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
Proposition 4.3 provides a characterization of solutions s on collections of upper
bounded subsets of R: there is an ´ [ 0, 1 `] such that s selects all ‘´-optimal’
elements. In order to get a nice characterization, which takes also the unbounded subsets of 6 into account, we can make use of the axiom CCA instead of TI.
Proposition 5.1. 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, NEM and CCA if and only if ]
s 5 s for some non-decreasing function a: R
→ R satisfying ak , k for every
a
k [ R .
Proof. We only prove the only-if-part. Suppose
s satisfies AC, NEM and CCA. By AC and NEM we may conclude, according to Proposition 4.2, that
s 5 s for some
a
] function a: R
→ R satisfying ak , k for every k [ R . In order to show that a is
non-decreasing let k, l [ R be such that k l. Suppose that ak . al . Then there is an s [
s2`, l such that s , ak , k. Since also s [ 2`, k we get, by CCA, s [
s2`, k, and hence s ak. Contradiction. h
One easily verifies that the solutions, characterized in Proposition 5.1 by AC, NEM and CCA, satisfy MON. However, since MON implies AC and CCA,
these solutions can also be characterized by MON and NEM. A result similar to that in Theorem 4.1 can be obtained by using CCA and WAC
instead of AC. Note that AC does not imply CCA [see, for example, Example
310 H
. Norde et al. Mathematical Social Sciences 40 2000 297 –311
3.1a] and that CCA and WAC does not imply AC [see, for example, Example 3.1b]. Of course, WAC, NEM, TI, MI and CCA do imply AC, according to
the following theorem.
Theorem 5.1. 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 WAC, NEM , TI , MI and CCA if and only if
s 5 s .
tot
Proof. Again, we only prove the only-if-part. Suppose
s satisfies WAC, NEM, TI, MI and CCA. Let k [ R and S [2`, k. By NEM we know that sS ± 5, so we
k k
can choose an s [ sS . Of course, s , k. We will show that sS 5 S . Therefore, let
k k
k
t [ S . Define l[k 2 t k 2 s . 0 and m[1 2 lk [ R. Then t 5 ls 1 m, S 5
k k
lS 1 m and hence, by MI and TI, t [ sS . Therefore, sS 5 S . By CCA we
k k
k k
get sS 5 S for every S [ 6 . By NEM and TI one easily infers that sR 5 R.
k
Hence, by CCA, we get that sS 5 S for every S [ 6
. h
1`
Remark. If the collection 6 is not complete, then Proposition 5.1 need not be valid
anymore, as the following example shows: consider the class 6 of all non-empty and upper bounded subsets 6 which satisfy the condition that there exists a t [ [0,1 such
that S t 1 Z. Define the solution s by:
a
k 2 22 if k [ Z
ak[
H
k 2 37 otherwise
Clearly s satisfies AC and NEM. It also satisfies CCA: this is due to the fact that
a
for S, T [ 6 with S T both are contained in the same t 1 Z for some t [ [0, 1. In fact, one can prove that any
s , with a feasible function a which is non-decreasing on t 1 Z
a
for every t [ [0, 1, satisfies AC , NEM and CCA. In fact, 6 can be partitioned into several subcollections such that sets belonging to different subcollections are not related
by inclusion. These problems do not occur if 6 is complete. A characterization of the solutions in Proposition
5.1 by MON and NEM on non-complete collections 6 can still be obtained by using an appropriate strengthening of MON , similar to the
strengthening SAC of AC , mentioned in the remark at the end of Section 4.
6. Conclusions
