Sets and Sequences

8/13/2019 Sets and Sequences

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Mathematical Foundations -1- Sets and Sequences

John Riley October 8, 2013

Sets and Sequences

Methods of proof 2

Sets and vectors 13

Planes and hyperplanes 18

Linearly independent vectors, vector spaces 20

Convex combinations of vectors 21

neighborhoods, boundary point and interior point of a set 24

Closed and open sets, convex sets, compact sets 28

Sequences and convergent sub-sequences 33

Reading: EM Appendix A.1, B.1


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Methods of proof

(i) direct proof,(ii) proof by induction,(iii)

proof of the contrapositive,

(iv) proof by contradiction


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Direct proof

Proposition: Let1

1 r

be the PV of 1 dollar a period into the future. Then the present value of an

investment yielding a constant payment of 1 dollar after each of nperiods is

1 1... [1 ... ]

1

n

n n

nV

.

The bracketed expression is an example of a geometric sum 2 11 ... nn

S a a a

. The well known

formula for this sum is 1 , 11

n

n

aS a

a

.

Proof: 2 ... nn

aS a a a

Hence, 2 11 (1 ... )n nn

aS a a a a

. But the term in the parentheses isn

S .

Therefore 1 nn n

aS S a . As long as 1a , we can solve forn

S rearrange to obtain1

1

n

n

aS

a

.

Q.E.D.


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Application: Prisoners Dilemma

Each firm chooses either a High or Low price.

Payoffs are as shown (row player indicated first)

Both firm follow the strategy of setting a high price

In the first period. However if cheated a firm

will follow the motto of my Graham ancestors.

Never forget a friend. Never forgive an enemy. That is they will evermore play low.

Suppose firm 2 cheats in the first period. Its payoff stream then has a PV of

2 2{5,1,1,1,...} 5 ( ....) 5 (1 ...)(1) 5 (1)1

PV

.

If firm 2 does not cheat his payoff is

2 2{2, 2, 2, 2,...} 25 ( ....) 5 (1 ...)(2) 2 (2)

1

PV

.

Therefore cheating is unprofitable if

5 (1) 2 (2)1 1

, that is if 31

, that 3 / 4

Firm 2

High Low

Firm 1High 2,2 0,5

Low 5,0 1,1


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Proof by induction

Proof by induction is often helpful if the goal is to establish that some propositionn

P is true for all

integers n= 1,2,.

Step 1: establish that the proposition is true for 1n .

Step 2: show that if the proposition holds for all integers up to k, it must hold for 1n k . If both can

be proved then we are finished because if1

Pis true then, by the second step2

P is true. Then, by the

second step, since1

Pand2

P are both true so is3

P . Since this argument can be repeated over and over

again it follows that for all nn

P is true.


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Geometric sum example

Define1

1

n

n

aS

a

Claim: 11 ... nn

a a S

.

Note that1

11

1

aS

a

so the claim is true for n=1.

Suppose true for n=k. Then

11

1

1 1 (1 ) 11 ...

1 1 1

k k k k

k k k k

k k

a a a a aa a a S a a S

a a a

.

Therefore true for 1n k


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Proof of the contrapositive

Suppose we would like to prove that if the

statementAis true then the statementB

must also be true. That is, any event in

whichAis true is also an event in whichB

is true. (B is necessary for A.)

Consider the three Venn diagrams. In each case the box is the set of all possible events. In the left

diagram the heavily shaded region is the set of events in which A is true. The set of events in whichA

is false is the set A. Similarly forB.

Note that A B is equivalent to the statement B A .

ifAis true thenBis true equivalently if B is true then A is true

(IfBis false thenAis false)

This statement is the contra positiveof the original statement.

A

B


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Thus instead of attempting a direct proof thatAimpliesB, we can appeal to the contrapositive and

attempt to prove that ifBis false, thenAmust also be false.

A good example of this approach can be found in section A.5 where we show how to construct a proof

of the following statement.

If the differentiable functionftakes on a maximum at x , then the slope is zero at x .

The contrapositive of this statement is that if the slope of a function is not zero at x , then the function

does not take on a maximum at x . The proof then follows from an examination of the formal

definition of the slope of a function.


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Proof by contradiction

Last but by no means least, we often prove that a statement is true by deriving the implications

that follow if the statement is false. Suppose, by a combination of luck and cunning, we find an

implication that is impossible. Then we know that the statement cannot be false and so it must be true!

To illustrate, we prove that the sum of two odd numbers must be even. Any even number can be

written as 2nwhere nis an integer and so any odd number can be written as 2n+1. Suppose that the

statement is false so that for some integers, , ,a b c

(2 1) (2 1) 2 1a b c . (*)

Rearranging this equation,

2( ) 1c a b .

The number 2( )c a b is divisible by 2 so it cannot be equal to 1. Hence equation (*) leads to a

contradiction. Thus there cannot be any such numbers a,b,cand so the statement must be true.


