Lecture 4

Econ 2001

2015 August 13

Lecture 4 Outline

1 Open and Closed Set2 Continuity

Announcements:- Tomorrow: first test at 3pm, in WWPH 4716. The exam will last an hour.- Tomorrow: recitation at 1pm, in WWPH 4716.

Open and Closed Sets

A set is open if at any point we can find a neighborhood of that pointcontained in the set.

DefinitionLet (X , d) be a metric space. A set A ⊂ X is open if

∀x ∈ A ∃ε > 0 such that Bε(x) ⊂ A

Remember that Bε(x) = {y ∈ X : d(y , x) < ε}... so openness depends on X .

DefinitionA set C ⊂ X is closed if X \ C is open.

Draw Pictures

Open and Closed Sets: Examples

Open Interval in R(a, b) is open in R (with the usual Euclidean metric).

Given x ∈ (a, b), a < x < b. Let

ε = min{x − a, b − x} > 0Then

y ∈ Bε(x)⇒ y ∈ (x − ε, x + ε) ⊂ (x − (x − a), x + (b − x)) = (a, b)

Hence Bε(x) ⊂ (a, b), so (a, b) is open.

ε must depend on x ; in particular, ε gets smaller as x nears the boundaries ofthe interval.

Closed Interval in R[a, b] is closed in R (with the usual Euclidean metric).

R \ [a, b] = (−∞, a) ∪ (b,∞) is the union of two open sets, which must beopen (prove this later).

Open and Closed Sets: Examples

An open ball is always an open setSuppose y ∈ Bε(x). Then d(x , y) < ε.

Letδ = ε− d(x , y) > 0

If d(z , y) < δ, then

d(z , x) ≤ d(z , y) + d(y , x)

< δ + d(x , y)

= ε− d(x , y) + d(x , y)

= ε

Hence Bδ(y) ⊂ Bε(x), so Bε(x) is open.

This is very usefuly since it holds for any X , d .

Open and Closed Sets: Examples

Openness and closedness depend on the underlying metric spaceIn the metric space X = [0, 1] (with standard metric), [0, 1] is open.

Since [0, 1] is the underlying metric space,

Bε(0) = {x ∈ X : d(x , 0) < ε} = {x ∈ [0, 1] : |x − 0| < ε} = [0, ε)

Most sets are neither open nor closed[0, 1] ∪ (2, 3) is neither open nor closed.

An open set may consist of a single pointIf X = N and d(m, n) = |m − n|, then

B1/2(1) = {m ∈ N : |m − 1| < 1/2} = {1}

Since 1 is the only element of the set {1} and B1/2(1) = {1} ⊂ {1}, the set{1} is open.

Open and Closed Sets: Examples

In any metric space (X , d), ∅ and X are both open... and closedTo see that ∅ is open, note that the statement

∀x ∈ ∅ ∃ε > 0 such that Bε(x) ⊂ ∅is vacuously true since there aren’t any x ∈ ∅ (any statement about points inthe empty set is true).

To see that X is open, note that Bε(x) = {z ∈ X : d(z , x) < ε} is triviallycontained in X .

Since ∅ is open, X is closed; since X is open, ∅ is closed.

Open and Closed Sets: Results

TheoremLet (X , d) be a metric space. Then

1 ∅ and X are both open and closed.2 The union of an arbitrary (finite, countable, or uncountable) collection ofopen sets is open.

3 The intersection of a finite collection of open sets is open.

Proof.1 Already done.2 Suppose {Aλ}λ∈Λ is a collection of open sets.

x ∈⋃λ∈Λ Aλ ⇒ ∃λ0 ∈ Λ such that x ∈ Aλ0

⇒ ∃ε > 0 such that Bε(x) ⊂ Aλ0 ⊂⋃λ∈Λ Aλ

so ∪λ∈ΛAλ is open.

Prove that the intersection of a finite collection of open sets is open.

Proof.Suppose A1, . . . ,An ⊂ X are open sets.

If x ∈ ∩ni=1Ai , thenx ∈ A1, x ∈ A2, . . . , x ∈ An

so∃ε1 > 0, . . . , εn > 0 such that Bε1(x) ⊂ A1, . . . ,Bεn (x) ⊂ An

Letε = min{ε1, . . . , εn} > 0

Then

Bε(x) ⊂ Bε1(x) ⊂ A1, . . . ,Bε(x) ⊂ Bεn (x) ⊂ An hence Bε(x) ⊂n⋂i=1

Ai

which proves that ∩ni=1Ai is open.

