H . Norde et al. Mathematical Social Sciences 40 2000 297 –311
305
MON AC
WAC NEM
TI MI
CCA s
2 1
1 1
2 2
2
mix
s 2
2 1
2 2
2 1
rat
s 1
1 1
1 1
1 1
tot
s 1
1 1
2 1
1 1
max
s 1
1 1
2 1
2 1
´ k
s 1
1 1
2 2
2 1
s 2
2 1
1 2
2 2
´,k
ˆ s
1 1
1 1
2 2
1
´,k
s 2
1 1
1 2
1 2
pro a, b
4. Characterizations for translation and multiplication invariant solutions
Let 6 be a collection of non-empty subsets of R . We write 6 5 6 where
k [R k
6 [hS [ 6 : sup S 5 kj for every k [ R . The collection 6 is complete if all intervals
k
belong to 6. Recall that elements of 6 can be considered as ranges uA, which are intervals if, for instance, A is connected and u continuous. So, 6 is complete if the
underlying collection of optimization problems 3 is ‘rich’ enough. A solution
s on 3 is closed if sS is a closed subset of S for every S [ 6. For a ]
function a: R →
R we define the closed solution s on 6 by:
a
s S[[ak, k] S
a
where k 5sup S in particular s S 5 5 if ak . k or ak 5 k and k [
⁄ S and s S 5 S if
a a
ak 5 2 `. So, s selects, for every S [ 6 , the elements s [ S with s ak. Clearly,
a k
s satisfies AC. The following proposition shows that the closed solutions satisfying
a
AC are precisely the solutions s , provided that 6 is complete. Note that the solutions,
a
defined in Example 3.1c–f, h, i, satisfy AC and are, in fact, s for some suitably
a
chosen a.
Proposition 4.1. Let 6 be a complete collection of non-empty subsets of R and let
s be a closed solution on 6. The solution
s satisfies AC if and only if s 5 s for some
a
function a.
Proof. We only prove the only-if-part. So, assume that
s satisfies AC. Let k [ R . Define S [ 6 by S [2`, k] if k [ R and S [R if k 5 1 `. Since, by AC and the
k k
k
] fact that
s is closed, sS is a closed interval there is an ak [ R such that
k
sS 5 hx [ S : x akj. By AC we get sS 5 [ak, k] S 5 s S for every
k k
a
S [ S . h
k
If we impose some feasibility condition upon the function a we get closed solutions which are characterized by AC and NEM.
306 H
. Norde et al. Mathematical Social Sciences 40 2000 297 –311
Proposition 4.2. Let 6 be a complete collection of non-empty subsets of R and let s be
a closed solution on 6. The solution s satisfies AC and NEM if and only if s 5 s
a
] for some function a: R
→ R satisfying ak , k for every k [ R .
Proof. Straightforward by using the fact that
s2`, k] ± 5 for every k [ R and
Proposition 4.1. h
The next theorem describes the class of solutions, which are characterized by AC, NEM and TI. It turns out that these solutions, restricted to the collection of upper
bounded sets, coincide with the collection of ‘ ´-optimal’ solutions for some ´ [
0, 1 `].
Proposition 4.3. Let 6 be a complete collection of non-empty subsets of R which satisfies CL 1 and let
s be a closed solution on 6. The solution s satisfies AC, NEM and TI if and only if there is an
´ [ 0, 1 `] such that s 5 s , where a:
a
] R
→ R is defined by:
ak[k 2 ´ for every k [ R
1
H
a 1 ` 5 2 `
Proof. Clearly,
s 5 s satisfies AC, NEM and TI if a is defined by 1. In order to
a
prove the only-if-part suppose that s satisfies AC, NEM and TI. By AC and
NEM we know, according to Proposition 4.2, that s 5 s , for some function a:
a
] R
→ R satisfying ak , k for every k [ R . Take
´ 5 2 a0. For every k [ R we have, by TI:
[ak, k] 5
s2`, k] 5
sk 1 2`, 0] 5 k 1
s2`, 0] 5 k 1 [a0, 0]
5 [k 1 a0, k] and hence ak 5 k 1 a0 5 k 2
´. The only thing which remains to be shown is that a 1 ` 5 2 `. Since
sR ± 0 we can choose s [ sR. Then for every t [ R we have t 1 R 5 R and hence, by TI, t 1 s [
st 1 R 5 sR. Therefore, sR 5 R and hence a 1 ` 5 2 `.
