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Mathematical Foundations -1- Mappings

© John Riley October 11, 2013

Mappings

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Mappings: transformations, correspondences, functions

Linear Transformationn m

11 1

1

. .. . . .

[ ]. . . .

. .

n

ij

m mn

a a

a

a a

A

Anm n

dimensional array. We call this array a matrix.

Special matrices:

1 n matrix 11 1[ ,..., ]na a a row vector 1m matrix

11

1

.

.

.

m

a

a

a column vector

It is often useful to think of an m n matrix as either a row of n column vectors or a column of m row

vectors

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Linear transformation

Linear combination of column vectors

11

1

.

.

.

m

a

a

,

12

2

.

.

.

m

a

a

,…,

1

.

.

.

n

mn

a

a

y =

11

1

1

.

.

.

m

a

x

a

+

12

2

2

.

.

.

m

a

x

a

+…+

1

.

.

.

n

n

mn

a

x

a

We write

y xA where1

n

i ij j

j

y a x

The inner product of the i-th row and the vector x

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Linear Transformations2 2

For any pair of column vectors

11

21

a

a

,

12

22

a

a

Let y be a linear combination of these two column vectors.

1 11 12 11 1 12 221

2 21 22 21 1 22 2

y a a a x a x x x

y a a a x a x

(*)

Graphically,

scale up the first vector by1

x ,

scale up the second by2

x

and then add the two resulting vectors as shown.

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For any set 2 X the linear combinations transform all the elements of X into a set 2

Y

Linear transformation

We write the linear transformation as follows.

y xA or1 11 12 1

2 21 22 2

y a a x

y a a x

Theni

y is the inner product of the row vector 1 2i ia a and the column vector

1

2

x

x

1 11 12 1 11 1 12 2

2 21 22 2 21 1 22 2

y a a x a x a x

y a a x a x a x

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Compound linear transformations

Consider the linear transformations y x A and z yB

1

1 2

2

i i i

y

z b b y

,

1 11 1 12 2

2 21 1 22 2

y a x a x

y a x a x

.

Then

11 1 12 2

1 2 1 11 2 21 1 1 12 2 22 2

21 1 22 2

( ) ( )i i i i i i i

a x a x z b b b a b a x b a b a x

a x a x

1

1 11 2 21 1 12 2 22

2

i i i i

xb a b a b a b a

x

Therefore

1 11 11 12 21 11 12 12 22 1

2 21 11 22 21 21 12 22 22 2

z b a b a b a b a x

z b a b a b a b a x

.

Thus the compound mapping is the linear transformation

z xC where

21

1 2

2 1

j

ij i i ik kj

j k

a

c b b a ba

.

Inner product of i-th row of B and j-th column of A

We write the compound mapping as C=BA.

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Linear Transformationsn n

Matrixes are square

Identity matrix n= 2

1 0

0 1

I

Note that, by the rules of matrix multiplication, the identity matrix maps x onto itself.

x xI

Transpose of a square matrix

Flip rows and columns

The ith row becomes the ith colum

11 12

21 22

a a

a a

A .

Transpose

11 21

12 22

a a

a a

A

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Determinant of a matrix

n=2

11 12

21 22

a a

a a

A

Delete a row and column and you get a numberij

m .

The matrix of these numbers is

11 12 22 21

21 22 12 11

m m a a

m m a a

Convert this into a matrix of cofactorsij

c where ( 1)i j

ij ijc m

2 311 12 22 2111 12

3 421 22 12 1121 22

( 1) ( 1)

( 1) ( 1)

c c a am m

c c a am m

Then the determinant of A is defined as follows:

1 1

n n

ij ij ij ij

j i

a c a c

A

It is a lovely theorem that it does not matter which row or column of A that we pick.

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Example:

6 2

8 3

A ,

11 12

21 22

3 8

2 6

m m

m m

2 311 12 11 12

3 421 22 21 22

3 8( 1) ( 1)

2 6( 1) ( 1)

c c m m

c c m m

C

1 1

2

n n

ij ij ij ij

j i

a c a c

A

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n=2

11 12

21 22

a a

a a

A

Eliminating a row and column yields a number so the minors are easily computed.

The matrix of minors is

11 12 22 21

21 22 12 11

m m a a

m m a a

The matrix of cofactors is

11 12 22 21

21 22 12 11

c c a a

c c a a

C .

