Lecture Notes MATHEMATICAL ECONOMICS

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1 Lecture Notes MATHEMATICAL ECONOMICS A. Neyman Fall 2006, Jerusalem The notes may/will be updated in few days. Please revisit to check for the updated version. Please comments to 1 Course Outline Exchange economies, efficient allocations, competitive equilibria, efficiency of competitive equilibria, existence of competitive equilibrium, generic finiteness of competitive equilibria, core and competitive equilibria. 2 References Debreu, G., Theory of Value. New York: Wiley (1959). Debreu, G., Mathematical Economics, Twenty papers of Gerard Debreu. Cambridge: Cambridge University Press (1983). Arrow, K. and Hann, F., General Competitive Analysis. San Francisco: Holden-Day (1971). 1

2 Arrow, K., and Intriligator, M., eds. Handbook of Mathematical Economics, vol. I. New york: North-Holland (1981). Arrow, K., and Intriligator, M., eds. Handbook of Mathematical Economics, vol. II. New york: North-Holland (1982). Mas-Colell, A., The theory of general economic equilibrium: A differentiable approach. Cambridge U.K.: Cambridge University Press (1985). Mas-Colell, A., M. D. Whinston and J. R. Green, Microeconomic Theory. New York Oxford: Oxford University Press (1995). 3 Exchange Economies An exchange economy consists of: 1) A set of traders T. 2) A vector space V - the space of goods. 3) For every t T a subset X(t) V - the consumption set of trader t. 4) For every t T, an element e(t) X(t)- the initial endowment of trader t. 2

3 5) For every t T a binary relation t on X(t) - the preference relation of trader t. For the mean time we assume that the trader set T consists of finitely many traders, i.e., T is a finite set. If there are l goods, V = IR l, and an element x V, x = (x 1,..., x l ) is interpreted as a bundle of x i units of commodity i. The set X(t) represents the consumption set of trader t, i.e. the set of all feasible consumption bundles of trader t. For instance, if one of the commodities is leisure in a given day, it is bounded by 24 hours. Restrictions of the consumption set may also result by interrelations of several commodities. The initial endowment e(t) represents the initial bundle of goods of trader t before trading takes place. The preference relation t represents trader s t preference over his feasible consumption. Given x, y X(t), x t y means that trader t prefers (strictly) bundle x over bundle y. Remarks The binary (preference) relation t is called transitive if x t y and y t z imply x t z. It is continuous if {(x, y) X(t) X(t) : x t y} is open in X(t) X(t). 3

4 A utility function on X(t) is a real valued function u : X(t) IR. The preference relation induced by a utility function u is the binary relation x y u(x) > u(y). A binary relation induced by a utility function is transitive, and a binary relation induced by a continuous utility function is continuous. An important implicit assumption in our present modeling of an exchange economy is that there are no externalities. Each individual cares about his consumption alone. Example of an exchange economy with 2 traders and 2 commodities: T = {1, 2} V = IR 2 IR 2 + = X(1) = X(2) e(1) = (4, 0) and e(2) = (0, 2) 1 is the preference relation induced by the utility function u(x 1, x 2 ) = x 1 x 2, and 2 is the preference relation induced by the utility function u(x 1, x 2 ) = min(x 1, x 2 ). Definition 1 An allocation in an exchange economy T ; V ; (X(t)) t T ; (e(t)) t T ; t T is a function z : T V such that 1. t T, z(t) X(t) 4

5 2. t T z(t) = t T e(t). Question 1 What is the set of all allocations in the above example? An allocation is a pair of 2-dimensional vectors, z(1) and z(2), in IR 2 + such that z(1) + z(2) = e(1) + e(2) = (4, 2). Therefore z(1) obeys (0, 0) z(1) (4, 2) (where for two finite dimensional vectors, x = (x 1,..., x n ) and y = (y 1,..., y n ), x y x i y i for every i = 1,..., n,) and z(2) = (4, 2) z(1). Therefore specifying z(1) determines the allocation z, thus the set of all allocation in this example is represented by the points in the rectangle {z IR 2 : (0, 0) z (4, 2)}. Note that whenever there are two traders and two commodities, and the consumption sets are the nonnegative orthant, there is a 1 1 correspondence between the rectangle {z IR 2 : (0, 0) z e(1) + e(2)} and the set of allocations. This rectangle is called the Edgeworth box. 5

6 (0,2) (4,2) (0,0) Edgeworth box initial endowment (4,0) Question 2 How can we describe the preferences in the above example? The consumption set of trader 1 is the positive orthant of IR 2 and his preference relation, 1 can be described by means of the indifference curves. E.g., consider all points x = (x 1, x 2 ) IR 2 + with x 1 x 2 = 1. The geometric location of all these points is called an indifference curve. Similarly, for any given α 0, the geometric location of all points x = (x 1, x 2 ) IR 2 + with x 1 x 2 = α is denoted Iα 1 and is an indifference curve. Trader 1 is indifferent between any two bundles in Iα, 1 and if x Iα 1 and y Iβ 1 then x 1 y α > β. Similarly, the consumption set of trader 2 is the positive orthant of IR 2 and his preference relation, 2 can be described by means of the indifference curves. E.g., consider all points y = (y 1, y 2 ) IR 2 + with min(y 1, y 2 ) = 1. The geometric location of all these points is an indifference curve of the 6

7 preference relation 2. Similarly, for any given α 0, the geometric location of all points x = (x 1, x 2 ) IR 2 + with min(x 1, x 2 ) = α is denoted Iα 2 and is an indifference curve of 2. Trader 2 is indifferent between any two bundles in Iα, 2 and if x Iα 2 and y Iβ 2 then x 2 y α > β. Definition 2 An allocation z is (weakly) efficient if there is no allocation y with y(t) t z(t) for every t T. Question 3 What are the (weakly) efficient allocations in the above example? Question 4 Are there efficient allocation in each exchange economy? Definition 3 A competitive equilibrium of an exchange economy E is an ordered pair (p, z) such that: p is a nontrivial linear functional on V. z is an allocation with p(z(t)) = p(e(t)) for every t T, and t T, {x X(t) : p(x) p(e(t)) and x t z(t)} =. Remarks If V = IR l then any linear functional over V is represented by a vector p = (p 1,..., p l ) where l p(x) = p i x i i=1 7

8 The vector p is called the vector of prices and p i is interpreted as the price of commodity i. Question 5 What are the competitive equilibria in the above example? Efficiency of Competitive Equilibrium Theorem 1 Let E = T ; V ; (X(t)) t T ; (e(t)) t T ; ( t ) t T be an exchange economy, and assume that (p, x) is a competitive equilibrium. Then the allocation x is weakly efficient. proof Assume that z is an allocation with z(t) t x(t) for every trader t T. As (p, x) is a competitive equilibrium, p(z(t)) > p(e(t)). As p is a linear functional, p( t T z(t)) = t T p(z(t)) and p( t T e(t)) = t T p(e(t)) and therefore p( t T z(t)) > p( t T e(t)) which contradicts the equality t T z(t) = t T e(t) 4.2 The Core A coalition is a subset of T. Let S T be a coalition. An S - allocation is a function x : S V such that for every t S x(t) X(t) and t S x(t) = 8

