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Unformatted text preview: X . By the above lemma, X 1 is not empty. Assume we have constructed the sets X 1 , . . . , X k . If X = X 1 X 2 . . . X k we are done. If not, deFne X k + 1 to be the set of minimal elements in X X 1 X 2 X k . By the lemma X k + 1 6= . Since X is Fnite we must be done after at most | X | steps. DeFne U ( x ) = k if x X k . Thus, U ( x ) is the step number at which x is eliminated. To verify that U represents % , let a % b . Then b / X X 1 X 2 X U ( a ) and thus U ( a ) U ( b ) . Without any further assumptions on the preferences, the exis-tence of a utility representation is guaranteed when the set X is...
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This note was uploaded on 12/29/2011 for the course ECO 443 taught by Professor Aswa during the Fall '10 term at SUNY Stony Brook.
- Fall '10