Try any row or column and you will be able to confirm that

11 22 21 12a a a a A

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n=3

Consider the ij-th minor matrix created by deleting the i-th row and j-th column of A. This is a 2 2

matrix. Letij

m be the determinant of this minor matrix.

Define the cofactor ( 1)ij ij

c m .

Then the determinant of A is defined as follows:

1 1

n n

ij ij ij ij

j i

a c a c

A

Again it does not matter which row or column of A that we pick.

A second feature of the matrix of cofactors is that if you compute the sum product of the k -th row

where k i and the ith row of cofactors, then the product is zero. The same is true for columns.

1

0,n

kj i j

j

a c k i

,

1

0,n

ik ij

i

a c k j

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Inverse of a matrix

C = matrix of cofactors

11 12 22 21

21 22 12 11

c c a a

c c a a

C .

Then the transposed matrix of cofactors is

11 21 22 11

12 22 21 11

c c a a

c c a a

C

Now consider the product of A and C

11 12 11 21 11 12 22 11

21 22 12 22 21 22 21 11

0

0

a a c c a a a a

a a c c a a a a

AAC

A

11 21 11 12 22 11 11 12

12 22 21 22 21 11 21 22

0

0

c c a a a a a a

c c a a a a a a

A

CA A

Finally, if 0A define22 111

21 11

1 1 a a

a a

A CA A

.

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22 111

21 11

1 1 a a

a a

A CA A

.

22 111

21 11

1 1 a a

a a

A C

A A .

Then

1 1 A A C A I

A and 1 1

AA AC IA

Mapping

y xA

Inverse mapping

1 1 y x x x

A A A I

We call 1 1A C

A the inverse matrix.

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Linear independence of the columns of an invertible matrix

Proof by contradiction.

11 1 1 11 1

1

1 1

. .

. . . . . . ....

. . . . . . .

. .

n n

n

n nn n n nn

a a x a a

x x x

a a x a a

A

If the columns are linearly dependent then, for some 0 x , 0 x A .

Since A is invertible, 1 10 0 x x

A A A . A contradiction.

Exercise:

0 1 1

1 0 1

1 1 0

A

(a) Compute the nine minor determinants of A and hence compute A .

(b) Use your results to obtain C and hence C and hence obtain the inverse matrix.

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Computing the inverse matrix

Example:

1 1 1

1 2 0

0 1 3

A ,

11

2 0

1 3m ,

12

1 0

0 3m ,

13

1 2

0 1m

21

1 1

1 3m , 22

1 1

0 3m , 23

1 1

0 1m

31

1 1

2 0m ,

32

1 1

1 0m ,

33

1 1

1 2m

6 3 1

[ ] 2 3 12 1 1

ijm

6 3 1

[ ] 2 3 12 1 1

ijc

C

6 2 2

3 3 11 1 1

C

([ th row of [ ],[ th row of [ ]) 4ij iji i sumproduct a c A

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Inverse of A

1 1 1

1 2 0

0 1 3

A

1

6 2 2

1 1 3 3 14

1 1 1

A CA

As may readily be checked,

-1A A = I

M h i l F d i M i

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Function 1

( ) ( ,..., )n

f x f x x

single valued mapping from nS

Increasing Function

The function ( ) f x is increasing over the set X if for any x and x X

( ) ( ) x x f x f x

Strictly increasing function

The function ( ) f x is increasing over the set X if for any x and x X

( ) ( ) x x f x f x

M h i l F d i M i

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Class Exercise:

(a) If a function : f

is differentiable and strictly increasing at 0 x , does it follow that the slope

of the function is strictly positive at 0 x ?

(b) If the function : n

f is increasing at 0 x in all of its arguments is it an increasing function?

**

M th ti l F d ti M i

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Class Exercise:






(a) Consider the following function

3 0( ) , 0 f x x x

M th ti l F d ti M i

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Class Exercise:






(a) Consider the following function

3 0( ) , 0 f x x x

(b) Consider the following function 2: f

4 2 2 4

1 1 2 2( ) 4 f x x x x x

Suppose

0

(0,0) x . Both

4

1 1( , 0) f x x

and

4

2 2(0, ) f x x

are strictly increasing functions on

2

.However if

1 2( , ) ( , ) x x z z then

4( , ) 2 f z z z . Thus the function f is definitely not an increasing

function.

