Abstract
We consider domains with a natural property called top-circularity. We show that if such a domain satisfies either the maximal conflict property or the weak conflict property, then it is dictatorial. We obtain the result in Sato (Rev Econ Des 14(3):331–342, 2010) as a corollary. Furthermore, it follows from our results that the union of a single-peaked domain and a single-dipped domain (with respect to a given ordering over the alternatives) is dictatorial.
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Notes
We denote by \(ab \ldots \) a preference which places a at the top and b at the second-ranked position.
Roy and Storcken (2016) shows that this property is necessary and sufficient for dictatorship on a large class of domains which they call short-path-connected domains. However, the domains that we consider are not short-path-connected.
The top-graph of a domain is defined as the graph where nodes are alternatives and there is an edge between two alternatives a, b if there are preferences \(ab\ldots \) and \(ba \ldots \) in the domain.
An undirected graph with nodes \(v_1,\ldots ,v_k\) is said to contain a maximal cycle if it has the following edges: \(\{v_1,v_2\},\{v_2,v_3\},\ldots ,\{v_{k-1},v_k\},\{v_k,v_1\}\).
Alternative models that consider similar practical situations exist in the literature. For instance, Thomson (2008) and Feigenbaum and Sethuraman (2014) partition the set of agents into those who can only have single-peaked preferences and those that can only have single-dipped preferences. On the other hand, Alcalde-Unzu and Vorsatz (2015) considers a situation where the social planner is informed about the location of the agents, but agents can have single-peaked preferences with the peak at her location or single-dipped preferences with the dip at her location. Though the domain restriction considered in the aforementioned models is close in spirit with ours, they admit non-dictatorial, unanimous, and strategy-proof SCFs.
We provide the technical definition of linked property in Remark 2.
They also provide two conditions T and \(T'\), and show that any domain satisfying those is dictatorial.
More formally, since P is a binary relation on X, xPy means that the pair \((x,y) \in P\).
A top-circular domain is minimal if none of its subsets is top-circular.
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Acknowledgements
The authors wish to thank an editor, an associate editor, and two anonymous referees for their insightful comments. The authors would also like to thank Madhuparna Karmakar, Manipushpak Mitra, Hans Peters, Soumyarup Sadhukhan, Arunava Sen, and Ton Storcken for their invaluable suggestions which helped improve this paper. The usual disclaimer holds.
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Appendix
Appendix
In this section, we prove Theorems 1 and 2. The following proposition in Aswal et al. (2003) allows us to restrict our attention to the case of two agents.
Proposition 1
(Aswal et al. 2003) Let \({\mathcal {D}}\) be a regular domain, such that every unanimous and strategy-proof SCF \(f:{\mathcal {D}}^{2}\rightarrow X\) is dictatorial. Then, every unanimous and strategy-proof SCF \(f:{\mathcal {D}}^{n}\rightarrow X\) is dictatorial.
The following proposition in Sanver (2007) allows us to restrict our attention to minimal top-circular domains satisfying either the maximal conflict or the weak conflict property.Footnote 11
Proposition 2
(Sanver 2007) A superset of a regular dictatorial domain is also dictatorial.
Now, we introduce the notion of option sets, which we use in our proofs.
Definition 13
Given an SCF \(f:{\mathcal {D}}^{2} \rightarrow X\), we define the option set of agent \(i \in \{1,2\}\) at preference \(P_j \in {\mathcal {D}}\) of agent \(j \in \{1,2\} {\setminus } i\), denoted by \(O_i(P_j)\), as \(O_i(P_j)= \bigcup _{P_i \in {\mathcal {D}}} f(P_i,P_j)\).
