Serial Dictatorship with Infinitely Many Agents∗
Shino Takayama†
Akira Yokotani
School of Economics
University of Queensland
This Version: February 17, 2014
Abstract. This paper studies social choice correspondences assigning a set of choices to each pair
consisting of a nonempty subset of the set of alternatives and a weak preference profile, which is
called an extended social choice correspondence (ESCC). The ESCC satisfies unanimity if, when
there is a weakly Pareto dominant alternative, the ESCC selects this alternative. Stability requires
that the ESCC is immune to manipulation through withdrawal of some alternatives. Independence
implies that the ESCC selects the same outcome from a subset of the set of alternatives for two
preference profiles that are the same on this set. We characterize the ESCC satisfying the three axioms, when the set of alternatives is finite but includes more than three alternatives, and the set of
voters can have any cardinality. Our main theorem establishes that the ESCC satisfying the three
axioms is a serial dictatorship à la Eraslan and McLennan (2004). Our second theorem shows
that a serial dictatorship includes “invisible serial dictators” à la Kirman and Sondermann (1972).
Key Words: Impossibility theorem, Social choice, Serial dictatorship, Ultrafilter.
JEL Classification Numbers: D71.
∗
We would like to thank Andrew McLennan for continuous discussion and many suggestions to improve this paper. We are also grateful to Simon Grant, Chiaki Hara, Atsushi Kajii, John Quah, Satoru Takahashi, Rabee Tourky,
and other participants in the Australasian Economics PhD Workshop, and the Econometric Society Australasian
Meeting held in Sydney. We similarly acknowledge the useful suggestions of seminar participants at Kyoto University and the National University of Singapore. Takayama also gratefully acknowledges the hospitality of the Kyoto
Institute of Economic Research and the University of British Columbia. All errors remaining are our own.
†
Contact: [email protected] (Takayama)
1
1
Introduction
In this paper, we consider social choice correspondences assigning a set of choices to each pair
consisting of a nonempty subset of the set of alternatives and a weak preference profile. We
call such an object an extended social choice correspondence (ESCC) and impose three axioms,
namely unanimity, stability, and independence. The ESCC satisfies unanimity if, when there is
a weakly Pareto dominant alternative, the ESCC selects this alternative. Stability requires that
the ESCC is immune to manipulation through withdrawal of some alternatives. Finally, independence implies that the ESCC selects the same outcome from a subset of the set of alternatives for
two preference profiles that are the same on this set. We characterize the ESCC satisfying the
three axioms, when the set of alternatives is finite but includes more than three alternatives, and
the set of voters can have any cardinality. Our main theorem establishes that the ESCC satisfying
the three axioms is a serial dictatorship à la Eraslan and McLennan (2004). Our second theorem
shows that a serial dictatorship includes “invisible serial dictators” à la Kirman and Sondermann
(1972).
Man and Takayama (2013) extend a social choice correspondence to an ESCC, characterize
the ESCC satisfying the three axioms in the case of finitely many agents and derive many wellknown impossibility theorems as corollaries to their main theorem, including Arrow’s Impossibility Theorem (Arrow, 1959), the Gibbard–Satterthwaite Theorem (Gibbard, 1973; Satterthwaite,
1975), the theorem of serial dictators (Grether and Plott, 1982; Dutta, Jackson, and Le Breton,
2001; Eraslan and McLennan, 2004), and the characterization of game-theoretic solutions implementing only dictatorial social choices by Jackson and Srivastava (1996). In each case, the
method of proof is to prove that the setting given by each theorem can be extended to, or used to
derive, a “desirable” ESCC satisfying the three axioms.
With infinitely many voters, the method taken by Man and Takayama (2013) to identify each
individual dictator does not work, because to identify an individual dictator, they use a Condorcet
cycle and shrink a group that contains a dictator to one individual by changing preference profiles. Given a social welfare function satisfying Arrow’s axioms of independence and unanimity,
Fishburn (1970) shows that when the number of agents is infinite, it is impossible to identify a
dictator whose preference dictates the outcome. In this context, Kirman and Sondermann (1972)
show that, even in the case of infinitely many agents, a set of decisive coalitions is an ultrafilter
and, in a sense, a dictatorship still persists. A filter on a ground set X is a collection of subsets
of X that satisfies the three conditions: (1) the empty set is not included in the collection; (2) a
superset of a set in the collection is also included; and (3) an intersection of finitely many sets in
the collection is also included. An ultrafilter is a largest filter, which means that, if a set is not a
member of the filter, then its complement is a member. An ultrafilter is called principal if it is
generated by one individual, otherwise, it is called non-principal. As in Kirman and Sondermann
2
(1972), a non-principal ultrafilter can be thought of as an invisible dictator.
Our main theorem extends the result of invisible dictators by Kirman and Sondermann (1972)
to a domain of weak preference profiles, and derives a serial dictatorship that generalizes the one
in Eraslan and McLennan (2004) by using a coherent hierarchy of ultrafilters.
The existence of non-principal ultrafilters is a nontrivial consequence of the axiom of choice.
In our analysis, the similar issue arises of how rich is the space of coherent hierarchies of ultrafilters. After the characterization of a serial dictatorship, the question remains if there exist
invisible serial dictators that correspond to a coherent hierarchy of non-principal ultrafilters. By
using Zorn’s lemma (equivalent to the axiom of choice), we show that a coherent hierarchy of
ultrafilters assigning non-principal ultrafilters to infinite sets exists. By our first main theorem,
this result implies that there exists an ESCC that induces invisible serial dictators.
Since Schmeidler and Sonnenschein (1974), there have been studies on the relationship between Arrow’s Impossibility Theorem and the Gibbard–Satterthwaite Theorem. Before the recent study by Man and Takayama (2013), Reny (2001) proves Arrow’s Impossibility Theorem
and the Gibbard–Satterthwaite Theorem in parallel arguments. Vohra (2011) proves Arrow’s
Impossibility Theorem, and then uses it to prove the Gibbard–Satterthwaite Theorem, and the
impossibility theorem of strategic candidacy, by adopting integer programming techniques. In
the case of finitely many agents, it is known that when preferences are strict, these two theorems share some common bases, whereas when preferences are weak, the Gibbard–Satterthwaite
Theorem does not hold1 . In this paper, we show that in the case of infinitely many agents, the
Gibbard–Satterthwaite Theorem does not hold, even when preferences are strict. This implies that
in the case of infinitely many agents, while a dictatorship persists when unanimity, independence,
and stability hold, the Gibbard–Satterthwaite Theorem does not hold, even in strict preferences.
Moreover, as our main theorem is not confined to the case of infinitely many agents, the direct
consequence is the existence of serial dictatorship in the case of finitely many agents. In the
literature, many preceding works (Geanakoplos, 2005; Yu, 2012; Barberà, 1980, 1983; Campbell
and Kelly, 2002, for a comprehensive survey) provide another proof of Arrow’s Impossibility
Theorem in the case of finitely many agents.
