Question

When there is a finite set of N = {1, ... n} as individuals, and there is n>X>=3 where X are possible alternatives. And R is the set of orderings of X and B is the set of binary relations of X. for all x, y in X and R in Rⁿ, define arrovian social welfare function f:Rⁿ -> B as

1) xPy IFF xPiy for at least n-1 individuals

2) xRy IFF not(yPx)

where R=f(R).

Prove that R=f(R) is acyclic for all R in Rⁿ

Answer 1

i don't known the exact answer to the question but I think that the "IFF" in the given question refers to the meaning of "if" then the answer maybe is:

N = {1, ... n}

R is R₁,R₂,R₃,R₄,R₅,R₆,.....Rn and R¹₁,R¹₂,R¹₃,R¹₄........R¹_n be two sets of individual orderings

n>X>=3

then X=3,...........n-1

Social Welfare Function is a function that ranks social states.

(1.)

xPy IFF xPiy for at least n-1 individuals

xPy ⇔ xPiy

the two profiles generate the same ranking of alternatives in set x,y until n-1.

When the set of alternativesis cartesian and the domain of the preferences is supersaturating an arrovian social welfare function.

f:Rⁿ -> B that is Rⁿ = Binary sets.

where R=f(R).

(2.)asymmetric part of R: xRy and not yPx

xRy $\neq$ yPx

where R=f(R).

R is transitive and reflective but not quasi-transitive then it doesn't imply to P

x₁Py_{2 ,} x₂Py_{3 ,} x₃Py_{4 ,} x₄Py_{5 ,............} xnPy_n-1 and x_nP₁

x_nP_{1 and} x₁P_n

If R is not quasi-transitive then P is acyclic for all R in Rn.

xRy $=$ yPx then P is cyclic.