
X
(yi2
−
x2i )
−
x22
+
y21
n
X
+
(yi1 − x1i ).
i=3
Now, since
the yij ’s are permutations
of xji ’s, 2m+x22 +x32 = y21 +2m+y23 and
P
P
x1i ) + ni=3 (yi2 − x2i ) + ni=3 (yi3 − x3i ) = 0. So, (17) is equivalent to
V (g 1 ) + y23 +
n
X
(17)
i=3
yi3 = V (x31 ) + x32 +
i=3
n
X
Pn
1
i=3 (yi −
x3i .
i=3
Since xji and yij , i ≥ 2, are multiples of 2,Pthere exists an integer
P k (positive or negative)
such that g 1 = x31 + 2k, and so g 1 + y23 + ni=3 yi3 = x31 + x32 + ni=3 x3i . But this equality
corresponds to the preference relation (g 1 , y23 , . . . , yn3 ) ∼ (x31 , x32 , . . . , x3n ). So (16) cannot
occur and, consequently, (C2 ) holds.
19
Third step: (Ci ), i ≥ 3, holds
According to theorem 2, it is sufficient to show that (Si ) holds for any i ≥ 3.
Let k ≤ (m + 1)/2 and assume that (S2k−1 ) does not hold, i.e., that there exist
elements xij , yji such that xi1 6= xp1 for all i ≤ k < p and either yji = xij or yji = xij + k,
and such that
(x11 , x12 , . . . , x1n )
(xi1 , xi2 , . . . , xin )
k+1 k+1
(x1 , x2 , . . . , xk+1
n )
k+j
k+j
(x1 , x2 , . . . , xk+j
n )
%
%
%
%
(xk1 , y21 , . . . , yn1 ),
i
i
(xi−1
1 , y2 , . . . , yn ), i ∈ {2, . . . , k},
k+1
2k
(x1 , y2 , . . . , ynk+1 ),
(xk+j−1
, y2k+j , . . . , ynk+j ), j ∈ {2, . . . , k},
1
(18)
with at least one strict preference relation . If there exists no j such that xj2 = 2m
or y2j = 2m, then (18) cannot occur because it would violate a cancellation axiom and
because % is represented on this subspace by an additive utility function. So, suppose
that there exists at least one index j ∈ {1, . . . , 2k} such that xj2 = 2m or y2j = 2m.
When xj2 = y2j , by independence substitute xj2 and y2j by 0’s. Now, by symmetry,
i
one can suppose that there exist p indices {i1 , i2 , . . . , ip }, ij ≤ k, such that y2j = 2m.
Using transformation (12) as in the proof of theorem 2, we can furthermore suppose
that i1 , . . . , ip are the first p indices. This transformation has no consequence on the
rest of the proof except to simplify notation. Now translating the k first preference
relations into their U counterpart, we get:
k X
n
X
xij
+ kH ≥
i=1 j=1
Thus,
k
X
xi2
p−1
X
V
(xi1 )
+V
(xk1 )
+
i=1
k
X
(x1i−1
i=p+1
+
y2i
+ H) +
k X
n
X
yji .
i=1 j=3
p−1
k
k X
n
i
X
X
X
h
i
i
k
k
i
y2 − pH +
(yji − xij ).
≥
V (x1 ) − x1 + V (x1 ) − x1 +
i=1
i=1
i=p+1
i=1 j=3
Now by (5), V (x) ≥ x − 0.5, so the inequality above implies that
k
X
xi2
≥ −0.5p +
i=1
But k ≤
k
X
y2i − pH − k(n − 2) ≥ −0.5p − pH − k(n − 2).
i=p+1
m+1
2 ,
so, expanding H, we get
k
X
m+1
(n − 2)
2
1
1
2
≥ pm + p −
mn + m + 3p −
n − 9p + 1.
2
2
xi2 ≥ −0.5p + p[m2 + mn + 3n − 8.5] −
i=1
Since m ≥ 3, n ≥ 2 and p ≥ 1,
k
X
i=1
xi2
1
1
≥ pm + 6 p −
+ 3 + 2 3p −
− 9p + 1 ≥ pm2 + 3p ≥ m2 + 3.
