Micro Barlo

7/29/2019 Micro Barlo

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Microeconomic Theory

ECON 501

Lecture Notes on Competitive Equilibrium in

Pure Exchange Economies1

Mehmet Barlo

Sabanc University

1These lecture notes are prepared from various sources (while honoring due credit) to be used in Eco-

nomics 501-502 course at the Sabanc University, and they are not intended for sale. Please contact the

author for permission to use these lecture notes elsewhere. Email [email protected] for additional

comments and questions.


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Contents

1 Pure Exchange Economies 1

2 Preferences 4

2.1 Completeness, Reflexivity and Transitivity . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 Continuity of Preferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.3 Debreus Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4 Desirability Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.5 Convexity of Preferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.6 Relation among the Assumptions on Preferences . . . . . . . . . . . . . . . . . . . . 15

3 Pareto Optimality and the Core 23

3.1 Pareto Optimality and Individual Rationality . . . . . . . . . . . . . . . . . . . . . 23

3.2 The Core . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3 Pareto optimality and the Planners Problem . . . . . . . . . . . . . . . . . . . . . . 27

4 The Price Mechanism 33

4.1 Budget Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.1.1 Properties of the Budget Correspondence . . . . . . . . . . . . . . . . . . . . 35

Continuity of the Budget Correspondence . . . . . . . . . . . . . . . . . . . 38

4.2 Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.2.1 Derivation of Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

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CONTENTS ii

Demand with Cobb-Douglas Preferences . . . . . . . . . . . . . . . . . . . . 55

Comparative Statics with Cobb-Douglas and Fixed Income . . . . . . . . . . 56

Further Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5 Competitive Equilibrium 60

5.1 Computing Competitive Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

5.1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.1.4 Example 4: Cobb-Douglas Utilities . . . . . . . . . . . . . . . . . . . . . . . 72

5.1.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.2 Existence of Competitive Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 76

5.2.1 The Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.3 Welfare Properties of Competitive Equilibria . . . . . . . . . . . . . . . . . . . . . . 94

5.3.1 First Fundamental Theorem of Welfare Economics . . . . . . . . . . . . . . . 94

5.3.2 Second Fundamental Theorem of Welfare Economics . . . . . . . . . . . . . 96

5.3.3 Competitive Equilibria and The Planners Problem . . . . . . . . . . . . . . 99

5.4 Competitive Equilibrium and The Core . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.4.1 Existence of The Core . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.5 Debreu Scarf Limit Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6 Appendix: Basic Properties of Topological and Metric Spaces 109

7 Appendix: Continuity of Correspondences 117

8 Appendix: Theorem of Maximum 123

9 Appendix: Fixed Point Theorems 125

9.1 Brouwers Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125


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CONTENTS iii

9.2 Kakutanis Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10 Appendix: Finite Normal Form Games and Nash Equilibrium 129

11 Appendix: Convexity and Separating Hyperplane Theorems 133


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1 Pure Exchange Economies

In this chapter we are going to concentrate on pure exchange economies. After describing the

environment, we will present the definition of general (competitive) equilibrium and analyze the

properties of its components. Finally, this chapter will establish its existence.

The notes presented here are based on Debreu (1959), and Hildenbrand and Kirman (1976).

Let the commodity space be given by RM, and the consumption space X by RM+ = {y RM :

yk 0, k = 1, . . . , M }.

This requirement restricts attention to situations where there are only M < goods each of

which are perfectly divisible. Hence, cases with infinitely many goods (thus, cases that involve non-

bounded time aspect) are ignored. Moreover, each good being assumed to be perfectly divisible,

eliminates some of quite interesting cases. Such a case could be tires, because half a tire is not good

for any kind of car, and may only be used in harbors.

We aim to describe agents behavior via their preferences over consumption bundles. A pref-

erence relation is a binary comparison, i.e. for each x and y a preference relation consists of a

comparison. (Note that this does not imply that each x and y must be comparable, indeed a

preference relation in which such a point occurs is allowed.)

The set of agents is given by N = {1, . . . , n}, and each of them have preferences (to be defined

below) given by i on X, where xi y for x, y X means that agent i thinks that consumption

bundle x is at least as good as consumption bundle y (weakly prefers x to y).

Definition 1 A preference relation, , is a binary relation defined on X, i.e. X X,

whose graph is given by Graph() = {(x, y) X X | x y}.

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Economics 501-2 by Mehmet Barlo: 1 Pure Exchange Economies 2

As an example consider X R+ and . Now we have Graph() = {(x, y) R2+ | x y}.

The following notation is needed for future analysis:

Definition 2 For x, y X let

1. x y if and only if x y and not yx, and we say x is strictly preferred to y;

2. x y if and only if x y and yx, and we say x is indifferent to y;

3. the upper-contour set of x is given by UCS(x) = {y X | yx}; and SUCS(x) = {y

X | y x} defines the strict upper-contour set of x.

4. the lower-contour set of x is given by LCS(x) = {y X | x y}; and SLCS(x) = {y

X | x y} defines the strict lower-contour set of x.

5. the indifference set of x is I(x) = {y X | x y}.

In words, the upper contour set of a consumption bundle x X RM+ given to player i is the

set of all consumption bundles each of which is weakly preferred to x. The strict upper contour set

is those in which the preference relation is strict. Similar considerations apply to the definitions of

the lower contour and strict lower contour sets.

Now we are ready to define a pure exchange economy:

Definition 3 (Pure Exchange Economy) A pure exchange economy E with consumption

space X and the set of agents N, is

E = {ei,i}iN ,

where ei X andi X X denote the agent is initial endowment and preference relation,

respectively.

That is, a pure exchange economy is defined with a finitely many and perfectly divisible goods,

finitely many agents each of whom possesses a preference relation and a given initial endowment

vector. That is summarized in the above definition.


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Economics 501-2 by Mehmet Barlo: 1 Pure Exchange Economies 3

In words, when we talk about pure exchange economies, we imagine an island in the Bahamas

which does not have any access to the outside world. Each agent possesses some goods (good or

bad depends on the tastes given by i) given by their initial endowments. If there were no trade,

each has to eat his own initial endowment by himself. But, with trade, he could exchange some

of his goods with others so that after the exchange everybody gets better off.

In formal economics, what we seek is an arrangement of allocations from which further trade

should not be beneficial to the agents. Because if it were, the point under question can hardly be

called a solution. Inherently, we believe in the rationality of agents, and therefore, we expect the

solution of such an exchange situation one with that particular feature, namely the requirement

that the solution must be one from which further trade should not beneficial to the agents.

These ideas, formalized in chapter 3, are captured by the notion of the core, which in essence

requires that the solution is one such that no coalitions of agents should be able to be strictly

better off by switching jointly to another feasible allocation. Consequently, whatever bargaining

is going on, whatever power distributions are in effect, we expect the solution to this economy be

an allocation in the core. Because if it is not, then that means there are some agents who are not

using an opportunity that they should have. And that is clearly against our belief that the solution

should not violate any rationality considerations.

Before going into the analysis of the core, we will first concentrate on assumptions on preferences

that will be needed in our analysis.


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2 Preferences

In order to employ various results from mathematics some assumptions (axioms) on preference

relations are needed. Some of those axioms might be stronger than desired, which would result that

the model we are going to analyze in this chapter not be suitable for situations for which some of

these axioms are violated.

2.1 Completeness, Reflexivity and Transitivity

Definition 4 is said to be reflexive if for all x X, xx.

Definition 5 is said to be transitive if for all x,y,x X,

x y and y z x z.

Definition 6 A binary relation on X X is a preorder if it is both reflexive and transitive.

Definition 7 is said to be complete if for all x, y X, either x y or yx, or both.

Note that sometimes the assumption that is a complete preorder is called rationality. I

personally dont agree with that definition, because after all I might not know my preferences

comparing Beatles St. Peppers Lonely Hearts Club Band and Help! (an example of incomplete

preferences), and that does not mean that I am indifferent between the two. It just means that I

do not prefer one over the other, and am not indifferent between them.

For a concrete example consider X = R2+ and =. It should be obvious to the reader that

this binary relation is transitive, reflexive but not complete. This is because of the following: For

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Economics 501-2 by Mehmet Barlo: 2 Preferences 5

any x, y X (recall that X is the same as RM+ ), x is greater or equal to x, thus, xx, hence these

preferences are reflexive. Moreover, for any x, y and z in X, x y and y z implies x z, leading

to the conclusion that for all x, y and z in X with x y and y z, we have x z. However, this

relation, , is not complete. To see this consider M = 2 and two points x = (1, 2) and y = (2, 1).

We cannot say that x y or x y.

Similar problems appear with transitivity as well. That is, it is not difficult to come up with

intuitive preference ordering that violate transitivity. Those examples generally involve preferences

over risky, or time dependent assets.

Analysis of preference relations when completeness (or transitivity) is dropped is a modern hot

research topic.