The organization of the paper is as follows. Section 2 presents the model. Section 3 defines
a generalized serial dictatorship. Section 4 provides the main theorem and proves it. Section 5
shows the existence of a hierarchy of non-principal ultrafilters. The last section concludes by
providing an example to show that the Gibbard–Satterthwaite Theorem does not hold.
1
An easy example is as follows. Let X = {a, b, c} and suppose that there are three agents. Let a social choice
function (SCF) be f : R3 → X. Suppose that f selects outcomes so that it meets the following requirement: (1)
if the set of most preferred alternatives for the first dictator 1 includes a, then agent 2 is the second dictator; (2)
otherwise, agent 3 is the second dictator. We can verify that f satisfies the axioms for the theorem (which are stated
in the last section of this paper), although this is not a serial dictatorship.
3
2
The Model
Let X be the set of potential alternatives. We assume that 3 ≤ |X | < ∞. Let N be the set of
agents. Let R be the space of weak preferences over X . Let %i ∈ R be agent i’s preference, and
let %∈ RN be a preference profile of all the agents in N . For each x, y ∈ X , we say that x i y
if x is strictly preferred to y, i.e., x %i y but not y %i x, and that x ∼i y if x is indifferent to y,
i.e., x %i y and y %i x.
For an arbitrary set Z, let P(Z) ≡ 2Z \{∅}. An environment is a pair (X, %) ∈ P(X )×RN .
An ESCC is a correspondence φ : P(X ) × RN → X such that, for each X ∈ P(X ),
φ(X, %) ∈ P(X).
Next, we define three properties that we impose on φ. Alternative x ∈ X is called weakly
Pareto dominant in X if, for every y ∈ X, x %i y for every i ∈ N , and x j y for at least one
j ∈ N.
Definition (Unanimity). An ESCC φ satisfies unanimity if φ(X, %) = {x} holds for all %∈ RN
such that x ∈ X is weakly Pareto dominant in X.
We say that % and %0 ∈ RN agree on X if % and %0 are the same on X, which we denote by
%=X %0 .
Definition (Independence). An ESCC φ satisfies independence if, for each X ∈ P(X ) and each
%, %0 ∈ RN , % =X %0 implies φ(X, %) = φ(X, %0 ).
Definition (Stability). An ESCC φ satisfies stability if, for each X, X 0 ∈ P(X ) and each %∈
RN , X 0 ⊂ X and φ(X, %) ∩ X 0 6= ∅ imply φ(X 0 , %) = φ(X, %) ∩ X 0 .
3
A Generalized Serial Dictatorship
In the literature of social choice theory, Eraslan and McLennan (2004) define a serial dictatorship
in the case of finitely many agents. To extend their results in the case where the number of voters
can be infinite, this section defines a “generalized serial dictatorship.” To do this, we start by
introducing our tool, an ultrafilter.
Definition (Ultrafilter). A family F ⊂ 2N is an ultrafilter of a set N ⊂ N if
(1) ∅ ∈
/ F.
(2) If S ∈ F , and S 0 ⊃ S, then S 0 ∈ F .
4
(3) If S, S 0 ∈ F , then S ∩ S 0 ∈ F .
(4) If S ⊂ N, then either S ∈ F or N \S ∈ F .
An ultrafilter is called principal or non-principal when the intersection of all its members is
nonempty or empty, respectively. If F is principal, then F = {S ⊂ N : i ∈ S} for some i ∈ N .
If N is finite, all ultrafilters are principal.
For each N ∈ P(N ) and %∈ RN , let %|N = (%i )i∈N ∈ RN . Let UN be an ultrafilter over
N . For each R ∈ R, let UR ≡ {i ∈ N | %i = R}, that is, UR is the set of agents holding the
preference R. As X is finite, there are finitely many possibilities for R, and so {UR ⊂ N | R ∈
R} is a finite partition of N . Then, by properties (2) and (4) of ultrafilters, exactly one set in the
partition belongs to UN . Thus, there exists a unique R∗ ∈ R such that UR∗ ∈ UN . Then, we can
use the preference of agents in UR∗ to represent ultrafilter UN ’s preference, which we denote by
%UN .
Definition. A coherent hierarchy of ultrafilters U assigns an ultrafilter UN over N to each N ∈
P(N ) such that, for all N, N 0 ∈ P(N ), if N ∈ UN 0 , then M ∈ UN 0 if and only if M ∩ N ∈ UN
for all M ∈ P(N ).
To explain the concept more clearly, we provide a simple example of violation of coherence.
Let N = {1, 2, 3, 4}. Suppose that we have identified a serial dictatorship and each number
i ∈ N corresponds to their ranking within the dictatorship. For example, agent 1 is the first
dictator in the serial dictatorship. In our definition, it implies UN = {U : 1 ∈ U }. Suppose that
there are two preference profiles % and %0 , each of which yields the following two cases:
• U1 = {1, 4} and U{2,3} = {U : 2 ∈ U }, and
• U10 = {1} and U{2,3,4} = {U : 3 ∈ U }.
Coherence is not satisfied by U{2,3} and U{2,3,4} , because {3, 4} ∈ U{2,3,4} but {3, 4}∩{2, 3} 6∈
U{2,3} . The problem with the serial dictatorship is that even if agent 2 has not yet been chosen
by the preceding ultrafilter UN , the next ultrafilter U{2,3,4} under %0 exclusively chooses a set
including agent 3. In this sense, agent 3 is chosen as a dictator before agent 2 in the second
dictatorship that U{2,3,4} defines.
To formally define a generalized serial dictatorship, we define a top set as follows: for each
R ∈ R and nonempty X ⊂ X , let
T op(X, R) ≡ {x ∈ X | for each y ∈ X, x R y}.
Finally, we provide a formal definition.
5
Definition. An ESCC φ is a generalized serial dictatorship if there exists a coherent hierarchy of
ultrafilters U and a tiebreaking rule ρ ∈ R such that, for each (X, %) ∈ P(X ) × RN , there
K(X,%)
exist K(X, %) ∈ N+ , and a series {Tk , Ck }k=0
satisfying:
1. T0 = X and C0 = N ,
2. for each k ∈ {0, · · · , K(X, %) − 1}, Ck 6= ∅,
3. for each k ∈ {1, · · · , K(X, %)}, Tk = T op(Tk−1 , %U (Ck−1 ) ),
4. for each k ∈ {1, · · · , K(X, %)}, Ck = {i ∈ N | Tk 6⊂ T op(Tk−1 , %i )},
5. CK(X,%) = ∅ and φ(X, %) = T op(TK(X,%) , ρ).
Having stated the definition of a generalized serial dictatorship in the environment where the
number of agents can be infinite, we consider the connection of our definition with the definition
of serial dictatorship for the case of finitely many agents found in Eraslan and McLennan (2004)
by applying a concept of a non-principal or principal ultrafilter.
