Social choice correspondences with infinitely many agents: serial dictatorship

Abstract

We study social choice correspondences (SCC) assigning a set of choices to each pair consisting of a nonempty subset of the set of alternatives and a weak preference profile. The SCC satisfies unanimity if when there is a weakly Pareto dominant alternative, the SCC selects this alternative. Stability requires that the SCC is unaffected by withdrawal of losing alternatives. Independence implies that the SCC 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 SCC satisfying the three axioms, when the set of alternatives is finite but includes more than three alternatives, and the set of agents can have any cardinality. We show that the SCC is a serial dictatorship à la Eraslan and McLennan (J Econ Theory 117:29–54, 2004) and that a serial dictatorship can include “invisible serial dictators” à la Kirman and Sondermann (J Econ Theory 5:267–277, 1972).

Appendices

Appendix I

The chain property is formally defined as follows. Denote the set of weak orderings on Y by f(Y), and the set of linear orderings on Y by l(Y). A social welfare function for outcome set X and domain $\mathcal {P}$ is a function from $\mathcal {P}$ into the set of complete binary relations on $\mathcal {X}$. Let $\mathcal {P}(Y)$ denote the set of profiles $r \in f(Y)^N$ such that there is some $R \in \mathcal {P}$ for which $r = R|_{Y}$. Given the domain $\mathcal {P}$, we say that a triple $\{x, y, z\}$ of alternatives is free if either

$$\begin{aligned} \mathcal {P}(\{x, y, z\}) = f(\{x, y, z\})^N \text { or } l(\{x, y, z\})^N. \end{aligned}$$

We say that the domain $\mathcal {P}$ has the chain property if $|\mathcal {X}| \ge 3$ and for every two ordered pairs (x, y) and (w, z) of alternatives in $\mathcal {X}$, there exists an integer k and a sequence $v_1, \ldots , v_k$ such that all of the triples $\{x, y, v_1\}, \{y, v_1, v_2\}, \{v_1, v_2, v_3\}, \ldots , \{v_k, w, z\}$ are free.

Proof of Lemma 9

The Ultrafilter Lemma of Campbell and Kelly (2002) shows that a domain of an SWF satisfies the chain property and the SWF satisfies unanimity and independence, $\mathcal {U}_N$ is an ultrafilter. In our environment, because there are finitely many alternatives in $\mathcal {X}$ and the domain is a full domain, $\mathcal {R}^{N}$ has the chain property, and by Lemma 8, $\mathbf {R}$ satisfies the two properties. Thus the Ultrafilter Lemma holds. $\square$

Appendix II

The “if” Part Proof of Proposition 4

Suppose that a coherent hierarchy of ultrafilters $\mathscr {U}$ and a tie-breaking rule $\rho$ induce a voting procedure V. Let $R \in \mathcal {R}^{\mathcal {N}}$. For all $A \in \mathcal {A}$, $T_{K(R)}(A, R) \subset A$ and thus V satisfies feasibility. Further, the unanimity and independence of V follow from the proofs of Lemma 5 and Lemma 6 by replacing X with $A \in \mathcal {A}$, and $\phi ^{\mathscr {U}, \rho }$ with V. To show that V is SCS, take $x \in \mathcal {X}$ and suppose $V(\mathcal {X},R) \not = \{x\}$. Then there is some $y \not = x$ such that $y \in V(\mathcal {X},R) {\setminus } \{x\}$. Then $y \in T_{K(R)}(\mathcal {X}, R)$, which implies that there is no $k = 1, \ldots , K(R)$ and $z \in \mathcal {X}{\setminus } \{x, y\}$ such that $z \succ _{R, k} y$. Thus, $y \in T_{K(R)}(\mathcal {X}{\setminus } \{x\}, R)$, and $y \in V(\mathcal {X}{\setminus } \{x\},R)$.