M th ti l F d ti 21 M i

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Mathematical Foundations 22 Mappings

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Special functions

Linear Function ( )l x a x

Quadratic form ( )q x x x A where A is a square matrix

WOLOG we may assume that A is symmetricij ji

a a

11 12 1 11 1 12 2 2 2 2 2

1 2 1 2 11 1 21 21 1 2 22 2 11 1 12 1 2 22 221 22 2 21 1 22 2

[ ] [ ] ( ) 2a a x a x a x

y x x x x a x a a x x a x a x a x x a xa a x a x a x

Quadratic Function ( ) f x k a x x x A

Equivalently, ( ) f x k a x x x A


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Special functions

Linear Function ( )l x a x

Quadratic form ( )q x x x A where A is a square matrix

WOLOG we may assume that A is symmetricij ji

a a

11 12 1 11 1 12 2 2 2 2 2

1 2 1 2 11 1 21 21 1 2 22 2 11 1 12 1 2 22 221 22 2 21 1 22 2[ ] [ ] ( ) 2

a a x a x a x

y x x x x a x a a x x a x a x a x x a xa a x a x a x

Quadratic Function ( ) f x k a x x x A

Equivalently, ( ) f x k a x x x A


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Negative definite matrix

A symmetric matrix A with the property that ( ) 'q x x x A is strictly negative for all 0. x

Consider n=2 2 2

11 1 12 1 2 22 2( ) 2q x x x a x a x x a x A

?

**


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Consider n=2 2 2

11 1 12 1 2 22 2( ) 2q x x x a x a x x a x A

Note first that11

0a (set2 0 x )

*


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Consider n=2 2 2

11 1 12 1 2 22 2( ) 2q x x x a x a x x a x A

Note first that11

0a (set2 0 x )

Complete the square 2 2 2

1 1 1( ) 2 x x x

22 2 2 212 12 12

11 1 1 2 2 22 2

11 1 1 11

( ) ( 2 ( ) ) ( )a a a

q x a x x x x a xa a a

2 2 21211 1 2 11 22 12 2

11 11

1( ) ( )

a

a x x a a a x

a a

Then ( ) 0q x for all 0 x if and only if11

0a and 2

11 22 12 0a a a


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Negative semi-definite matrix

A symmetric matrix A with the property that ( ) 'q x x x A is negative for all x .

n=2: ( ) 0q x for all x if and only if11 22

0, 0a a and 2

11 22 12 0a a a


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0, 0a a and 2

11 22 12 0a a a

Case (i)11

0a

2 2 21211 1 2 11 22 12 2

11 11

1( ) ( ) ( )

aq x a x x a a a x

a a

Then a necessary and sufficient condition is11

0a and 2

11 22 12 0a a a


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0, 0a a and 2

11 22 12 0a a a

Case (i)11 0a

2 2 21211 1 2 11 22 12 2

11 11

1( ) ( ) ( )

aq x a x x a a a x

a a

Then a necessary and sufficient condition is 11 0a and2

11 22 12 0a a a

Case (ii) 11

0a

Exercise: Complete the proof.


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Limit of a function

The function : n f S has a limit L at 0 x S if, for any 0 , there exists a deleted

neighborhood 0( , )

D N x such that if 0

( , ) D

x N x S then ( ) f x L .

Exercise: Suppose2

2

1 , 1( )

2 , 1

x x f x

x x

.

(a) Does the function have a limit at 0 1 x if [0,2]S ?

(b) What if [0,1]S ?

Continuous function

Let f be a function defined on nS and suppose 0

x S . Then f is continuous at 0 x if it has a

limit at 0 x equal to 0( ) f x .


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3 pp g


Example 1: A function : f with no limit at 02 x

Consider some small .

For all x in the interval (2 ,2) , ( ) f x is close to 2.

For all x in the interval (2,2 ) , ( ) f x is close to 4.

Thus the function does not approach a limit L as

x approaches 02 x .

2

4


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pp g


Example 2: A function :[0,2) (2,4] f with a limit point at 02 x

Note that even though the function is not defined

at 0 2 x , for all x close to 2, ( ) f x is close to 4.