From now on unless otherwise stated we will keep the following assumption:

Assumption 1 Preference relation given by X X is a complete preorder. That is, is

reflexive, complete and transitive.

2.2 Continuity of Preferences

The next assumption will require that preferences should not feature any jumps.

In other words, continuity eliminates cases when an agents preferences may involve a drastic

change. In order to see that, consider the following example. Suppose that there are 3 goods, the

first being coffee, the second tea and the last sugar. Now assume that the preferences of the agent

under analysis is such that he prefers coffee to tea as long as none of it has any sugar. Yet, he

cannot take any sugar in his coffee no matter what happens, because then he hates it. (Of course

we also need to assume that this agent is like a tasting machine where he can detect very tiny

amounts of sugar in his coffee.) Therefore, for any natural number n our agent strictly prefers yn

given by (0, 1, 1/n) to xn = (1, 0, 1/n) (that is he strictly prefers getting one tea with 1/n sugar to

one coffee with the same amount of sugar). It should be obvious that xn converges to x = (1, 0, 0)


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and yn to y = (0, 0, 1). Finally, even though for any natural number n, we have yn xn, in the limit

this relation is reversed so that x y. Thus, these preferences are not continuous.

Definition 8 is continuous if for all x X, both UCS(x) and LCS(x) are closed.

The following Proposition will establish equivalent definitions for continuity based on closedness

and openness of sets and sequences, concepts that the reader is assumed to be familiar with from

textbooks such as Kolmogorov and Fomin (1970) and Rudin (1976):

Proposition 1 The following statements are equivalent:

1. is continuous;

2. for all x X, both SUCS(x) and SLCS(x) are open;

3. Graph() is closed;

4. for all x, y X with y x, there exists 1, 2 > 0 with

|x x

| < 1 and |y y

| < 2 y

x

;

5. for any sequence {xn, yn}nN X X with xn yn for all n N, and (xn, yn) (x, y), we

have that x y.

Proof. The proof will involve the following steps:

Step 1. is continuous if and only if for all x X, both SUCS(x) and SLCS(x) are open.

Proof. From basic mathematical analysis, it is well known that a set A X is open if and only

ifX\ A is closed. Noting that for any x X, X\SUCS(x) = LCS(x) and X\SLCS(x) = UCS(x),

finishes the proof.

Step 2. is continuous if and only if Graph() is closed.

Proof. Recall that Graph() = {(x, y) X X | x y}. Suppose that is continuous. Then

for any y X, {x X : x y} = UCS(y) is closed. Moreover, for any x X, {y X : x y} =

LCS(x) is closed. Thus, because that x and y were arbitrary, Graph() is closed.


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Now, suppose that Graph() is closed. Then for any y X, UCS(y) = {x : x y} is closed.

Moreover, for any x X, LCS(x) = {y : x y} is also closed. Thus, is continuous.

Step 3. is continuous if and only if for all x, y X with y x, there exists 1, 2 > 0 with

|x x| < 1 and |y y| < 2 y

x.

Proof. Suppose is continuous. Then for any x, y with y x, it must be that y SUCS(x),

and x SLCS(y). Because of the second step, we know that both SUCS(x) and SLCS(y) are open.

Hence, again from basic mathematical analysis we know that there exists 1 > 0 and 2 > 0, such

that B1(x) = {x X : |xx| < 1} is contained in SLCS(y), and B2(y) = {y

X : |yy| < 2}

is contained in SUCS(x). Thus, for all |x x| < 1 and |y y| < 2 we know that y x

and y x, because y B2(y) SUCS(x) and x B1(x) SLCS(y). Let 1, 2 > 0 be fixed.

Repeating the same argument for any y with fixed 2, and thus y x, we obtain the existence of

2,1 with 2 > 2,1 > 0 such that y x for any x B2,1(x); and, for any x

with fixed 1, and

thus y x, we obtain the existence of 1,2 with 1 > 1,2 > 0 such that y x for any y B1,2(y).

Letting 1 = 1,2, and 2 = 2,1 delivers the required result.1

Suppose that there exists x, y with y x, but there are no 1, 2 > 0 such that for all x and y,

|x x| < 1 and |y y| < 2, implies y x. Then, using the above technique, it can be shown

that these would contradict with the openness of SUCS(x) or SLCS(y).

Step 4. is continuous if and only if for any sequence {xn, yn}nN X X with xn yn for all

n N, and (xn, yn) (x, y), we have that x y.

Proof. Suppose for a contradiction that is continuous, but there exists a sequence {xn, yn}nN

X X with xn yn for all n N, and (xn, yn) (x, y), we have that y x. Therefore, by step 3

we know that there exists 1 > 0 and 2 > 0 such that x and y, |x x| < 1 and |y y| < 2,

implies y x. Let N be high enough so that for all n > N, xn B1(x) and yn B2(y). Thus, by

step 3 we know that yn xn for all n > N, a contradiction to xn yn for all n N.

1Generally, such a tedious check is not needed because this technique presented above is quite standard, and is

implied by the openness of sets.


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Conversely, assume that for any sequence {xn, yn}nN X X with xn yn for all n N, and

(xn, yn) (x, y), we have that x y. Take any x X, and we will show that LCS(x) is closed.

From basic mathematics, recall that a set is closed if and only if every convergent sequence in that

set, converges in that set. Take any sequence yn in LCS(x) converging to y. I need to show that

y LCS(x). Define a sequence by xn = x and yn with yn LCS(x), and limn yn = y. Because of

our hypothesis, we have that x y, y LCS(x), hence LCS(x) is closed. Closedness of the upper

contour sets can easily be shown using a similar argument.

Instead of continuity we could have worked with less restrictive notions:

Definition 9 is upper-semi continuous if for all x X, UCS(x) is closed (alternatively

SLCS(x) is open). And is lower-semi continuous if for all x X, LCS(x) is closed (alter-

natively SUCS(x) is open).

The reader should note that a preference ordering is continuous if and only if it is both upper-

semi continuous and lower-semi continuous.

2.3 Debreus Representation Theorem

Our next task is to present the phenomenal result known as Debreus representation theorem, which

establishes that under very weak assumptions, there exists a continuous real valued function that

represents an agents preferences.

Definition 10 Given preference ordering onXX we say that a real valued function u : X R

represents if and only if

x y u(x) u(y).

Proposition 2 If a real valued function u : X R represents some preference relation, and

f : R R is a strictly increasing function2, then v : X R defined by v = f u also represents

2We say that a function f : R R is a strictly increasing if x > y f(x) > f(y), x, y R.


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the same preferences. The assertion fails if f is a non-decreasing function3.

Exercise 1 Prove Proposition 2.

Now we are ready for the famous Representation Theorem by Debreu (1959).

Theorem 1 (Debreu (1959)) Suppose that preferences X X are continuous complete

preorders. Then there exists a continuous real valued function u : X R which represents.

Proof. The proof of this Theorem is beyond the scope of this course, and hence, is omitted.

We refer the reader to page 56 of Debreu (1959).

Under the light of the following definition, the next Proposition is easier to be obtained:

Definition 11 A function f : X R is upper-semi continuous if for all R, {x X |

f(x) } is a closed set in X. Similarly, it is lower-semi continuous if for all R,

{x X | f(x) } is a closed set in X.

Lemma 1 A function f : X R, where (X, ) is a metric space, is continuous4

if and only if it

is both upper-semi continuous and lower-semi continuous.

Exercise 2 Prove this lemma.

Proposition 3 Suppose that u : X R represents a given preference relation . Then the

following must hold:

1. is a complete preorder;

2. if u : X R representing is continuous so is.

3. if u : X R representing is upper-semi continuous so is.

4. if u : X R representing is lower-semi continuous so is.

3We say that a function f : R R is a non-decreasing function if x y f(x) f(y), x, y R.4A function f : X R is continuous if for all {xn}nN X with xn x, we have f(xn) f(x).


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Exercise 3 Prove the previous Proposition.

Exercise 4 Suppose that the preference ordering on R2+ is given by the following: For any

x, y R2

+, x y if either x1 > y1, or x1 = y1 and x2 > y2. This relation is called the lexicographicorder, and is basically the simplified version of the alphabetical order.

1. Is this relation reflexive, complete and transitive? Is this relation continuous? Prove your

answers.

2. For any given x X find I(x) with this order.

3. Prove that there is no a continuous utility function u(x1, x2) representing this preference re-

lation.

The utility function that Theorem 1 implies is ordinal. That is what matters is the ranking and

the particular value of the utility function does not have any particular meaning apart from it being

employed to rank alternatives.

The following Proposition elaborates more on that point:

Proposition 4 Suppose a functionu : X R represents preferences given by, and letf : R R

be a strictly increasing function (i.e. r > r for r, r R if and only if f(r) > f(r)). Then

v : X R defined by v = f u (i.e. v(x) = f(u(x)) for all x X) also represents.