Conversely, suppose $y \in V(\mathcal {X}{\setminus } \{x\},R)$. Then $y \in T_{K(R)}(\mathcal {X}{\setminus } \{x\}, R)$, which implies that there is no $k = 1, \ldots , K(R)$ and $z \in \mathcal {X}{\setminus } \{x\}$ such that $z \succ _{R, k} y$. Then $y \in T_{K(R)}(\mathcal {X}, R) {\setminus } \{x\}$ or $\{x\} = T_{K(R)}(\mathcal {X}, R)$. Thus $y \in V(\mathcal {X},R) {\setminus } \{x\}$ or otherwise, $\{x\} = V(\mathcal {X},R)$. By the initial assumption, $V(\mathcal {X},R) \not = \{x\}$ and thus we conclude that $y \in V(\mathcal {X},R) {\setminus } \{x\}$. Thus, we obtain $V(\mathcal {X}{\setminus } \{x\},R) = V(\mathcal {X},R) {\setminus } \{x\}$. $\square$

Now suppose that a voting procedure V satisfies feasibility, unanimity, IIA and SCS. Our strategy for proving the “only-if” part of Proposition 4 is as follows. For all $X \in \mathscr {P}(\mathcal {X})$ and all $R \in \mathcal {R}^{\mathcal {N}}$, we define $\phi$ to be:

$$\begin{aligned} \phi (X, R) = V(\mathcal {X}, R^X), \end{aligned}$$

(14)

where $R^X$ is the preference profile that takes X to the top from R. We will show that $\phi$ satisfies unanimity, independence and stability. Then by Theorem 1, a coherent hierarchy of ultrafilters and a tie-breaking rule induce $\phi$, and $\phi$ is a serial dictatorship. Then we show that the coherent hierarchy of ultrafilters and tie-breaking rule which induce $\phi$ also induce V defined in (14).

Fix $R \in \mathcal {R}^{\mathcal {N}}$. The proof of the following lemma, which guarantees that $V(\mathcal {X}, R^X) \subset X$ occupies the rest of the subsection. Eraslan and McLennan (2004) prove a similar result for the case of finite $\mathcal {N}$ as their Lemma 1 (in pages 41 – 42). A similar logic works in our case, even when the case of infinitely many agents is included.

Lemma 16

If $x \in V(\mathcal {X},R)$, then x is not strongly Pareto dominated at R.

Lemma 16 implies that $V(\mathcal {X}, R^X) \subset X$ for every $X \in \mathscr {P}(\mathcal {X})$. Let $x \in \mathscr {P}(\mathcal {X})$. We say that $R'$ is an x -modification of $R$ if $R|_{\mathcal {X}\backslash \{x\}}=R'|_{\mathcal {X}\backslash \{x\}}$ holds. Several results prepare the proof of Lemma 16.

Lemma 17

Suppose that $R'$ is an x-modification of $R$, and $V(\mathcal {X}, R') \not = \{x\} \not = V(\mathcal {X}, R)$. Then $V(\mathcal {X}, R) \backslash \{x\} = V(\mathcal {X}, R') \backslash \{x\}$.

Proof

We have

$$\begin{aligned} \begin{array} {r c l l} V(\mathcal {X}, R) \backslash \{x\} &{} = &{} V(\mathcal {X}\backslash \{x\}, R) &{} \quad \text {(SCS and }V(\mathcal {X}, R) \not = \{x\}) \\ &{} = &{} V(\mathcal {X}\backslash \{x\}, R') &{} \quad \text {(IIA)} \\ &{} = &{} V(\mathcal {X}, R') \backslash \{x\} \text {.} &{} \quad \text {(SCS and }V(\mathcal {X}, R') \not = \{x\}) \end{array} \end{aligned}$$

$\square$

Let $X = \{x_1, \ldots , x_K\}$ denote a set of K alternatives. Take a sequence of preference profiles $\{R^k\}_{k=0}^{K}$ such that each $R^k$ is an $x_{k}$-modification of $R^{k-1}$.

Lemma 18

For all $k \ge 1$, $V(\mathcal {X}, R^k) \subset V(\mathcal {X}, R^{0}) \cup \{x_{1}, \ldots , x_{k}\}$.