Remark 5
Note that if an SCF \(f:{\mathcal {D}}^{2} \rightarrow X\) is unanimous, then \(r_1(P_j) \in O_i(P_j)\) for all \(P_j \in {\mathcal {D}}\). Furthermore, if f is strategy-proof, then for all \(i,j \in \{1,2\}; i \ne j\) and all \((P_1,P_2) \in {\mathcal {D}}^2\), \( f(P_1,P_2) = \max _{P_i}O_i(P_j)\), where \(\max _{P_i}O_i(P_j)=x\) if and only if \(x \in O_i(P_j)\) and \(xP_iy\) for all \(y \in O_i(P_j) {\setminus } x\).
Remark 6
Note that an SCF \(f:{\mathcal {D}}^{2} \rightarrow X\) is dictatorial if and only if there is \(i \in \{1,2\}\), such that \(O_i(P_j)=\{r_1(P_j)\}\) for all \(P_j \in {\mathcal {D}}\).
For all the subsequent results, let \({\mathcal {C}}\) be a minimal top-circular domain. Suppose that \(f:{\mathcal {C}}^2 \rightarrow X\) is a unanimous and strategy-proof SCF and \(O_i(P_j)\) is the corresponding option set of agent i at a preference \(P_j\) of agent \(j \in \{1,2\} {\setminus } i\). We prove a sequence of lemmas that we use in the proofs of Theorems 1 and 2.
The following lemma establishes a property of a minimal top-circular domain. We assume for this lemma that \(0 \equiv m\) and \(m+1 \equiv 1\).
Lemma 1
Let \(P_2,P_2' \in {{\mathcal {C}}}\) be such that \(r_1(P_2)=r_1(P_2')=x_k\). Then, for all \(j \in \{k-1,k+1\}\), \(x_{j} \in O_1(P_2)\) if and only if \(x_{j} \in O_1(P'_2)\).
Proof
Assume for contradiction that there exist \(P_2,P'_2 \in {{\mathcal {C}}}\) with \(r_1(P_2)=r_1(P'_2)=x_k\), such that \(x_{j} \in O_1(P_2)\) and \(x_{j} \not \in O_1(P'_2)\) for some \(j \in \{k-1,k+1\}\). Consider \(P_1 \in {{\mathcal {C}}}\), such that \(r_1(P_1)=x_{j}\) and \(r_2(P_1)=x_{k}\). Such a preference exists in \({\mathcal {C}}\) as \(|j-k| = 1\). Then, by the strategy-proofness of f, \(f(P_1,P_2)=x_{j}\) and \(f(P_1,P'_2)=x_{k}\). This means that agent 2 manipulates at \((P_1,P_2)\) via \(P'_2\), a contradiction. This completes the proof of the lemma. \(\square \)
The subsequent lemmas establish a few crucial properties of a minimal top-circular domain \({\mathcal {C}}\), such that \(\{x_1x_2\ldots x_m, x_mx_{m-1}\ldots x_1 \} \subseteq {\mathcal {C}}\). Note that if a minimal top-circular domain \({\mathcal {C}}\) satisfies either the maximal conflict property or the weak conflict property, then such two preferences are there in \({\mathcal {C}}\).
Lemma 2
Let \(\{x_1x_2\ldots x_m,x_mx_{m-1}\ldots x_1 \} \subseteq {\mathcal {C}}\). Then, for all \(P_2 \in \{x_1x_2\ldots x_m,x_mx_{m-1}\ldots x_1\}\), \(r_m({P}_2) \notin O_1({P}_2)\) implies that \(O_1({P}_2)= \{r_1(P_2)\}\).
