Utility Representation of Lower Separable Preferences
Özgür Yılmaz∗
June 2008
Abstract
Topological separability is crucial for the utility representation of a complete preference
relation. When preferences are incomplete, this axiom has suitably defined counterparts:
upper separability and lower separability (Ok (2002)). We consider the problem of representing an incomplete preference relation by means of a vector-valued utility function; we
obtain representation results under the lower separability assumption. Our results extend
the main representation theorems by Ok (2002) in terms of the separability axioms.
Keywords: Incomplete Preferences, Utility Representation, Upper/Lower Separability,
Near-completeness
Journal of Economic Literature Classification Number: D11
1
Introduction
Preference relations are usually assumed to satisfy completeness and transitivity. These two
properties have been the core axioms, tying self-interest to ‘rationality’. They are also useful
for analytical tractability. Particularly, the possibility of representation by real-valued functions
owes much to the completeness axiom. However, for most problems of individual decision making, it seems unrealistic to assume completeness. Our focus is the utility representation without
the completeness axiom.
∗
College of Administrative Science and Economics, Koç University, Sarıyer, İstanbul, Turkey 34450; Fax:+90
212 338 1653; E-mail address: [email protected]
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Obviously, an incomplete preorder can not be represented by a real-valued utility function.
The scope of this limitation of analytical tractability is extensively discussed in the literature.1
Basically, in the case of incomplete preferences, while utility representation is not possible, one
may still find a utility function that represents the preference order in a weaker sense: as before,
if x is preferred to y then u(x) > u(y), but the opposite implication is no longer true. This
approach is explored by Richter (1966) and Peleg (1970). Clearly, this ‘quasi-representation’ is
very much problematic, because the preorder cannot be recovered from the utility function u.
An alternative approach is representation by means of a vector-valued utility function.2 The
question is whether, given a preorder %, it is possible to find a function u : X → Rn for some
positive integer n such that, for each x, y ∈ X,
x % y if and only if u(x) ≥ u(y).
As a natural extension of the classical utility representation, multi-criteria decision making is not only intuitive but also analytically convenient. Ok (2002) shows that, under certain
topological separability assumptions, a not-too-incomplete preference relation, a preference relation such that any subset of mutually incomparable alternatives is finite, can be represented
by a vector-valued utility function. First, he suggests two counterparts of the usual separability axiom, upper separability and lower separability, which are suitably defined for incomplete
preferences. Then, he shows that upper separability is sufficient for the vector-valued utility
representation; also, he shows that upper and lower separability together imply the existence of
an upper semicontinuous vector-valued utility representation. We extend these results in terms
of the separability axioms: we show that lower separability is sufficient for representability by
means of a vector-valued utility function (Theorem 3); it follows from this result that lower separability is also the key axiom for the existence of such a function that is upper semicontinuous
(Theorem 4).
1
2
See Aumann (1962), Richter (1966), Peleg (1970), and Sondermann (1980).
For a discussion of the multi-attribute utility representation, see Ok (2002) and Eliaz and Ok (2006).
2
2
Preliminary Definitions and Existing Results
Let < be a binary relation on a set of alternatives X. Two alternatives x and y are <-comparable
if either x < y or y < x holds. They are <-incomparable if they are not <-comparable. For each
<-incomparable pair of alternatives x and y, we write x o
n y. A binary relation < is complete if
and only if each x and y in X are <-comparable. The strict (asymmetric) part of <, denoted
by , is the relation on X defined as x y if and only if x < y and ¬(y < x). The symmetric
part of <, denoted by ∼, is the relation on X defined as x ∼ y if and only if x < y and y < x.
For each Y ⊆ X, the relation induced by < on Y, denoted <|Y , is defined as <|Y ≡< ∩(Y × Y ).
The relation <0 is an extension of <, if <⊆<0 and ⊆0 . A relation < is a preorder if < is
reflexive and transitive, a partial order if it is an antisymmetric preorder, a linear order if it is
a complete partial order.
A preordered set is a pair (X, <) such that X is a nonempty set and < is a preorder on X.
Let (X, <) be a preordered set. A subset Y of X is <-dense if, for each pair x, y ∈ X with
x y, there is z ∈ Y such that x z y. The preorder < is weakly separable if there is a
countable <-dense set in X. Clearly, if < is weakly separable, then X is infinite.
A preordered set (X, <) is a poset (partially ordered set) if < is a partial order on X.
Given a poset (X, <), the partial order < is spacious if x y implies that {z : z ∈ X and
x z} ⊇ Closure{z : z ∈ X and y z}.
Theorem 1 Peleg (1970) Let < be a partial order on a nonempty set X. If for each x ∈ X,
{y : y ∈ X and x y} is open and < is weakly separable and spacious, then there exists a
continuous function u : X → [0, 1] such that for each x, y ∈ X, x y implies u(x) > u(y) .