9 t S e(t). Definition 4 The core of an exchange economy E is the set of all allocations y for which there is no coalition S and an S - allocation x with x(t) t y(t) for every t S. Definition 5 A competitive allocation is an allocation z for which there is a competitive equilibrium (p, z). Consider the following exchange economy E. The set of traders is T = {1, 2, 3, 4}. There are two goods in the market; the space of goods is IR 2. The consumption sets of each trader t T is X(t) = IR 2 +. The initial endowments are: e(1) = e(2) = (1, 0) and e(2) = e(4) = (0, 1). The preference relation of each trader t T is induced by the utility function u t (x 1, x 2 ) = x 1 x 2. Consider the allocation x where x(1) = x(3) = (1/4, 1/4) and x(2) = x(4) = (3/4, 3/4). Prove that x is an efficient allocation. Question 6 Find all efficient allocations y with y(1) = y(3) and y(2) = y(4) of the above exchange economy. Consider the above allocation x. Show that it is not a core allocation by exhibiting a {1, 2, 3}-allocation y where y(t) x(t) for every 1 t 3. 9

10 Question 7 Find all core allocations y with y(1) = y(3) and y(2) = y(4) of the above exchange economy. Show that every core allocation y of the above exchange economy satisfies y(1) = y(3) and y(2) = y(4). Therefore the solution of the previous question describes all allocations in the core of E. Theorem 2 Any competitive allocation is in the core. Proof. Assume that (p, z) is a competitive equilibrium, S a nonempty subset of T, and x : S V with x(t) t z(t) for every t S. As (p, z) is a competitive equilibrium, px(t) > pz(t) for every t S, and therefore t S px(t) > t S pz(t) = t S pe(t), i.e., p t S x(t) > p t S z(t) = p t S e(t), implying that t S x(t) t S e(t). Thus x is not an S-allocation. We have thus proved that there is no coalition S and an S-allocation x with x(t) z(t) for every t S, i.e., that z is in the core of the exchange economy. 4.3 Preferences, Utility, and Demand A binary relation R on a set X is a subset R X X. We write xry whenever (x, y) R. The binary relation R is reflexive if xrx for every x X. It is transitive if for every x, y, z X, xry and yrz implies 10

11 xrz. A binary relation R on X is complete if for every x, y X either xry or yrx. 4.4 Brouwer s Fixed Point Theorem Theorem 3 (Brouwer s fixed point theorem): Let C be a nonempty convex compact subset of IR n and let f : C C be a continuous function. Then f has a fixed point, i.e., there is a point x C such that f(x) = x. Remarks The proof of Brouwer s fixed point theorem in the 1-dimensional case is equivalent to the mean value theorem of continuous functions. Indeed, a convex compact subset of IR is a closed interval [a, b], and a continuous function f : [a, b] [a, b] satisfies f(a) a and f(b) b. Therefore, the continuous function g : [a, b] IR defined by g(x) = f(x) x obeys g(a) 0 and g(b) 0 and therefore there is x [a, b] such that g(x) = 0 i.e. f(x) = x. Each one of the assumptions in Brouwer s fixed point theorem necessary. Indeed, if S 1 = {(x, y) IR 2 : x 2 +y 2 = 1}, the continuous function f : S 1 S 1 defined by f(x, y) = (x, y) does not have a fixed point. The continuous function f : (0, 1] (0, 1] defined by f(x) = x/2 does not have a fixed point. The continuous function f : IR IR defined by f(x) = x + 1 does not have a fixed point. The function f : [0, 1] [0, 1] defined by f(x) = x/2 if x > 0 11

12 and f(0) does not have a fixed point. The graph of a function f : C D is the subset {(x, y) : y = f(x)} of C D, and is denoted Γ f. When D IR n is a compact subset, the function f is continuous if and only if Γ f is a closed subset of C D. Indeed, assume that f is continuous and that the sequence (x k, y k ) Γ f converges as k to a point (x, y) C D. Then x k x as k and y k = f(x k ). As f is continuous y k = f(x k ) f(x) and therefore y = f(x), i.e., (x, y) Γ f. In the other direction, assume that Γ f is closed in C D. Let x k, x C and assume that x k x. We have to prove that y k = f(x k ) f(x) as k. Otherwise, as D is compact, there is a subsequence (y km ) which converges to a point y D with y f(x); therefore, (x km, y km ) (x, y ) C D. As (x km, y km ) Γ f and Γ f is closed in C D, (x, y ) C D, i.e y = f(x) a contradiction. 4.5 Set-Valued Functions A set-valued function from a set X to a set Y is a function which maps points in X to subsets of Y. A fixed point of a set-valued function f from X to X 12

13 is a point x X such that x f(x). Consider the following set-valued function from [0, 1] to [0, 1]. f(x) = {x + 1/2} if x < 1/2 {0, 1} if x = 1/2 {x 1/2} if x > 1/2 The graph of a set valued function f from X to Y is the subset {(x, y) : x X, y = f(x)} of X Y. The graph of the above set valued function is depicted in the following figure: y 1 1 Figure 1 (1,1) x Note that this set-valued function does not have any fixed point. We describe next another set-valued function from [0, 1] to [0, 1]. 13

14 f(x) = {x + 1/2} if x < 1/2 [0, 1] if x = 1/2 {x 1/2} if x > 1/2 The graph of this set-valued function is depicted in the following figure: y 1 1 Figure 2 (1,1) x 5 Kaukutani s Fixed point Theorem Definition 6 A set-valued function f from X to Y is called uppersemicontinuous if x n x X, y n y Y, and y n f(x n ) for all n imply y f(x). The graph of a set-valued function f from C to D is the subset {(x, y) : y f(x)} of C D, and is denoted Γ f. 14

15 The function f is uppersemicontinuous if and only if Γ f is a closed subset of C D. Indeed, let f be a set-valued function from C to D and assume that (x n, y n ) Γ f C D is a sequence that converges to (x, y) C D. If f is uppersemicontinuous then y f(x) and therefore Γ f is closed in C D, and if Γ f is closed in C D then (x, y) Γ f, i.e., y f(x) and thus f is uppersemicontinuous. If f is an uppersemicontinuous set-valued function from C to D, then, for any x C, f(x) is a closed subset of D. If f is a function from C to D, we may identify the function f with the set valued function which maps a point x C to the subset {f(x)} of D. Note that if f is continuous then the associated set-valued function is uppersemicontinuous, and if D is compact and the associated set-valued function is uppersemicontinuous then f is continuous. Given a subset A of IR k and a point x IR k we define the distance of x to A, d(x, A), as inf{ x y : y A}. Claim 1: Let f be an uppersemicontinuous set-valued function from a 15

