Lecture 3

Econ 2001

2015 August 12

Lecture 3 Outline

1 Metric and Metric Spaces2 Norm and Normed Spaces3 Sequences and Subsequences4 Convergence5 Monotone and Bounded Sequences

Announcements:- Friday’s exam will be at 3pm, in WWPH 4716; recitation will be at 1pm. Theexam will last an hour.

Vector Space Over the Reals

DefinitionA vector space V over R is a 4-tuple (V ,R,+, ·) where V is a set, + : V ×V → Vis vector addition, · : R× V → V is scalar multiplication, satisfying:

1 Closure under addition: ∀x, y ∈ V , (x+ y) ∈ V2 Associativity of +: ∀x, y, z ∈ V , (x+ y) + z = x+ (y + z)3 Commutativity of +: ∀x, y ∈ V , x+ y = y + x4 Existence of vector additive identity: ∃!0 ∈ V s.t. ∀x ∈ V , x+ 0 = 0+ x = x5 Existence of vector additive inverse: ∀x ∈ V ∃!y ∈ V s.t. x+ y = y + x = 0.Denote this unique y as (−x), and define x− y as x+ (−y).

6 Distributivity of scalar multiplication over vector addition: ∀α ∈ R ∀x, y ∈ V ,α · (x+ y) = α · x+ α · y

7 Distributivity of scalar multiplication over scalar addition: ∀α, β ∈ R ∀x ∈ V ,(α+ β) · x = α · x+ β · x

8 Associativity of · : ∀α, β ∈ R,∀x ∈ V , (α · β) · x = α · (β · x)9 Multiplicative identity: ∀x ∈ V , 1 · x = x

Fun With Vector Spaces

Supose we have a vector space V over R and x ∈ V .What is 0 · x =?

0·x = (0+0)·xdistributivity of scalar muliplication︷︸︸︷

= 0·x+0·xexistence of vector additive identity︷︸︸︷

= 0where the first zero is a scalar while the last is a vector.

Using this result:

0 = 0 · x = (1− 1) · x = 1 · x+ (−1) · x = x+ (−1) · x

So we conclude(−1) · x = (−x).

which is not something implied by the nine properties of a vector space.

DistanceWe work with abstract sets (typically vector spaces) and our main objective isto be able to talk about their properties as well as properties of functionsdefined over them.

For example, we would like to define notions like limits and convergence,continuity of a function from one of these sets to another, and so on.

Continuity of a function f : R→ R is defined as∀ε > 0 ∃δ > 0 such that |x − y | < δ ⇒ |f (x)− f (y)| < ε

How can we construct a similar definition when the domain is some abstractset X?

ε, δ, and f (·) are scalars, but we do not know how to measure the distancebetween two points.

So, next we define function that measures “distance”between any two pointsin a set.

using this function, one can express the idea that two points are “close” (saywithin δ of each other).Notice that the even more general case in which f is a function from X to Ydoes not complicate things.

First, we define distance for abstract settings; next, specialize to vector spaces.

Metric Spaces and Metrics

DefinitionA metric d on a set X is a function d : X × X → R+ satisfying

1 d(x, y) ≥ 0, with d(x, y) = 0⇔ x = y ∀x, y ∈ X .2 d(x, y) = d(y, x) ∀x, y ∈ X (symmetry).3 triangle inequality:

d(x, z) ≤ d(x, y) + d(y, z) ∀x, y, z ∈ X

A metric space (X , d) is a set X with a metric d defined on it.

The value d(x, y) is interpreted as the distance between x and y.This definition generalizes the concept of distance to abstract spaces.

Why “triangle inequality”? y↗ ↘

x → z

Notice that a metric space need not be a vector space.

Metrics: Examples

ExampleThe discrete metric on any given set X is defined as

d(x , y) ={0 if x = y1 if x 6= y

It trivially satisfies 1. and 2., and as for triangle inequality: 1 ≤ 2.