A poset (X, <) is a chain if < is a linear order, and an antichain if = ∅. Clearly, if (X, <)
is a chain, then o
n= ∅, and if it is an antichain, then o
n= {(x, y) : x 6= y}. A chain in a poset
(X, <) is a set Y ⊆ X such that the relation <|Y is complete. Similarly, a set Y ⊆ X is an
antichain in a poset (X, <) if |Y = ∅. An antichain Y in (X, <) is of maximal cardinality if its
cardinality is at least as large as that of each antichain in (X, <). In this case, the cardinality
of Y is the width of the poset (X, <). It is denoted by w(X, <). Clearly, the width of each chain
3
is zero, while the width of an antichain (X, <) is |X| . An important result which we will use
extensively in the following analysis is Dilworth’s Decomposition Theorem (1950): If (X, <) is
a poset with finite width w, then X is the union of w chains in (X, <). For each poset (X, <),
let L(X, <) be the set of all extensions of < that are linear orders. By the classical Szpilrajn’s
Theorem (1930), each partial order can be extended to a linear order. Thus, for each poset
(X, <), L(X, <) 6= ∅. From this result, it follows that, for each poset (X, <),
<=
R,
R∈L(X,<)
that is, each partial order is the intersection of all of its linear extensions. The order dimension
of a poset (X, <), denoted as dim(X, <), is the minimum number of linear extensions of <, the
intersection of which is <, provided that this number is finite, and is ∞, otherwise. That is,
dim(X, <) ≡ min{k ∈ N : Ri ∈ L(X, <), i = 1, ..., k, and <=
n Ri }.
i=1
For each poset (X, <), w(X, <) ≥ dim(X, <) (Hiragushi (1955)).
A partial order < is representable if there is a positive integer n and a function u : X → Rn
such that, for each x, y ∈ X,
x < y ⇔ u(x) ≥ u(y).
Proposition 1 Ok (2002) Let < be a partial order on a nonempty set X. If < is representable,
then dim(X, <) < ∞. Moreover, if X is countable and dim(X, <) < ∞, then < is representable.
Since w(X, <) ≥ dim(X, <), it implies, that if X is countable and w(X, <) is finite, then
< is representable. Our focus, on the other hand, is on representability without the restriction
that X is a countable set. We analyze this problem in the following section.
3
Representation of Near-Complete Preferences
A natural conception of the ‘degree of incompleteness’ is near-completeness: A preorder is nearcomplete, if each subset A of X, with x o
n y for each distinct x, y in A, is finite (Ok (2002)).
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Thus, a partial order < on X is near-complete if and only if w(X, <) < ∞.
Let (X, <) be any preordered set. A subset Y of X is upper <-dense if, for each x, y ∈ X
such that x o
n y, there is z ∈ Y such that x z o
n y. It is lower <-dense if, for each x, y ∈ X
such that x o
n y, there is z ∈ Y such that x o
n z y. A preorder < is upper (lower) separable
if there is a countable set in X which is both <-dense and upper (lower) <-dense. Finally, < is
separable if it is both upper and lower separable.
Theorem 2 Ok (2002) Let X be any nonempty set and let < be a near-complete and upper
separable partial order on X. Then, < is representable.
Our first result is the counterpart of this theorem in terms of the separability assumption.
We replace upper separability with lower separability and obtain a representability result.
Theorem 3 Let X be any nonempty set and let < be a near-complete and lower separable partial
order on X. Then, < is representable.
Proof. Let (X, <) be a poset such that its width is n < ∞, and < is lower separable. By
Dilworth’s Decomposition Theorem, there is a partition of X, say X1 , X2 , ..., Xn such that, for
each k, <|Xk is a linear order on Xk . For each k, define Rk and Rk0 as follows:
x Rk y ⇔ x 6∈ Xk and ∃z ∈ Xk s.t. x z o
n y,
n y.
x Rk0 y ⇔ x ∈ Xk and x o
Then define the relation Dk = < ∪ Rk ∪ Rk0 .
Step 1: Dk is a partial order.
• Reflexivity of Dk : It follows from reflexivity of < .
• Antisymmetry of Dk : Let x Dk y Dk x. Suppose x 6= y. There are three cases:
i) x y.
Thus, ¬(y o
n x), and it implies ¬(y Rk0 x). Since < is antisymmetric, y Rk x. By definition of
Rk , there is z ∈ Xk such that y z o
n x. Since x y, transitivity of < implies that x z. It
contradicts with z o
n x.
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ii) x Rk y.