16 subset C of a Euclidean space into a compact subset D of a Euclidean space. Then, for every x C and ε > 0 there is δ = δ(x, ε) > 0 such that if x, x C with x x < δ and z f(x ), then d(f(x), z) < ε. Proof. Otherwise, there are x C, ε > 0, and two sequences (x n ) n=1, and (z n ) n=1 such that x n x 0 as n, z n f(x n ) and d(z n, f(x)) > ε. W.l.o.g. we assume that z n z D as n (otherwise we take a subsequence). As f is uppersemicontinuous, z f(x). Therefore, d(z n, f(x)) z n z 0 as n. A contradiction. Claim 2: Let A be a convex subset of IR m, y 1,..., y m IR m and α i 0 with m i=1 α i = 1. Then m d( α i y i, A) max m d(y i, A). i=1 i=1 Proof. Let ε > 0. For every 1 i m there is x i A with x i y i < d(y i, A) + ε. Note that A is convex and therefore m i=1 α i x i A thus d( m i=1 α i y i, A) m i=1 α i y i m i=1 α i x i which by the triangle inequality is at most m i=1 α i y i x i which is bounded by max m i=1 y i x i max m i=1 d(y i, A) + ε. We are ready now to state Kakutani s fixed point theorem which is a 16

17 generalization of Brouwer s fixed point theorem. Theorem 4 (Kakutani s fixed point theorem) Let f be an uppersemicontinuous set-valued function from a nonempty convex compact subset C of IR n into itself (i.e., f(x) C for all x C) such that for all x C f(x) is a nonempty convex subset of C. Then f has a fixed point, i.e., there is a point x C with x f(x). Proof We prove Kakutani s fixed point theorem from Brouwer s fixed point theorem. The idea of the proof is as follows: we construct a sequence of continuous functions g n : C C such that any limit point of a sequence of points (x n, y n ) Γ gn is in the graph of f. By Brouwer s fixed point theorem it follows that g n has a fixed point x n, i.e., a point x n C with x n = g n (x n ). Let (x, y) C C be a limit point of (x n, g n (x n )) = (x n, x n ). As the diagonal, {(x, x) : x C}, is closed in C C, y = x and thus x is a fixed point of f. (Alternatively, w.l.o.g. we assume that x n x C as n, (otherwise we replace the sequence (n) n=1 with a subsequence (n k ) k=1), and therefore (x n, x n ) (x, x) Γ f i.e., x is a fixed point of the set-valued function f.) Construction of the functions g n, n = 1,

18 Fix n > 0 and for every x C let B(x, 1/n) denote the open ball of radius 1/n and center x, i.e., B(x, 1/n) = {y C : y x < 1/n}. Recall that C is compact, and observe that C = x C B(x, 1/n). Therefore there is a finite subset of these balls that covers C, i.e., for any n there is a list x n 1,..., x n m (m = m(n)) of points in C such that C = m i=1b(x n i, 1/n). Define the real valued functions δi n : C IR by δi n (y) = max(0, 1/n y x i ). Note that δi n (y) > 0 if x i y < 1/n and δi n (y) = 0 if y x i 1/n, and that m j=1 δi n (x) > 0 for every x C. Therefore, for every i = 1,..., m, δ n i (x) m j=1 δn j (x) is a continuous positive (real valued) function defined on C, and m i=1 δ n i (x) mj=1 δ n j (x) = 1. For any i = 1,..., m let y n i f(x n i ) and define g n : C C by g n (x) = m i=1 δ n i (x) mj=1 δ n j (x) yn i It follows that g n is a continuous function from C to C. Therefore, by Brouwer s fixed-point theorem, g n has a fixed point x n, i.e., g n ( x n ) = x n. Let x be a limit point of the sequence ( x n ) n=1. We will show that d(x, f(x)) = 0. As f(x) is closed, it will follow that x f(x) i.e., x is a fixed point of f. 18

19 Fix ε > 0 and let δ = δ(x, ε) be defined as in claim 1. As x is a limit point of the sequence ( x n ) n=1, there is n > 2/δ with x n x < δ/2. For every i with x n i x n δ/2 > 1/n, δ n i ( x n ) = 0, and therefore g n ( x n ) is a convex combination of finitely many points y j with y j f(x n j ) and x n j x x n j x n + x n x < δ, implying that d(y j, f(x)) < ε. Thus x n = g n ( x n ) = j: x n j x <δ with d(y j, f(x)) < ε and thus by claim 2, δn j ( x n ) j δ n j ( x n ) yn j d( x n, f(x)) j: x n j x <δ δn j ( x n ) j δ n j ( x n ) d(yn j, f(x)) ε, implying that d(x, f(x)) x x n + d( x n, f(x)) < 2ε. As this last inequality holds for every ε > 0, d(x, f(x)) = 0, and as f(x) is closed, x f(x). 6 Applications of Kaukutani s f.p. theorem We start with few remarks: (i) If F i is an upper semicontinuous set-valued function from X to Y i, i = 1,..., n, then (F 1,..., F n ) is an upper semicontiuous set-valued function from 19

20 X to Y 1... Y n. (ii) If Y i is a convex subset of a (real) vector space and F i is a convex valued set-valued function from X to Y i, i = 1,..., n, then (F 1,..., F n ) is a convex valued set-valued function from X to Y 1... Y n. (iii) If g : X Y is a continuous function, and F is an upper semicontinuous set-valued function from Y to Z, then F g is an uppersemicontinuous set-valued function from X to Z. (iv) If F i is an uppersemicontiuous set-valued function from Y i to X i, i = 1,..., n, and π i is a continuous function from X to X i, then (F 1 π 1,..., F n π n ) is an uppersemicontinuous set-valued function from X to X 1,..., X n. (v) If F i is a convex valued set-valued function from Y i to X i, where X i is a convex subset of a real vector space, i = 1,..., n, and π i is a continuous function from X to Y i, then (F 1 π 1,..., F n π n ) is a convex valued set-valued function from X to X 1,..., X n. Theorem 5 Let X 1,..., X n be compact convex subsets of Euclidean spaces, and assume that f i are nonempty convex valued and uppersemicontinuous set-valued functions from n i=1x i X i. The (f 1,..., f n ) has a fixed point. 20

21 Proof. By remarks (i) and (ii), F = (f 1,..., f n ) is a convex valued uppersemicontinuous set-valued function. As each f i in nonempty valued, F is nonempty valued. As each X i is convex and compact, so is n i=1x i. Therefore the set-valued function F from n i=1x i to itself obey all the assumptions of Kakutani s fixed point theorem, and thus has a fixed point. Lemma 1 Let X and Y be compact subsets of Euclidean spaces, and let u : X Y IR be a continuous function. Then the set valued function F from X to Y defined by F (x) = {y Y : u(x, y) = max u(x, z)}, z Y is uppersemicontinuous. Proof. Assume that x n X and y n Y, n = 1,... are sequences that converge to x X and y Y respectively, and that y n F (x n ). Let z Y. As y n F (x n ), u(x n, y n ) u(x n, z). As u is continuous, u(x n, y n ) u(x, y) and u(x n, z) u(x, z) as n, and therefore u(x, y) = lim n u(x n, y n ) u(x, z) = lim n u(x n, z), 21