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Proposition:If the demand function 2 ( )q p is strictly more elastic than 1( )q p then the demand curves

have at most one intersection.

2 12 1

2 1

( ( ), ) ( ( ), )dq dqp p

q p p q p p

q dp q dp

**


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Proposition:If the demand function 2 ( )q p is strictly more elastic than 1( )q p then the demand curves

have at most one intersection.

2 12 1

2 1

( ( ), ) ( ( ), )dq dqp p

q p p q p p

q dp q dp

If the demand curves intersect at p .

Then 1 2( ) ( )q p q p and so2 1( ) ( )

dq dqp p

dp dp . (*)

Therefore for some prices greater then p , 2 1( ) ( )q p q p .

*

1( )q p


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Proposition:If the demand function 1( )q p is strictly less elastic than 2 ( )q p

then the demand curves have at most one intersection.

2 1

2 12 1

( ( ), ) ( ( ), )

dq dqp p

q p p q p pq dp q dp

If the demand curves intersect at p .

Then 1 2( ) ( )q p q p and so2 1( ) ( )

dq dqp p

dp dp . (*)

Therefore for some prices greater then p , 2 1( ) ( )q p q p

.

Let1

p be the lowest price at which the curves intersect

And let2

p be the second lowest such price.

Then, as depicted 2 12 2( ) ( )dq dq

p pdp dp

But at any intersection (*) must hold so we have a contradiction.


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Sets and Vectors

Set=collection of objects

Ifxis in the set we write x S

xbelongs to S

Ifxis not in the set we write x S

xdoes not belong to S

Subset, S T

superset S T

Union, S T x S orx T (shaded)

Intersection S T x S andx T

Set difference \S T { | }x S x T


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Vector

an ordered n-tuple1 2( , ,..., )nx x x x where ix (a real number)

The set of all such vectors is the Euclidean vector space of dimension n. nx

Economists typically consider subsets of n ( nS )

Length of a vector

2 2

1

n

j

j

x x


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Distance between 2 vectors

If 1n , a natural measure is the absolute value of the difference so that ( , )d y z y z .

Similarly, with2n

, a natural measure is the Euclidean distance y z

between the two vectors.

Appealing to Pythagoras Theorem

22 2 21 1 2 2( , ) ( ) ( )d x y y z y z y z

( , )d y z y z

Note that the distance betweenyandzis the length of the vectory-z.

Extending this to n-dimensions, the square of

Euclidean distance between ordered n-tuples is defined as follows.

2 2

1

( )n

j j

j

y z y z


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Vector Product

The product (inner product or sumproduct) of two n-dimensional vectorsyandzis.

1

n

i ii

y z y z

The square of Euclidean distance between two vectors can then be rewritten as follows.

22( , ) ( ) ( )d y z y z y z y z . (*)

Properties of vector products

1 1

n n

i i i i

i i

a b a b b a b a

( )a b c a b a c and hence ( )a b d a b a d

M h i l F d i S d S


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Orthogonal Vectors

Two vectors,xandpare depicted

in 3-dimensional space.

xlies in the plane perpendicular top.

Question: What is the equation of the

plane of vectors perpendicular top?

Pythagoras Theorem

2 2 2x p p x

O

M h i l F d i S d S


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Planes and hyperplanes

2 2 2p x p x

From (*) on page 16, we can rewrite this as follows.

( ) ( )p x p x p p x x

Applying the rules of vector multiplication,

( ) ( ) ( ) ( )p x p x p x p p x x

p p x p p x x x

2p p p x x x

Appealing to Pythagoras Theorem,

2p p p x x x p p x x

Hence 0p x .

We generalize perpendicularity to higher dimensions as follows.

Definition: Orthogonal Vectors

The vectorsxandpare orthogonal if their product1

0n

i i

i

p x p x

.

M th ti l F d ti S t d S


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(Hyper)plane through 0x orthogonal top

Set of vectorsxsuch that

0x x is orthogonal top.

Example: Budget set

Price vector and income

np

,I

Consumption vector nx

Suppose that 0p x I

For feasibility

0p x I p x

Equivalently,

0( ) 0p x x

Geometry of the budget set

Set of positive consumption vectors bounded by the hyperplane through 0x orthogonal to p

O

M th ti l F d ti S t d S


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Linear combination of vectors 1 1 ... m my v v where , 1,...,i n

v i m

Linearly independent vectors

1{ ,..., }mv v such that 1 1 ... 0m mv v if and only if 1 ... 0m

Euclidean Vector Space

Set of vectors such that each vector can be represented as a combination of linearly independent

vectors 1{ ,..., }mv v known as a basis for the vector space

Dimension of a vector space

number of vectors in a basis

Class Exercise: 1 2 3(1, 2,1), (5, 2, 3), (6, 0, 2)v v v .

(a) What is the dimension of the vector space of all linear combinations of these vectors?(b) The vector space of linear combinations of 1v and 2v is a two dimensional vector space (a

plane .) What is the equation of this plane?