2
2
2
20
But this is impossible because k ≤
k
X
i=1
xi2 ≤
m+1
2
i
and xi2 ≤ 2(m − 1) since xi2 6= y2j = 2m, so that
m+1
· 2(m − 1) = m2 − 1.
2
Hence (18) cannot hold. Thus, (Si ) holds for any i ∈ {3, . . . , m}, and consequently
(Cm ) also holds. Since m is odd, by theorem 2, (Cm+1 ) also holds.
In order to prove lemma 6, I introduce the following lemma:
Lemma 7 Let % be a weak order on X = X1 × X2 . Assume restricted solvability
w.r.t. X1 (axiom 6), independence (axiom 1) and the Archimedean axiom (axiom 12).
Suppose that % is representable by a utility function, and that Card(X2 ) = 2. Then %
is representable by an additive utility function.
Proof of lemma 7: Let X2 = {a, b}. If a ∼2 b, the proof is obvious. Assume henceforth that b 2 a. Let X̃1 be the set of equivalent classes of %1 . The existence of
an additive utility function on X1 × {a, b} is equivalent to the existence of an additive
utility function on X̃1 × {a, b}. The advantage of working in the latter set rather than
in X1 × {a, b} is that the utility function is one-to-one in this space. So, let us work in
X̃1 × {a, b}.
By hypothesis, % is representable by a utility function, say u(·, ·). Let ua (·) = u(·, a)
and ub (·) = u(·, b). To prove the existence of an additive utility, it is sufficient to
prove that there exists a strictly increasing function, say ϕ(·), transforming u(·) into an
additive function. In other words, there exists α ∈ R∗+ such that
ϕ ◦ ub (x) = ϕ ◦ ua (x) + α, for all x ∈ X̃1 .
(19)
Suppose that (x1 , a) ≺ (y1 , b) for all x1 , y1 ∈ X̃1 . Then, in terms of utility functions,
ua (X̃1 ) ∩ ub (X̃1 ) = ∅. Let ϕ(·) be defined by:
arctan(x)
if x ∈ ua (X̃1 ),
ϕ(x) =
−1
arctan(ua (ub (x))) + π if x ∈ ub (X̃1 ).
Then it is obvious that ϕ ◦ u(·, ·) is an additive utility representing % on X̃1 × {a, b}.
Now, suppose that ua (X̃1 )∩ub (X̃1 ) 6= ∅. (19) is equivalent to the following equation:
ϕ ◦ (ub ◦ u−1
a )(x) = ϕ(x) + α, for all x ∈ ua (X̃1 ).
(20)
Note that the value of α is not important, as long as it remains strictly positive. Indeed,
if it were replaced by β > 0, ϕ(·) would be replaced by αβ ϕ(·). Let x0 ∈ ua (X̃1 ).
The value of ϕ(x0 ) can be taken arbitrarily; indeed, adding a constant to ϕ(·) has no
consequence on the validity of (20). Now, let us construct a function ϕ(·) satisfying (20).
0
We first define ϕ(·) on [x0 , ub ◦ u−1
a (x )] as:
ϕ(x) =
α(x − x0 )
0
+ ϕ(x0 ), for all x ∈ [x0 , ub ◦ u−1
a (x )].
−1 0
0
ub ◦ ua (x ) − x
21
(21)
0
0
Note that since (x, a) ≺ (x, b) for every x ∈ X̃1 , ub ◦ u−1
a (x ) > x , so that ϕ(·), as
defined by equation (21), is well defined and is strictly increasing.
0
Extend ϕ(·) as follows: for all x ∈ [x0 , ub ◦ u−1
a (x )] ∩ ua (X̃1 ), equation (20) can be
0
applied. Let x1 = sup{x : x ∈ [x0 , ub ◦ u−1
a (x )] ∩ ua (X̃1 )}. Then, applying (20) for all
0
1
1
x ∈ [x , x [∩ua (X̃1 ) if x is not reached (resp. [x0 , x1 ] ∩ ua (X̃1 ) otherwise), we define
0
−1 1
−1 0
−1 1
ϕ(·) on [ub ◦ u−1
a (x ), ub ◦ ua (x )[∩ub (X̃1 ) (resp. [ub ◦ ua (x ), ub ◦ ua (x )] ∩ ub (X̃1 )).