Proof
We prove the lemma for the case where \(P_2=x_1x_2\ldots x_m \in {\mathcal {C}}\), the proof of the same for the other case can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. Let \(P_2=x_1x_2\ldots x_m \in {\mathcal {C}}\) and let \(r_m({P}_2)=x_m \notin O_1({P}_2)\). We show \(O_1({P}_2)= \{r_1(P_2)\}\). Assume for contradiction that \(x_j \in O_1(P_2)\) for some \(j \ne 1,m\). Let \(P'_2 \in {\mathcal {C}} \) be such that \(r_1(P'_2)=x_1\) and \(r_2(P'_2)=x_m\). Since \(x_m \notin O_{1}(P_{2})\), by Lemma 1, \(x_m \notin O_1(P'_2)\). Let \(P_1=x_mx_{m-1}\ldots x_1\). By unanimity and strategy-proofness, we must have \(f(P_1,P'_2) \in \{x_1,x_m\}\); as otherwise, agent 2 manipulates at \((P_1,P'_2)\) via a preference which places \(x_m\) at the top. In addition, since \(x_m \notin O_1(P_2')\), we have \(f(P_1,P'_2) =x_1\). However, since \(x_j \in O_1(P_2)\) and \(x_j P_1 x_1\), it must be that \(f(P_1,P_2)\ne x_1\). Because \(r_1(P_2)=x_1=r_1(P'_2)\), this means that agent 2 manipulates at \((P_1,P_2)\) via \(P_2'\), a contradiction. This completes the proof of the lemma. \(\square \)
Lemma 3
Let \(\{x_1x_2\ldots x_m,x_mx_{m-1}\ldots x_1\} \subseteq {\mathcal {C}}\) and let \(O_1(P_2) \in \{\{r_1(P_2)\},X\}\) for all \(P_2 \in \{x_1x_2\ldots x_m,x_mx_{m-1}\ldots x_1\}\). Suppose \(\hat{P}_2,\bar{P}_2 \in {\mathcal {C}}\) are such that \(r_1(\hat{P}_2)=x_1\) and \(r_1(\bar{P}_2)=x_{m}\). Then, \(O_1(\hat{P}_2)=\{x_1\}\) if and only if \(O_1(\bar{P}_2)=\{x_{m}\}\).
Proof
Let \(\hat{P}_2,\bar{P}_2 \in {\mathcal {C}}\) be such that \(r_1(\hat{P}_2) = x_1\) and \(r_1(\bar{P}_2) = x_{m}\). It is sufficient to show that \(O_1(\hat{P}_2)=\{x_1\}\) implies \(O_1(\bar{P}_2)=\{x_{m}\}\). The proof of the fact that \(O_1(\bar{P}_2)=\{x_{m}\}\) implies \(O_1(\hat{P}_2)=\{x_1\}\) can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. By strategy-proofness, it is enough to show that \(O_1(\bar{P}_2) = \{x_{m}\}\) where \(\bar{P}_{2} = x_{m}x_{m-1}\ldots x_{1}\).
Assume for contradiction that \(O_1(\hat{P}_2) = \{x_{1}\}\) and \(O_1(\bar{P}_2) \ne \{x_{m}\}\). By the assumption of the lemma, \(O_1(\bar{P}_2) \ne \{x_{m}\}\) implies that \(O_1(\bar{P}_2)=X\). Consider \(\bar{P}'_2 \in {\mathcal {C}}\), such that \(r_1(\bar{P}'_2) = x_{m}\) and \(r_2(\bar{P}'_2) = x_{1}\). Since \(O_1(\bar{P}_2) \ne \{x_m\}\), it follows from strategy-proofness that \(O_1(\bar{P}'_2) \ne \{x_m\}\). We show that \( x_j \not \in O_1(\bar{P}_2')\) for all \(j \ne 1,m\). Suppose not. Then, \(f(P_1,\bar{P}_2')=x_j\) for some \(P_1 \in {\mathcal {C}}\) with \(x_j\) at the top. However, because \(O_1(\hat{P}_2)=\{x_{1}\}\), agent 2 manipulates at \((P_1,\bar{P}_2')\) via \(\hat{P}_2\). Since \( O_1(\bar{P}'_2) \ne \{x_m\}\) and \(x_j \notin O_1(\bar{P}'_2)\) for all \(j \ne 1,m\), it must be that \(O_1(\bar{P}'_2) = \{x_1,x_m\}\). However, since \( O_1(\bar{P}_2)=X\), which in turn means \(x_{m-1} \in O_1(\bar{P}_2)\), by Lemma 1, we must have \(x_{m-1} \in O_1(\bar{P}'_2)\), a contradiction. This completes the proof of the lemma. \(\square \)
1.1 Proof of Theorem 1
In this section, we provide a Proof of Theorem 1. First, we establish a few properties of a top-circular domain satisfying the maximal conflict property.