22 and as this holds for any z Y, y F (x), and thus F is uppersemicontinuous. Definition 7 Let X be a convex subset of a Euclidean space. A real valued function f : X IR is called quasi concave if for every α IR the set {x X : f(x) α} is convex. If Y is a set, a real valued function f : X Y IR is called quasi concave in x if for every α IR and every y Y, the set {x X : f(x, y) α} is convex. Theorem 6 Let X 1,..., X n be compact convex subsets of Euclidean spaces, and assume that u 1,..., u n are real valued quasi concave continuous function defined on n i=1x i. Then there is a point x = (x 1,..., x n ) n i=1x i such that for every 1 i n and every y i X i, u i (x 1,..., y i,..., x n ) u i (x 1,..., x i,..., x n ) = u i (x). Proof. For every i = 1,..., n define the set valued function F i from n i=1x i to X i by F i (x 1,..., x n ) = {y i X i : u i (x 1,..., y i,..., x n ) = max z i X i u i (x 1,..., z i,..., x n )}. Then, as X 1,..., X n are compact and u i is continuous, F i is non empty valued and uppersemicontinuous (Lemma 1). As X 1,..., X n are convex and u i is 22

23 quasi concave, F i is convex valued. By remarks (i) and (ii) F = (F 1,..., F n ) is a convex valued and uppersemicontinuous, and as each F i is nonempty valued F is also non empty valued. Therefore, by Theorem 4 F has a fixed point, i.e., there is a point x = (x 1,..., x n ) n i=1x i such that for every i, x i F i (x), i.e., for every y i X i, u i (x 1,..., y i,..., x n ) u i (x 1,..., x i,..., x n ) = u i (x). Definition 8 A set valued function f from X to Y is called lowersemicontiuous if for every sequence x n X which converges to a point x X and every y f(x), there is a sequence y n f(x n ) which converges to y. It is called continuous if it is uppersemicontinuous and lowersemicontinuous. Lemma 2 Let X and Y be compact subsets of Euclidean spaces, and let u : X Y IR be a continuous function, and G a continuous set-valued function from X to Y. Then the set valued function F from X to Y defined by is uppersemicontinuous. F (x) = {y G(x) : u(x, y) = max u(x, z)}, z G(x) 23

24 Proof. Assume that x n X and y n F (x n ) Y, n = 1,..., are sequences that converge to x X and y Y respectively. Recall that F (x n ) G(x n ) and G is uppersemicontinuous, and therefore y G(x). Let z G(x). As G is a continuous set valued function, there is a sequence z n G(x n ) that converges to z. As y n F (x n ), u(x n, y n ) u(x n, z n ). As u is continuous, u(x n, y n ) u(x, y) and u(x n, z n ) u(x, z) as n, and therefore u(x, y) u(x, z), and as this holds for any z G(x), y F (x), and thus F is uppersemicontinuous. Given sets X 1,..., X n, an index 1 i n, an element x = (x 1,..., x n ) in n i=1x i, and an element y i in X i, we denote by (x i, y i ) or by (x y i ) the element (x 1,..., x i 1, y i, x i+1,..., x n ) in n i=1x i. Theorem 7 Let X 1,..., X n be non-empty convex compact subsets of Euclidean spaces, X = n i=1x i, and assume that ϕ 1,..., ϕ n are continuous set-valued functions from X to X 1,..., X n respectively such that ϕ i (x) is non-empty and convex for every x X and 1 i n. If u i : X IR are continuous and quasi concave in x i, and F i is the set-valued function from X to X i defined by F i (x) = {y i ϕ i (x) : u i (x i, y i ) u i (x i, z i ) for every z i ϕ i (x)}, 24

25 then, the set-valued function (F 1,..., F n ) from X to X has a fixed point. Proof. Note that F i is an uppersemicontinuous set-valued function from X to X i with non-empty convex values. Therefore, (F 1,..., F n ) is an uppersemicontinuous set-valued function with non-empty convex values and thus by Kakutani s fixed point theorem has a fixed point. Lemma 3 Let = {p = (p 1,..., p n ) : p i 0 and n i=1 p i = 1}, e IR n + with e i > 0, and let Y be a convex subset of IR n with 0 Y. Then the set-valued functions ϕ from to Y that is given by n n ϕ(p) = {x Y : px = p i x i p i e i = pe} i=1 i=1 is continuous and has non-empty convex values. Proof. Assume that p k p and x k x Y as k and that p k x k = ni=1 p k i x k i n i=1 p k i e i = p k e. Then it follows that lim k ni=1 p k i x k i = ni=1 p i x i and lim k ni=1 p k i e i = n i=1 p i e i and therefore n n px = p i x i p i e i = pe, i.e., x ϕ(p), i=1 i=1 proving that ϕ is uppersemecontinuous. Assume next that p k k p and x ϕ(p). If n i=1 p i x i = px < pe = n i=1 p i e i, then for sufficiently large k, n i=1 p k i x i = p k x p k e = n i=1 p k i e i, i.e., x ϕ(p k ), and therefore 25

26 by setting x k = x we deduce that x k ϕ(p k ) and x k k x. Otherwise, if px = pe, p k x k pe > 0. Therefore, if ε k is defined by ε k = max(0, ( n i=1 p k i x i n i=1 p k i e i ))/ n i=1 p k i x i, 0 ε k k 0. As Y is convex and contains the points x and 0, x k = (1 ε k )x Y. As ε k k 0, x k k x and by the definition of ε k, x k p k p k e and thus x k ϕ(p k ). Thus ϕ is lowersemicontinuous. Lemma 4 Let = {p = (p 1,..., p n ) : p i 0 and n i=1 p i = 1}, e IR n + with e i > 0, and let Y be a convex subset of IR n that contains a neighborhood of e and with 0 Y. Then the set-valued functions ϕ from to Y that is given by n n ϕ(p) = {x Y : px = p i x i = p i e i = pe} i=1 i=1 is continuous and has non-empty convex values. Proof. Assume that p k p and x k x Y as k and that p k x k = ni=1 p k i x k i = n i=1 p k i e i = p k e. Then it follows that lim k ni=1 p k i x k i = ni=1 p i x i and lim k ni=1 p k i e i = n i=1 p i e i and therefore n n px = p i x i = p i e i = pe, i.e., x ϕ(p), i=1 i=1 proving that ϕ is uppersemecontinuous. Assume next that p k k p and x ϕ(p). Then, p k x k px = pe > 0. Y contains a neighborhood 26

27 of e. Therefore there is α > 0 such that (1 + α)e Y. Let θ k be defined as inf{θ 0 p k (θ(1 + α)e + (1 θ)x k ) p k e}. For every θ > 0 we have p k (θ(1 + α)e + (1 θ)x) k pe(1 + α) > pe. Therefore, θ k 0 as k. Set y k = θ k (1 + α)e + (1 θ k )x. (In other words, y k is the point on the interval [x, (1+α)e] with p k y k p k e that is the closest point to x.) Note that y k k x. Let ε k be defined by ε k = ( n i=1 p k i yi k n i=1 p k i e i )/ n i=1 p k i yi k, 0 ε k k 0. As Y is convex and contains the points x and 0, x k = (1 ε k )y k Y. As ε k k 0 and y k k x, x k k x and by the definition of ε k, x k p k p k e and thus x k ϕ(p k ). Thus ϕ is lowersemicontinuous. 27