This needs finite intersection as the infimum of an infinite set of positivenumbers could be zero and the proof would fail.

The intersection of an infinite collection of open sets need not be open.

Interior, Closure, Exterior and BoundaryLet (X , d) be a metric space and A ⊂ X .

DefinitionThe interior of A, denoted intA, is the largest open set contained in A(alternatively, the union of all open sets contained in A).

DefinitionThe closure of A, denoted A, is the smallest closed set containing A(alternatively, the intersection of all closed sets containing A).

DefinitionThe exterior of A, denoted extA, is the largest open set contained in X \ A.

Note that extA = intX \ A.

Definition

The boundary of A, denoted ∂A is equal to (X \ A) ∩ A.

Interior, Closure, Exterior and Boundary

Let (X , d) be a metric space and A ⊂ X .

FACTSA point is interior if and only if it has an open ball that is a subset of the set

x ∈ intA ⇔ ∃ε > 0,Bε(x) ⊂ AA point is in the closure if and only if any open ball around it intersects the set

x ∈ A ⇔ ∀ε > 0,Bε(x) ∩ A 6= ∅A point is exterior if and only if an open ball around it is entirely outside theset

x ∈ extA ⇔ ∃ε > 0,Bε(x) ⊂ X \ AA point is on the boundary if any open ball around it intersects the set andintersects the outside of the set

x ∈ ∂A⇔ ∀ε > 0,Bε(x) ∩ A 6= ∅

andBε(x) ∩ X \ A 6= ∅

Interior, Closure, Exterior and Boundary

Interior, Closure, Exterior and Boundary ExampleLet A = [0, 1] ∪ (2, 3). Then

intA = (0, 1) ∪ (2, 3)

A = [0, 1] ∪ [2, 3]

extA = int (X \ A)

= int ((−∞, 0) ∪ (1, 2] ∪ [3,+∞))

= (−∞, 0) ∪ (1, 2) ∪ (3,+∞)

∂A = (X \ A) ∩ A= ((−∞, 0] ∪ [1, 2] ∪ [3,+∞)) ∩ ([0, 1] ∪ [2, 3])

= {0, 1, 2, 3}

Sequences and Closed Sets

We can characterize closedness also using sequences: a set is closed if itcontains the limit of any convergent sequence within it, and a set thatcontains the limit of any sequence within it must be closed.

TheoremA set A in a metric space (X , d) is closed if and only if

{xn} ⊂ Aand

xn → x ∈ X⇒ x ∈ A

We will prove the two directions in turn.

A set is closed if it contains the limit of any convergent sequence within it.

Proof.Let A be closed. Then X \ A is open.

Consider a convergent sequence xn → x ∈ X , with xn ∈ A for all n.We need to show that x ∈ A. Suppose not.If x 6∈ A, then x ∈ X \ A, so there is some ε > 0 such that Bε(x) ⊂ X \ A (bythe definition of open set).

Since xn → x , there exists N(ε) such that

n > N(ε) ⇒ xn ∈ Bε(x)

⇒ xn ∈ X \ A⇒ xn 6∈ A

This is a contradiction. Therefore,

{xn} ⊂ A, xn → x ∈ X ⇒ x ∈ Aas desired.

A set that contains the limit of any sequence within it must be closed

Proof.

Suppose{xn} ⊂ Aand

xn → x ∈ X⇒ x ∈ A.

We need to show that A is closed (or equivalently, that X \ A is open).Suppose not: X \ A is not open. Hence, there exists x ∈ X \ A such that forevery ε > 0,

Bε(x) 6⊆ X \ ASo there exists y ∈ Bε(x) such that y 6∈ X \ A. Then y ∈ A, hence

Bε(x)⋂A 6= ∅

Construct a sequence {xn} as follows: for each n, choosexn ∈ B 1

n(x) ∩ A

Given ε > 0, by the Archimedean Property we can find N(ε) such thatN(ε) > 1

ε , so n > N(ε)⇒ 1n <

1N(ε) < ε, therefore xn → x .