Lemma 4
Let \({\mathcal {C}}\) satisfy the maximal conflict property. Let \(P,P' \in {\mathcal {C}}\) be such that \(r_k(P)=r_{m-k+1}(P')=x_k\) for all \(k=1,\ldots ,m\). Then, for all \({P}_2 \in \{P,P'\}\), \(r_m({P}_2) \in O_1({P}_2)\) implies \(O_1({P}_2)=X\).
Proof
We prove this lemma for the case where \(P_2=P\), the proof of the same for the other case can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. Let \(P_2 = P\). Suppose that \(x_m \in O_1(P_2)\). We show \(O_1(P_2)=X\). We prove this by induction. Since \(x_m \in O_1(P_2)\), it is sufficient to show that for all \(1 < k \le m\), \(x_k \in O_1(P_2)\) implies \(x_{k-1} \in O_1(P_2)\). Assume for contradiction that \(x_k \in O_1(P_2)\), but \(x_{k-1} \notin O_1(P_2)\) for some \(1 < k \le m\). Consider \(P_1 = x_{k-1}x_{k} \ldots \in {\mathcal {C}}\). Since \(x_k \in O_1(P_2)\) and \(x_{k-1} \notin O_1(P_2)\), \(f(P_{1},P_{2})=x_k\). However, this means that agent 2 manipulates at \((P_{1},P_{2})\) via a preference which places \(x_{k-1}\) at the top, a contradiction. This completes the proof of the lemma. \(\square \)
Remark 7
Let \({\mathcal {C}}\) be a minimal top-circular domain satisfying the maximal conflict property, and let \(P,P' \in {\mathcal {C}}\) be such that \(r_k(P)=r_{m-k+1}(P')=x_k\) for all \(k=1,\ldots ,m\). Then, it follows from Lemma 2 that for all \(P_2 \in \{P,P'\}\), \(r_m(P_2) \notin O_1(P_2)\) implies that \(O_1(P_2)=\{r_1(P_2)\}\). Again, it follows from Lemma 4 that for all \(P_2 \in \{P,P'\}\), \(r_m(P_2) \in O_1(P_2)\) implies \(O_1(P_2)=X\). Thus, for all \(P_2 \in \{P,P'\}\), we have \( O_1(P_2) \in \{\{r_1(P_2)\},X \}\).
Lemma 5
Let \({\mathcal {C}}\) satisfy the maximal conflict property. Furthermore, let \(P,P' \in {\mathcal {C}}\) be such that \(r_k(P)=r_{m-k+1}(P')=x_{k}\) for all \(k=1,\ldots ,m\). Then, for all \(P_2 \in \{P,P' \}\), \(O_1(P_2)=\{r_1({P}_2)\}\) implies that \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2) \}\) for all \(\bar{P}_{2} \in {\mathcal {C}}\).
Proof
It is enough to prove the lemma for the case where \(P_2=P\), the proof for the other case can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. Let \(P_2=P\). Suppose that \(O_1(P_2)=\{r_1({P}_2)\}\). We show \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2)\}\) for all \(\bar{P}_{2} \in {\mathcal {C}}\). By strategy-proofness, we have \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2)\}\) for all \(\bar{P}_2 \in {\mathcal {C}}\) with \(r_{1}(\bar{P}_{2}) = x_{1}\). Moreover, by Lemma 3 and Remark 7, we have \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2)\}\) for all \(\bar{P}_{2} \in {\mathcal {C}}\) with \(r_{1}(\bar{P}_{2}) = x_{m}\). Take \(j \ne 1,m\) and \(\hat{P}_{2} \in {\mathcal {C}}\) with \(r_{1}(\hat{P}_{2}) = x_j\). We show \(O_1(\hat{P}_2)=\{r_1(\hat{P}_2)\}\).