ExampleThe l1 metric (taxicab) on R2 is defined as d : R2 × R2 → R+:

d(x , y) = |x1 − y1|+ |x2 − y2|

ExampleThe l∞ metric on R2 is defined as d : R2 × R2 → R+:

d(x , y) = max {|x1 − y1| , |x2 − y2|}

Metrics and ApplicationsDefinitionIn a metric space (X , d), a subset S ⊆ X is bounded if

∃x ∈ X and β ∈ R such that ∀s ∈ S , d(s, x) ≤ β

Given a metric space, boundedness is a propety of subsets (it is relative!).

DefinitionIn a metric space (X , d), we define the diameter of a subset S ⊆ X by

diam (S) = sup{d(s, s′) : s, s′ ∈ S}

A metric can be used to define a “ball of a given radius around a point ”.

DefinitionIn a metric space (X , d), we define an open ball of center x and radius ε as

Bε(x) = {y ∈ X : d(y, x) < ε}and a closed ball of and center x and radius ε as

Bε[x] = {y ∈ X : d(y, x) ≤ ε}

Distance Between Sets

We can also define the distance from a point to a set:

d(A, x) = infa∈A

d(a, x)

and the distance between sets:

d(A,B) = infa∈A

d(B, a)

= inf{d(a,b) : a ∈ A,b ∈ B}

Note that d(A,B) is not a metric (why?).

Norms and Normed SpacesThe following defines a vector’s length (its size).

DefinitionLet V be a vector space. A norm on V is a function ‖ · ‖ : V → R+ satisfying

1 ‖x‖ ≥ 0 ∀x ∈ V .2 ‖x‖ = 0⇔ x = 0 ∀x ∈ V .3 triangle inequality: ‖x+ y‖ ≤ ‖x‖+ ‖y‖ ∀x, y ∈ V4 ‖αx‖ = |α|‖x‖ ∀α ∈ R, ∀x ∈ V

Why is this also called “triangle” inequality?

xx↗ ↘ y0 → x+ yy↘ ↗ x

y

DefinitionA normed vector space is a vector space over R equipped with a norm.

There are many different norms: if ‖ · ‖ is a norm on V , so are 2‖ · ‖ and3‖ · ‖ and k‖ · ‖ for any k > 0.

Normed Spaces

Can one relate norm and distance on a given normed vector space?In Rn , the standard norm is defined as

‖z‖ =(

n∑i=1

z2i

) 12

while the Euclidean distance between two vectors x and y

d(x, y) =√(x1 − y1)2 + (x2 − y2)2 + ...+ (xn − yn)2 =

√√√√ n∑i=1

(xi − yi )2

Norm and Euclidean distanceHere, the distance between x and y is the length of the vector given by theirdifference:

d(x, y) = ‖x− y‖

This idea generalizes.

Norm and Distance

In any normed vector space, the norm can be used to define a distance.

TheoremLet (V , ‖ · ‖) be a normed vector space. Let d : V × V ⇒ R+ be defined by

d(v,w) = ‖v −w‖Then (V , d) is a metric space.

In a normed vector space, we define a metric using the norm.

Remember that ‖x‖ : V → R+ is the lenght of a vector x ∈ V .Here x = v −w.Hence, the distance between two vectors is the length of the difference vector.

Proof that (V , d) is a Metric Space I

We must verify that the d defined satisfies the three properties of a metric.

First property: d(x, y) ≥ 0,with d(x, y) = 0⇔ x = y ∀x, y ∈ X

Proof.Let v,w ∈ V . By definition, d(v,w) = ‖v −w‖ ≥ 0. Moreover

d(v,w) = 0 ⇔ ‖v −w‖ = 0⇔ v −w = 0 by def of ‖ · ‖⇔ (v + (−w)) +w = w by vector additive inverse

⇔ v + ((−w) +w) = w by associative property

⇔ v + 0 = w by vector additive inverse

⇔ v = w

Proof that (V , d) is a Metric Space II

For symmetry, we need to show that d(x, y) = d(y, x) ∀x, y ∈ X

Proof.Let v,w ∈ V .

d(v,w) = ‖v −w‖= | − 1|‖v −w‖= ‖(−1)(v + (−w))‖ by def of ‖ · ‖= ‖(−1)v + (−1)(−w)‖= ‖ − v +w‖ by the first slide

= ‖w + (−v)‖= ‖w − v‖= d(w, v)

Proof that (V , d) is a Metric Space III

The triangle inequality is d(x, z) ≤ d(x, y) + d(y, z) ∀x, y, z ∈ X .