Proof. We need to prove that v = f u represents . In order to do that take any x, y X

and without loss of generality suppose that x y. (Thus, we need to show that v(x) v(y).) We

already know that u(x) u(y) because u represents . Since f is strictly increasing, we must have

v(x) = f(u(x)) f(u(y)) = v(y), which delivers the required conclusion.

Therefore, the ranking given by and represented by u is preserved under a strictly increasing

transformation. This, in turn, means that the utility figure does not mean anything apart from being

a tool for ranking. In order to see that consider the following: Suppose M = 1 and u is an ordinal

utility function with u(100) = 1, u(50) = 1/2 and finally u(0) = 0. Now consider f defined by


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f(r) = r1000 for each r R+. Because of the above Proposition we now that v = fu also represents

the same preferences and v(100) = 1, v(50) =12

1000, and v(0) = 0. Note that the ranking is still

the same. But as it should be obvious, the particular value of u(x) does not mean much because

if it were the following would hold: under v the difference between v(100) v(50) < v(50) v(0)

implies going from 50 to 100 is more desirable than going from 0 to 50. But under u because that

u(100) u(50) = u(50) u(0) going from 50 to 100 is seen the same as going from 0 to 50. Thus,

u and v do not represent the same cardinal preferences.

2.4 Desirability Assumptions

Now we are going to consider assumptions onwhich will relate the physical amount of consumption

to preferences.

Definition 12 A preference relation X X is locally non-satiated if for allx X and all

> 0, there exists y X with y x < and y x.

The important thing to note regarding local non-satiation is that satiated preferences and thick

indifference curves are ruled out. Because with thick indifference curves, any point inside a thick

indifference curve would have a close by neighborhood so that for this agent each point in that

neighborhood is indifferent to the others.

Definition 13 is monotone if x y implies x y. is weakly monotone if for all x =

y X with x y, we have x

y. Moreover,

is strongly monotone if for all x, y X withx y and x = y, we have x y.

Strong monotonicity, a rather strong assumption, says that no matter what ones level of con-

sumption, a bundle with slightly more of anything is preferred to what one has. Whereas, weak

monotonicity says that in such cases the agent should not be worse off.

The following easier Representation Theorem can be found in standard microeconomics text-

books, and hence, is left as an exercise.


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Proposition 5 Suppose that preferences are continuous complete preorders and monotone. Then

there exists a continuous non-decreasing utility function representing them.

Exercise 5 Prove this Proposition. (Hint: Check Mas-Colell, Whinston, and Green (1995).)

2.5 Convexity of Preferences

These assumptions about convexity of preferences will entail the idea that individuals prefer bundles

in which commodities are fairly evenly distributed to those which are concentrated on a few goods.

That is why an agent with convex preferences should prefer orta sekerli kahve.

We will have three sets of assumptions for convexity, weak convexity, convexity and strong

convexity of preferences. The definitions and results are due to section 7 of chapter 4 of Debreu

(1959).

Definition 14 (Weak Convexity) If x2x1 then x2 + (1 )x1x1 for any (0, 1).

Note that this definition allows for thick indifference curves, hence a weakly convex preference

relation might be violating local non-satiation. In fact an agent with weakly convex (and with

convex) preferences might be indifferent between all the feasible consumption bundles. But these

preferences would not be strictly convex with our definitions.

Proposition 6 is weakly convex if and only if for every x X, UCS(x) is convex.

Proof. Suppose that is weakly convex, and I aim to prove that for all x X, UCS(x)

is convex. That is, given any x X, for any y, z in UCS(x), y + (1 )z UCS(x), i.e.

y + (1 )zx for all [0, 1]. Note that due to weak convexity of, for any y, z with (without

loss of generality) y z, we already have y + (1 )z z. Due to transitivity, we can conclude

that because z UCS(x), y + (1 )z zx, hence, y + (1 )z UCS(x).

Conversely, suppose that for all x X UCS(x) is convex, and I will prove that is weakly

convex. Pick any x, y with yx. Thus, y UCS(x). Also note that x is also trivially in UCS(x).


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Hence, because that UCS(x) is convex, for any [0, 1], y + (1 )x UCS(x), thus, y + (1

)xx. Hence, is weakly convex.

Definition 15 (Convexity) If x2 x1 then x2 + (1 )x1 x1 for any (0, 1).

In other words, is convex if a possible consumption bundle x2 is strictly preferred to another

x1, then their weighted average with arbitrary positive weights in (0 , 1) is strictly preferred to x1.

Finally the last notion of convexity is the following:

Definition 16 (Strong-Convexity) If x2 x1 with x1 = x2, then x2 + (1 )x1 x1 for any

(0, 1).

In other words, is strongly convex if two possible consumption bundles x1, x2 are indifferent,

then their weighted average with arbitrary positive weights in (0, 1) is strictly preferred to x1.

The following gives us a related notion in terms of functions.

Definition 17 A function f : X R is said to be:

1. concave if for all [0, 1] and every x, y X, f( x + (1 )y) f(x) + (1 )f(y).

2. strictly concave if for all (0, 1) and every x, y X, f( x + (1 )y) > f(x) + (1

)f(y)

3. quasi-concave if for all [0, 1] and every x, y X, f( x + (1 )y) min{f(x), f(y)}.

4. strictly quasi-concave if for all (0, 1) and every x, y X, f( x + (1 )y) >

min{f(x), f(y)}.

Proposition 7 Suppose that is a continuous complete preorder and is represented by a continuous

utility function u. Then:

1. If u is concave, then is weakly convex;

2. If u is strictly concave, then is strictly convex;


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3. Every concave function is also quasi-concave, but the reverse is not true;

4. Every strictly concave function is also strictly quasi-concave, but the reverse is not true.

5. is weakly convex if and only if u is quasi-concave;

6. is strictly convex if and only if u is strictly quasi-concave;

Proof. The proof will have the following steps:

Step 1. If u is concave, then is weakly convex.

Proof. Let x, y X with x y. Need to show that x + (1 )y y for all [0, 1]. By

hypothesis (i.e. concavity of u), for any [0, 1] we have u( x + (1 )y) u(x) + ( 1 )u(y),

and because x y, u(x) u(y). Thus, u(x) + ( 1 )u(y) u(y) + ( 1 )u(y) = u(y). Thence,

u( x + (1 )y) u(y) implying that (due to representation) x + (1 )y y for all [0, 1].

Step 2. If u is strictly concave, then is strictly convex.

Proof. Let x, y Xbe such that x y. I need to show that x+(1)y y x. By hypothesis,

u(x) = u(y), and because of strict concavity ofu, for any (0, 1) it must be that u( x+(1)y) >

u(x) + (1 )u(y) = u(x) = u(y). Thus, due to representation, x + (1 )y yx.

Step 3. Every concave function is also quasi-concave, but the reverse is not true.

Proof. Suppose u is concave. Thus, for any [0, 1], and for any x, y X, u( x + (1 )y)

u(x) + (1 )u(y) min{u(x), u(y)} + (1 )min{u(x), u(y)} = min{u(x), u(y)}. Hence, u is

quasi-concave.

For a counter example consider X = R2

+, and u(x1, x2) = x10

1 x10

2 . This function is clearly not

concave, yet is quasi-concave.

Step 4. Every strictly concave function is also strictly quasi-concave, but the reverse is not true.

Proof. The proof is left to the reader, because it involves the same argument as in the proof of

step 3. Moreover, the example given there also works for this situation.

Step 5. is weakly convex if and only if u is quasi-concave.


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Proof. Assume that is weakly convex, and without loss of generality pick any x, y X with

x y, thus, min{u(x), u(y)} = u(y). I need to prove that for all [0, 1], u( x + (1 )y)

min{u(x), u(y)} = u(y). Because that is weakly convex, for all [0, 1], x + (1 )y y, thus,

by representation, u( x + (1 )y) u(y) = min{u(x), u(y)}, implying that u is quasi concave.

For the reverse direction, assume that u is quasi concave, and consider any x, y with x y.

I need to show that for all [0, 1], x + (1 )y y. Because that u is quasi concave, and

x y implies min{u(x), u(y)} = u(y), we have u( x + (1 )y) min{u(x), u(y)} = u(y), and by

representation this means that x + (1 )y y.

Step 6. is strictly convex if and only if u is strictly quasi-concave.

Proof. The proof is left to the reader, because it involves the same argument used in the proof

of step 5.

2.6 Relation among the Assumptions on Preferences

First of all it should be mentioned that a preference relation generally is thought of being complete,

reflexive and transitive. While preferences not satisfying one or more of these assumptions are

definitely quite interesting, it needs to be noted that dealing with them becomes technically very

difficult. That is why in this section (and in the course) you will not be able to go deeper into that

subject. For the curious reader should want to read the introductions of Ok (2002) and Dubra,

Maccheroni, and Ok (2004).