We say that there exists a hierarchy of dictators {i1 , . . . , i|N | }, which is a permutation on N
with |N | < ∞, such that for all (X, %) ∈ P(X ) × RN , φ(X, %) ⊂ T op(Tn−1 , %in ) where
T0 = X and Tn = T op(Tn−1 , %in ) for n = 1, . . . , |N |, and φ(X, %) = T op(Ti|N | , ρ).
T
Next, let π1 = {C : C ∈ U (C 0 )}. When the ground set is finite, an ultrafilter is principal.
Thus, when an ESCC φN is a generalized serial dictatorship, π1 is a singleton set and only contains i1 , who is the first dictator in the definition of Eraslan and McLennan (2004). Recursively,
we can obtain a permutation of individuals.
4
The Main Theorem
Here, we state our main theorem.
Theorem 1. An ESCC φ satisfies unanimity, independence, and stability if and only if φ is a
generalized serial dictatorship.
4.1
The Proof of the “If” Part
In this subsection, we assume that φ is a generalized serial dictatorship associated with a coherent
hierarchy of ultrafilters U . In each of the following three lemmas, we establish that φ satisfies
one of the axioms.
Lemma 1. If φ is a generalized serial dictatorship, then φ satisfies unanimity.
6
Proof. Let X ∈ P(X ) and %∈ RN . Suppose that x is weakly Pareto dominant in X. Then, for
each k ∈ {0, · · · , K(X, %)}, we have that x ∈ Tk . On the other hand, for each y 6= x, let iy ∈ N
be such that x iy y. Then, there exists ky ∈ {0, · · · , K(X, %) − 1} such that %UCk =%iy . It
y
implies that y ∈
/ Tky +1 . Therefore, TK(X,%) = {x}. Thus, we have that
φ(X, %) = T op(TK(X,%) , ρ) = {x}. Lemma 2. If φ is a generalized serial dictatorship, then φ satisfies independence.
Proof. The result is direct, because T op(X, %) depends only on %|X . Lemma 3. If φ is a generalized serial dictatorship, then φ satisfies stability.
0 L
0
Proof. Let X ⊂ X 0 ⊂ X and %∈ RN . Let {Tm , Cm }M
m=0 and {Tl , Cl }l=0 be two sequences
defined by a generalized serial dictatorship for (X, %) and (X 0 , %), respectively. For each m ∈
{1, · · · , M } and l ∈ {1, · · · , L}, let
0
Um = {i ∈ N | Tm = T op(Tm−1 , %i )} and Ul0 = {i ∈ N | Tl0 = T op(Tl−1
, %i )}.
(1)
Suppose that φ(X 0 , %) ∩ X 6= ∅. We want to show that TM = TL0 ∩ X. Because C0 = C00 ,
%UC0 =%UC 0 and T0 = T00 ∩ X. By induction, take m ∈ {1, · · · , M } and suppose that there exists
0
an l∗ ∈ {1, · · · , L} such that Tm−1 = Tl0∗ ∩ X, Cm−1 ⊂ Cl0∗ , and %UCm−1 =%UC 0 .
l∗
Because Tm = T op(Tm−1 , %UCm−1 ) and Tl0∗ +1 = T op(Tl0∗ , %UC 0 ), we have that Tm = Tl0∗ +1 ∩
l∗
X. By (1) and the definition of a generalized serial dictatorship,
Ul0∗ +1 = {i ∈ N | Tl0∗ +1 = T op(Tl0∗ , %i )} ∈ UCl0∗ .
(2)
For each i ∈ Ul0∗ +1 , as Tm = Tl0∗ +1 ∩ X,
T op(Tm−1 , %UCm−1 ) = T op(Tl0∗ , %UC 0 ) ∩ X
(%UCm−1 =%UC 0 )
l∗
=
T op(Tl0∗ , %i )
∩X
= T op(Tm−1 , %i ),
l∗
(i ∈
Ul0∗ +1
and (2))
(Tm−1 = Tl0∗ ∩ X),
which implies that Ul0∗ +1 ⊂ Um by (1). As Cm = Cm−1 \Um , Cl0∗ +1 = Cl0∗ \Ul0∗ +1 and Cm−1 ⊂ Cl0∗ ,
Cm ⊂ Cl0∗ +1 . Then,
{i ∈ N | %i =%UCm } ∈ UCm and {i ∈ N | %i =%UC 0
l∗ +1
} ∈ UCl0∗ +1 .
Now, we have to consider two cases: (1) Cl0∗ +1 \Cm ∈
/ UCl0∗ +1 ; and (2) Cl0∗ +1 \Cm ∈ UCl0∗ +1 .
(Case 1) Suppose that Cl0∗ +1 \Cm ∈
/ UCl0∗ +1 . By coherence, {i ∈ N | %i =%UC 0 } ∩ Cm ∈ UCm .
l∗ +1
It implies that {i ∈ N | %i =%UC 0
l∗ +1
} ∩ {i ∈ N | %i =%UCm } =
6 ∅. Hence, %UCm =%UC 0
l∗ +1
7
.
(Case 2) Suppose that Cl0∗ +1 \Cm ∈ UCl0∗ +1 . Then, there exists an i∗ ∈ Cl0∗ +1 \Cm such that
%i∗ =%UC 0 . Therefore, %Cm 6=%UC 0 . Note that
lm
lm
Tl0∗ +2 = T op(Tl0∗ +1 , %UC 0
l∗ +1
) = T op(Tl0∗ +1 , %i∗ ).
(3)
On the other hand, because i∗ ∈ Um , we have that
Tm = T op(Tm−1 , %i∗ ).
(4)
Because Tm ⊂ Tl0∗ +1 because Tm = Tl0∗ +1 ∩ X, we have that
Tm = T op(Tm−1 , %i∗ ) ⊂ T op(Tl0∗ +1 , %i∗ ) = Tl0∗ +2 .
(5)
Let Ul0∗ +2 = {i ∈ N | Tl0∗ +2 = T op(Tl0∗ +1 , %i )}. Because Tm ⊂ Tl0∗ +2 ⊂ Tl0∗ +1 , we have that
Ul0∗ +2 ⊂ Um , because if i ∈ Ul0∗ +2 and a ∈ Tm , then a ∈ Tl0∗ +2 , which indicates i ∈ Um by (5).
By definition, Cm ⊂ Cl0∗ +2 = Cl0∗ +1 \Ul0∗ +2 . Because there are finitely many different preferences
among agents in Cl0∗ +2 \Cm , we can repeat this elimination of Cl0∗ +j for j ∈ {1, · · · , J} at most
finite J times until Cl0∗ +J \Cm ∈
/ UCl0∗ +1 . Therefore, there exists an l∗∗ ∈ {1, · · · , L} such that
Tm = Tl0∗∗ ∩ X, Cm ⊂ Cl0∗∗ , and %UCm =%UC 0 . Hence, by induction, for each m ∈ {0, · · · , M },
l∗∗
there exists an lm ∈ {0, · · · , L} such that Tm = Tl0m ∩ X, Cm ⊂ Cl0m , and %UCm =%UC 0 .
lm
0 L
0
Then, take a subsequence {Tl0m , Cl0m }M
m=0 of the sequence {Tl , Cl }l=0 such that, for each
m ∈ {0, · · · , M }, Tm = Tl0m ∩ X, Cm ⊂ Cl0m , and %UCm =%UC 0 . If Cl0M = ∅, then Cl0M = CL0 ,
lm
which indicates TM = TL0 ∩ X and φ(X, %) = φ(X 0 , %) ∩ X. If Cl0M 6= ∅, by the same argument
with (5),
TM ⊂ Tl0M +1 = T op(Tl0M , %Cl0 ).