0
In fact, restricted solvability w.r.t. X1 ensures that we define ϕ(·) on [ub ◦ u−1
a (x ), ub ◦
−1
1
−1
0
−1
1
ua (x )[∩(ua (X̃1 ) ∪ ub (X̃1 )) (resp. [ub ◦ ua (x ), ub ◦ ua (x )] ∩ (ua (X̃1 ) ∪ ub (X̃1 ))).
Moreover, ϕ(·) strictly increases because, by independence, ub ◦ u−1
a (x) > x for all
x ∈ ua (X̃1 ) and α > 0.
0
1
Now, two cases can occur. In the first one, x1 = ub ◦ u−1
a (x ) but x 6∈ ua (X̃1 ), or
0
x1 < ub ◦ u−1
a (x ). Then, by restricted solvability, there exists no element, say y ∈ X̃1 ,
0
such that ua (y) > x1 — otherwise, by definition of x1 , ua (y) > ub ◦u−1
a (x ), which would
0
−1 0
imply that (y, a) (u−1
a (x ), b) (ua (x ), a) and would ensure the existence of t ∈ X̃1
−1
0
such that (t, a) ∼ (ua (x ), b). Therefore, ϕ(·) has been defined on {x ∈ ua (X̃1 )∪ub (X̃1 )
such that x ≥ x0 }.
0
1
In the second case, x1 = ub ◦ u−1
a (x ) and x ∈ ua (X̃1 ). Then ϕ(·) has been defined
0
−1
2
0
2 0
on [x , (ub ◦ ua ) (x )] ∩ (ua (X̃1 ) ∪ ub (X̃1 )). Let x2 = sup{x : x ∈ [x0 , (ub ◦ u−1
a ) (x )] ∩
ua (X̃1 )}. A process similar to that of the preceding paragraphs can be applied with the
set [x1 , x2 ] instead of [x0 , x1 ]. By continuing the same process, we construct a sequence
of points (xi )ni=0 , where n is either an integer or the infinity. ϕ(·) is then defined either
on [x0 , xn ] ∩ (ua (X̃1 ) ∪ ub (X̃1 )) or on [x0 , xn [∩(ua (X̃1 ) ∪ ub (X̃1 )), depending on whether
xn is reached or not. If n is finite, by restricted solvability w.r.t. the first component,
ϕ(·) has been defined on the set {x ∈ ua (X̃1 ) ∪ ub (X̃1 ) : x ≥ x0 }.
Problems arise when n is infinite. In such a case, ϕ(·) has been defined on [x0 , (ub ◦
−1
ua )n (x0 )[∩(ua (X̃1 ) ∪ ub (X̃1 )). Now, suppose that there exists x ∈ ub (X̃1 ) such that,
i 0
−1 i 0
0
for every integer i, (ub ◦u−1
a ) (x ) < x. By construction, ϕ((ub ◦ua ) (x )) = ϕ(x )+iα;
hence, this should tend toward +∞ when i tends toward +∞. But, then, the problem
is that ϕ(x) cannot be finite, and so ϕ(·) is not well defined on ua (X̃1 ) ∪ ub (X̃1 ).
i
Fortunately, this case cannot arise. For all integers i, i + 1 ≥ 0, xi+1 = ub ◦ u−1
a (x ).
i
i+1
Now, by the process of construction, x , x
∈ ua (X̃1 ); so, there exists a sequence of
i
i
elements of X̃1 , say (y )i≥0 , such that x = ua (y i ) for all i. But, then, ua (y i+1 ) =
ub (y i ). In terms of preference relations, (y i+1 , a) ∼ (y i , b). Therefore, (y i ) is a standard
sequence. So, by the Archimedean axiom, if x existed such that x is greater than
i 0
(ub ◦ u−1
a ) (x ) for all i, then {i} would be finite. Hence, the case described in the
previous paragraph would not arise. Consequently, ϕ(·) has been well defined on {x ∈
ua (X̃1 ) ∪ ub (X̃1 ) : x ≥ x0 }.
