A -Theoretical Proof of Hartogs Extension Theorem on Stein by Ruppenthal J.

By Ruppenthal J.

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V Then the preference ordering produced by δ is obtained by similarly permuting the preference ordering produced from ρ. That, is, for any B, C ∈ A, ρ B ⊒ C ⇐⇒ δ σ(B) ⊒ σ(C) . 2) between three or more alternatives does not satisfy the axiom (N). 9 Check that the following voting procedures do satisfy axiom (N): 1. 1(a)). 2. 1(b)). 3. 1(c). 10 Show that, if a voting procedure satisfies the Monotonicity axiom (M) and the Neutrality axiom (N), then it must satisfy the Pareto axiom (P). 3(b) on page 34.

First if each voter casts a single approval vote, so that we basically have a traditional plurality competition, then Arianne wins, with 4 votes: # Preferences Approval Points A B C 4 A ≻ D ≻ C ≻ B 1 0 0 D 0 0 1 0 0 2 B ≻ A ≻ C ≻ D 0 1 0 0 3 C ≻ B ≻ D ≻ A 0 0 1 0 2 D ≻ B ≻ C ≻ A 0 0 0 1 1 D ≻ C ≻ B ≻ A 0 0 0 1 Total score: 4 3 3 3 1 B ≻ A ≻ D ≻ C Next, if each voter casts two approval vote, then Bryn wins, with 8 points: # Preferences Approval Points A B C 4 A ≻ D ≻ C ≻ B 1 0 0 D 1 1 1 0 0 2 B ≻ A ≻ C ≻ D 1 1 0 0 3 C ≻ B ≻ D ≻ A 0 1 1 0 2 D ≻ B ≻ C ≻ A 0 1 0 1 1 D ≻ C ≻ B ≻ A 0 0 1 1 Total score: 7 8 4 7 1 B ≻ A ≻ D ≻ C However, if each voter casts three approval vote, then Chloe wins, with 12 points: # Preferences Approval Points A B 1 0 C 1 D 1 1 1 0 1 2 B ≻ A ≻ C ≻ D 1 1 1 0 3 C ≻ B ≻ D ≻ A 0 1 1 1 2 D ≻ B ≻ C ≻ A 0 1 1 1 1 D ≻ C ≻ B ≻ A 0 1 1 1 4 A ≻ D ≻ C ≻ B 1 B ≻ A ≻ D ≻ C Total score: 7 9 12 11 2C.

We want to show that A ⊒ B ⇐⇒ A ρ B (regardless of m what the other voters think). Suppose that ρ is some profile. 8) To do this, we will construct a new profile δ, so that each voter’s ρ-preferences concerning {A, B} are identical to her δ-preferences. Thus, by (IIA), ρ ⇐⇒ A ⊒ B δ A ⊒ B . δ We will build δ so that it is clear that A ⊒ B. To do this, we introduce a third alternative, C ∈ A. By axiom (IIA), the position of C in the δ-preferences of voters v1 , . . , vn has no δ δ effect on whether A ⊒ B or B ⊒ A.

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