28 7 Existence of Competitive Equilibria Theorem 8 Let E = T ; IR l ; (X(t)) t T ; (e(t)) t T ; t T be an exchange economy such that: 1. t T, X(t) = [0, M] l where M > t T e i (t) 2. t T and 1 i l, e i (t) > 0 3. t T, the preference relation of trader t, t, is represented by a continuous and monotonic quasi concave utility function u t : IR l IR. Then, there exists a competitive equilibrium (p, x) = (p, x(1),..., x(n)) with p i > 0 for every 1 i l. Proof. Assume that T = {1,..., n}. Set X 0 = = {p IR l + l i=1 p i = 1}, and for t T, set X t = X(t). The sets X i, i = 0, 1,..., n, are non-empty convex and compact subsets of Euclidean spaces. For every 0 t n we define a set-valued function ϕ t from X = n i=0x i to X t as follows: ϕ 0 (p, x(1),..., x(n)) =, and for 1 t n, ϕ t (p, x(1),..., x(n)) = {y X t py pe(t)}. By Lemma 3 the set valued functions ϕ t, 1 t n, are continuous and with non-empty and convex values. Obviously, the set-valued function ϕ 0 28

29 is continuous and with non-empty and convex values. Define the set-valued functions F i : X X i, i = 0, 1,..., n, by: n n F 0 (p, x(1),..., x(n)) = {q q (x(t) e(t)) r (x(t) e(t)) r } i=1 t=1 = arg max q n (x(t) e(t)), q i=1 and for 1 i n, F i (p, x(1),..., x(n)) = {y ϕ i (p, x(1),..., x(n)) u i (y) u i (z) z ϕ i (p, x(1),..., x(n))}. Note that by Theorem 7 the set-valued function F = (F 0,..., F n ) from X to X has a fixed point. Let (p, x(1),..., x(n)) be a fixed point of F. We will show that it is a competitive euilibrium. As x(i) F i (p, x(1),..., x(n)) ϕ i (p, x), px(i) pe(i) for every 1 i n, (1) and therefore n p( (x(i) e(i))) 0. (2) i=1 Let f(i) denote the i-th unit vector in IR l. As p F 0 (p, x(1),..., x(n)), we deduce that p( n i=1 (x(i) e(i))) f(j)( n i=1 (x(i) e(i))) = n i=1 (x j (i) e j (i)) for every 1 j l. Therefore, using (5), for every 1 j l, ni=1 (x j (i) e j (i)) 0. In particular, if ε > 0 with ε < M n t=1 e j (t), then 29

30 x(t)+εf(j) X t. By the monotonicity of u t we deduce that u t (x(t)+εf(j)) > u t (x(t)). Therefore, as x(t) F t (p, x), we deduce that px(t) + εp j = p(x(t) + εf(j)) > pe(t) px(t) (3) and thus we deduce that p j > 0, and as (6) holds for all sufficiently small ε > 0, we deduce also that px(t) pe(t) for every t T which together with (4) implies that for every t T, px(t) = pe(t). Therefore, using p > 0 and the inequality t T x(t) t T e(t), we deduce that t T x(t) = t T e(t). As x(i) F i (p, x(1),..., x(n)) we conclude that (p, x(1),..., x(n)) is a competitive equilibrium. The assumption that for every t T and every 1 i l we have e i (t) > 0 is restrictive. It assumes that every trader has a strictly positive quantity of every commodity. The next Theorem replaces this assumption by the assumption that every commodity is present in the market, namely, that for every 1 i l we have t T e i (t) > 0. Theorem 9 Let E = T ; IR l ; (X(t)) t T ; (e(t)) t T ; t T be an exchange economy such that: 1. t T, X(t) = [0, M] l where M > t T e i (t) 30

31 2. 1 j l, t T e j (t) > 0 3. t T, the preference relation of trader t, t, is represented by a continuous and monotonic quasi concave utility function u t : IR l IR. Then, there exists a competitive equilibrium (p, x) = (p, x(1),..., x(n)) with p j > 0 for every 1 j l. Proof. Assume that T = {1,..., n}. Set X 0 = = {p IR l + l i=1 p i = 1}, and for t T, set X t = [0, M] l. The sets X i, i = 0, 1,..., n, are non-empty convex and compact subsets of Euclidean spaces. Set X = 0 i n X i, and B i (p) = {y [0, M] l : py pe(t)}. Define the set-valued functions F i : X X i, i = 0, 1,..., n, by: n n F 0 (p, x(1),..., x(n)) = {q q (x(t) e(t)) r (x(t) e(t)) r } i=1 t=1 = arg max q n (x(t) e(t)), q i=1 and for 1 i n, if pe(i) > 0 then F i (p, x(1),..., x(n)) = {y B i (p) u i (y) u i (z) z B i (p)} and F i (p, x(1),..., x(n)) = B i (p) 31

32 if pe(i) = 0. The quasi concavity of u t implies that F t is convex valued for 1 t n. Next we show that F i are upper semi continuous with nonempty values. Assume (p k, x k (1),..., x k (n)) k (p, x(1),..., x(n)). We have to prove that p F 0 (p, x(1),..., x(n)) and that for every 1 t n we have x(t) F t (p, x(1),..., x(n)). Fix q. Then q( x(t) e(t)) = lim k q( x k (t) e(t)) lim k p k ( x k (t) e(t)) = p( x(t) e(t)) where the equalities follow from the continuity of the inner product and from the assumptions that p k p and x k (t) x(t), and the inequality follows from the fact that p k F 0 (p, x(1),..., x(n)). Thus, p F 0 (p, x(1),..., x(n)). Next we show that x(t) F t (p, x(1),..., x(n)). Note that px(t) = lim p k x k (t) lim p k e(t) = pe(t) and thus x(t) B t (p) and thus pe(t) = 0 implies that x(t) F t (p,...). If pe(t) > 0, set y k = y if p k y p k e(t) and y k = pk e(t) y if p k y pk y > p k e(t). It follows that p k y k p k e(t) and y k y. Therefore, for sufficiently large k we have u t (y k ) > u t (x k (t)) and p k y k p k e(t) contradicting the assumption that x k (t) F t (p,...). Thus F t is upper semi continuous. Note that by Theorem 7 the set-valued function F = (F 0,..., F n ) from X to X has a fixed point. Let (p, x(1),..., x(n)) be a fixed point of F. We will 32

33 show that it is a competitive euilibrium. As x(i) F i (p, x(1),..., x(n)) B i (p), px(i) pe(i) for every 1 i n, (4) and therefore n p( (x(i) e(i))) 0. (5) i=1 Let f(i) denote the i-th unit vector in IR l. As p F 0 (p, x(1),..., x(n)), we deduce that p( n i=1 (x(i) e(i))) f(j)( n i=1 (x(i) e(i))) = n i=1 (x j (i) e j (i)) for every 1 j l. Therefore, using (5), for every 1 j l, ni=1 (x j (i) e j (i)) 0. In particular, if ε > 0 with ε < M n t=1 e j (t), then x(t) + εf(j) X t. Assume that 1 t n with pe(t) > 0. By the monotonicity of u t we deduce that u t (x(t) + εf(j)) > u t (x(t)). Therefore, as x(t) F t (p, x), we deduce that px(t) + εp j = p(x(t) + εf(j)) > pe(t) > 0 (6) and thus we deduce that for every 1 j l we have p j > 0. Therefore (w.l.o.g.) pe(t) > 0 for every t. Thus (6) holds for all ε > 0. 33