HINT: Find a vectorpthat is orthogonal (perpendicular) to both 1v and 2v .

M th ti l F d ti 21 S t d S


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Convex combination of vectors

For any 0 1,x x X the vector y is a

convex combination of 0x and 1x if for some (0,1)

0 1(1 )y x x .

**

Mathematical Foundations 22 Sets and Sequences


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0 1(1 )y x x .

It is often helpful to write a convex combination as

0 1(1 )x x x .

Note that 0 1 0( )x x x x .

Then 0 1 0( )x x x x

Hence 0 1 0x x x x

The convex combination with weight is a fraction of the way down the line segment

beginning at0

x and ending at1

x .

*



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0 1(1 )y x x .

It is often helpful to write a convex combination as

0 1(1 )x x x .

Note that 0 1 0( )x x x x .

Then 0 1 0( )x x x x

Hence 0 1 0x x x x

The convex combination with weight is a fraction of the way down the line segment

beginning at0

x and ending at1

x .

Exercise: xand x are both convex combinations of 0x and 1x . Suppose that (so x is closer

to 0x

(a) Depict the 4 vectors in a diagram.

(b) Prove that x is a convex combination of 0x and x.



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Neighborhood and deleted (punctured) neighborhood of 0x

neighborhood of 0x 0 0( , ) { | ( , ) }N x x d x x

Deleted Neighborhood of a vector0 0 0( , ) { | ( , ) , }DN x x d x x x x

Class Exercise:

Define a Local Maximum for the function(i) :f (ii) : nf



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Henceforth focus on sets that are subsets of Euclidean space

Boundary point of a set

0x is on the boundary of nS if for any 0 the neighborhood 0( , )N x contains some

andy S z S

**



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andy S z S

Consider a sequence of neighborhoods 0( , )tN x where 0t

. Then there must be a corresponding

sequence of points { }

t

y in Sand a second sequence { }

tz

not in Sthat both converge to

0x

*



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andy S z S

Consider a sequence of neighborhoods 0( , )tN x where 0t

. Then there must be a corresponding

sequence of points { }

t

y in Sand a second sequence { }

tz

not in Sthat both converge to

0x

Interior point of S.

Any point in S that is not a boundary point

The set of interior points is written as intS



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Closed Set in Euclidean SpacenS is closed if it contains all its boundary points.

Example: [ , ] { | }a b x a x b

Note that the set of real numbers is closed as it has no boundary points.

In economics we are often interested in the set of positive real vectors

{ | 0}n nx x

This is also a closed set.

*



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Closed Set in Euclidean SpacenS is closed if it contains all its boundary points.

Example: [ , ] { | }a b x a x b

In economics we are often interested in the set of positive real vectors

{ | 0}n nx x

This is also a closed set.

Open Set in Euclidean Space

nS is open if every point is S is an interior point. That is, for any 0x S there exists 0 such

that the neighborhood 0( , )N x S

Example: ( , ) { | }a b x a x b

Class Exercise:

Which statements are correct?

The set of real numbers is (a) open (b) closed (c) neither open nor closed.



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Bounded set: The set nX is bounded if for somen

a , a x a for all x X

Compact set: A set that is both closed and bounded

**



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3 q





Convex set: Sis convex if, for any 0 1,x x S and any (0,1) ,

the convex combination 0 1(1 )x x x S

Strictly Convex Set Sis strictly convex if, for any 0 1,x x S and any (0,1) ,

0 1(1 ) intx x x S

*



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q





Convex set: Sis convex if, for any 0 1,x x S and any (0,1) ,

the convex combination 0 1(1 )x x x S

Strictly Convex Set Sis strictly convex if, for any 0 1,x x S and any (0,1) ,

0 1(1 ) intx x x S

Examples of strictly convex sets2 2 2

1 2{ | 25}S x x x ,2

1 2{ | 25}S x x x ,2

1 2{ | 25}S x x x

Exercise

(a) Suppose that a b c . What are the boundary points of { | or }S x a x b b x c ?

(b) Is Sopen?



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q


Infinite sequence of vectors 1{ }t

tx

Convergent sequence 01{ }t

tx x

For any0

there exists some ( )T

such that for all

0

( ), ( , )

tt T x N x

The vector 0x is the limit point of this sequence.

Question: Does every infinite sequence have a limit point?

Question: Does every bounded infinite sequence have a limit point?



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q

Convergent subsequence

Bolanzo-Weierstrass Theorem: Any bounded sequence

of vectors 1{ }t

tx

has a convergent subsequence.

Proof: 2x

Sequence is bounded so for some a

ta x a

infinite sequence in a square of side 2a

Step 1:must be an infinite subsequence in a square of side a

Step 2:

must be an infinite subsequence in a square of side / 2a

.

Step n:must be an infinite subsequence in a square of side /a n

.

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