Second, the assumption of continuity is a very convenient one when it comes to maximization

issues. Because after all continuity is one of the most important ingredients to make sure that an

optimum is obtained. Indeed, we will employ the following Theorem on more than a few instances:

every continuous function on a compact set achieves its maximum. Dealing with continuity requires

tools that are covered in a principles of mathematical analysis course. There are lots of great books

on that subject, and the following two are widely regarded as being two of the best introductory

textbooks into that subject: Rudin (1976), Kolmogorov and Fomin (1970).


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In this course we can handle the relation between the desirability assumptions that we have

seen so far. Later when we go into convexity and strict convexity, we also will be dealing with their

relation to the existing desirability assumptions.

The student should understand the requirement of these assumptions very well in order to deal

with them in a proper manner.

The assumption of local non-satiation says that every consumption bundle has to possess a

strictly better rival in any of the neighborhoods one can consider. In order to understand an

assumption I always find it helpful to consider the case when that assumption is not satisfied.

In this particular case, if a preference relation is not locally non-satiated, then there must be a

consumption bundle which does not have a strictly better rival in any neighborhoods you can

consider. Therefore, if the preferences are given so that they involve a thick indifference curve, then

any consumption bundle residing in the thick indifference curve (that is that point must be strictly

inside, and not on the boundary) would not have a strictly better rival when the neighborhoods

one considers is sufficiently close to that point.

Similarly, the assumption of monotonicity requires that when an agent gets strictly more of

all the goods, he should get strictly better off. Note that monotonicity does not say about what

happens between two consumption bundles (that is, it does not make any claims about which

one should be chosen) when the agent does not get more from all the goods. Again in order to

understand this assumption consider a preference relation what is not monotone. Then, it must be

the case that there exists at least two consumption bundles one providing strictly more in all the

coordinates than the other; and the agent does not strictly prefer the one that gives strictly more

consumption.

Finally, the assumption of strict monotonicity implies that each good is valuable (which generally

is referred to as each good being good!). In particular it says that when the agent is confronted with

a choice between a consumption bundle that provides at least as much as another but more in

one coordinate, the agent has to choose the first one. Thus, if a preference relation is not strictly


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monotone, then there exists two different consumption bundles first one greater equal to the second,

and the agent does not strictly prefer the first one over the second.

As a concrete example to see these consider the following preference relation:

Example 1 LetM = 2 and the preferences are given as follows: When x1, x2, y1, y2 > 3

x y x1x2 > y1y2.

On the other hand, if the assumption of x1, x2 > 3, and not y1, y2 > 3, then no matter what the

values are x y. Finally, when x1, x2, y1, y2 3, then x y.

In this example I wish to over the assumptions we have seen so far. First of all, for any x, the

definition of the preferences implies that x x, thus, this relation is reflexive.

How about completeness? Do we have that for all x, y in R2+, either x y or yx or both. Let

us start by saying yes. Thus, take any x and y. Thus, without loss of generality either one of the

following 3 cases can happen: (1) x1, x2, y1, y2 > 3; or (2) x1, x2 > 3, and not y1, y2 > 3 (Actually,

symmetrically we also have the case where y1, y2 > 3, and not x1, x2 > 3, but changing x and y

handles this case with the second one we have already written. That is why I wrote without loss ofgenerality in the previous sentence.); or (3) x1, x2, y1, y2 3. If case 1 were to happen, then since

x1 x2 and y1 y2 are both real numbers, it is either x1 x2 y1 y2, or x1 x2 y1 y2, or both.

Thus, for all x, y satisfying x1, x2, y1, y2 > 3 these preference relation is complete. Next consider the

case when x, y are such that x1, x2 > 3, and not y1, y2 > 3. In this case we already know that x y,

thus, x y and not yx. Hence, when x, y satisfy x1, x2 > 3, and not y1, y2 > 3, the preference

relation is complete. Finally, for the case when x, y are such that x1, x2, y1, y2 3, we already know

that x y. Thus, x y and yx. Consequently, this preference relation is complete in that case

as well.

When we check for transitivity, we need to consider any x,y,zin R2+ such that x y and y z.

From this we need to conclude that no matter what the levels of x,y,zare that x z. Thus, take

any x,y,z with x y and y z. The rest of the argument should follow as demonstrated above

checking cases.


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This preference relation is continuous. In order to see this take any x and y, and any two

sequences {xn}nN, and {yn}nN with xn x and yn y and xn yn for all n N. In order

to prove continuity we need to show that x y. Without loss of generality, again there are three

cases as in the previous paragraphs: (1) x1, x2, y1, y2 > 3; or (2) x1, x2 > 3, and not y1, y2 > 3;

or (3) x1, x2, y1, y2 3. In the first case, for n high enough since xn and yn converges to x and y

respectively, xn and yn are both such that xn1 , x

n2 , y

n1 , y

n2 > 3. Thus, because that x

n yn for n high

enough we know that xn1 xn2 y

n1 y

n2 . Thus in the limit x1 x2 y1 y2, enabling the conclusion

that x y. In the second case x1, x2 > 3, and not y1, y2 > 3, again for n high enough xn1 , x

n2 > 3,

and not yn1 , yn2 > 3. By the definition of the preferences, then, it is easy to see that x y. Finally

for the third case when x1, x2, y1, y2 3, we already know that x y (because x y means x y

and yx). Thus, in all these cases we have x y, enabling the conclusion that these preferences

are continuous.

These preferences are not locally non-satiated: In order to see that consider x = (1, 1). Now

any point y in its close by neighborhoods (with a radius less than or equal to 2 actually) will have

y1, y2 3. Thus, by the definition of these preferences, there is no y in a close by neighborhood of

x = (1, 1) such that y x. Hence, these preferences are not locally non-satiated.

Because of the following two propositions these preference are not monotone and strictly mono-

tone.

Proposition 8 Suppose that a preference relation defined onRM+ is monotone. Then it is locally

non-satiated.

Proof. Suppose is monotone. Take any x, and consider any close by neighborhoods of

it. It is easy to see that any close by neighborhoods of x will contain y with ym > xm for all

m = 1, . . . , M . Because of monotonicity we know that y x. This, in turn, establishes that is

locally non-satiated, because any close by neighborhood of x contains a y with y x.

Proposition 9 Suppose that a preference relation defined onRM+ is strictly monotone. Then it

is both locally non-satiated and monotone. But the reverse conclusions do not hold.


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Proof. Suppose is strictly monotone. Thus, for any x and y with x y, it is true that

x y and x = y. Because of strict monotonicity we know that x y. Thus, these preferences are

monotone, because for all x and y with x y, we have x y. Now the conclusion that any strictly

monotone preference relation has to be locally non-satiated follows from the above and previous

Proposition.

The example to show that monotonicity does not imply strict monotonicity is the Leontieff

preferences that we have done in the lectures.

For the rest of the section we assume that preferences are complete preorders. But not necessarily

continuous. We will consider more technicalities and some difficult situations.

Let us start with continuity and monotonicity.

Obviously, weak monotonicity (monotonicity) does not imply monotonicity (strong monotonic-

ity). In order to see the first relation consider a preference ordering under which all x, y X

are x y. This relation is weakly monotonic, but is neither locally non-satiated nor monotonic

(strongly monotonic). In order to see the second relation consider Leontieff preferences, which are

monotone, but not strongly monotone. As noted above, it is clear that strong monotonicity implies

monotonicity: In order to see this consider x, y X with x y. Since the preference ordering is

strongly monotone, and x y implies x y and x = y, we have x y, thus is monotone. By

the same logic strong monotonicity implies weak monotonicity, which is left to verify by the reader.

Whether or not monotonicity implies weak monotonicity is a nontrivial question which is not that

interesting. In fact, when a preference relation does not satisfy continuity one can find examples of

preferences where they are not weakly monotone, but monotone. The reader is asked to be aware of

this complication, which could make a nice bonus question. However, if preferences are continuous,

monotonicity implies weak monotonicity. To see this, consider x = y, x y. Since preferences are

monotone for each n N, xn defined by xn = (1 +1n

)x x y, is strictly preferred to y. Since

the preference ordering is continuous, x y thus, is weakly monotone.

For convexity assumptions same kind of complications due to continuity might arise: Although


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at first reading it looks like convexity implies weak convexity, that is not true in general. However,

that relation is restored if is continuous.

The following example will display that convexity does not imply weak convexity.

Example 2 Consider a preference relation R2+ R2+ defined by: x y if and only if either

x1 > y1, or x1 = y1 and x2 = y2 = k, k > 0 (k is fixed, say k = 4); x y if and only if either

x1 = y1 and x2, y2 = k, or x1 = y1 and x2 = y2 = k.