M
Tl0M +1
Therefore, TM =
∩ X. Recursively, as lM ∈ {0, · · · , L}, we obtain TM = TL0 ∩ X and
φ(X, %) = φ(X 0 , %) ∩ X. 4.2
The Proof of the “Only If” Part
Because Kirman and Sondermann (1972) only deal with the strict preferences and use a social
welfare function, there are two issues that we need to take care of. First, we show that there exists
a social preference whose top set on an arbitrary nonempty set X selects the same set chosen by
the ESCC satisfying the three axioms. Then, we extend the proof in Kirman and Sondermann
(1972) to the full domain of preferences.
An environment for N is a pair (X, %|N ) ∈ P(X )×RN . An ESCC for N is a correspondence
φN : P(X )×RN → X such that, for each nonempty X ⊂ X , φN (X, %|N ) ∈ P(X). As before,
we simply denote an ESCC for N by φ.
8
Lemma 4. If φN satisfies unanimity, independence, and stability, then, for each %|N ∈ RN , there
exists an R ∈ R such that, for each X ⊂ X with |X| ≥ 2, φN (X, %|N ) = T op(X, R).
Proof. Let %|N ∈ RN . We define a binary relationship R on X by xRy if x ∈ φN ({x, y}, %|N ).
Then, because φN ({x, y}, %|N ) 6= ∅, R is complete.
Next, we show transitivity. Suppose that x, y, z ∈ X satisfy that xRy and yRz. Then
x ∈ φN ({x, y}, %|N ) and y ∈ φN ({y, z}, %|N ).
We need to show x R z, namely x ∈ φN ({x, z}, %|N ). Suppose that this is not true. Then, by
stability,
x∈
/ φN ({x, y, z}, %|N ).
Hence, y ∈
/ φN ({x, y, z}, % |N ), because otherwise φN ({x, y}, % |N ) = {y} would hold by
stability. Hence, φN ({x, y, z}, %|N ) = {z}, which implies that y ∈
/ φN ({y, z}, %|N ) by stability.
This is a contradiction. Thus, R is complete and transitive, that is, a weak preference.
Finally, we show that φN is the top set of R. Let X ⊂ X with |X| ≥ 2. First, suppose
that there exists x ∈ φN (X, %|N ) but x ∈
/ T op(X, R). Then, there exists a y ∈ X such that
y ∈ φN ({x, y}, % |N ) but x ∈
/ φN ({x, y}, % |N ) by the definition of R. However, because
{x, y} ⊂ X, x ∈ φN ({x, y}, %|N ) by stability. This is a contradiction. Thus, φN (X, %|N ) ⊂
T op(X, R). Next, suppose that x ∈ T op(X, R). Then, for each y ∈ X, xRy, which implies
that, for each y ∈ X, x ∈ φN ({x, y}, % |N ) and, thus, x ∈ φN (X, % |N ) by stability. Hence,
T op(X, R) ⊂ φN (X, %|N ). By Lemma 4, we can define a social welfare function R : RN → R such that, for each
X ∈ P(X ) with |X| ≥ 2,
φN (X, %|N ) = T op(X, R(%|N )).
(6)
We write that xP%|N y if x is strictly preferred to y, and that xI%|N y if x is indifferent to y under
R(%|N ), respectively.
For each U ⊂ N ∈ P(N ), x, y ∈ X and each %|N ∈ RN , we denote x %|U y if, for each
i ∈ U , x %|i∈N y. We consider three families of groups of agents.
• FN is the set of subsets U ⊂ N such that, for all x, y ∈ X , and for all % |N ∈ RN ,
x |U y and y %|N \U x imply xP%|N y;
• FN0 is the set of subsets U ⊂ N such that there exist distinct x, y ∈ X where, for all
%|N ∈ RN , x |U y and y %|N \U x imply xP%|N y;
9
• FN00 is the set of subsets U ⊂ N such that there exist distinct x, y ∈ X and an %|N ∈ RN
where the following hold: (1) x |U y, (2) y ∼|Ũ x, (3) y |N \(U ∪Ũ ) x, and (4) xP%|N y for
all Ũ ⊂ N \U .
Note that each member of FN is a decisive coalition in the sense that as long as they strictly
prefer some alternatives to others, those unfavored alternatives are never chosen.
We now show that FN is an ultrafilter on N . We prove it by showing that (1) FN = FN00 and
(2) FN00 is an ultrafilter. By definition, it is clear that
FN ⊂ FN0 ⊂ FN00 .
(7)
The following two lemmas show that the inverse inclusions also hold and these sets are not
empty.
Lemma 5. N ∈ FN00 .
Proof. In the definition of FN00 , let U = N and suppose that (1), (2), and (3) hold for some
x, y ∈ X and %|N ∈ RN . Then, by unanimity, we obtain φN (X, %|N ) = {x}, which implies (4)
xP%|N y by (6). Thus, we obtain N ∈ FN00 .
Lemma 6. FN00 ⊂ FN0 .
Proof. Let U ∈ FN00 and U 0 ≡ N \U . Let %|N ∈ RN . Suppose that
x |U y and y %|U 0 x
(8)
at %|N . Now, let Ũ ≡ {i ∈ U 0 | x ∼i y}, and let U 00 ≡ U 0 \Ũ . Then, because U ∈ FN00 , there
exist x, y ∈ X and %α ∈ RN such that x αU y, y ∼αŨ x, y αU 00 x, and xP%α y, which implies
%α ={x,y} % |N by (8). Thus, by independence, φN ({x, y}, % |N ) = φN ({x, y}, %α ). Because
xP%α y, φN ({x, y}, %|N ) = φN ({x, y}, %α ) = {x}. This implies that xP%|N y. Thus, U ∈ FN0 .
Lemma 7. FN0 ⊂ FN .
Proof. Let U ∈ FN0 , and let U 0 ≡ N \U . Then, there exist distinct x, y ∈ X such that
for all %|N ∈ RN , x |U y and y %|U 0 x imply xP%|N y.
Let z ∈ X \{x, y} and %|N ∈ RN . Suppose that z |U y and y %|U 0 z. Now, we want to
show that z P%|N y. Let %β ∈ RN such that z βU x βU y, y %βU 0 z βU 0 x, and, for each
i ∈ U 0 , %i ={y,z} %βi . Then, by unanimity, φN ({x, z}, %β ) = {z}. This implies that zP%β x. Now
x βU y and y βU 0 x. Because U ∈ FN0 , we have xP%β y. Then, we have zP%β y by transitivity.
