ECON
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2 example 1 ε ε 2 ε 2 ε 2 ε 2 1 ε ε 2 1 0 ε 2

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2 Example: (1 + ε - ε 2 , ε - 2 ε 2 , - ε 2 ) = (1 + ε - ε 2 )(1 , 0 , 0) + ( ε - 2 ε 2 )(0 , 1 , 0) + ε 2 (0 , 0 , - 1) . If the relations (2)–(3) are true, we say that a hierarchic price ( q 1 , . . . , q k ) represents a non-standard price vector p and denote it by q p . Conversely, if q = ( q 1 , . . . , q k ) is a hierarchic price, one may consider for some ε 0 , ε > 0 a vector of non-standard prices p ( q, ε ) = q 1 + εq 2 + · · · + ε k - 1 q k such that q p ( q,ε ) = q. Evidently, there is much more than one non-standard vector that satisfies this property, in particular q p ( q,ε ) = q p ( q,ε 2 ) . Furthermore, if p = 0 , then put q p = { 0 } . The next proposition gives a characterization of the set ¯ B ( p ) and is crucial for our analysis. It asserts that if X is a polyhedral set then there is a number m ∈ { 1 , . . . , k } such that the set ¯ B ( p ) consists of elements x such that the vector ( q j x ) j =1 ,...,m is lexicographically less than zero. Remind that a subset P IR l is called a polyhedral set or a polyhedron if it is the intersection of a finite number of closed half-spaces. Each polyhedral set is closed and convex but not necessarily compact. Denote the set { x X | q 1 x = 0 , . . . , q m x = 0 } by X ( q 1 , . . . , q m ) and put X ( ) = X. Consider the sets B m ( p ) = { x X ( q 1 , . . . , q m - 1 ) | q m x 0 } , m k, (5) and B k +1 ( p ) = X ( q 1 , . . . , q k ) . (6) Proposition 3.2 Suppose that p * IR l , and at least one of the following alternatives is true: (a) The set X is a polyhedron; (b) The set X is closed, star-shaped with respect to 0 and locally polyhedral at 0 , that is for some small positive γ IR ++ the set P = { x X : | x j | ≤ γ, j L } is a polyhedron. 11

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Then there exists a natural number m ∈ { 1 , . . . , k + 1 } such that ¯ B ( p ) = B m ( p ) . Moreover, for m k there exists y ¯ B ( p ) such that q m y < 0 . The proof is based on the following lemma. Lemma 3.3 Suppose that X IR l is a polyhedral set and p * IR l . Then pX 0 implies p ( * X ) 0 . Note first that the conclusion of the lemma does not hold if X is not polyhedral. Take for instance X = { x IR 2 + | x 2 ( x 1 ) 2 } and p = ( - ε, 1) , ε 0 , ε > 0 . Then for each x X px = - εx 1 + x 2 0 . But once an element ˜ x = ( ε/ 2 , ε 2 / 4) * X is taken, p ˜ x = - ε 2 / 2 + ε 2 / 4 < 0 . Proof. The set X consists of all vectors in IR l that satisfy some system of linear inequalities: X = { x IR l : d α x g α , α A } , where d α IR l \{ 0 } , g α IR , and A is finite. It follows from the Fundamental Theorem of Linear Inequalities (see, for example, Schrijver (1986), Theorem 7.16) that X can be represented as a sum of a compact polyhedron Y and a convex cone Z with a finite number of generators. In other words, X = Y + Z, and for some finite sets B IR m , C IR m Y = conv B = { X b B β b b | β b IR + & X b B β b = 1 } , and Z = con C = { X c C γ c c | γ c IR + } . Then by definition pX 0 ⇐⇒ pb 0 & pc 0 b B, c C. By the Transfer Principle, * X = * Y + * Z, where * Y and * Z are defined by substitution IR + for * IR + in the definitions of Y and Z, respectively. Therefore, pb 0 & pc 0 b B, c C ⇐⇒ p ( * X ) 0 .
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• Spring '16
• Equilibrium, Economic equilibrium, General equilibrium theory, Non-standard analysis, Florig

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