We should note that this relation is not continuous. Moreover, take x, y such that x1 = y1,

and x2 > k > y2, thus, x y. There is a (0, 1) such that x2 + (1 )y2 = k. Hence,

x, y x + (1 )y, thus, this is not weakly convex (and not strictly convex). On the other

hand, it is an easy exercise to show that this is convex. Indeed, for any x, y, x y can be satisfied

only when x1 > y1, or x1 = y1 but y2 = k. For any (0, 1), in the first case x + (1 )y

will be such that x1 + (1 )y1 > y1, thus, by the definition of these preferences we must have

x + (1 )y y. In the second case, for any (0, 1), we must have x1 + (1 )y1 = x1 = y1,

and x2 + (1 )y2 = k. Thus, x + (1 )y y.

Proposition 10 Suppose that is continuous and convex. Then it is weakly convex.

Proof. Suppose not. Let x1, x2 X be such that x2 x1. It must be shown that A = {x

[x1, x2] | x1 x} is empty. Note that A cannot contain a single point, since its complement in

[x1, x2] is the set {x [x1, x2] : xx1} closed by the assumption of continuity. Thus, if A were

not empty it would own at least two different points x, x. However, x1 x implies by convexity

x x, and x2 x implies again by convexity x x. A contradiction would thus obtain.

As stated before weak convexity allows for preferences that violate local non-satiation (even

when continuity is satisfied). In that sense convexity is stronger, that is convex and continuous

preferences must obey local non-satiation.

Proposition 11 Suppose that is convex and continuous. Then it is locally non-satiated.


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Proof. Without loss of generality assume that there is not a global satiation point, i.e. for

every x X there is y X such that y x. Because, otherwise, trivially local non-satiation cannot

be obtained. Take any x X. We need to show that there exits a x in the neighborhood of x

such that x x. Now pick any y X with y x. Thus as shown above all the points in the line

segment (x, y) are strictly preferred to x. Thus, pick x such that it is (1) close to x; and (2) is in

(x, y).

Similar complications do arise in the interaction between convexity and strong convexity, that

is strong convexity does not necessarily imply convexity. That holds only whenever the preferences

are continuous.

Proposition 12 Suppose that is continuous and strongly convex. Then it is convex.

Proof. We refer the reader to page 61 of Debreu (1959).

The following is an explicitly defined example of which are strongly convex but not convex:

Example 3 Consider a preference relation R2+ R2+ defined by: x y if and only if either

x1 > y1; or x1 = y1 and x2, y2 2, and |x2 1| < |y2 1|; or x1 = y1 and x2, y2 > 2 and

x2 > y2. That is x y if and only if either x1 = y1 and x2 = y2, or x1 = y1 and x2, y2 are such that

|x2 1| = |y2 1|.

We should note that this relation is not continuous.

Therefore, in the first case the hypothesis of strict convexity is not satisfied at all (that is there

are no x, y with x = y and ...). In the second case, the mixture with (0, 1) when x1 = y1 would

have |( x2 + (1 )y2) 1| < |x2 1| = |y2 1|. Thus this relation is strictly convex.

Moreover, take x, y such that x1 = y1, and x2 = 3, y2 = 1, thus, x y (therefore x y). Let

= 1/2. Note that z = 1/2x + 1/2y involves z1 = x1 = y1 and z2 = 2. By the definition of these

preferences it should be clear that y z because z1 = y1 and |z2 1| = 1 > 0 = |y2 1|. This shows

that these preferences are not convex.

It is worthwhile to see that strong convexity, unlike convexity, does not allow for linear segments

in indifference curves. Moreover, it is intuitively hard to justify strong convexity, but not convexity.


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Another important relation one has to keep in mind is obtaining strict monotonicity with the

use of monotonicity, strict convexity of preferences which need to be continuous complete preorders.

Proposition 13 Suppose that is continuous complete preorders satisfying monotonicity and

strict convexity. Then is strongly monotone.

Proof. Take any x, y X with x y and x = y. I need to show that under the hypothesis of

the Proposition, x y in order to obtain strong monotonicity.

Recall that for any x, y X, x y means that xk > yk for all k = 1, . . . , M .

Clearly, y x is not possible. Because that would violate monotonicity, since due to continuity

we would have the existence of > 0 but small, and (1 )y x. Then, x y, and y x, a

contradiction to monotonicity.

Thus, if x y is not true, the only situation to be worried about is that x y. But then, for

any (0, 1), x + (1 )y x due to strict convexity of the preferences. Consequently, because

that x y and x = y, for any 1 > 0, x (1 )( x + (1 )y). Thus, we reach the

required contradiction to monotonicity, because (by continuity) there exists > 0, strictly positive

but small, such that x (1 )( x + (1 )y) and (1 )( x + (1 )y) x.


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3 Pareto Optimality and the Core

First we will introduce notions of welfare for pure exchange economies.

In this chapter preferences are assumed to be complete preorders.

For notational purposes for any given pure exchange economy E = i, eiiN, let the set of

feasible allocations be given by F(E), i.e.

F(E) =

x Xn :

iN

(xi ei) 0

.

3.1 Pareto Optimality and Individual Rationality

The first notion is individual rationality. In general it says that an allocation is individually rational

if it does not make none of the agents strictly worse off than the initial situation. Otherwise, as

economists we should expect agents to oppose to such kind of allocations.

Definition 18 Given a pure exchange economy E = i, eiiN, an allocation x Xn is individ-

ually rational if for all players i = 1, . . . , n we have

xii ei. (3.1)

We denote the set of individually rational allocations of a pure exchange economy E by IR(E).

The reader is encouraged to visualize in the Edgeworth box the representing a pure exchange

economy, the set of feasible and individually rational points.

The second notion of welfare that we are going to analyze is due to an Italian economist, Pareto.

We will present 2 versions of this notion. The regular one, in words, will say that none of the agents

23


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Economics 501-2 by Mehmet Barlo: 3 Pareto Optimality and the Core 24

can be made strictly better off without harming some other agent(s). The weak one, on the other

hand, will tell us that the test will involve trying to determine whether or not all the agents can

be made strictly better off. Note that as economists, we would not think that non-Pareto optimal

allocations as nice because by its definition, this allocation is not likely to be observed, because

it is in the interests of all the agents to change it.

Definition 19 Given a pure exchange economy E = i, eiiN, a feasible allocation x Xn is

1. Pareto optimal if there is no allocation y F(E) such that for all i N

yii xi,

and for some j N

yj j xj,

2. weakly Pareto optimal if there is no other allocation y F(E) such that for all i N

yi xi.

We denote the set of Pareto optimal and weakly Pareto optimal allocations of a pure exchange

economy E byPO(E) andWPO(E), respectively.

It is clear that if an allocation x Xn is Pareto optimal, then it is weakly Pareto optimal. But

the reverse is not true. This is to be verified by the reader, as required in the following exercise:

Exercise 6 Show that for all pure exchange economies E, where preference are complete preorders,

PO(E) WPO(E). Prove by giving a concisely specified example that the converse does not hold

even if preferences are assumed to be monotone, continuous and weakly convex.

The following Proposition will be quite useful in the future, and it gives us a sufficient condition

when for a given E, we have PO(E) = WPO(E).


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Proposition 14 When preferences are strongly monotone and continuous complete preorders then

PO(E) = WPO(E).

Proof. I will prove that under these hypothesis, for any exchange economy E, WPO(E)

PO(E), because then this would imply PO(E) = WPO(E) (we already know from the above that

PO(E) WPO(E)). That is, I will prove that any weakly Pareto optimal allocation is also Pareto

optimal. In particular, I will show that if an allocation is not Pareto optimal, then it cannot be

weakly Pareto optimal (that is, the proof will be done by counter positive).

Consider an allocation x XN and assume that it is not Pareto optimal. Thus, there exists

y F(E) such that yii xi for all i N and yj j xj for at least 1 j N.

Let S N be given by {j N : yj j xj}, i.e. the set of agents who are strictly better off under

y. Note that this set is non-empty, because x is not Pareto optimal.

Consider the following allocation y XN: yj = (1)yj for all j S; and yi = yi+

#N\S

jSyj

for all i N \ S. Due to continuity, there exists (0, 1] but small, so that yj j xj for all j S.

Moreover, for any i N \ S, due to strict monotonicity and transitivity, yi yii xi. Thus, y is

such that for every player i N, yi i xi.

The only step left to finish to show that x is not weakly Pareto optimal is to prove that y F(E),

i.e. y is feasible. This follows because:

iN

yi =jS

yj +iN\S

yi = (1 )jS

yj +

iN\S

yi + #N \ S

#N \ S

jS

yj =iN

yi iN

ei,

where the last inequality follows from y F(E).

Exercise 7 (Bonus) Give an explicitly constructed example of a pure exchange economy with at

least 2 goods with strictly positive total endowments and preferences given by complete preorders,

for which:

1. there are no Pareto optimal al locations;

2. there is only one Pareto optimal allocation.


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3.2 The Core

In a pure exchange economy Pareto optimality answers what kind of allocations are plausible when

the group of all of the agents considers possible deviations. On the other hand, individual rationality

considers only single agent deviations, that is such allocations cannot be improved upon by a

deviation done by a single agent.