Proof.Let u,w, v ∈ V .

d(u,w) = ‖u−w‖= ‖u+ (−v + v)−w‖= ‖u− v + v −w‖≤ ‖u− v‖+ ‖v −w‖ by triangle for ‖ · ‖= d(u, v) + d(v,w)

SummarySince we have shown that d(·) = ‖ · ‖ satisfies the three properites of a metric onV , we conclude that (V , d) is a metric space.

Normed Vector Spaces: Examples

Rn admits many normsThe n-dimensional Euclidean space Rn supports the following norms

the standard (Euclidean) norm:

‖x‖2 =

√√√√ n∑i=1

(xi )2 =√x · x

where · (the “dot product”of vectors) is defined as:

x · y = x1y1 + x2y2 + ...+ xnyn =n∑i=1

xiyi

the L1 norm: ‖x‖1 =n∑i=1

|xi |

the Lp norm: ‖x‖∞ = (|x1|p + . . .+ |xn |p)1p

the maximum norm, or sup norm, or L∞ norm:

‖x‖∞ = max{|x1|, . . . , |xn |}

Normed Vector Spaces: Results

Theorem (Cauchy-Schwarz Inequality)

If v,w ∈ Rn , then (n∑i=1

viwi

)2≤(

n∑i=1

v 2i

)(n∑i=1

w 2i

)

This can also be written as

‖v ·w‖2 ≤ ‖v‖2‖w‖2

or‖v ·w‖ ≤ ‖v‖‖w‖

where ‖ · ‖ is the standard norm.Prove this as an exercise.

Different NormsDifferent norms on a given vector space yield different geometric properties.

In Rn , a closed ball around the origin of radius ε is:Bε [0] = {x ∈ Rn : d(x, 0) ≤ ε} = {0 ∈ X : ‖x‖ ≤ ε}

Draw a ball in R2 ({x ∈ R2 : ‖x‖ = 1}) in 3 different cases:standard norm (‖·‖2)

‖x‖2 =√x21 + x

22 = 1,

L1 norm (‖·‖1)

‖x‖1 = |x1|+ |x2| = 1,

sup-norm (‖·‖∞)

‖x‖∞ = max{|x1| , |x2|} = 1

Equivalent Norms

Do different norm imply different results when applied to ideas likeconvergence, continuity and so on? Hopefully not.

DefinitionTwo norms ‖ · ‖ and ‖ · ‖∗ on the same vector space V are said to beLipschitz-equivalent (or equivalent) if ∃m,M > 0 such that ∀x ∈ V ,

m‖x‖ ≤ ‖x‖∗ ≤ M‖x‖Equivalently, ∃m,M > 0 such that ∀x ∈ V , x 6= 0,

m ≤ ‖x‖∗

‖x‖ ≤ M

Equivalent Norms

Equivalent norms define the same notions of convergence and continuity.

For topological purposes, equivalent norms are indistinguishable.

Compare Different NormsLet ‖ · ‖ and ‖ · ‖∗ be equivalent norms on the vector space V , and fix x ∈ V .Let

Bε(x, ‖·‖) = {y ∈ V : ‖x−y‖ < ε} and Bε(x, ‖·‖∗) = {y ∈ V : ‖x−y‖∗ < ε}Then, for any ε > 0,

B εM(x, ‖ · ‖) ⊆ Bε(x, ‖ · ‖∗) ⊆ B ε

m(x, ‖ · ‖)

ε balls are ‘close’to each other.

ResultIn Rn (or any finite-dimensional normed vector space), all norms areequivalent.

Roughly, up to a difference in scaling, for topological purposes there is a uniquenorm in Rn .