First, we show \(O_{2}(P_{1})=O_{2}(P'_{1}) = X\), where \(r_{m-k+1}(P_1) = r_{k}(P'_1) = x_{k}\) for all \(k = 1,\ldots ,m\). We show that, for \(P_1\), the proof of the same for \(P_1'\) can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. Since \(O_1(P_2) = \{x_1\}\), we have \(f(P_{1},P_{2}) = x_{1}\). Because \(r_m(P_1) = x_1\), this means that \(r_m(P_1) \in O_{2}(P_{1})\). By interchanging the roles of the agents in Lemma 4, this means \(O_{2}(P_{1}) = X\).
Now, we complete the proof of the lemma. Assume, for contradiction, that \(x_{l} \in O_{1}(\hat{P}_{2})\) for some \(x_{l} \ne r_{1}(\hat{P}_{2}) = x_{j}\). Since \(r_{m-k+1}(P_{1}) =r_{k}(P'_{1}) = x_{k}\) for all \(k = 1,\ldots ,m\), we must have either \(x_{l} P_{1} x_{j}\) or \(x_{l} P'_{1} x_{j}\). Assume without loss of generality that \(x_{l} P_{1} x_{j}\). Since \(O_{2}(P_{1}) = X\) and \(r_1(\hat{P}_2)=x_j\), \(f(P_{1},\hat{P}_{2}) = x_{j}\). Let \(\hat{P}_{1} \in {\mathcal {C}}\), such that \(r_{1}(\hat{P}_{1}) = x_{l}\). Since \(x_{l} \in O_{1}(\hat{P}_{2})\) and \(r_{1}(\hat{P}_{1}) = x_{l}\), we have \(f(\hat{P}_{1},\hat{P}_{2}) = x_{l}\). This means that agent 1 manipulates at \((P_{1},\hat{P}_{2})\) via \(\hat{P}_{1}\), a contradiction. Therefore, \(O_{1}(\hat{P}_{2}) = \{r_{1}(\hat{P}_{2})\}\), which completes the proof of the lemma. \(\square \)
Now, we are ready to prove Theorem 1.
Proof of Theorem 1
In view of Propositions 1 and 2, it sufficient to show that a minimal top-circular domain with the maximal conflict property is dictatorial for two agents. Consider \(P_2 \in {\mathcal {C}}\), such that \(r_k(P_2)=x_k\) for all \(1 \le k \le m\). By Remark 7, we have \(O_1(P_2) \in \{\{r_{1}(P_{2})\},X\}\). Suppose \(O_1(P_2) = \{r_{1}(P_{2})\}\). Then, by Lemma 5, it follows that \(O_1(P'_2) = \{r_{1}(P'_{2})\}\) for all \(P'_2 \in {\mathcal {C}}\), which implies that agent 2 is the dictator.
Now, suppose \(O_1(P_2) = X\). Consider \(P_{1} \in {\mathcal {C}}\), such that \(r_1(P_1) = x_m\). Since \(O_1(P_2) = X\), we have \(f(P_1,P_2) = x_m\). We claim \(O_2(P_1) = \{r_{1}(P_{1})\}\). Assume for contradiction that \(x_{j} \in O_{2}(P_{1})\) for some \(j \ne m\). Since \(r_{m}(P_{2}) = x_{m}\), we have \(x_{j} P_{2} x_{m}\). However, since \(x_{j} \in O_{2}(P_{1})\), agent 2 manipulates at \((P_{1},P_{2})\) via some preference \(\bar{P}_{2}\) with \(r_1(\bar{P}_{2})=x_j\). Therefore, \(O_2(P_{1}) = \{r_{1}(P_{1})\}\). By interchanging the roles of the agents in Lemma 5, this means that \(O_2(P_1) = \{r_{1}(P_{1})\}\) for all \(P_1 \in {\mathcal {C}}\), which implies agent 1 is the dictator. This completes the proof of the theorem. \(\square \)
1.2 Proof of Theorem 2
In this section, we provide a Proof of Theorem 2. First, we establish a few properties of a top-circular domain satisfying the weak conflict property.