The consideration of group deviations (not necessarily the whole group) leads us to the notion of

the core. In words, a feasible allocation will be in the core of an economy, if there are no coalitions

with a feasible group deviation opportunity.

Definition 20 The core of an pure exchange economy E = i, eiiN, denoted byC(E), is the set

of feasible allocations x Xn such that there is no coalition S 2N \ {} and yS X|S| such that

1. yS is resource feasible for S, i.e.

iS(ySi ei) 0; and

2. ySi i xi for all i S.

For notational convenience let S 2N \ {}. It is simply the set of all coalitions that can be

formed out of the society given by N. Again for notational convenience for any S S let

F(E | S) =

yS X|S| :

jS

(ySj ej) 0

.

Clearly, F(E | N) = F(E).

Whenever an allocation x Xn is not in the core, there must be a coalition S S, that we refer

to as the blocking coalition and an allocation yS X|S| which is feasible for S, and renders strictly

higher utility for all the members in the blocking coalition. In other words, for any x / C(E) with

S S blocking it via the use of allocation yS X|S|, we say that x can be improved upon by

coalition S S via yS X|S|.

Therefore, for any x C(E), the coalitions S = N, and S = {i} for i N, are not blocking.

Consequently, any allocation in the core is both weakly Pareto optimal and individually rational.


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However, the reverse direction does not hold, that is there might be individually rational and weakly

Pareto optimal allocations that are not in the core. These are summarized in the following Theorem:

Theorem 2 For all exchange economies E, C(E) IR(E) WPO(E). Moreover, the containment

might be strictly whenever n > 2. In the case of n = 2, if agents preferences are given by continuous

complete preorders satisfying strong monotonicity, thenC(E) = IR(E) PO(E).

Exercise 8 Prove this Theorem (also by identifying a concrete, well-defined example).

Whether or not the core of an exchange economy is empty is an important question that we

need to answer. Answering this question directly, on the other hand, would make us go deeper into

cooperative game theory. Therefore, we will deal with that question and show the existence of the

core later in the course.

The following exercise is for the motivated student:

Exercise 9 (Bonus) Give explicitly specified examples of 2 agent 2 good pure exchange economies

with strictly positive total endowments and preferences given by complete preorders, satisfying the

following properties:

1. the core of the economy is empty;

2. even though there exists a Pareto optimal allocation and a core allocation, there are no al lo-

cations that are both Pareto optimal and in the core.

3.3 Pareto optimality and the Planners Problem

In this section, we will be describing an alternative way of finding the set of Pareto optimal allo-

cations. Indeed, the set of Pareto optimal allocations, under some assumptions, will be given as

maximizers of a certain program, which is call the Social Planners Problem.


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I have to stress that this formulation is used extensively in modern macroeconomics, and provides

not only important insight, but also a very useful tool for us to identify the set of Pareto optimal

allocations.

The result that we wish to obtain is as follows: 1

Consider a pure exchange economy E. Then, an allocation x Xn is Pareto optimal if and

only if there exists (N) (i.e. i [0, 1] for all i N and

iN i = 1) and x

solves the

social planners problem for E at , where it is is given by

maxxF(E)iN

i ui(xi). (SP(E | ))

Of course, this result does not hold in general. The assumptions that we need for this result to hold

(the proofs are given later in this section) is as follows: Every agents preferences are represented

by a continuous, strictly increasing, and concave utility function ui : X R.

The proof of this result involves the use of an important Theorem (Minkowskis Separating

Hyperplane Theorem, Theorem 3 given below, and it is also used in the proof of the Second Fun-

damental Theorem of Welfare Economics, which we will see later in the course).

Theorem 3 Let C be a convex set ofR, < , and y /

C, where

C denotes the interior of C.

Then there exists R \ {0} such that for all x C

y x.

Notice that, the requirement for this Proposition is not the convexity of preferences, but the con-

cavity of the utility function representing these preferences. This is rather not a usual requirement,

and is due to the techniques that must be used in the proof of this result.

1Note that for any finite set E = {a1, a2, . . . , a|E|}, (E) denotes the simplex formed on E, i.e. the set of all

probability distributions on E. Formally,

(E) =

p R|E| : pk [0, 1], and

|E|k=1

pk = 1

.

pk, k = 1, . . . , |E|, denotes the weight (or probability) that p assigns to the kth component of E.


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Given weights = (i)iN (N), where (N) denotes the set of probability distributions on

the finite set of players N, i [0, 1] is often referred to as the bargaining weight of player i in the

Social Planners Problem.

Let us start with the formalities: For the rest of this section, for convenience assume that agents

preferences are continuous complete preorders, satisfying strict monotonicity.

For what follows, the following set will be quite useful:

UE = {u Rn : ui [ui(0), ui(xi)], i N for some x PO(E)}. (3.2)

Note that UE is well defined because ui(0) exists. (ln utility function for example does not satisfy

this assumption.)

Under these assumptions (strict monotonicity in particular), PO(E) is non-empty, thus, so is

UE for all E. To see that why PO(E) is non-empty, consider the following: For any given exchange

economy E, we can define the following allocation x by xi = 0 for all i = 1 and x1 =

jN ej. Such

an allocation is clearly Pareto optimal (due to strict monotonicity), since all the goods belong to

agent 1.

In words, for a given economy E, UE contains the set of all utility vectors that can be obtained

from all distributions of total endowments. Thus, it can be called utility possibility set. That is,

due to continuity of the utility functions u(F(E)) = UE, where

u(F(E)) = {u Rn : there exists x F(E) with ui = ui(xi) for all i N}.

Let us have a simple example to understand what that set is: Say there are 2 agents and 2 goods,

and both agents have a utility function given by ui(xi,1, xi,2) = 2 (xi,1 + xi,2). Say the amount of

endowments is e1 = (1, 0) and e2 = (0, 1). Here, any allocation that does not waste any good, i.e.

any x R22+ with x1 + x2 = (1, 1), is Pareto optimal (note that there is no difference between weak

Pareto optimality and the regular one, because these preference are strictly monotone). Hence, UE

is simply given by {(u1, u2) : u1 + u2 4}. The reader is asked to either visualize or draw this

particular UE.


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Now the following Lemma will establish that UE is a convex set whenever agents preferences

are weakly convex.

Lemma 2 Suppose that agents preferences are represented by a continuous, strictly increasing2

and concave utility functions. Then for every exchange economy E, UE is a convex subset ofRn.

Proof. Pick u and u both in UE. I will show that for any [0, 1], u + (1 )u UE.

Let x and x both in PO(E) be the associated Pareto optimal allocations (x the Pareto optimal

allocation require for u, and x the one for u), and fix [0, 1]. Since for all i N

ui + (1 )ui ui(xi) + (1 )ui(x

i) ui(xi + (1 )x

i),

for the proof to finish, it suffices to show that there exits x PO(E) such that ui(xi) ui(xi +(1

)xi) for all i N. If the allocation x + (1 )x is Pareto optimal, we are done. If it is not (recall

that due to strict monotonicity, weak Pareto optimality and Pareto optimality coincide), define x

as follows: x solves maxxF(E) u1(x1) subject to ui(xi) = ui(xi + (1 )xi) for all i = 1. Note that,

the constraint set is non-empty because x + (1 )x is an element. Because that the constraint

set is compact and the the utilities are continuous, there exists a solution x. Note that player 1 is

obtaining at least as much utility as under x+(1)x (but maybe more), while the others are not

worse off. Hence, by strict monotonicity x is Pareto optimal. Thus, since ui + (1 )ui ui(xi)

for all i N and x PO(E), ui + (1 )ui UE.

We are ready for the main result of this section:

Proposition 15 Given a pure exchange economy E where every agents preferences are strictly

monotone, convex, continuous and complete preorders represented by a continuous and concave

utility function ui : X R; an allocation x Xn is Pareto optimal if and only if there exists

(N) and x solves the social planners problem for E at , where it is is given by

maxxF(E)

iN

i ui(xi). (SP(E | ))

2Remember that a function f : RM R is strictly increasing if for all x, y RM with x y and x = y we have

f(x) > f(y).


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Proof. First off, for the converse direction in order to render a proof by counter-positive, fix

E suppose that x / PO(E). Thus, there is y F(E) with ui(yi) > ui(xi ) for all i N since

under these assumptions by Proposition 14 we have PO(E) = WPO(E). Thus, for all (N),

ui(yi) > ui(xi ), rendering the conclusion that there is no (N) such that x solves

SP(E | ). Thus we have proven that if x solves SP(E | ) at some (N), then x PO(E).