h ˆ
The solution s
of Example 3.1 satisfies AC and NEM but not TI, the solution
´,k
s satisfies AC and TI but not NEM and one easily constructs a solution satisfying
´
NEM and TI but not AC simply by defining the solution for S with sup S [ h0, 1 `j in an arbitrary but not approximation consistent way and extending this
solution by translation invariance. Therefore, the axioms AC, NEM and TI are logically independent.
In the following proposition we characterize the ‘proportional’ solutions by AC, NEM and MI.
Proposition 4.4. Let 6 be a complete collection of non-empty subsets of R which satisfies CL and let
s be a closed solution on 6. The solution s satisfies AC,
H . Norde et al. Mathematical Social Sciences 40 2000 297 –311
307
NEM and MI if and only if there are a . 1 and b , 1 such that s 5 s , where a:
a
] R
→ R is defined by:
ak[ak for every k , 0
a0 5 2 ` 2
ak[bk for every k [ 0, `
5
a 1 ` [ h 2 `, 0j
Proof. Clearly, s 5 s satisfies AC, NEM and MI if a is defined by 2. In order to
a
prove the only-if-part suppose that s satisfies AC, NEM and MI. By AC and
NEM we know, according to Proposition 4.2, that s 5 s , for some function a:
a
] R
→ R satisfying ak , k for every k [ R . Take
a 5 2 a21 and b 5 a1. For every k , 0 we have, by MI:
[ak, k] 5
s2`, k] 5
s2k2`, 2 1] 5 2 k
s2`, 2 1] 5 2 k[a21, 2 1]
5 [2ka21, k] and hence ak 5 2 ka21 5
ak. In the same way one can prove that ak 5 bk for every k [ 0, 1 `. For every
l [ 0, ` we have: [a0, 0]
5 s2`, 0]
5 sl2`, 0]
5 ls2`, 0]
5 l[a0, 0]
5 [ la0, 0]
Therefore, a0 5 la0 for every l [ 0, 1 `. Since a0 , 0 we get a0 5 2 `. In a
similar way one can prove that a 1 ` [ h 2 `, 0j. h
ˆ The solution
s of Example 3.1 satisfies AC and NEM but not MI, the solution
´,k
s satisfies AC and MI but not NEM and one easily constructs a solution
max
satisfying NEM and MI but not AC simply by defining the solution for S with sup S [
h21, 0, 1, 1`j in an arbitrary but not approximation consistent way and extending this solution by multiplication invariance. Therefore, the axioms AC, NEM and MI
are logically independent. Clearly, the trivial solution
s satisfies AC, NEM, TI and MI. In the following
tot
proposition we show the impossibility of finding another solution, satisfying these four properties.
Theorem 4.1. Let 6 be a complete collection of non-empty subsets of R, which satisfies CL 1 and CL Let
s be a closed solution on 6. The solution s satisfies AC, NEM , TI and MI if and only if
s 5 s .
tot
Proof. Again we only prove the only-if-part. Suppose
s satisfies AC, NEM, TI and MI. By Proposition 4.4 we get
s2`, 0] 5 2`, 0] and hence, by TI: s2`, k] 5 k 1 s2`, 0] 5 k 1 2`, 0] 5 2`, k]
308 H
. Norde et al. Mathematical Social Sciences 40 2000 297 –311
for every k [ R. Moreover, by Proposition 4.3, we get sR 5 R. Using AC we may
conclude that sS 5 S for every S [ 6.
h Let us notice that, in the context of decision making under risk, u and v are von
Neumann–Morgenstern utility functions representing the same preferences iff v 5 cu 1 d, with c . 0 and d [ R. So, if one wants to stress the point of view that only
preferences have a true meaning, one should use a ‘solution rule’ for optimization problems that takes this fact into account. But Theorem 4.1 just shows that it is
impossible to do this in a non-trivial way. Stated otherwise: for von Neumann– Morgenstern preferences there is no sensible concept of approximate optimum If one
wants to talk in a meaningful way of approximate optimization, an escape route could be the addition of further details that allow for some ‘absolute’ reference point for
example: how do we decide whether the oscillations of last week at the New York Stock Exchange were wild or not? Maybe we refer to the previous history as a benchmark.