10
Because %|N ={y,z} %β , we have that φN ({y, z}, %|N ) = φN ({y, z}, %β ). Therefore, zP%|N y,
which shows that the preference of U dictates the social welfare between arbitrary z and y.
Finally, we show that it also dictates the social welfare between any arbitrary z and w. Let
0
0
0
w ∈ X \{x, y, z}. Suppose that z |U w and w %|U 0 z. Let %β ∈ RN such that z βU y βU w,
0
0
0
0
w %βU 0 z βU 0 y, and, for each i ∈ U 0 , %i ={w,z} %βi . By unanimity, φN ({y, z}, %β ) = {z}. So,
0
0
zP%β0 y. Now, y βU w and w βU 0 y. By the first part of this proof, we have yP%β0 w, which
implies zP%β0 w by transitivity. The rest follows in the same way as for the first part of this proof,
by replacing y with w, and we obtain zP%|N w. Because %|N is arbitrary, we obtain U ∈ FN . Together with (7), Lemma 6 and Lemma 7 establish the following relationship.
Corollary 1. FN = FN0 = FN00
Finally, we show that FN is an ultrafilter.
Proposition 1. FN is an ultrafilter.
Proof. By Corollary 1, it is enough to show that FN00 is an ultrafilter.
(Property 1) On the contrary, suppose ∅ ∈ FN00 . By taking U = ∅ and Ũ 6= N , there exist
x, y ∈ X and %|N ∈ RN such that y %|Ũ x, y |N \Ũ x, and x P%|N y. This contradicts unanimity.
(Property 3) Let W1 , W2 ∈ FN00 . We separate N into four disjoint subsets:
(1) V1 ≡ W1 ∩ W2 ; (2) V2 ≡ W1 \V1 ; (3) V3 ≡ W2 \V1 ; (4) V4 ≡ N \(W1 ∪ W2 ).
Let {a, b, c} ⊂ X and Ṽ1 ⊂ N \V1 . Then, we can find %γ ∈ RN such that (1) c γV1 a γV1 b;
(2) a γV2 b %γV2 c; (3) b %γV3 c γV3 a; (4) b %γV4 a %γV4 c; (5) b ∼γṼ1 c; (6) b γN \(V ∪Ṽ1 ) c.
Note that a γW1 b, and b %γN \W1 a. By Corollary 1, we have W1 ∈ FN00 = FN . Therefore,
aP%γ b. In the same way, W2 ∈ FN , c γW2 a, and a %γN \W2 c. This implies that cP%γ a. By
transitivity, we have cP%γ b. Now, we also have that c γV1 b, b ∼γṼ1 c, and b γN \(V ∪Ṽ1 ) c. Thus,
by the definition of FN00 , V1 = W1 ∩ W2 ∈ FN00 .
(Property 4) Next, we show that for each V ⊂ N , V ∈ FN00 or N \V ∈ FN00 . If V ∈ FN00 , then
the result is immediate. Hence, we suppose that V ∈
/ FN00 . Then, as FN00 6= ∅ by Lemma 5 and
FN00 = FN0 by Corollary 1, for some {a, b, c} ⊂ X , there exists a Ṽ ⊂ N \V with Ṽ ∈ FN00 such
that for each %|N ∈ RN , if b |V a, a ∼|Ṽ b, and a |N \(V ∪Ṽ ) b, then aP%|N b. Now, let V̂ ⊂ V .
Then, we can find %δ ∈ RN such that (1) b δV̂ c ∼δV̂ a; (2) b δV \V̂ c δV \V̂ a; (3) a ∼δṼ b δṼ c;
(4) a δN \(V ∪Ṽ ) b δN \(V ∪Ṽ ) c.
By unanimity, bP%δ c. Note that b δV a, a ∼δṼ b, and a δN \(V ∪Ṽ ) b. By our initial assumption
of V ∈
/ FN00 and Ṽ ∈ FN00 , we have that aP%δ b. Thus, aP%δ c. At the same time, we have that
a δN \V b, c ∼δV̂ a, and c δV \V̂ a. Because V̂ is an arbitrary subset of V , by the definition of FN00 ,
we have that N \V ∈ FN00 .
11
(Property 2) On the contrary, suppose that there exist W, U ⊂ N with U ⊂ W such that U ∈ FN00
but W ∈
/ FN00 . Then, by Property 4, N \W ∈ FN00 . By Property 3, we have that U ∩(N \W ) ∈ FN00 .
However, U ∩ (N \W ) = ∅, which contradicts Property 1. The following lemma completes the proof of the “only if” part.
Lemma 8. If φ satisfies unanimity, independence, and stability, then there exists a coherent
hierarchy of ultrafilters U such that, for each (X, %) ∈ P(X )×RN , there exist K(X, %) ∈ N+
K(X,%)
and {Tk , Ck }k=0
satisfying 1-4 in the definition of the generalized serial dictatorship.
Proof. Suppose that φ satisfies unanimity, independence, and stability. For each N ∈ P(N ) and
%|N ∈ RN , let %|N = (%|1∈N , · · · , %|i∈N , · · · , %||N |∈N ), and define %∈ RN by
for each i ∈ N, %i =%|i∈N ,
for each i ∈ N \N, for each x, y ∈ X , x ∼i y.
Let ψN be an ESCC such that, for each (X, %|N ) ∈ P(X ) × RN , ψN (X, %|N ) ≡ φ(X, %).
We show that ψN satisfies unanimity, independence, and stability. Let x ∈ X and %|N ∈ RN .
(Unanimity) Suppose that for each i ∈ N and y ∈ X, x %|i∈N y, and there exists i∗ ∈ N such
that x %|i∗ ∈N y. Then, by construction, we have that for each i ∈ N and y ∈ X, x %i y. Because
i∗ ∈ N , we have that %|i∗ ∈N =%i∗ . Therefore, x i∗ y. By unanimity of φ, φ(X, %) = {x}.
This implies that ψN (X, %|N ) = {x}.
(Independence) Let X ⊂ X , and %|N , %̃|N ∈ RN be such that %|N =X %̃|N . Then, for each
i ∈ N , we have that %|i∈N =X %̃|i∈N . For each i ∈ N \N , i is indifferent among X in both
cases. Therefore, %|N =X %δ . Thus, we have that
ψN (X, %|N ) = φ(X, %) = φN (X, %δ ) = ψN (X, %̃|N ).
The second equality is due to independence of φ.
(Stability) Let Y ⊂ X ⊂ X with Y 6= ∅. Let %|N ∈ RN . Suppose that ψN (X, %|N ) ∩ Y 6= ∅.
Then, φ(X, %) ∩ Y 6= ∅. Because φ satisfies stability,
φ(Y, %) = φ(X, %) ∩ Y.
Therefore, ψN (Y, %|N ) = ψN (X, %|N ) ∩ Y .
