Rubinstein2005-page137

# Rubinstein2005-page137 - x ± ² i y for all i ∈ G 1 and...

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October 21, 2005 12:18 master Sheet number 135 Page number 119 Social Choice 119 Let c be a third alternative. Let ( ± i ) i N be a proﬁle satisfying a ± i b iff a ± ² i b for all i , and for all i G , a ± i c ± i b ± i x for every x X { a , b , c } and for all j / G , c ± j y ± j x for every y ∈{ a , b } and for every x X −{ a , b , c } . Since G is almost decisive, a ± c .By Par , c ± b , therefore, a ± b by transitivity. Group Contraction Lemma: If G is decisive and | G |≥ 2, then there exists G ² G such that G ² is decisive. Proof: Let G = G 1 G 2 , where G 1 and G 2 are nonempty and G 1 G 2 =∅ . By the Field Expansion Lemma it is enough to show that G 1 or G 2 is almost decisive. Take three alternatives a , b , and c and a proﬁle of preference rela- tions ( ± i ) i N satisfying for all i G 1 , c ± i a ± i b , and for all i G 2 , a ± i b ± i c , and for all other i , b ± i c ± i a . If G 1 is not almost decisive, then there are x and y and a proﬁle ( ± ² i ) i N such that
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Unformatted text preview: x ± ² i y for all i ∈ G 1 and y ± ² i x for all i / ∈ G 1 , such that F ( ± ² 1 , . . . , ± ² n ) determines y to be at least as preferable as x . Therefore, by I ∗ , b ± c . Similarly, if G 2 is not almost decisive, then c ± a . Thus, by transi-tivity b ± a , but since G is decisive, a ± b , a contradiction. Thus, G 1 or G 2 is almost decisive. Proof of the Theorem: By Par , the set N is decisive. By the Group Contraction Lemma, every decisive set that includes more than one member has a proper subset that is decisive. Thus, there is a set { i ∗ } that is decisive, namely, F ( ± 1 , . . . , ± n ) ≡± i ∗ ....
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