Lemma 6
Let \({\mathcal {C}}\) satisfy the weak conflict property. Suppose that \(P_2 \in \{x_1x_2\ldots x_m,x_mx_{m-1}\ldots x_1\} \subseteq {\mathcal {C}}\). Then, \(r_m(P_2) \in O_1(P_2)\) implies that \(O_1(P_2)=X\).
Proof
It is enough to prove the lemma for \(P_2=x_1x_2\ldots x_m \in {\mathcal {C}}\), the proof for the other case can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. Suppose \(x_m \in O_1(P_2)\). We show \(O_1(P_2)=X\). We prove this by induction. By unanimity, \(x_1 \in O_1(P_2)\). Therefore, it is sufficient to show that for all \(1\le k < m\), \(x_k \in O_1(P_2)\) implies \(x_{k+1} \in O_1(P_2)\). Assume for contradiction that \(x_k \in O_1(P_2)\) and \(x_{k+1} \notin O_1(P_2)\) for some \(1\le k < m\). Let \(\hat{P}_2 = x_kx_{k+1}\ldots x_m \ldots x_{k-1} \ldots \in {\mathcal {C}}\). Note that since \(x_m \in O_1(P_2)\) and \(r_m(P_2)=x_m\), by strategy-proofness, it must be that \(x_m \in O_1(\hat{P}_2)\). Let \(P_1 = x_{k+1}x_{k+2}\ldots x_m \ldots x_k \ldots \in {\mathcal {C}}\). By unanimity and strategy-proofness, \(f(P_1,\hat{P}_2) \in \{x_k,x_{k+1}\}\), as otherwise agent 2 manipulates at \((P_1,\hat{P}_2)\) via some preference with \(x_{k+1}\) at the top. Suppose \(f(P_1,\hat{P}_2)=x_k\). Since \(x_mP_1 x_k\) and \(x_m \in O_1(\hat{P}_2)\), this means that agent 1 manipulates at \((P_1,\hat{P}_2)\) via some preference with \(x_m\) at the top. Therefore, we have \(f(P_1,\hat{P}_2)=x_{k+1}\). Now, let \({P}'_1 =x_{k+1}x_k \ldots \in {\mathcal {C}}\). Then, since \(f(P_1,\hat{P}_2)=x_{k+1}\) and \(r_1(P_1)=r_1(P_1')=x_{k+1}\), by strategy-proofness, \(f(P'_1,\hat{P}_2)=x_{k+1}\). In addition, because \(x_{k} \in O_1(P_2)\) and \(x_{k+1} \notin O_1(P_2)\), we have \(f(P'_1,P_2)=x_k\). Therefore, agent 2 manipulates at \((P'_1,\hat{P}_2)\) via \(P_2\), a contradiction. This completes the proof of the lemma. \(\square \)
Remark 8
Let \({\mathcal {C}}\) satisfy the weak conflict property. Then, using arguments similar to the ones employed in Remark 7, it follows from Lemmas 2 and 6 that for all \(P_2 \in \{x_1x_2\ldots x_m, x_mx_{m-1}\ldots x_1\}\), \(O_1(P_2) \in \{\{r_1(P_2)\},X \}\).
Lemma 7
Let \({\mathcal {C}}\) satisfy the weak conflict property. Furthermore, let \(P_2 \in \{x_1x_2 \ldots x_m,x_mx_{m-1} \ldots x_1\}\). Then, \(O_1(P_2)=\{r_1({P}_2)\}\) implies that \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2) \}\) for all \(\bar{P}_{2} \in {\mathcal {C}}\).