Thus the limit of the function is 4 L .

KEY POINT: The function itself need not be defined at 0 x to have a limit point at 0

x .

4


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pp g


Example 3: A discontinuous function : f with a limit point

Arguing as in Example 2, the function has a

limit 2 L at 02 x . Also 0

( ) f x a .

Therefore the function is discontinuous at 0 x unless 2a .

a

2


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pp g


Rules of limits: : , :n n f S f S

Rule 1: Limit of sum = sum of limits

Rule 2: Limit of product = product of limits

We will prove Rule 1 this assuming that1 2, 0 L L .

We know that for any1 2, there exist deleted delta neighborhoods 0 0

1 2( , ), ( , ) D D

N x N x such that

0

1 1 1 1 1( ) , ( , ) D L f x L x N x S , 0

2 2 2 2 2( ) , ( , ) D

L g x L x N x S

We wish to prove that for any there exists a deleted delta neighborhood 0( , ) D

N x such that

0

1 2 1 2( ) ( ) , ( , ) D L L f x g x L L x N x S

?


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pp g


We will prove Rule 1 this assuming that1 2

, 0 L L .

We know that for any1 2

, there exist deleted delta neighborhoods 0 0

1 2( , ), ( , ) D D

N x N x such that

01 1 1 1 1( ) , ( , ) D L f x L x N x S , 0

2 2 2 2 2( ) , ( , ) D L g x L x N x S

We wish to prove that for any there exists a deleted delta neighborhood 0( , )

D N x such that

0

1 2 1 2( ) ( ) , ( , ) D L L f x g x L L x N x S

*


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Proof:

We will prove Rule 1

0

1 1 1 1 1( ) , ( , )

D

L f x L x N x S

,

0

2 2 2 2 2( ) , ( , )

D

L g x L x N x S

Then

1 1 2 2 1 1 2 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( ) L L f x g x L L L L ,

for all 0( , ) D x N x S where1 2{ , } Min


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Proof:


0

1 1 1 1 1( ) , ( , )

D

L f x L x N x S

,

0

2 2 2 2 2( ) , ( , )

D

L g x L x N x S

Then

1 1 2 2 1 1 2 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( ) L L f x g x L L L L ,


That is,

1 2 1 2 1 2 1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) L L f x g x L L L L ,

for all 0( , ) D

x N x S where1 2

{ , } Min


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Proof:


0

1 1 1 1 1( ) , ( , )

D

L f x L x N x S

,

0

2 2 2 2 2( ) , ( , )

D

L g x L x N x S

Then

1 1 2 2 1 1 2 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( ) L L f x g x L L L L ,


That is,

1 2 1 2 1 2 1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) L L f x g x L L L L ,

for all 0( , ) D

x N x S where1 2

{ , } Min

Choose1

1 2 and1

2 2 .

Then1 2 1 2( ) ( ) L L f x g x L L

QED


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Properties of functions on a set

Some properties of functions must be defined point-wise. When this is the case and the property P

holds for the function f at every point in a set S , the function f is said to have the property P on S.

Example 1:1 2

( ) f x x x is continuous on 2

(the set of all positive elements of 2 ) because it is

continuous at each point in 2

.

Example 2:1 2( ) ln f x x x is continuously differentiable on 2

(the set of strictly positive elements of

2 ) because the partial derivatives exist and are continuous at each point in 2

.

Note: If the partial derivatives of a function : n f S are continuous we write 1 f C .


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However if a property need not be defined point-wise, it is best defined on a set directly.

An increasing function can defined point-wise on a set or directly on the set. To see why

mathematicians take the latter approach consider the following example.

Example 1:

[0,1] [2,3]S ,, [0,1]

( )5, [2,3]

x x f x

x x

.

Note that the function is strictly increasing at each point in S however the function is not strictly

increasing on S .


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Key foundational theorem for economics

Extreme Value Theorem

If the function f is continuous over a compact (i.e. closed and bounded) set nS , then f attains its

maximizing and minimizing value at some point in S .

Class exercise:

To see that there may be no maximizing value if S is not compact and f is not continuous,

consider the following examples.