34 As (6) holds for all sufficiently small ε > 0, we deduce also that px(t) pe(t) for every t T which together with (4) implies that for every t T, px(t) = pe(t). Therefore, using p > 0 and the inequality t T x(t) t T e(t), we deduce that t T x(t) = t T e(t). As x(i) F i (p, x(1),..., x(n)) we conclude that (p, x(1),..., x(n)) is a competitive equilibrium of the exchange economy E = T ; IR l ; (X(t)) t T ; (e(t)) t T ; t T. The next theorem replaces the assumption that X(t) = [0, M] l where M > t T e i (t) with the assumption that t T, X(t) is convex and IR l + X(t) [0, M] l where M > t T e i (t). We state and prove the result for the the special case X(t) = IR l +. However, the same proof applies to the general case. Theorem 10 Let E = T ; (X(t)) t T ; (e(t)) t T ; ( t ) t T be an exchange economy with l commodities such that: 1. t T, X(t) = IR l + 2. t T e i (t) > 0 for every 1 i l 3. t T, the preference relation of trader t, t, is represented by a con- 34

35 tinuous and monotonic quasi concave utility function u t : IR l IR. Then, there exists a competitive equilibrium (p, x) with p i > 0 for every 1 i l. Proof. Let E be the economy obtained from E by modifying the consumption set IR l of each trader to the consumption set [0, M] l where M = 2 max i t T e i (t). By the previous theorem, E has a competitive equilibrium (p, x) with p > 0. We claim that the competitive equilibrium (p, x) of E is a competitive equilibrium of E. As x is an allocation of E it is also an allocation of E. It remains to show that if t T and y IR l + with u t (y) > u t (x(t)) then py > pe(t). Indeed, if u t (y) > u t (e(t)), the continuity of u t implies that there is a sufficiently small 1 > ε > 0 such that u t ((1 ε)y) > u t (x(t)). Set y ε = (1 ε)y. If py pe(t) then py ε = (1 ε)py < pe(t) and for any 1 δ > 0 the point y ε,δ = δy ε + (1 δ)x(t) obeys py ε,δ < pe(t), and if δ > 0 is sufficiently small, y ε,δ i < M. By the quasi concavity of u t, u t (y ε,δ ) u t (x(t)). As u t is monotonic u t (y ε,δ + θe j ) > u t (y ε,δ ) u t (x(t)) for every θ > 0 and every 1 j l. On the other hand, for sufficiently small θ > 0, y ε,δ +θe j [0, M] l contradicting the assumption that (p, x) is a competitive equilibrium of E. Consider the following examples of exchange economies. The set of traders is T = {1, 2} and the number of commodities is l = 2. 35

36 The utilities of the traders are u 1 (x, y) = x + y and u 2 (x, y) = x + 2 y. What is the set of competitive equilibrium when e(1) = (1, 2) and e(2) = (2, 1)? Show that when e(1) = (1, 0) = e(2) there is no competitive equilibrium. Show that there is no competitive equilibrium in the case e(1) = e(2) = (1, 1), u 1 (x, y) = x 2 + y 2, and u 2 (x, y) = x + y. 8 Definition 9 An abstract social system consists of: 1. A finite set N 2. i N a nonempty set A i ; set A = i N A i and A i = j i A j 3. i N a set-valued function ϕ i : A A i 4. i N a set-valued function P i : A A i. The set A i is interpreted as the possible actions of agent i, (e.g., in an exchange economy it may stand for the consumption set). The set-valued function ϕ i represents the constrained feasible actions of agent i as a function of the actions of all agents or all the other agents, (e.g., if we associate to 36

37 an exchange economy an additional agent whose action set is the simplex of prices, the feasible actions of a trader is a consumption bundle in his budget set). The set-valued function P i is interpreted as the wishful actions of agent i as a function of the N-tuple of actions. Definition 10 An equilibrium of an abstract social system is an element a A such that P i (a) ϕ i (a) = Theorem 11 Let N; (A i ) i N ; (ϕ i ) i N ; (P i ) i N be an abstract social system such that: 1. i N, A i is a non-empty convex compact subset of a Euclidean space. 2. i N, ϕ i is a continuous set-valued function with nonempty convex values. 3. a A, a i / cop i (a), and 4. i N, the graph of P i, Γ(P i ) = {(a, b) A A i b P i (a)} is an open subset of A A i. Then the abstract social system has an equilibrium. We start with a lemma. 37

38 Lemma 5 Let K IR m. Then any point in cok is a convex combination of m + 1 points in K. proof. Assume that x cok. Then x is a convex combination of finitely many points in K, i.e., x can be written as x = k i=1 a i x i where x i K, a i > 0 and k i=1 a i = 1. Let k be the smallest positive integer for which x can be written as a convex combination of k points from K. Assume that k > n+1, and let x = k i=1 a i x i where x i K, a i > 0 and k i=1 a i = 1. It is sufficient to prove that any other representation of x of the form, x = k i=1 c i x i where x i K, c i 0 and k i=1 c i = 1, obeys c i > 0 for every 1 i k. Define the linear transformation T : IR k IR m+1 by k k T (b 1,..., b k ) = ( b i x i, b i ). i=1 i=1 The kernel of T, {b IR k T b = 0}, has dimension at least k (m + 1) 1. Therefore there is b IR k \ {0} with k i=1 b i x i = 0 and k i=1 b i = 0. Set a = (a 1,..., a k ) and k = {c IR k k i=1 c i = 1 and i c i 0}. Set J = {j b j < 0} and note that J is not empty. For every j J set t j = a j /b j. Then t j > 0, a j + t j b j = 0, and for every 0 < t < t j, a j + tb j > a j + t j b j = 0. Therefore, if t = min t j, a + tb k, at least one of the coordinates of a + tb vanishes, and k i=1 (a + tb) i x i = x contradicting 38

39 the minimality of k. (Alternatively, consider the half line starting in a in the direction of b and let t = sup{s 0 a + sb k }. Note that k is convex and bounded and therefore t <. As k is closed a + tb k and as a i > 0 for every i it follows that t > 0. If (a + tb) i > 0 for every 1 i k then there is t > t with a + t b k contradicting the definition of t. Thus ki=1 (a + tb) i x i is a representation of x as a convex combination of less than k points from K.) Lemma 6 Let X IR l and Y IR m be convex compact subsets, and assume that F is an upper semicontinuous set-valued function from X to Y. Then the set-valued function G from X to Y defined by G(x) = cof (x) is upper semicontinuous. Proof. Assume that x n x X, y n G(x n ) and y n y. We have to prove that y G(x). By the previous lemma, each y n is a convex combination of m + 1 elements of F (x n ). Therefore, for every n we can write y n = m+1 i=1 a i (n)z i (n) where a i (n) 0, m+1 i=1 a i (n) = 1 and z i (n) F (x n ). The sequence (a 1 (n),..., a m+1 (n), z 1 (n),..., z m+1 (n)), n = 1,..., is a bounded sequence in IR m+1+(m+1)m, and therefore has a subsequence that converges, i.e. there is a sequence n k k such that lim k a i (n k ) = a i and lim k z i (n k ) = z i. It follows that a i 0 and m+1 i=1 a i = 1, and 39