By definition of Rk , x 6∈ Xk and there is z ∈ Xk such that x z o
n y. Suppose y Rk x. Then,
there is z 0 ∈ Xk such that x z o
n y z0 o
n x. Clearly, z = z 0 is impossible. If z z 0 , then
transitivity of < implies x z 0 . It contradicts with z 0 o
n x. If z 0 z, transitivity of < implies
y z. It contradicts with z o
n y. Note that since z and z 0 are in Xk , and Xk is a chain, z and z 0
are <-comparable. Suppose y x. It implies x z o
n y x. This contradicts with transitivity
of <. Finally, since z ∈ Xk , y 6∈ Xk and ¬(y Rk0 x).
iii) x Rk0 y.
By definition of Rk0 , x ∈ Xk and x o
n y. Since Xk is a chain, y 6∈ Xk and ¬(y Rk0 x). Thus,
y Rk x, and there is z ∈ Xk such that y z o
n x. But, since x, z ∈ Xk , it is not possible. Thus,
x Dk y Dk x implies x = y. Thus, Dk is antisymmetric.
• Transitivity of Dk : Let x Dk y Dk z. If x < y < z, then, by transitivity of <, x < z. Note that,
x Rk0 y implies y 6∈ Xk and ¬(y Rk0 z) for each z ∈ X. Thus, x Rk0 y Rk0 z is not possible. Also,
since Xk is a chain and x a o
n y for some a ∈ Xk , y 6∈ Xk . Then, for each z ∈ X, ¬(y Rk0 z).
Thus, x Rk y Rk0 z is not possible. There are six cases left to be considered:
i) x < y Rk z
By definition of Rk , y 6∈ Xk and there is a ∈ Xk such that x < y a o
n z. If x 6∈ Xk , then
transitivity of < implies x Rk z. Suppose x ∈ Xk . Since x < y a o
n z, either x < z or x o
n z.
If x o
n z, then x Rk0 z. In either case, x Dk z.
ii) x Rk y < z
There is a ∈ Xk such that x a o
n y < z, where x 6∈ Xk . Since a and y are <-incomparable,
either a z or a o
n z. If a z, then x z. If a o
n z, then, since x 6∈ Xk and a ∈ Xk , x a o
nz
implies x Rk z. Thus, x Dk z.
iii) x Rk y Rk z
There are a, b ∈ Xk such that x a o
nybo
n z where x, y 6∈ Xk . Since Xk is a chain, a and b
are <-comparable. Since x a o
n y b, a 6= b. If b a, then y a. It is a contradiction. Thus,
a b. By transitivity of <, x b o
n z. Thus, x Dk z.
iv) x < y Rk0 z
6
If x 6∈ Xk , then, since y ∈ Xk and x y o
n z, x Rk z. Suppose x ∈ Xk . If z < x, then z < y. It
is a contradiction. Thus, either x o
n z or x z. If x o
n z, then, by definition of Rk0 , x Rk0 z. In
both cases, x Dk z.
v) x Rk0 y < z
Clearly, z < x is not possible. Thus, either x < z or x o
n z. Since x ∈ Xk , the latter implies
x Rk0 z . Thus, x Dk z.
vi) x Rk0 y Rk z
There is a ∈ Xk such that x o
nyao
n z, where x ∈ Xk . Clearly, a < x is not possible. Since
Xk is a chain, and both x and a are in Xk , x a. If z < x, then z a. It is a contradiction.
Thus, either x z or x o
n z. Since x ∈ Xk , the latter implies x Rk0 z. In both cases, x Dk z.
Thus, Dk is transitive. Thus, Dk is a partial order.
Step 2: Dk is weakly separable.
Since < is lower separable, there is a countable set Y in X which is both <-dense and lower
<-dense. We claim that Y is Dk -dense in X. Let x, y ∈ X such that x Bk y, where Bk is the
strict part of Dk .
i) x y
Since Y is <-dense, there is z ∈ Y such that x z y. Thus, there is z ∈ Y such that
x Bk z Bk y.
ii) x Rk0 y
n y. By lower <-denseness of Y, there is z ∈ Y such that
By definition of Rk0 , x ∈ Xk and x o
xo
n z y. Since x ∈ Xk , x Rk0 z. Clearly, x, y, z are all distinct. Thus, x Bk z Bk y.
iii) x Rk y
There is a ∈ Xk such that x a o
n y, where x 6∈ Xk . Since Y is lower <-dense, there is z ∈ Y
such that x a o
n z y. Then, x Rk z. Since x and z are distinct, x Bk z Bk y. We conclude
that there is a countable set Y in X, which is Dk -dense in X. Thus, Dk is a weakly separable
partial order on X.
Step 3: Constructing multi-vector utility indices that represent < .