To prove that if x PO(E), then there exists a (N) such that x solves SP(E | ), we

will employ Lemma 2 and the hyperplane separation result of Theorem 3. Let x PO(E). Note

that x /

UE, because if it were, by the defining property of UE, there is another Pareto optimal

allocation that provides strictly higher utilities to all the agents, a contradiction to the (weak) Pareto

optimality of x. Moreover, by Lemma 2, UE is convex and non-empty. Thus, by the hyperplane

separating Theorem, Theorem 3, there exists Rn with = 0 such that u(x) u for

all u UE. Moreover, 0 (i.e. i 0 for all i N), and in order to provide a continuous

reading the particular proof of this step is given in the footnote. 3 Thus, letting (N) be

defined by j =ij

jfinishes the proof.

The following notion of welfare has will bring individual rationality and Pareto optimality under

the same umbrella. In fact, the reader should note that these notions are pretty similar as long as

deviations from a given allocation is considered. Given an allocation for a pure exchange economy,

E, considering individual deviations (and the objection of an agent to his portion of that allocation

3Due to Proposition 4, which says that increasing transformation of utility functions does not matter for repre-

sentation of preferences, we may assume that ui(0) > 0 for all i N. If you will, just add a sufficiently high real

number to the lowest level of utilities given by the utility of consuming 0, and due to strict monotonicity, all other

consumption bundles would give utilities that are higher. If 0 was not the case, that means there exists at least

one agent j N with j < 0. Fix such an agent j, and consider u defined by uj = uj(0) > 0, and ui = ui(xi ) for

all i = j. By the defining property of UE, because that u(x) u and x being Pareto optimal, u UE. We already

know that by the separating hyperplane Theorem, Theorem 3, that u(x) u because u UE. Doing simple

arithmetics then shows:

u(x) u = u(x) j uj(0),

hence, because that uj(0) > 0, we have j 0, a clear contradiction to j < 0.


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based on the comparison of his welfare under that allocation with that under his initial endowment)

leads us to the notion of individual rationality. On the other hand, if the society as a whole is

considered (and objections to that allocation is solely are based on what the whole society could have

done under the restriction of feasibility) leads us to the notion of Pareto optimality. Moreover, we

could consider these simultaneously, and identify the set of Pareto optimal and individually rational

allocations by the following optimization problem (of course under the hypothesis of Proposition 15)

as follows: An allocation x X is Pareto optimal and individually rational, if there is (N)

such that x solves the following problem:

maxiN

i ui(xi) (3.3)

s.t. x F(E) IR(E).

The proof of this observation is omitted since it consists of repeating the same arguments.


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4 The Price Mechanism

Walras (1874-1877) formulated a solution to the trade problem that prevails in a pure exchange

economy not by considering a result emerging from a bargaining structure (which is exactly what

was done by Edgeworth (1881)), but instead, looking at a certain mechanism, which we call the

price mechanism.

Every agent in a pure exchange economy observes the same level of prices, and it will be assumed

that none of the agents are able or aware of their ability to influence these prices. This is called the

price taking behavior assumption.

Therefore, each agent will go to the market with all their endowments (leaving or hiding some

of them at home in order to consume after they return from the market is not allowed), sell them

using the price given to them, and then identify the optimal consumption bundles they can afford.

It is often thought that the price taking behavior assumption is justified when there are many

consumers in the economy. While having some merit, this is not sufficient to justify this assump-

tion. Because even when there are many consumers, if the economy has a family called the House

of Saud (and the King of Saudi Arabia, the largest oil producer in the world, comes from that

family), or Bill Gates, or J. P. Morgan inside, and these agents are aware of their power and

can contemplate manipulating the market, the assumption of many agents does not suffice. In-

deed, mathematical economists have shown such counterexamples even with infinitely many agents.

Therefore, what we need is to have many alike agents, none of which has a critical control on a

critical resource. And to me that assumption is far from being realistic, especially when one sees

that %99 of the wealth of U.S.A. is in the hands of %1 of the population in that country.

33


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Economics 501-2 by Mehmet Barlo: 4 The Price Mechanism 34

Leaving these observations aside, and pretending to believe in the price taking assumption, lead

to the following considerations: The budget set, and the demand.

4.1 Budget Set

In this section we are going to treat the budgetary constraints the agents face, as a set (which also

can be viewed as a set-valued function, i.e. a correspondence; which will be handled in the further

parts of this section). For the demand and competitive equilibrium, the result we are going to derive

in this section that says that the budget correspondence behaves well, is an essential one.

Moreover, note that as far as the analysis of general equilibrium is concerned, the level of

prices does not matter in a pure exchange economy. What matters is the relative prices. That is

established in the following discussion.

Definition 21 The budget set is of player i in a pure exchange economy E with prices p RM+ is

Bi(p, ei) = {xi X : p (xi ei) 0}.

It is an easy exercise to see that for p, p RM+ , with p = p, where R++, Bi(p, ei) =

Bi(p, ei). In other words, the budget set is homogeneous of degree 0 in prices. That is why in our

analysis we may restrict attention to p , where is the M 1 dimensional simplex (that is

pk [0, 1] for all k = 1, . . . , M , andM

k=1pk = 1), because we can always let =1

Mk=1pk. Thus, for

all k = 1, . . . , M , (1) pk [0, 1] and (2) Mk=1kpk = 1; consequently, p .

The following Proposition is essential for future results:

Proposition 16 For every pure exchange economy E, Bi(p, ei) is a non-empty, compact (closed

and bounded) subset of X for every pure exchange economy E with strictly positive prices p 0.

Moreover, it is also convex, i.e. for every xi, yi Bi(p, ei), xi + (1 )yi Bi(p, ei) for every

[0, 1].

Proof. Because that for all p 0, 0 is in Bi(p, ei), Bi(p, ei) is non-empty for every p 0.


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Let p 0, and let

Kp = maxk=1,...,M

p eipk

.

Then a circle with its center at the origin, 0, and radius given by Kp contains Bi(p, ei), thus, Bi(p, ei)

is bounded.

It is closed because for any convergent sequence {xn}nN Bi(p, ei) with limn xn = x, it must

be that p xn p ei implies in the limit, p x p ei since the dot product and greater or equal

to operations are both continuous. Thus, x Bi(p, ei), and hence, Bi(p, ei) is closed.

Because Bi(p, ei) is both bounded and closed, it is compact.

Finally, for convexity, assume that xi, yi Bi(p, ei). Then, p ( xi + (1 )yi) = (p xi) + ( 1

)(p yi) p ei for every [0, 1], because xi, yi Bi(p, ei). Thus, ( xi + (1 )yi) Bi(p, ei),

as was to be shown.

This finishes the proof of Proposition 16

4.1.1 Properties of the Budget Correspondence

In this section we are going to treat the budgetary constraints the agents face, as a set-valued

function (i.e. a correspondence). It is appropriate to remind the reader that for the existence of

equilibrium, the results we are going to derive in this section (saying that the budget correspondence

behaves well) are essential.

First let me introduce the notion of correspondences (set-valued functions) to the reader. The

reader should already be aware that a function is a mapping which maps every point in the domain

to a single point in the range. A correspondence, on the other hand, is a mapping which maps every

point in the domain to a subset of the range. That is, trivially every function is a singleton-valued

correspondence.

As an example consider the following correspondence F that maps X into subsets ofY, and the

notation we are going to abide by is: F : X Y. So, every point x X is mapped to F(x) Y.

Let X = [0, 1] and Y = [0, 1]. As a first example consider F : X Y defined by F(x) = [0,12x].


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Thus, for any x [0, 1], F(x) = [0, x/2]. Another example is G : X Y defined by

G(x) =

x if x = 1/2,

{0, 1} otherwise.

Note that apart from the exception of x = 1/2, G(x) is given by x. But when x = 1/2, there are

two points in G, 0 and 1.

The following definition will be used for the rest of the course:

Definition 22 LetF be a correspondence from X into Y, i.e. F : X Y. We say that, F is:

1. non-empty valued, if for all x X, F(x) = ;

2. closed-valued, if for all x X, F(x) Y is a closed set in Y;

3. compact-valued, if for all x X, F(x) Y is a compact set in Y;

4. convex-valued, if for all x X, F(x) Y is a convex set in Y.

The definition of continuity of correspondences requires more care, and will be dealt later in this

section. Moreover, these notes contain an appendix for the motivated and interested reader.

One possible interpretation of the budget is to see it as a correspondence, where the arguments

are p, e. I.e., B(p, e) : RM+ X X, where for all (p, e) RM+ X, B(p, e) X.

Note that because for any > 0, Bi(p) = Bi(p), we can always let =1

Mk=1pk. Thus, instead

of using a price vector p RM+ \ {0}, we could use p =1

Mk=1pkp, and p . Hence, the domain of

the budget correspondence (the set of prices to be considered) can be restricted (without any loss

of generality) to .

Definition 23 For a given economy E = (i, ei)iN, the budget correspondence of player i for a

given endowment ei is a set-valued function, Bi : X X defined by

Bi(p | e) = {x X : p (x e) 0}. (4.1)


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Whenever we do not need to index across the players and keep e Xn fixed we are going to

abuse notation and let the budget correspondence be given by B(p).