Proof

By induction suppose that $V(\mathcal {X}, R^{k-1}) \subset V(\mathcal {X}, R^{0}) \cup \{x_1, \ldots , x_{k-1}\}$ for some $k \ge 1$. Then by Lemma 17, $V(\mathcal {X}, R^{k}) = \{x_{k}\}$ or $V(\mathcal {X}, R^{k}) \backslash \{x_{k}\} = V(\mathcal {X}, R^{k-1}) \backslash \{x_{k}\}$. $\square$

Lemma 19

Suppose that $V(\mathcal {X}, R^0) \not \subset X$. Then for every k, $V(\mathcal {X}, R^k) \backslash X = V(\mathcal {X}, R^0) \backslash X$.

Proof

This follows from repeated applications of Lemma 17 unless $V(\mathcal {X}, R^k) = \{x_k\}$ for some k. In this case applying Lemma 18 to the sequence from $R^k$ to $R^0$ implies that $V(\mathcal {X}, R^0) \subset \{x_{k}, \ldots , x_{1}\} \subset X$. $\square$

Proof of Lemma 16

Let $x \in V(\mathcal {X},R)$. On the contrary suppose that there exists a $y \in \mathcal {X}$ such that $y P_i x$ for all $i \in \mathcal {N}$. Consider $R'$ such that at $R'$, $y P'_i z$ for all $z \in \mathcal {X}$ and all $i \in \mathcal {N}$, and $R'|_{\mathcal {X}{\setminus } \{y\}}=R|_{\mathcal {X}{\setminus } \{y\}}$. Then y is strongly Pareto dominant at $R'$ and so unanimity implies that $V(\mathcal {X},R')=\{y\}$. Since R is a y-modification of $R'$, by Lemma 18, $\{x, y\} \subset V(\mathcal {X},R) \subset V(\mathcal {X},R') = \{y\}$, which is a contradiction. $\square$

Lemma 20

The SCC $\phi$ defined in (14) satisfies unanimity, independence and stability.

Proof

• Unanimity: If y is weakly Pareto dominant in $Y \in \mathscr {P}(\mathcal {X})$ at some $R \in \mathcal {R}^{\mathcal {N}}$, then y is weakly Pareto dominant in $\mathcal {X}$ at $R^Y$. By the unanimity of V and (14), $\phi (Y, R) = V(\mathcal {X}, R^Y) = \{y\}$.

• Independence: Let $Y \in \mathscr {P}(\mathcal {X})$, $R, R' \in \mathcal {R}^{\mathcal {N}}$ and $R|_Y = R'|_Y$. By the definition of $\phi$ in (14), $\phi (Y, R) = V(\mathcal {X}, R^Y)$ and $\phi (Y, R') = V(\mathcal {X}, {R'}^Y)$. We can construct ${R'}^Y$ from $R^Y$ through finitely many interim preference profiles $\{R^k\}$ by repeatedly moving each element $x^k$ so that for each $x^k \not \in Y$ and $y \in Y$ with $R^Y|_{\{x^k, y\}} \not = {R'}^Y|_{\{x^k, y\}}$, $R^k|_{\{x^k, y\}} = {R'}^Y|_{\{x^k, y\}}$ holds. By Lemma 16 and Lemma 19, we obtain $V(\mathcal {X}, R^Y) = V(\mathcal {X}, {R'}^Y)$. Thus, $\phi (Y, R) = \phi (Y, R')$.

• Stability: Fix $X, Y \in \mathscr {P}(\mathcal {X})$ where $Y \subset X$, and $R \in \mathcal {R}^{\mathcal {N}}$ such that $V(\mathcal {X}, R^X) \cap Y \ne \varnothing$. Let $Z = X {\setminus } Y = \{z_1, \ldots , z_K\}$. Define $Z_0 = \varnothing$ and $Z_k = \{z_1, \ldots , z_k \}$. Construct $R^0 = R^X$ and for all $k>0$, a preference profile $R^k$ such that $R^k$ is the $z_k$-modification of $R^{k-1}$ such that $z_k$ is strongly Pareto dominated by alternatives in $X {\setminus } \{z_k\}$ at $R^k$.