The interesting question is whether it can be done in a consistent way, without resorting to an ‘absolute’ utility function.
Remark. If a solution
s on 6 satisfies AC then, for every S [ 6, sS can be described as
hs [ S: s g j or hs [ S: s . g j for some g depending on S. However, we can have strict or weak inequality, depending on the value of sup S, as can be seen in
Example 3.1a. In order to get rid of these kind of approximate solutions we have
added the requirement that sS is a closed subset of S. However, this addition does not
‘force’ the parentheses to be closed. Consider, for example, 6 5 hh0, 1jj h[a, 1]:
a [ 0, 1 j and let s be the solution on 6, defined by sh0, 1j[h1j and s[a, 1][
[ a, 1] for every a [ 0, 1. The solution s satisfies AC and sS is a closed subset of
] S for every S [ 6, but there is no
g [ R such that sS 5 hs [ S: s g j for every
S [ 6. The problems in this example are due to the fact that the collection 6 is rather ‘poor’. For this reason we added the requirement that 6 should be complete, i.e.
contains all intervals. However, the Propositions
4.1 –4.4 and Theorem 4.1 can also be shown to be true for collections 6 which are not complete. In this situation however, the proofs become a
little bit more technical and an appropriate strengthening of AC is needed: a solution s on 6 satisfies SAC strong approximation consistency if for every S, S ,S , . . . [ 6
1 2
with sup S 5 sup S for every i [ N the following statement is true:
i
if s [ sS for every i [ N and s [ S is such that s lim inf
s then s [ sS
i i
i →
` i
One easily verifies that SAC induces AC . Moreover, SAC implies that sS is a
closed subset of S for every S [ 6. In fact, if 6 is a collection of intervals, then s
satisfies SAC if and only if s satisfies AC and sS is a closed subset of S for every
S [ 6. The impossibility result of Theorem 4.1 is caused by the fact that the domain S
contains elements without maximum. Let us notice, in particular, that whenever we have at least a couple of problems in 6 with the same supremum, but one of which has a
maximum and the other not, the AC axiom prevents the possibility of choosing the
H . Norde et al. Mathematical Social Sciences 40 2000 297 –311
309
maximum only, whenever it exists. In fact, a solution like s
in Example 3.1g, which
´,k
tries to capture this kind of idea, violates AC, as can easily be checked, whenever in 6 there are sets with maximum and others without. If we restrict our attention to domains
6 containing only elements S for which max S is well-defined then it turns out that s
max
is the only non-trivial solution satisfying AC, NEM, TI and MI.
Theorem 4.2. Let 6 be a collection of non-empty subsets of R such that every element of 6 has a maximum. Suppose, moreover, that 6 satisfies CL 1 and CL. Let
s be a solution on 6. The solution
s satisfies AC, NEM, TI and MI if and only if s 5 s
or s 5 s
.
tot max
Proof. Again we only prove the only-if-part. Suppose
s satisfies AC, NEM, TI and MI and suppose that
s ± s . Then there is some S [ 6 with
sS ± s S 5
max max
hmax Sj. By NEM we may conclude that there is an s [ sS such that s , max S. In order to show that
s 5 s take S [ 6 and s [ S. It suffices to prove that
tot
s [ sS. Choose a . 0 such that as 2 max S 1 max S s and define b[2a max
S 1 max S. Define, moreover, S9[aS 1 b. By CL1 and CL we have S9 [ 6 and by TI and MI we have
as 1 b [ sS9. Since as 1 b 5 as 2 max S 1 max S s and max S9 5 max S we infer by AC that s [
sS. h
5. Characterizations under CCA