40 lim k i a i (n k )z i (n k ) = i a i z i. As F is upper semicontinuous z i F (x) and therefore i a i z i G(x). 9 We return now to the proof of the theorem. For every i N we define a continuous function g i : A A i IR + such that: x P i (a), g i (a, x) > 0 x / P i (a), g i (a, x) = 0 E.g. we define g i (a, x) = dist((a, x), A A i \ Γ(P i )). Note that the graph of P i is open and thus its complement in A A i is closed and therefore the distance of (a, x) to A A i \ Γ(P i ) is positive (> 0) iff x P i (a). For every i N define a set-valued function F i from A to A i by: F i (a) = {x ϕ i (a) g i (a, x) = max{g i (a, y) y ϕ i (a)}}. By lemma 2, F i is uppersemicontinuous. Next we define a set-valued function G i from A to A i by G i (a) = cof i (a). Again by a previous result G i is upper semicontinuous. Also for every a A, F i (a) is nonempty and thus also G i (a)is nonempty. Therefore letting G(a) = i N G i (a), G is an upper 40

41 semicontinuous set-valued function with non empty convex values from A to itself and therefore has (by Kakutani s fixed point theorem) a fixed point, i.e., there is a A with a G(a ). We will show next that this point a is an equilibrium of the abstract social system. Thus we have to prove that a i ϕ i (a ) and that ϕ i (a ) P i (a ) =. Indeed, a i G i (a ) = cof i (a ) co ϕ i (a ) = ϕ i (a ) and thus a i ϕ i (a ). If ϕ i (a ) P i (a ), there is x ϕ i (a ) P i (a ) and therefore g i (a, x) > 0 implying that F i (a ) P i (a ) and therefore G i (a ) cop i (a ) which together with a i G i (a ) imply that a i cop i (a ) which contradicts the assumption of the theorem. Definition 11 An exchange economy with externalities consists of: 1. A set of traders T, 2. A positive integer l, representing the number of commodities. 3. For every t T a subset X(t) IR l the consumption set of trader t. 4. For every t T, a element e(t) X(t) the initial endowment of trader t. 5. For every t T a set-valued function P t from t T X(t) to X(t) the preference of trader t given the vector of prices and the consumption of 41

42 each agent. Definition 12 A competitive equilibrium of an exchange economy with externalities, E = T, (X(t)) t T, (e(t)) t T, (P t ) t T, is an ordered pair (x, p ) where p and x : T IR l are such that: 1) t T, x (t) X(t), 2) t T x (t) e(t), 3) t T, p x (t) = p e(t), 4) t T P t (x, p) {y X(t) p y = p e(t)} =. Theorem 12 Let E = T, (X(t)) t T, (e(t)) t T, (P t ) t T be an exchange economy with externalities, such that: (0) T is finite, (1) X(t) = IR l +, (2) 0 < e(t) IR l +, with e i (t) > 0 1 i l, (3)the graph of the set-valued function P t : t T X(t) X(t) is open, (4) ((x(t)) t T, p) j T X(t), x(t) / cop t ((x(t)), p). Then there is p and x X = i T X(t) such that: (a) x(t) e(t), (b) t T, p x (t) = p e(t), 42

43 (c) t T, P t (x, p) {y X(t) p y = p e(t)} =. Proof. We construct an abstract social system whose equilibrium points correspond to those of the economy E. Assume that T = {1,..., n}, then N = {0, 1,..., n}, where we imagine player 0 as the price maker. A 0 =, and for 1 i n, A i = [0, M] l where M is a sufficiently large constant, e.g., M > max l ni=1 j=1 e j (i). For i = 0 we define ϕ 0 by ϕ 0 ( ) =, and for i = 1,..., n, ϕ i (x 1,..., x n, p) = {y A i py = pe(i)}. Define ˆP i, i = 0, 1,..., n, by ˆP 0 (x 1,..., x n, p) = {q q( x i e(i)) > p( x i e(i))}, and for i = 1,..., n, ˆP i (x 1,..., x n, p) = P i (x 1,..., x n, p) A i. The abstract social system S(M) := N; (A i ) i N ; (ϕ i ); ( ˆP i ) satisfies: a) N is finite, b) 0 i n, A i is a convex and compact subset of a Euclidean space, 43

44 c.1) ϕ 0 is a set-valued function with a constant nonempty compact convex value ( ), and thus is a continuous set-valued function with convex values, c.2) 1 i n, ϕ i is a continuous set-valued function with nonempty convex values (by lemma 4) d.1) as ˆP i (a) P i (a) and a i / cop i (a), we have a i / co ˆP i (a) d.2) as ˆP 0 (x 1,..., x n, p) is convex and p / ˆP 0 (x 1,..., x n, p) we deduce that p / co ˆP 0 (x 1,..., x n, p), e) 0 i n, Γ( ˆP i ) is open in A A i (in the relative topology). Therefore, the abstract social system N; (A i ) i N ; (ϕ i ); ( ˆP i ) satisfies all the conditions of the previous theorem and thus has a fixed point (x, p ). We show that this is an equilibrium of S(M) = T, ([0, M] l ), (e i ), (P i ) i T. As x (i) ϕ i (x, p ), p x i = p e(i). Therefore i p x i = i p e(i) and thus p ( i x i i e(i)) = 0. If there is j with ( i x i ) j > ( i e i ) j, setting δ j = (0,..., 1, 0..., 0), δ j ( i x i i e i ) > 0 = p ( i x i i e i ) and therefore δ j ϕ 0 (x, p ) P 0 (x, p ) =, a contradiction. Therefore, i x i i e(i). Obviously, x i ϕ i (x, p) implies that x i [0, M] l, and as ϕ i (x, p) ˆP i (X ) = we deduce that for every i T, P i (x, p ) {y [0, M] l p y = p e(i)} =. Thus (x, p ) is an equilibrium of E(M). 44

45 Let (x(m), p(m)) be competitive equilibria of the economy E(M). Then x(m) = (x 1 (M),..., x n (M)) [0, M] ln and p(m). As both sets, [0, M] ln and are compact, there is a sequence (M k ) k=1 such that M k as k, x(m k ) x IR ln as k, and p(m k ) p as k. As n i=1 e(i) n i=1 x i (M k ) k ni=1 x i, x is a suballocation. As px i = lim k p(m k )x i (M k ) = lim k p(m k )e(i) = pe(i), we deduce that px i = pe(i). In order to prove that indeed (x, p) is a competitive equilibrium of E we have to show further that there is no i T and y i IR l + such that py i = pe(i) and y i P i (x, p). Assume that y i IR l + with py i = pe(i) and y i P i (x, p). Let y i (k) be a sequence such that y i (k) y i as k and p(m k )y i (k) = p(m k )e(i). E.g., y i (k) = p(m k)e(i) p(m k y i. As P )y i i is continuous, i.e., Γ(P i ) is open, it will follow that there is a sufficiently large k, such that y i (k) P i (x(m k ), p(m k )) [0, M k ] l = ˆP i (x(m k ), p(m k )) contradicting the fact that (x(m k ), p(m k )) is an equilibrium of S(M k ). 45