By Proposition 1, the set of decisive coalitions is an ultrafilter over N . We denote the ultrafilter as UN . Define a correspondence U : P(N ) → P(N ) such that, for every N ∈ P(N ),
U (N ) = UN , that is, U assigns a dictatorial ultrafilter UN , which is the set of decisive coalitions,
to each nonempty N . We show that U is a coherent hierarchy of ultrafilters.
Let N, N 0 ∈ P(N ) with N ⊂ N 0 . For each %|N ∈ R, let %|N 0 be such that (1) for each
i ∈ N , % |i∈N =% |i∈N 0 , and (2) for each i ∈ N 0 \N and x, y ∈ X , x ∼ |N 0 y. Then, we
12
have that for each (X, %|N ) ∈ P(X ) × RN , ψN (X, %|N ) = ψN 0 (X, %|N 0 ). Suppose that for
U ∈ UN 0 , it holds that U ⊂ N . By construction, for each %∈ RN and x, y ∈ X , if x U y,
ψN 0 ({x, y}, % |N 0 ) = {x}. This implies that, for each % |N ∈ RN and x, y ∈ X , if x U y,
ψN ({x, y}, %|N ) = {x}. Thus, U ∈ UN .
Now, suppose that N 0 \N ∈
/ UN 0 . Because UN 0 is an ultrafilter, N ∈ UN 0 . If M 0 ∈ UN 0 , then
N ∩ M 0 ∈ UN 0 . The above argument implies that N ∩ M 0 ∈ UN 0 . If N ∩ M 0 ∈ UN , we want to
show that N ∩M 0 ∈ UN 0 . Suppose this is not true. Because N ∈ UN 0 , we have that N \M 0 ∈ UN 0 .
Because N \M 0 ⊂ N , we have that N \M 0 ∈ UN . However, we assume that N ∩ M 0 ∈ UN . This
is a contradiction. Thus, N ∩ M 0 ∈ UN 0 . Because N ∩ M 0 ⊂ M 0 , we have that M 0 ∈ UN 0 .
It remains to show that φ is a generalized serial dictatorship. Let (X, %) ∈ P(X ) × RN . Let
T0 ≡ X and C0 ≡ N . Because ultrafilter UC0 picks out only one set from the equivalent classes
with respect to their preferences, there exists a unique %◦ ∈ R such that {i ∈ N | %i =%◦ } ∈
UC0 . We denote this %◦ as %UC0 . Then, we have that φ(X, %) ⊂ T1 ≡ T op(T0 , %UC0 ).
Next, let C1 ≡ {i ∈ N | T1 6⊂ T op(T0 , %i )}. Let %C1 ∈ RC1 be such that, for each
C1
1
i ∈ C 1 , %C
i =%i . Then, by construction, ψC1 (T1 , % ) = φ(T1 , %). By stability, we have
that ψC1 (T1 , %C1 ) = φ(X, %). In a similar fashion to %UC0 , we define %UC1 so that we obtain
ψC1 (T1 , %C1 ) ⊂ T2 ≡ T op(T1 , %UC1 ). Because the equivalent classes with respect to the agents’
preferences are only finite, this process ends at finite K(X , %) steps. Thus, we construct our
K(X ,%)
desired sequence, {Tk , Ck }k=0 . 4.2.1
A Tiebreaking Rule
Finally, we show the existence of a tiebreaking rule.
Lemma 9. There exists a ρ ∈ R, such that, for each X ∈ P(X ) and %∈ RN , φ(X, %) =
T op(TK(X,%) , ρ).
Proof. Let ρ be a binary relation on X such that, for each x, y ∈ X , xρy if and only if there
exists a %∈ RN such that {x, y} ⊂ TK(X ,%) and x ∈ φ(X , %). First, we show that ρ is a weak
preference.
Let x, y ∈ X . Then, we can find %̃ ∈ RN such that, for each i ∈ N and z ∈ X \{x, y},
˜ i z. Then, TK(X ,%̃) = {x, y}. Because φ(X , %̃) ⊂ TK(X ,%̃) and φ(X , %̃) 6= ∅, we
x∼
˜ i y and x
have x ∈ φ(X , %̃) or y ∈ φ(X , %̃), that is, xρy or yρx. Therefore, ρ is complete.
Next, we show transitivity. Let x, y, z ∈ X be such that xρy and yρz. Then, there exists a %∈ RN such that {x, y} ⊂ TK(X ,%) and x ∈ φ(X , %). In the same way, there also
exists a %0 ∈ RN such that {y, z} ⊂ TK(X ,%0 ) and y ∈ φ(X , %0 ). Then, by stability, x ∈
φ({x, y}, %) and y ∈ φ({y, z}, %0 ). Now, let %∗ ∈ RN be such that for each i ∈ N and each
w ∈ X \{x, y, z}, x ∼∗ y ∼∗i z and z ∗i w. Then, TK(X ,%∗ ) = {x, y, z}, which implies that
13
φ(X , %∗ ) ⊂ {x, y, z}. By stability, we have that φ(X , %∗ ) = φ({x, y, z}, %∗ ). Thus, it is enough
to show that x ∈ φ({x, y, z}, %∗ ). Suppose that x ∈
/ φ({x, y, z}, %∗ ). Because %∗ ={y,z} %0 ,
we have that y ∈ φ({y, z}, %0 ) = φ({y, z}, %∗ ). Because x ∈
/ φ({x, y, z}, %∗ ), stability implies that y ∈ φ({x, y, z}, %∗ ). Otherwise, φ({y, z}, %∗ ) = {z}, which is a contradiction. By
stability again, we have that φ({x, y}, %∗ ) = φ({x, y, z}, %∗ ) ∩ {x, y} = {y}. However, note
that %∗ ={x,y} %. Thus, we have that φ({x, y}, %) = φ({x, y}, %∗ ). Because xρy, we have that
x ∈ φ({x, y}, %) = φ({x, y}, %∗ ), which is a contradiction.
It remains to show that, for each X ∈ P(X ) and %∈ RN , φ(X, %) = T op(TK(X,%) , ρ). First,
we show that, for each X ∈ P(X ) and each %∈ RN , φ(X, %) ⊂ T op(TK(X,%) , ρ). Let x ∈
φ(X, %) and y ∈ T op(TK(X,%) , ρ). Then, {x, y} ⊂ TK(X,%) and, by stability, x ∈ φ({x, y}, %).
This implies that xρy. Because y ∈ T op(TK(X,%) , ρ), we have x ∈ T op(TK(X,%) , ρ).
Next, we show that, for each X ∈ P(X ) and each %∈ RN , T op(TK(X,%) , ρ) ⊂ φ(X, %).
Let x ∈ T op(TK(X,%) , ρ) and y ∈ φ(X, %). Because x ρ y, there exists %0 ∈ RN such that
{x, y} ⊂ TK(X ,%0 ) and x ∈ φ(X , %0 ). Because {x, y} ⊂ TK(X,%) , we have that % ={x,y} %0 .