Now, using (20) conversely, i.e.,
ϕ ◦ (ua ◦ u−1
b )(x) = ϕ(x) − α, for all x ∈ ub (X̃1 ),
and applying a process similar to that of the preceding paragraphs, staring with the set
0
0
[x0 , (ub ◦ u−1
a )(x )], one can construct ϕ(·) on {x ∈ ua (X̃1 ) ∪ ub (X̃1 ) : x ≥ x}. And, of
course, as shown above, ϕ(·) is strictly increasing. Hence, ϕ(·) exists on ua (X̃1 )∪ub (X̃1 ).
Note that ϕ(·) is not unique up to positive affine transformations because ϕ(·) can be
defined with a certain degree of freedom in equation (21).
22
Proof of lemma 6: Suppose that a utility function exists, representing % on X. Consider arbitrary elements i, i + 1 ∈ N . By lemma 7, there exists an additive utility, say
u, representing % on X1 × {xi2 , xi+1
2 }. This means that there exists a constant, say α,
i+1
i
such that u(·, x2 ) = u(·, x2 ) + α. Suppose that i + 2 ∈ N . Let A = {y ∈ X1 : there
i+2
i
exists x ∈ X1 such that (y, xi+1
2 ) ∼ (x, x2 )}. (6) defines uniquely u(·, x2 ) on A because
i+1
i+1
i+1
i
i
i
(x, x2 ) ∼ (y, x2 ) ⇔ u(x, x2 ) = u(y, x2 ) = u(y, x2 ) + α ⇔ (x, x2 ) ∼ (y, xi+2
2 ) ⇔
i+2
i+1
i
i
u(y, x2 ) = u(x, x2 ) = u(x, x2 ) + α = u(y, x2 ) + 2α. Obviously, u, as defined above,
i+2
is a utility function over A × {xi2 , xi+1
2 , x2 }.
i+2
Now, there remains to extend u(·, x2 ) on B = {y ∈ X1 : there exists no x ∈ X1 such
i+1
i
i
that (y, xi+1
2 ) ∼ (x, x2 )}. But, by restricted solvability, B = {y ∈ X1 : (y, x2 ) (x, x2 )
i+2
i+1
for all x ∈ X1 } and, by definition 2, B = {y ∈ X1 : (y, x2 ) (x, x2 ) for all x ∈ X1 }.
i+1
So, u(·, xi+2
2 ) defined on B as u(·, x2 ) + α fits as a utility function. Indeed, let
β = sup u(x, xi2 ) − inf u(y, xi2 ).
y∈B
x∈B
Then β < α if both the sup and the inf are reached, else β ≤ α. If this were not true,
there would exists (x, y) ∈ B × B such that u(x, xi2 ) − u(y, xi2 ) ≥ α, or equivalently
such that (x, xi2 ) % (y, xi+1
2 ). But this leads to a contradiction because y ∈ B. Since
i+1
i+1
u is additive on X1 × {xi2 , xi+1
2 }, β = supx∈B u(x, x2 ) − inf y∈B u(y, x2 ). Hence
i+2
i+1
i+2
u(x, x2 ) = u(x, x2 ) + α for all x ∈ B guarranties that (x, x2 ) (y, xi+1
2 ) for all
i+2
i+1
i+1
x, y ∈ B × X1 . Moreover, u(x, xi+2
)
≥
u(y,
x
)
⇔
u(x,
x
)
≥
u(y,
x
2
2
2
2 ) for all
x, y ∈ B, so u, as defined above, preserves the ordering.
i+2
Hence, there exists an additive utility function on X1 × {xi2 , xi+1
2 , x2 }. By induction, we construct the utility function on X1 × {xj2 : j ∈ N and j ≥ i}. With a similar
process, we construct u on X1 × {xj2 : j ∈ N and j ≤ i}. So, an additive utility function
exists, representing % on X.