46 10 Sard s Lemma A rectangular parallelepiped in IR n is a set of the form I = n i=1[a i, b i ]. Its volume, vol I, is defined as Π n i=1(b i a i ). A subset V of IR n is a set of measure 0 if for every ε > 0 there is a sequence of rectangular parallelopipeds, I k = n i=1[a k i, b k i ] which cover V, i.e., such that V k=1i k and the sum of their volumes is less then ε, i.e., k=1 Π n i=1(b k i a k i ) < ε. A countable union of sets of measure 0 is a set of measure 0. Indeed, if V j, j = 1,... is a sequence of subsets of IR n having measure 0, for every ε > 0 and every j 1, there is a sequence of rectangular parallelopiped I j,k such that V j k=1i j,k and k=1 voli j,k = k=1 Π n i=1(b j,k i a j,k i ) < ε2 j. Then j=1v j j=1 k=1 I j,k and k=1 j=1 voli j,k = k=1 j=1 Π n i=1(b j,k i a j,k i ) < ε, and therefore j=1v j is a set of measure 0. A function f : U IR n where U is an open subset of IR n is called C 1 if it has continuous partial derivatives. The Jacobian of f at x U is the determinant of the n n matrix J(x) where J i,j (x) = f i x j (x). A critical point (of f) is a point x U with det J(x) = 0. A critical value is a point of the form f(x) where x is a critical point. A regular value is a point y IR n which is not a critical value. 46

47 Theorem 13 (Sard s Lemma). Let f : U IR n be a C 1 function where U is an open subset of IR n. Then the set of critical values of f has measure 0. Proof. For every x IR n we denote max n i=1 x i by x. Assume first that [0, 1] n U, and we will prove that the set, {f(x) : x [0, 1] n with det J(x) = 0} has measure 0. For every x IR n and α > 0 we denote by I(x, α) the cube {y IR n : 1 i n, y i x i α}. Fix ε > 0. For every 1 i n, f i (y) f i (x) = j J ij (x + θ i (y x))(y j x j ) for some 0 < θ i < 1 and therefore, as J(x) is continuous in the compact set [0, 1] n it is uniformly continuous there, and thus there is δ i > 0 such that J ij (x + θ(y x)) J ij (x) < ε/n whenever x y < δ i and 0 < θ < 1 and therefore f i (y) f i (x) J i (x)(y x) < ε y x whenever x y < δ i. Therefore, there is δ > 0 (e.g. δ = min δ i ) such that for every x, y [0, 1] n with x y < δ, f(x) f(y) J(y)(x y) < ε x y. There is a sufficiently large constant K, such that for every x, y [0, 1] n, f(x) f(y) < K x y. Let k and l be two positive integers with k > δ 1 and K/ε < l 1 + K/ε. Note that for any two points x, y [0, 1] n 47

48 with x y 1/(kl), f(x) f(y) < ε/k. Define for every 1 i n, A i = {x [0, 1] n : kx i ZZ and 1 j n, klx j ZZ} and A = A i. Note that the number of points in A i is (k + 1)(kl + 1) n 1 2 n l n 1 k n and thus the number of points in A is n(k + 1)(kl + 1) n 1 n2 n l n 1 k n. Assume that x 0 [0, 1] n with det J(x 0 ) = 0. As det J(x 0 ) = 0, the kernel of the linear transformation J(x 0 ) : IR n IR n is nonempty, i.e., there is z IR n with z 0 and J(x 0 )z = 0. W.l.o.g. assume that x 0 j 1/k. Therefore for any point y on the line {x 0 + tz : t IR}, J(x 0 )(y x 0 ) = 0 and therefore, there is x 1 in the boundary of [0, 1/k] n with J(x 0 )(x 1 x 0 ) = 0, and therefore in particular f(x 1 ) f(x 0 ) ε/k, and there is x 2 A with x 2 x 1 1/(kl), and therefore in particular f(x 2 ) f(x 1 ) K/(kl). Therefore, f(x 0 ) f(x 2 ) f(x 0 ) f(x 1 ) + f(x 1 ) f(x 2 ) ε/k +K/(kl) < 2ε/k. Therefore, f(x 0 ) I(f(x 2 ), 2ε/k). Therefore the set of critical values (in f([0, 1] n )) is contained in y A I(y, 2ε/k) and y A voli(y, 2ε/k) A (4ε/k) n n2 n l n 1 k n (4ε/k) n n8 n l n 1 (ε) n < n8 n K n ε. Next assume that x U and δ > 0 are such that I(x, δ) = {y IR n : y x δ} U. The function g : U IR n defined by g i (z) = (z x)/2δ + 1/2 maps I(x, δ) onto [0, 1] n and has a nonvanishing Jaco- 48

49 bian. Therefore, {f(z) : z I(x, δ) with det J(z) = 0} = {f g 1 (y) : y [0, 1] n with det J(f g 1 )(y) = 0} has measure zero by the first part of our proof. Next we show that any open set U is the (countable) union of all (rational) cubes I(x, δ) with x IQ n and δ IQ which are subsets of U. Indeed, given x U there is α > 0 such that I(x, α) U. Let δ IQ with δ < α/2 and y IQ n with y x < δ. Then x I(y, δ) I(x, α) U which proves our claim. As the set of all rational cubes which are subsets of U, is a countable collection of sets which cover U, the set of critical values has measure Generic finiteness of Competitive Equilibria In the previous sections the models of an exchange economy included as a basic ingredient the preference relations of the individuals, and the resulting income as a function of prices (pe(t)) and the demand set-valued functions of each individual were derived. In the present section we consider a model in which demand functions are part of the description of the exchange economy. We consider here a model of an exchange economy which consists of: 49

50 1. A finite set of traders, T = {1,..., n}, 2. A finite number of commodities, l, 3. For every t T, an initial bundle e(t) IR l ++, 4. For every t T, a demand function δ t : IR ++ IR l ++ which satisfies pδ t (p, y) = y. The demand function δ t represents the demand of trader t as a function of the price vector p and his income y IR ++. The condition pδ t (p, y) = y asserts that trader t balances his budget. A Competitive equilibrium price vector, or an equilibrium price vector for short, of the economy E = (T ; l; (e(t)) t T ; (δ t ) t T ) is a price vector p such that δ t (p, pe(t)) = e(t). t T t T We will state and prove a result that asserts that given demand functions which satisfy suitable conditions (e.g. continuously differentiable), for almost all initial endowments, (e(t)) t T, the exchange economy has finitely many competitive equilibria. Theorem 14 Assume that the demand functions δ t are continuously differentiable and for every y 0 > 0, δ 1 (p, y) as p uniformly in y y 0, i.e., for every K > 0 there is ε > 0 such that for every y y 0 and ev- 50