Therefore, x ∈ φ({x, y}, %0 ) = φ({x, y}, %). By stability, we have that x ∈ φ(X , %). 5
A Coherent Hierarchy of Non-principal Ultrafilters
In this section, we show that there exists a coherent hierarchy of ultrafilters that assigns nonprincipal ultrafilters to infinite sets. To do so, we define a consistent ultrafilter mapping that
assigns a coherent ultrafilter to a collection of nonempty subsets of N .
Definition. Let M ⊂ P(N ). A correspondence F : M → P(N ) is a consistent ultrafilter
mapping if (1) for each N ∈ M, F (N ) is an ultrafilter on N , and (2) for each N, N 0 ∈ M, if
N ∈ F (N 0 ), then
F (N 0 ) = {Z ⊂ N 0 : Z ∩ N ∈ F (N )}.
Suppose that F is a consistent ultrafilter mapping. Suppose N ∈ F (N 0 ) and M ∈ F (N 0 ).
Then, M ∩ N ∈ F (N ). In a reverse way, if M ⊂ N and M ∩ N ∈ F (N ), then M ∈ F (N 0 ).
By replacing F (N ) and F (N 0 ) with UN and UN 0 in the definition of coherence, we see that a
consistent ultrafilter mapping, F , can yield a family of coherent ultrafilters.
Proposition 2. There exists a coherent hierarchy of ultrafilters that assigns a non-principal ultrafilter to each infinite set N ∈ P(N ).
Proof. Let F be the class of pairs (M, F ) satisfying: (i) M ⊂ 2N is a set of infinite sets and
F is a consistent ultrafilter mapping whose domain is M; (ii) if there exists a Z ∈ M and
Q ∈ F (Z), then Q ∈ M; (iii) if Q ∈ M and Q ⊂ Z, then Z ∈ M; and (iv) F (Q) is a
non-principal ultrafilter for every infinite set Q ∈ M. Then, we can define a partial order ≤ on F
14
such that, for each (M, F ) and (M0 , F 0 ) ∈ F, (M, F ) ≤ (M0 , F 0 ) if M ⊂ M0 and, for each
M ∈ M, F (M ) = F 0 (M ).
To begin with, we show that F is not empty. Let F (N ) be a non-principal ultrafilter on N .
Then, let M = F (N ) and for each N ∈ M, F (N ) = 2N ∩ F (N ). Then, F is consistent.
In addition, condition (ii) is satisfied by construction of F for each Q ∈ M. Moreover, by the
second property of ultrafilters, condition (iii) is satisfied. Finally, (iv) is satisfied, because there
exists a non-principal ultrafilter when its ground set is infinite.
We now show that there exists a maximal element in F, and consistency is also satisfied by
S
the maximal element. Let C be a chain in F. Let M̄ ≡ M∈C M and F¯ : M̄ → 2N be as
follows:
for each M ∈ M̄, F¯ (M ) ≡ F (M ), where (M, F ) ∈ C and M ∈ M.
We show that (M̄, F¯ ) ∈ F.
Condition (i). Suppose that there exist L, M ∈ M̄ such that L ⊂ M and L ∈ F¯ (M ). Then,
there exists (M, F ) ∈ F such that L, M ∈ M. Note that F (L) = F¯ (L) and F (M ) = F¯ (M ).
Thus, F¯ is a consistent ultrafilter mapping.
Condition (ii). Suppose that Z ∈ M̄ and Q ∈ F¯ (Z) for some Z. Then, there are M and F
such that Z ∈ M and Q ∈ F (Z). Thus, Q ∈ M̄.
Condition (iii). Suppose that Q ∈ M̄ and Q ⊂ Z. Then, there is an M such that Q ∈ M. Thus,
Z ∈ M. Hence, Z ∈ M̄.
Condition (iv). When each F (Q) is non-principal for each infinite set Q, clearly F¯ (Q) is a
non-principal ultrafilter.
Finally, we conclude that (M̄, F¯ ) ∈ F and it is an upper bound of C . Therefore, by Zorn’s
lemma, there exists a maximal element in F. Let (M∗ , F ∗ ) ∈ F be a maximal element. To show
that it is a coherent hierarchy of ultrafilters, it is enough to show that M∗ = 2N . Suppose that
M∗ 6= 2N . Then, there exists Q ∈ 2N \M∗ . Because F ’s domain is assumed to be a set of
infinite sets, we can assume that Q is infinite. Because condition (ii) of F is satisfied by M∗ ,
there exists no Z ∈ M∗ such that Q ∈ F ∗ (Z).
Let UQ∗ denote the set of non-principal ultrafilters on Q. Choose U ∈ UQ∗ arbitrarily. Let
χ denote the set of supersets of elements in U that are not yet assigned any ultrafilters. Define
Fˆ : M∗ ∪ χ → 2N as follows. Let Fˆ (Q) = U and
for each M ∈ M∗ , Fˆ (M ) = F ∗ (M ), and
for each M ∈ χ, Fˆ (M ) = {Z ⊂ M : Z ∩ Q ∈ Fˆ (Q)}.
We show that Fˆ is a consistent ultrafilter mapping. Take L, M ∈ M∗ ∪ χ and L ∈ Fˆ (M ).
Then, L ⊂ M . By condition (ii), if M ∈ M∗ , then L ∈ M∗ . By condition (iii), if L ∈ M∗ , then
15
M ∈ M∗ . Thus, there are two possible cases: (Case I) L, M ∈ M∗ ; (Case II) L, M ∈ χ. We
need to show Fˆ (M ) = {N ⊂ M : N ∩ L ∈ Fˆ (L)}.
(Case I) L, M ∈ M∗ . First, let S ∈ Fˆ (M ) and N ⊂ M . Then, S ∈ M∗ by condition (ii). Then,
because L ∈ F ∗ (M ) and S ∈ F ∗ (M ), L ∩ S ∈ F ∗ (M ) by the second property of ultrafilters.
Because L ⊂ M , and F ∗ is consistent, S ∩ L ∈ F ∗ (L). Because L ∈ M∗ , S ∩ L ∈ Fˆ (L).
Thus, we conclude
S ∈ {Z ⊆ M : Z ∩ L ∈ Fˆ (L)}.
Second, suppose that S ⊂ M and S ∩ L ∈ Fˆ (L). Then, because L ∈ M∗ , S ∩ L ∈ F ∗ (L).
Because F ∗ is consistent, we obtain S ∈ F ∗ (M ), which implies S ∈ Fˆ (M ).
(Case II) L, M ∈ χ. First, suppose that S ∈ Fˆ (M ). Then, S ⊂ M and S ∩ Q ∈ Fˆ (Q).
Because L ∈ Fˆ (M ), L ⊂ M and L ∩ Q ∈ Fˆ (Q). Then, because the intersection of two sets
in an ultrafilter is also in an ultrafilter, (S ∩ L) ∩ Q ∈ Fˆ (Q). Because S ∩ L ⊂ L, we obtain
S ∩ L ∈ Fˆ (L). Thus, we conclude
S ∈ {Z ⊂ M : Z ∩ L ∈ Fˆ (L)}.