We have shown that (X, Dk ) is a poset and for each k = 1, ..., n, Dk is weakly separable. By
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Theorem 1 above, there is a function ϕk : X → [0, 1] such that, for each x, y ∈ X and for each k,
x Bk y implies ϕk (x) > ϕk (y). Note that, for each k, x y implies x Bk y and ϕk (x) > ϕk (y).
Suppose there are x, y ∈ X such that x o
n y. Since X1 , ..., Xn is a partition of X, and for each
k, <|Xk is complete, there are i, j with i 6= j, such that x ∈ Xi and y ∈ Xj . By definition of Rk0 ,
x Rk0 y and y Rk0 x. Thus, x Bi y and y Bj x. Thus, ϕi (x) > ϕi (y) and ϕj (y) > ϕj (x). Finally,
for each x ∈ X, define u(x) = (ϕ1 (x), ..., ϕn (x)). This completes the proof.
Our next result is on the upper semicontinuous representation of < . This is an important
result because upper semicontinuity is often sufficient for maximization of utility functions.
Theorem 4 Let X be a topological space and let < be a near-complete and lower separable
partial order on X such that {y : x y} is open for each x ∈ X. Then there exists an upper
semicontinuous mapping u :X → [0, 1]n with n being the width of (X, <), such that x < y if and
only if u(x) ≥ u(y) for each x, y ∈ X.
Proof. Following Ok (2002), we define the mapping uk : X → [0, 1] as
uk (a) =


 ϕk (x)
if U (a) = ∅

 Infb∈U (a) Supt∈L(b) ϕk (t)
otherwise
where ϕk is the utility function in Theorem 3, for each a ∈ X, U (a) = {b : b a} and
L(a) = {t : a t}. Let x, y ∈ X. If x y, then, by weak separability, there is z ∈ U (y) such
that x z. By Theorem 1 above, ϕk (x) > ϕk (z). Since U (y) 6= ∅ and
uk (y) = Infb∈U (y) Supt∈L(b) ϕk (t) ≤ Supt∈L(z) ϕk (t),
where z ∈ U (y), Supt∈L(z) ϕk (t) ≥ uk (y). Thus, by definition of uk ,
uk (x) ≥ ϕk (x) > ϕk (z) ≥ Supt∈L(z) ϕk (t) ≥ uk (y).
Thus, x y implies that, for each k, uk (x) > uk (y). Now, suppose x o
n y. By lower separability,
there is z ∈ X such that x o
n z y. Since X1 , ..., Xn is a partition of X and each set in this
8
partition is a chain, x ∈ Xk and z 6∈ Xk for some k. Thus, x Bk z and ϕk (x) > ϕk (z). Since
z y, the inequalities above apply. Thus,
uk (x) ≥ ϕk (x) > ϕk (z) ≥ Supt∈L(z) ϕk (t) ≥ uk (y).
Since o
n is symmetric, for some l 6= k, ul (y) > ul (x). Thus, x (=)y holds if and only if,
for each k, uk (x) > (=)uk (y) Thus < is represented by u = (u1 , . . . , un ). Finally, we need to
show that, for each k, uk is upper semicontinuous. To this end, we repeat the argument in the
proof of Theorem 2 in Ok (2002): First, note u−1
k [θ, 1] is closed for θ ≤ 0 or θ ≥ 1. For upper
semicontinuity, let θ ∈ (0, 1). Let (xα )α∈A be a sequence in X such that, for each α ∈ A, xα → a
and uk (xα ) ≥ θ. If, for some α ∈ A, a < xα , then uk (a) ≥ uk (xα ) ≥ θ. Now, assume that,
for each α ∈ A, either xα a or xα o
n a. Then, lower separability of < implies U (a) 6= ∅. Let
b ∈ U (a), and observe that, since L(b) is open, there is α(b) ∈ A such that xα(b) ∈ L(b). Thus,
for each b ∈ U (a), Supt∈L(b) ϕk (t) ≥ uk (xα(b) ) ≥ θ. It implies
uk (a) = Infb∈U (a) Supt∈L(b) ϕk (t) ≥ θ.
Thus, uk is upper semicontinuous. Since k is chosen arbitrarily, u is upper semicontinuous. This
completes the proof.
Ok (2002) proves a corollary of this theorem; the difference is that he assumes separability
instead of lower separability. Thus, Theorem 4 extends his result to lower separable partial
orders.
4
Conclusion
We extend some of the existing results on the multi-vector utility representation by Ok (2002)
in terms of separability assumptions. In its technicality, it is restricted to near-complete preferences. We showed that a lower separable partial order is representable by means of a multi-vector
utility function. We also show that a lower separable partial order can be represented by an
upper semicontinuous vector-valued utility function.
9
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