The following Theorem is essential, and lists the properties of the budget correspondence: (Recall

that

int(), and thus, when the domain of the budget correspondence is restricted to only

strictly positive prices, there is no loss of generality to consider

as the set of prices that are

allowed.)

Theorem 4 B :

X is non-empty valued, homogeneous of degree 0 in prices 1, compact-valued,

convex valued, and continuous.

Proof. We are going to present the proof using separate lemmas in order for the result be more

tractable.

Lemma 3 B :

X is non-empty valued.

Proof. The origin, 0 X, is in B(p) for all p , thus the result follows.

Lemma 4 B :

X is homogeneous of degree 0 in prices.

Proof. Follows from the discussion following definition 21.

Lemma 5 B :

X is convex valued.

Proof. Let p

and x, x B(p). Then for any [0, 1], and x = x + (1 )x

p (x e) = p(( x + (1 )x) e)

= p ((x e) + (1 )x e) = p (x e) + (1 )p (x e) 0,

since p (x e) 0 and p (x e) 0. Thus, x B(p).

Lemma 6 B :

X is compact valued.

1

I.e. for any > 0, Bi(p) = Bi(p).


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Proof. Pick p

, and let (p) = mink=1,...,Mpk, and define the (p)-constraint simplex

(p)

by

(p) = {p : pk (p), for all k = 1, . . . , M }. (4.2)

The reader is asked to verify that for any p

, (p) is non-empty2, convex, and compact.

Now we can define K(p) R++ by

K(p) = 2 maxk=1,...,M

p e

pk, (4.3)

since pk (p) for all k = 1, . . . , M . Thus, for p

, B(p) K(p), where K(p) = {x X : xk

K(p)

, k}, is a compact subset of X,3

since it is both closed and bounded. Thus, the remaining

task to complete to proof is to show that B(p) K(p) is closed. Take any sequence {xn}n B(p),

with xn x. Thus, for all n N, p (xn e) 0. Since all the operations involved (the dot

product, and the operation) are continuous, p (x e) 0, thus, x B(p), showing that B(p)

is closed. As p

was arbitrary, result follows.

Finally, continuity of the budget correspondence follows from Lemmas 7 and 8 which will be

presented after the introduction and discussion of continuity of correspondences.

Continuity of the Budget Correspondence

Because that our attention is restricted to finite dimensional Euclidian spaces, namely RM, the

continuity of functions is easy: We know from basic mathematical analysis that a function f :

RM RK is continuous if and only if: (1) for every sequence {xn} RM with xn x (i.e. for any

> 0, there exists N N such that for all n > N we have xn x < ), we have f(xn) f(x);

OR (2) for any open set E RK, f1(E) = {x RM : f(x) E} RM is open in RM.

Let us analyze if we can modify these definitions (even with being restricted to finite dimensional

Euclidian spaces) and obtain a nice working definition for the continuity of correspondences.

2Because for any p

, p defined by pk =1M

is in (p).3The curious reader should prove this step by using Theorem 26 from appendix 6.


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In fact, let us start with the first definition, the one that involves sequences in the domain of

the correspondence. That definition would then read: F : RM RK is continuous if and only if for

every sequence {xn} RM with xn x, we have F(xn) F(x). The problem with that definition

is that the convergence of F(xn) to F(x) is unfortunately not trivial at all. This is because, we are

talking about two sets. In fact, F(xn) F(x), for F(xn), F(x) RK means that for any > 0,

there exists N N such that for all n > N we have d(F(xn), F(x)) < . So we need to have a

distance notion on subsets ofRK, because without it, F(xn) F(x) does not make any meaning.

In mathematics, when attention is restricted to finite dimensional Euclidian spaces, to my knowl-

edge the only distance notion used is the Hausdorff distance (metric), which will be defined below:

(The formalities presented below are taken from Berge (1963).) Let A and B be two non-empty

sets in RK, and write

(A, B) = supxA

infyB

d(x, y),

(B, A) = supyB

infxA

d(x, y).

The numerical function defined by

(A, B) = max{(A, B), (B, A)}

is called a Hausdorff Metric.

As an example consider the following two sets: Let A = [0, 1] and B = [3, 5]. Then in order

to find (A, B), consider any x A, and solve infyB d(x, y) for that given x. It should be clear

that no matter what x A is, the closest member of B is 3, thus, infyB d(x, y) = 3 x. Then,

supxA 3 x is clearly given by 3 when x = 0 A. Thus, (A, B) = 3. Now, in order to compute

(B, A), fix any y B, and and solve infxA d(x, y) for that given x. Clearly, the closest member

of A for any y B is 1, thus, infxA d(x, y) = y 1. Hence, supyB infxA d(x, y) = supyB(y 1),

thus is equal to 4, and the supremum is given by y = 5. Thus, (B, A) = 4. Thence, (A, B) = 4.

This metric, while being very useful in many situations, is not well suited for our analysis because

of the following reason: This metric can only be used on compact sets (not even closed ones; to see


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this let A = {0} and B = R, it can be easily shown that (A, B) does not exists even though both

of these sets are closed): To see this, let A = (0, 1) and B = [0, 1]. Clearly, because that A B we

have (A, B) = 0. Moreover, for (B, A) fix any y B. Ify A, clearly we have infxA d(x, y) = 0.

Ify / A, then y is equal to either 1 or 0. In any case, infxA d(x, y) = 0. Thus, (B, A) = 0. This is

a contradiction to being a metric on all (bounded) subsets ofRK, because A = B but (A, B) = 0.

The following presented and proven in Lemma 9 of appendix 7 makes this observation precise:

as defined above is a metric for the family of non-empty and closed subsets of a compact metric

space X.

Therefore, because that the budget correspondence is a mapping from into X, it is not

necessarily compact valued on . To see this, consider 2 goods with prices p = (0, 1), and it is

easy to see that the resulting budget set (therefore, the budget correspondence for these prices) is

not compact. Thus, this distance notion as it is is not well suited for our purposes. But, it is a

very useful way of defining continuity of correspondences whenever the correspondence is question

is compact-valued, because then the Hausdorff distance is well defined.

Next, consider a modification to the second definition of continuous functions: F : RM R

K

is continuous if and only if for any open set E RK, F1(E) RM is open in RM. That again

sounds promising, yet we have to ask ourselves about the meaning of F1(E). When attention

is restricted to functions (singleton-valued correspondences) its meaning is clear: F1(E) = {x

RM : F(x) E}. But when F(x) is not a single point but a set, would the following be sufficient?

F1(E) = {x RM : F(x) E}. To my knowledge, the answer is affirmative, even though I have

not personally seen a proof. That is why, it would be great if any one of you could provide a clean

proof of this (the details about what I mean by being sufficient, will be given later).

In the literature, there are two notions of continuity: upper-hemi continuity and lower-hemi

continuity, and they involve a definition of continuity via open sets. More technical details can be

found in the appendix 7.

First, the upper-hemi continuity:


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Figure 4.1: The Graph of F

Definition 24 Let F : RM RK, M , K < , be a correspondence mappingRM into RK. We

say that F is upper-hemi continuous if for all x0 RM, for each open set G RK containing

F(x0) RK, there exists a neighborhood Ux0 R

M such that

x U(x0) F(x) G.

Consider the following example (remember that I think that in order to understand a notion, it is

best to see a counter example first): Let M = K = 1 and consider a correspondence F : [0, 1] [0, 2]

defined by:

F(x) =

[1/2, 1] if x < 1/2,

[3/4, 1] otherwise.

The graph ofF is given in figure 4.1. To see why this correspondence is not upper-hemi continuous,


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consider x0 = 1/2, and G = (0.70, 1.1) and note that F(1/2) = [3/4, 1] G. Now, for any

neighborhood of x0 = 1/2, denoted by U(x0), it must be the case that there are x U(x0) with

x < 1/2. Thus, for all such x we have F(x) = [1/2, 1], hence, it is not the case that F(x) G.

Moreover, the reader should note that even though F is not upper-hemi continuous, it is compact-

valued.

An important result, which is not that difficult to establish (and the reader is asked to do it) is:

Proposition 17 Every non-empty valued, singleton-valued, upper-semi continuous correspondence

is a continuous function.

Next, I define lower-hemi continuity:

Definition 25 Let F : RM RK, M , K < , be a correspondence mappingRM into RK. We

say that F is lower-hemi continuous if for all x0 RM, for each open set G RK with

G F(x0) = , there exists a neighborhood Ux0 RM such that

x U(x0) F(x) G = .

Again, let me consider the following counter-example: Let M = K = 1 and consider a corre-

spondence F : [0, 1] [0, 2] defined by:

F(x) =

{3/4} if x < 1/2,

[1/2, 1] otherwise.

The graph ofF is given in figure 4.2. To see why this correspondence is not lower-hemi continuous,

consider x0

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