Second, suppose that S ⊂ M and S ∩ L ∈ Fˆ (L). Because L ∈ Fˆ (M ), L ⊂ M and
L ∩ Q ∈ Fˆ (Q). Also, because N ∩ L ∈ Fˆ (L), (S ∩ L) ∩ Q ∈ Fˆ (Q). Because L ∈ χ, Q ⊂ L
and so (S ∩ L) ∩ Q = S ∩ Q. Hence, we conclude that S ∩ Q ∈ Fˆ (Q). Then, because S ⊂ M ,
S ∈ Fˆ (M ).
So, we conclude that Fˆ is a consistent ultrafilter mapping and, thus, condition (i) is satisfied.
Also, condition (iv) is satisfied by Fˆ . Thus, it remains to show that conditions (ii) and (iii) are
satisfied. Suppose that Z ∈ M∗ ∪ χ and S ∈ Fˆ (Z). If Z ∈ M∗ , then obviously Q ∈ M∗ by
condition (ii) of F ∗ . If Z ∈ χ, then S ∩ Q ∈ Fˆ (Q). If S is assigned any ultrafilter, S ∈ M∗ .
Otherwise, as χ includes all the supersets of elements in U = Fˆ (Q) that are not yet assigned any
ultrafilter, S ∈ χ. Thus, condition (ii) is satisfied.
Next, suppose that S ∈ M∗ ∪ χ and S ⊂ Z. If S ∈ M∗ , then there exists S 0 with S ∈
F ∗ (N 0 ). If Z ⊂ S 0 , then Z ∈ F ∗ (S 0 ) by the third property of ultrafilters and so Z ∈ M∗ .
Otherwise, because S ⊂ S 0 ∩ Z, S 0 ∩ Z ∈ F ∗ (S 0 ) by the third property of ultrafilters. Then,
condition (ii) implies that S 0 ∩ Z ∈ M∗ and because S 0 ∩ Z ⊂ Z, condition (iii) indicates that
Z ∈ M∗ . If S ∈ χ, then obviously Z ∈ χ. Hence, we obtain Z ∈ M∗ ∪ χ. Therefore, we
conclude that the four conditions are satisfied by (M∗ ∪ χ, Fˆ ) and (M∗ ∪ χ, Fˆ ) ∈ F. By
construction, we have (M∗ , F ∗ ) < (M∗ ∪ χ, Fˆ ). However, this is a contradiction with the
maximality of (M∗ , F ∗ ).
Theorem 2. There exists a hierarchy of ultrafilters such that (i) all the ultrafilters satisfy coherence and (ii) all the infinite sets are assigned non-principal ultrafilters.
16
Proof. By Proposition 2, there exists a consistent ultrafilter mapping F that satisfies the second
condition required in this theorem. Now, suppose that N0 , N ⊂ N , N0 ⊂ N is a finite set and N
is an infinite set. Then, F (N ) only includes infinite sets and by the fourth property of ultrafilters,
N0 6∈ F (N 0 ). Thus, we can see that coherence is not required between F (N0 ) and F (N ).
Now, by Zorn’s lemma, there is a complete strict ordering for N . There is a minimal element
in this ordering for any finite set N ⊂ N , which we denote by m̄N . For any finite set N ⊂ N ,
define F (N ) by
F (N ) = {U ⊂ N : m̄N ∈ U }.
We show that F satisfies consistency. Take two finite sets N0 and N1 with N0 ⊂ N1 ⊂ N .
Suppose that N0 ∈ F (N1 ). Then, m̄N1 ∈ N0 . Because N0 ⊂ N1 , m̄N1 = m̄N0 . If M ∈ F (N1 ),
then m̄N0 ∈ M and, thus, m̄N0 ∈ M ∩ N0 , which implies M ∩ N0 ∈ F (N0 ). On the other hand,
take M ⊂ N1 and let M ∩ N0 ∈ F (N0 ). Then, m̄N0 ∈ M ∩ N0 . Thus, because m̄N1 = m̄N0 ,
m̄N1 ∈ M . Thus, M ∈ F (N1 ).
Therefore, we conclude that there exists a hierarchy of ultrafilters for which condition (i) is
satisfied for any finite sets. This completes the proof.
6
Concluding Remarks
To conclude, we discuss the Gibbard–Satterthwaite Theorem with infinitely many agents. In this
example, we restrict our attention to strict preference profiles2 . Let P be the entire space of strict
preferences over X . Let i ∈ P be agent i’s strict preference. An SCF is a mapping from P N to
X . By an abuse of notation, let (0i , −i ) denote a preference profile such that only i is replaced
by 0i while holding everything else the same as in . Now, we introduce two properties about
SCFs.
Definition (Truthfully Implementable in Dominant Strategies). An SCF f : P N → X is truthfully implementable in dominant strategies if, for each i ∈ N , each i , 0i ∈ P, and each
−i ∈ P N −1 , f (i , −i ) %i f (0i , −i ).
Note that f (i , −i ) ∼i f (0i , −i ) if and only if f (i , −i ) = f (0i , −i ). This indicates
that it is always best for the agents to report their true preferences to the social planner.
Definition (Dictatorial). An SCF f : P N → X is dictatorial if there exists i∗ ∈ N such that, for
each ∈ P N and each x ∈ X , f () %i∗ x.
2
In the case of weak preference profiles, the Gibbard–Satterthwaite Theorem does not hold even in the case of
finitely many agents, as described in the Introduction. Man and Takayama (2013) also provide an example of a SCF
that does not satisfy a dictatorship when it does not satisfy the set of axioms for the Gibbard–Satterthwaite Theorem.
17
The Gibbard–Satterthwaite Theorem states the following.
Theorem (Gibbard–Satterthwaite). Let f : P N → X . Suppose that (i) N is finite; (ii) for every
x ∈ X , there is a ∈ P N with f () = x; and (iii) f is truthfully implementable in dominant
strategies. Then f is dictatorial.
We say that is a modification of 0 if i 6=0i for only finitely many i ∈ N . We say that
0
and 0 are close if there exists a finite sequence {m }M
m=0 such that 0 =, M = and for
each m ∈ {0, · · · , M − 1}, m+1 is a modification of m .
We denote a set of preference profiles that are close to by A(). Then, A() is an equivalence class of . The axiom of choice implies that there is an SCF f with the following properties:
(I) f satisfies unanimity;
(II) for every ∈ P N and 0 ∈ A(), f () = f (0 ).
Then, f is truthfully implementable in dominant strategies, because (0i , −i ) ∈ A() and
f () ∼i f (0i , −i )
for all ∈ P N , all i ∈ N and all 0i ∈ P.
However, because of (II) in the definition of f , a dictatorship does not hold with f . Therefore,
we can say that the Gibbard–Satterthwaite Theorem does not hold when there are infinitely many
agents. Note that, when there are only finitely many agents, there is only one set of preference
profiles that is close to . Thus, unanimity makes (II) impossible. Therefore, a dictatorship can
be proved as in Geanakoplos (2005).
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