References
Adams, E. W.: 1965, Elements of a theory of inexact measurement, Philosophy of
Science 32, 205–228.
Debreu, G.: 1954, Representation of a preference ordering by a numerical function, in
R. Thrall, C. Coombs and R. Davis (eds), Decision Processes, Wiley, New York,
pp. 159–165.
Debreu, G.: 1960, Topological methods in cardinal utility theory, in K. J. Arrow, S. Karlin and P. Suppes (eds), Mathematical Methods in the Social Sciences, Stanford
University Press, Stanford, CA, pp. 16–26.
Debreu, G.: 1964, Continuity properties of paretian utility, International Economic
Review 5(3), 285–293.
Fishburn, P. C.: 1970, Utility Theory for Decision Making, Wiley, NewYork.
Fishburn, P. C.: 1981, Uniqueness properties in finite-continuous additive measurement,
Mathematical Social Sciences 1, 145–153.
23
Fishburn, P. C.: 1992, A general axiomatization of additive measurement with applications, Naval Research Logistics 39, 741–755.
Fishburn, P. C.: 1997a, Cancellation conditions for finite two-dimensional additive measurement, communicated by the author.
Fishburn, P. C.: 1997b, Cancellation conditions for multi-attribute preferences on finite
sets, in M. H. Karwan, J. Spronk and J. Wallenius (eds), Essays on Decision
Making, Springer, Berlin, pp. 157–167.
Fishburn, P. C.: 1997c, Failure of cancellation conditions for additive linear orders,
Journal of Combinatorial Designs 5, 353–365.
Fuhrken, G. and Richter, M. K.: 1991, Additive Utility, Economic Theory 1, 83–105.
Gonzales, C.: 1996a, Additive utilities when some components are solvable and others
are not, Journal of Mathematical Psychology 40(2), 141–151.
Gonzales, C.: 1996b, Utilités additives : existence et construction, PhD thesis, Université Paris VI.
Gonzales, C.: 1998, Additive utility without restricted solvability on all components,
working paper.
Gonzales, C. and Jaffray, J.-Y.: 1997, Imprecise sampling and direct decision making,
to appear in Annals of Operations Research.
Jaffray, J.-Y.: 1974a, Existence, Propriétés de Continuité, Additivité de Fonctions
d’Utilité sur un Espace Partiellement ou Totalement Ordonné, Thèse d’Etat, Université Paris VI.
Jaffray, J.-Y.: 1974b, On the extension of additive utilities to infinite sets, Journal of
Mathematical Psychology 11(4), 431–452.
Jaffray, J.-Y.: 1992, Préférences représentables par une fonction d’utilité additivement
séparable, mimeograph, DEA Informatique et Recherche Opérationelle, Université
Paris 6, Paris.
Karni, E. and Safra, Z.: 1997, Hexagon condition and additive representation of preferences: An algebraic approach, to appear in Journal of Mathematical Psychology.
Krantz, D. H.: 1964, Conjoint measurement: The Luce-Tukey axiomatization and some
extensions, Journal of Mathematical Psychology 1, 248–277.
Krantz, D. H., Luce, R. D., Suppes, P. and Tversky, A.: 1971, Foundations of Measurement (Additive and Polynomial Representations), Vol. 1, Academic Press, New
York.
Luce, R. D.: 1966, Two extensions of conjoint measurement, Journal of Mathematical
Psychology 3, 348–370.
24
Luce, R. D. and Tukey, J. W.: 1964, Simultaneous conjoint measurement : A new type
of fondamental measurement, Journal of Mathematical Psychology 1, 1–27.
Scott, D.: 1964, Measurement structures and linear inequalities, Journal of Mathematical Psychology 1, 233–247.
Scott, D. and Suppes, P.: 1958, Foundational aspects of theories of measurement, Journal of Symbolic Logic 23(2), 113–128.
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 Representations of Preferences, A New Foundation of
Decision Analysis, Kluwer Academic Publishers, Dordrecht.
Wakker, P. P.: 1991, Additive representation for equally spaced structures, Journal of
Mathematical Psychology 35, 260–266.
25