Because $V(\mathcal {X}, R^X) \cap Y \ne \varnothing$, $V(\mathcal {X}, R^X) \not \subset Z$. By Lemma 19, $V(\mathcal {X}, R^0) {\setminus } Z = V(\mathcal {X}, R^K) {\setminus } Z$. At the same time, at $R^K$, alternatives in Z are strongly Pareto dominated by alternatives in Y. Thus, by Lemma 16, $V(\mathcal {X}, R^K) {\setminus } Z = V(\mathcal {X}, R^K)$. Because $R^K|_Y = R^Y|_Y$, by the same argument as in the proof of independence, we obtain $V(\mathcal {X}, R^K) = V(\mathcal {X}, R^Y)$ and hence

$$\begin{aligned} V(\mathcal {X}, R^Y) = V(\mathcal {X}, R^X) {\setminus } Z \text {.} \end{aligned}$$

(15)

Then by the definition of $\phi$ in (14), $V(\mathcal {X}, R^X) \cap Y \ne \varnothing$ implies $\phi (X, R) \cap Y \ne \varnothing$ and at the same time,

$$\begin{aligned} \phi (Y, R) = \phi (X, R) {\setminus } Z, \end{aligned}$$

(16)

which implies $\phi (Y, R) = \phi (X, R) \cap Y$. Thus, stability follows. $\square$

The “only-if” part proof of Proposition 4

By Lemma 20, the SCC $\phi$ defined in (14) satisfies unanimity, independence and stability. Then by Theorem 1, $\phi$ is a serial dictatorship. Thus, a coherent hierarchy of ultrafilters $\mathscr {U}$ and a tie-breaking rule $\rho$ induce $\phi$. Then by Definition 6, for all $R \in \mathcal {R}^{\mathcal {N}}$,

$$\begin{aligned} \phi (\mathcal {X}, R)=\mathrm {Top}(T_{K(R)}(\mathcal {X}, R), \rho ). \end{aligned}$$

(17)

By replacing X with $\mathcal {X}$ in (14), for all $R \in \mathcal {R}^{\mathcal {N}}$,

$$\begin{aligned} V(\mathcal {X}, R)=\mathrm {Top}(T_{K(R)}(\mathcal {X}, R), \rho ). \end{aligned}$$

Therefore, a coherent hierarchy of ultrafilters $\mathscr {U}$ and a tie-breaking rule $\rho$ also induce V for $\mathcal {X}$. To show the result for any agenda domain, fix $x \in \mathcal {X}$. Let $A = \mathcal {X}{\setminus } \{x\}$. Then similarly to (17), for all $R \in \mathcal {R}^{\mathcal {N}}$,

$$\begin{aligned} \phi (A, R)=\mathrm {Top}(T_{K(R)}(A, R), \rho ). \end{aligned}$$

By replacing X with A in (14), for all $R \in \mathcal {R}^{\mathcal {N}}$,

$$\begin{aligned} V(A, R^{A})=\mathrm {Top}(T_{K(R)}(A, R), \rho ). \end{aligned}$$

By SCS, for all $R \in \mathcal {R}^{\mathcal {N}}$, $V(A, R) = V(\mathcal {X}, R) {\setminus } \{x\}$ and $V(A, R^A) = V(\mathcal {X}, R^A) {\setminus } \{x\}$. By IIA, $V(A, R) = V(A, R^A)$ for all $R \in \mathcal {R}^{\mathcal {N}}$. Finally, for all $R \in \mathcal {R}^{\mathcal {N}}$, $V(A, R^A) = V(A, R)$, and thus

$$\begin{aligned} V(A, R)=\mathrm {Top}(T_{K(R)}(A, R), \rho ). \end{aligned}$$

Therefore, a coherent hierarchy of ultrafilters $\mathscr {U}$ and a tie-breaking rule $\rho$ also induce V. $\square$