Rubinstein2005-page32 - X . By the above lemma, X 1 is not...

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October 21, 2005 12:18 master Sheet number 30 Page number 14 14 Lecture Two start with a lemma regarding the existence of minimal elements (an element a X is minimal if a - x for any x X ). Lemma: In any Fnite set A X there is a minimal element (similarly, there is also a maximal element). Proof: By induction on the size of A .I f A is a singleton, then by complete- ness its only element is minimal. ±or the inductive step, let A be of cardinality n + 1 and let x A . The set A −{ x } is of cardinality n and by the inductive assumption has a minimal element denoted by y .I f x % y , then y is minimal in A .I f y % x , then by transitivity z % x for all z A −{ x } and thus x is minimal. Claim: If % is a preference relation on a Fnite set X , then % has a utility representation with values being natural numbers. Proof: We will construct a sequence of sets inductively. Let X
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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.

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