Proof
We prove this lemma for the case where \(P_2 =x_1x_2 \ldots x_m\), the proof for the case where \(P_2 =x_mx_{m-1}\ldots x_1\) can be obtained by employing the same set of arguments after renaming \(x_1,x_2,\ldots ,x_m\) as \(x_m,x_{m-1}, \ldots ,x_1\), respectively. Let \(P_2=x_1x_2 \ldots x_m\). Suppose \(O_{1}(P_{2}) = \{x_{1}\}\). We show \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2)\}\) for all \(\bar{P}_{2} \in {\mathcal {C}}\). By strategy-proofness, we have \(O_1(\bar{P}_2)=\{r_1(\bar{P}_2)\}\) for all \(\bar{P}_2 \in {\mathcal {C}}\) with \(r_{1}(\bar{P}_{2}) = x_{1}\). By Lemma 3 and Remark 8, \(O_1(P_2) =\{x_1\}\) implies \(O_1(\bar{P}_2) =\{x_m\}\) for all \(\bar{P}_2 \in {\mathcal {C}}\) with \(r_1(\bar{P}_2) =x_m\). We prove the lemma using induction. Take \(1 \le j <m\). Suppose \(O_1(\bar{P}_2) =\{x_j\}\) for all \(\bar{P}_2 \in {\mathcal {C}}\) with \(r_1(\bar{P}_2) =x_j\). We show \(O_1(\hat{P}_2) =\{x_{j+1}\}\) for all \(\hat{P}_2 \in {\mathcal {C}}\) with \(r_1(\hat{P}_2) =x_{j+1}\). Take \(\hat{P}_2 \in {\mathcal {C}}\) with \(r_1(\hat{P}_2) =x_{j+1}\). We show \(O_1(\hat{P}_2) = \{x_{j+1}\}\). By strategy-proofness, it is enough to show this for \(\hat{P}_2=x_{j+1}x_j \ldots \).
First, we claim \(x_k \notin O_1(\hat{P}_2)\) for all \(k \ne j,j+1\). Assume for contradiction that \(x_k \in O_1(\hat{P}_2)\) for some \(k \ne j,j+1\). Then, \(f(P_1,\hat{P}_2)=x_k\) for some \(P_1 \in {\mathcal {C}}\) with \(r_1(P_1)=x_k\). However, since \(O_1(\bar{P}_2)=\{x_j\}\) for all \(\bar{P}_2 \in {\mathcal {C}}\) with \(r_1(\bar{P}_2)=x_j\), agent 2 manipulates at \((P_1,\hat{P}_2)\) via some preference \(\bar{P}_2\) with \(r_1(\bar{P}_2)=x_j\).
Now, we show that \(x_j \notin O_1(\hat{P}_2)\). Assume for contradiction that \(x_j \in O_1(\hat{P}_2)\). Let \(\hat{P}_2'=x_{j+1}x_{j+2} \ldots x_m \ldots x_j \ldots \). Then, by Lemma 1, \(x_j \in O_1(\hat{P}'_2)\). Take \(P_1 \in {\mathcal {C}}\), such that \(r_1(P_1)=x_j\). Then, because \(x_j \in O_1(\hat{P}'_2)\), \(f(P_1,\hat{P}'_2)=x_j\). Now, take \(P_2 \in {\mathcal {C}}\) with \(r_1(P_2)=x_m\). Since \(O_1(P_2) = \{x_m\}\), we have \(f(P_1,P_2)=x_m\). This means that agent 2 manipulates at \((P_1,\hat{P}'_2)\) via \(P_2\). This completes the proof of the lemma. \(\square \)
Proof of Theorem 2
The Proof of Theorem 2 follows using the same arguments as for the proof of Theorem 1 with the exception that Lemma 5 should be used in place of Lemma 7.\(\square \)
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Achuthankutty, G., Roy, S. Dictatorship on top-circular domains. Theory Decis 85, 479–493 (2018). https://doi.org/10.1007/s11238-018-9667-7
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DOI: https://doi.org/10.1007/s11238-018-9667-7