Sequences of Real Numbers

DefinitionA sequence of real numbers is a function f : N −→ R.

NotationIf f (n) = an , for n ∈ N, we denote the sequence by the symbol {an}∞n=1, orsometimes by {a1, a2, a3, . . . }.The values of f , that is, the elements an , are called the terms of the sequence.

ExamplesLet an = n then {an}∞n=1 = {1, 2, 3, . . . }Let a1 = 1, an+1 = an + 2,∀ n > 1 then {an}∞n=1 = {1, 3, 5, . . . }Let an = (−1)n , n ≥ 1 then {an}∞n=1 = {−1, 1,−1, 1, . . . }

Limit of a Sequence of Real Numbers

DefinitionA sequence of real numbers {an} is said to converge to a point a ∈ R if

∀ε > 0 there exists N(ε) ∈ N such that n ≥ N ⇒ |an − a| < ε

The sequence converges to a limit point if for every strictly positive epsilon,there exists a natural number (possibly dependent on epsilon) such that allthe points in the sequence on and after that number are at most epsilon awayfrom the limit point.

NotationWhen {an} converges to a we usually write

an −→ a

orlimn→∞

an = a

Generalize to arbitrary sequences next.

Sequences

The objects along the sequence are now elements of some set.

DefinitionLet X be a set. A sequence is a function f : N −→ X .

NotationWe denote the sequence by the symbols {xn}∞n=1, or sometimes by {xn}.

When X = Rn each term in the sequence is a vector in Rn .

ConvergenceTo define convergence, X must be a metric space so that one talk aboutpoints being close to each other using the metric to measure distance.

DefinitionLet (X , d) be a metric space. A sequence {xn} converges to x ∈ X if

∀ε > 0 ∃N(ε) ∈ N such that n > N(ε)⇒ d(xn , x) < ε

NotationWhen {xn} converges to x we write

xn → x or limn→∞

xn = x

For any strictly positive epsilon, there exists a number such that all points inthe sequence from then on are at most epsilon away from the limit point.

Similar to real numbers: we replaced the standard distance with a metric.

RemarkThe limit point must be an element of X .

Uniqueness of Limits for Real Numbers

TheoremLet {an} be a sequence of real numbers that converges to two real numbers b andb′; then, b = b′.

Proof.By contradiction. Assume (wlog) that b > b′ and let

ε =12(b − b′) > 0.

By definition, there are N(ε) and N ′(ε) such that

∀ n ≥ N(ε) |b − an | < ε and ∀n ≥ N ′(ε) |b′ − an | < ε

But this is impossible since it implies that

2ε = b − b′ ≤ |b − an |+ |b′ − an | < 2εfor n > max{N(ε),N ′(ε)}.

Uniqueness of Limits

Theorem (Uniqueness of Limits)In a metric space (X , d), if xn → x and xn → x′, then x = x′.

Proof.By contradiction. Suppose {xn} is a sequence in X , xn → x, xn → x′, and x 6= x′.Since x 6= x′, d(x, x′) > 0. Let ε = d (x,x′)

2

Then, there exist N(ε) and N ′(ε) such thatn > N(ε)⇒ d(xn , x) < εn > N ′(ε)⇒ d(xn , x′) < ε

.

Choose n > max{N(ε),N ′(ε)} Thend(x, x′) ≤ d(x, xn) + d(xn , x′)

< ε+ ε

= 2ε

= d(x, x′)d(x, x′) < d(x, x′)

a contradiction.

Subsequences

If {xn} is a sequence and n1 < n2 < n3 < · · · then {xnk } is called a subsequence.

DefinitionA subsequence of the sequence described by f from N to X is the sequence givenby a composite function

f ◦ g : N −→ X

whereg : N −→ N

and g(m) > g(n) whenever m > n.

A subsequence is formed by taking some of the elements of the parentsequence in the same order.

Example

xn = 1n , so {xn} =

(1, 12 ,

13 ,

14 ,

15 ,

16 , . . .

). If nk = 2k , then {xnk } =

(12 ,

14 ,

16 , . . .