(i) ( ) , [0, 2) f x x S

(ii) ( ) , f x x S

(the set of positive real numbers)

(iii) , 1( ) , [0,1]

0, 1

x x f x S

x


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Exercise: Suppose that : f S is continuous and for every 0 x S there exists a 0 such that for

all 0( , ) x N x ,0

0

( ) ( )0

f x f x

x x

. If S is a closed interval, show that f is strictly increasing on S .

HINT: If not, then for some 0 x and 1 0

x x , (a) 1 0( ) ( ) f x f x or (b) 1 0( ) ( ) f x f x . First consider

case (a) and appeal to the Extreme Value theorem for the interval 0 1[ , ] x x . Use your conclusion to

obtain a contradiction.


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Correspondence:

Multi valued mapping

Each point n x X is mapped into a subset ( )

m y x

Example 1: Supply correspondence

6 , [0,10]

( ) 60 12( 10), (10, 20]

, (20, )

q q

C q q q

q

The firm solves the following maximization problem.

{ ( ) | }q

Max pq C q q

The set of quantities that solve this problem is written as follows.

( ) arg { ( ) | }q

y p Max pq C q q

.

Note that ( ) y p is not single valued for all prices.

q


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Example 2: Demand correspondence

A consumer enjoys ice cream but really loves raspberry ripple.

Each scoop of raspberry ripple (commodity 2) is as good as two scoops of vanilla (commodity 3)

1 2 3 1 2 3( , , ) ln ln(2 )U x x x x x x

The marginal utilities of the three commodities are

1 1

1U

x x

,2 2 3

2

2

U

x x x

,3 2 3

1

2

U

x x x

Hence the marginal utilities per dollar are

1 1 1 1

1 1U

p x p x

,

2 2 2 2 3

1 2 1( )

2

U

p x p x x

,

3 3 3 3

1 1U

p x p x

.


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Case (i)2 3{ | 2 }

r p P p p p

Then the marginal utility per dollar is higher for scoops of raspberry so*

3 ( ) 0 x p

Equating marginal utilities per dollar

1 1 1 1 2 3 2 2

1 1 1 2 1( )

2

U U

p x p x p x p x

.

Then spending on each of the other two commodities is half total spending so

*

1 2

( ) ( , ,0)2 2

I I x p

p p .

Case (ii) 2 3{ | 2 }v

p P p p p

By an almost identical argument

*

1 3

( ) ( ,0, )2 2

I I x p

p p .


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Case (iii) 3 \b r v p P P P

2 2 3 3

1 1U U

p x p x

Consumer is indifferent between

0

1 2

( ) ( , ,0)2 2

I I x p

p p , 1

1 3

( ) ( ,0, )2 2

I I x p

p p and any convex combination.

Thus for b p P there is a set of maximizing consumption bundles

* 0 1( ) (1 ) ( ) ( ), 0 1 x p x p x p .


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Correspondence

A non-empty set valued mapping ( ) X of each value of into a subset of n

.

Upper hemi-continuous correspondence

The set valued mapping ( ) X is upper hemi-continuous at 0 if for any open neighborhood, V of

0( ) X there exists a neighborhood of 0

, 0( , ) N , such that ( ) X V ,for all 0

( , ) N

Exercise: Explain why the following correspondence is upper hemi-continuous on [0,4]S .

[2 ,3 ], 0 2( )[ , 4 ], 2 4

X

Lower hemi-continuous correspondence

The set valued mapping ( ) X is lower hemi-continuous at 0 if for any open set V that intersects

0( ) X , there exists a neighborhood of 0

, 0( , ) N , such that ( ) X intersects V for all

0( , ) N


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Consider the examples below. In both cases ( ) X is an interval. Note that the left hand mapping is

continuous from the right 0 and the second is continuous from the left but neither is continuous.

However the mapping on the left satisfies the conditions for upper hemi-continuity and the one on the

right does not.

The right-hand diagram does, however, satisfy the conditions for lower hemi-continuity while the one

on the left does not.

Upper hemi-continuity Lower hemi-continuity


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Continuous correspondence

A mapping is continuous if it is both upper and lower hemi-continuous.

Exercise: Supply correspondence

Explain why the supply correspondence of Example 1 is upper hemi-continuous.

Exercise: Demand correspondence

For the demand correspondence of Example 2, explain why * ( ) x p is upper hemi-continuous but not

lower hemi-continuous

mapping s

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