).

Monotone Sequences of Reals

DefinitionsA sequence of real numbers {xn} is increasing if xn+1 ≥ xn for all n.A sequence of real numbers {xn} is decreasing if xn+1 ≤ xn for all n.

We say strictly increasing or decreasing if the above inequalities are strict.

DefinitionA sequence of real numbers is monotone if it is either increasing or decreasing.

DefinitionIf {xn} is a sequence of real numbers, {xn} tends to infinity (written xn →∞ orlimn→∞ xn =∞) if

∀K ∈ R ∃N(K ) such that n > N(K )⇒ xn > K

Similarly define xn → −∞ or lim xn = −∞.

Facts about Convergence of Sequences of Reals

Convergence and Subsequences{an} converges to b if and only if every subsequence of {an} converges to b.

Convergence OperationsLet {an} and {bn} be sequences such that lim

n→∞an = a and lim

n→∞bn = b. Then

1 limn→∞

(an + bn) = a+ b

2 for c ∈ R, limn→∞

can = ca,

3 limn→∞

anbn = ab

4 limn→∞

(an)k = ak

5 limn→∞

1an= 1

a provided an 6= 0 (n = 1, 2, 3, . . . ), and a 6= 0

6 limn→∞

anbn= a

b provided bn 6= 0 (n = 1, 2, 3, . . . ), and b 6= 0

Boundedness for Sequences of Real Numbers

DefinitionThe sequence of real numbers {an} is bounded above if

∃m ∈ R such that an ≤ m ∀ n ∈ N

DefinitionThe sequence of real numbers {an} is bounded below if

∃m ∈ R such that an ≥ m ∀ n ∈ N

Boundedness and Convergence

TheoremIf a sequence of real numbers {an} converges then it is bounded.

Proof.Suppose an −→ a.

Then, there is an integer N such that n > N implies |an − a| < 1.Define r as follows:

r = max{1, |a1 − a|, . . . , |aN − a|}Then |an − a| ≤ r for n = 1, 2, 3, . . . .

Boundedness and Convergence

TheoremLet {an}, {bn}, and {cn} be sequences of real numbers such that

an ≤ bn ≤ cn , ∀ n ∈ Nand we have that both

an −→ a, cn −→ a

Thenbn −→ a

Proof in the Problem Set.

Monotonicity and ConvergenceTheoremA monotone sequence of real numbers is convergent if and only if it is bounded.

Proof.We have already shown that a convergent sequence is bounded (monotone or not),so we need only prove that a bounded and monotone sequence converges.Suppose {an} is bounded and increasing. Let a = sup{an : n ∈ N}.We want a to be the limit of {an}; so, we need to show that

∀ε > 0, ∃N ∈ N such that |a− an | < ε, ∀ n ≥ NFor any ε > 0, a− ε is not an upper bound of {an}; thus ∃N such thataN > a− ε .Since {an} is increasing, for every n ≥ N we have aN ≤ anPutting these observations together

a− ε < aN ≤ an ≤ awhere the last inequality follows from an ≤ a, ∀ n.Thus,

|an − a| < ε, ∀ n ≥ NProve the case in which {an} is bounded and decreasing as an exercise.

A Monotone Subsequence Always Exist

Theorem (Rising Sun Lemma)Every sequence of real numbers has a monotone subsequence.

Why the name?◦ ← ← ← ← ← ← ← ← ← ← ← ← S

• • • • ◦ ← ← ← ← ← ← ← ← U• • • • • ◦ ← ← ← N

• • ••

Proof.Exercise

Bolzano-Weierstrass Theorem

Theorem (Bolzano-Weierstrass)Every sequence of real numbers that is bounded has a convergent subsequence.

Proof.This is done by using the Rising Sun Lemma and the previous theorem:every real sequence has a monotone subsequence and

a monotone sequence is convergent if and only if it is bounded

Tomorrow

We start defining open and closed sets, then we move to functions and definelimits and continuity.

1 Open and Closed Set2 Limits of functions3 Continuity