Then {xn} ⊆ A, xn → x , so x ∈ A, a contradiction. Therefore, X \ A is open,and A is closed.

Familiar (maybe) Terminology

DefinitionsLet (X , d) be a metric space and E ⊂ X .

A point x is a limit point of E if every Bε(x) contains a point y 6= x such thaty ∈ E .If x ∈ E and x is not a limit point of E , then x is called an isolated point of E .E is dense in X if every point of X is a limit point of E , or a point of E (orboth).

ResultsE is closed if every limit point of E is a point of E .

E is open if every point of E is an interior point of E .

Limits of Functions in Metric SpacesYesterday we defined the limit of a sequence, and now we extend those ideasto functions from one metric space to another.

For functions from reals to reals: f : (c , d)→ R, y is the limit of f at x0 iffor each ε > 0 there is a δ(ε) > 0 such that 0 < |x−x0| < δ(ε)⇒ |f (x)−y | < ε

We extend this to metric spaces by replacing each absolute value with ametric.

DefinitionLet (X , d) and (Y , ρ) be metric spaces with A ⊂ X , f : A→ Y , and x0 ∈ A. Wesay that f has limit y0 as x→ x0 if

∀ε > 0 ∃δ(ε) > 0 such that 0 < d(x, x0) < δ(ε)⇒ ρ(f (x), y0) < ε

NotationWhen f has limit y0 as x goes to x0 we write

f (x)→ y0 as x→ x0 or limx→x0

f (x) = y0

Notice the different metrics.

Limits of Functions and Sequences

Standard results like the uniqueness of limits theorem hold. Also, the resultswe saw on sequences about sums, multiplication by a scalar, products, anddivision extend to functions (when appropriate).

The limit of a function can also be characterized using sequences.

TheoremLet (X , d) and (Y , ρ) be metric spaces, f : X → Y , and x0 ∈ X. Then

limx→x0

f (x) = y0

if and only iffor each {xn} → x0 in (X , d) the sequence {f (xn)}

with xn 6= x0 converges to y0 in (Y , ρ)

For every sequence that converges in the domain, the corresponding sequencegiven by the function converges to in the range.

Prove these results as part of Problem Set 4.

Continuity in Metric SpacesFor functions from reals to reals: f : (a, b) −→ R is continuous at x0 meansfor each ε > 0 there is a δ(ε) > 0 such that |x−x0| < δ(ε)⇒ |f (x)−f (x0)| < ε

which is easy to generalize.

DefinitionLet (X , d) and (Y , ρ) be metric spaces. A function f : X → Y is continuous at apoint x0 ∈ X if

∀ε > 0 ∃δ(x0, ε) > 0 such that d(x, x0) < δ(x0, ε)⇒ ρ(f (x), f (x0)) < ε

Notice that different metrics are used when appropriate.Continuity at x0 requires:

f (x0) is defined;and either

x0 is an isolated point of X (i.e. ∃ε > 0 such that Bε(x0) = {x0}); orlimx→x0

f (x) exists and equals f (x0)

Definitionf is continuous if it is continuous at every element of its domain.

δ(x0, ε) means that δ can depend on x0. What if it does not? Later.

Continuity in Metric SpacesRemember, given f : X → Y and A ⊂ Y the inverse image is a subset of Xdefined as:

f −1(A) = {x ∈ X : f (x) ∈ A} ⊂ XThe inverse image is used to provide a characterization of continuousfunctions.

TheoremLet (X , d) and (Y , ρ) be metric spaces, and f : X → Y . Then

f is continuous

if and only if

f −1(A) is open in X for every A ⊂ Y such that A is open in Y

Continuity is equivalent to the fact that the inverse image of every open set inthe range is an open set in the domain.

NOTE: an equivalent statement of the theoremf is continuous if and only if f −1(C ) is closed in X for every closed C ⊂ Y .

f is continuous ⇒ f −1(A) is open in X for every A ⊂ Y such that A is open in Y .

Proof.Suppose f is continuous.Given A ⊂ Y , with A open, we must show that f −1(A) is open in X .

Suppose x0 ∈ f −1(A). Let y0 = f (x0) ∈ A.Since A is open, we can find ε > 0 such that Bε(y0) ⊂ A.Since f is continuous, there exists δ > 0 such that

d(x, x0) < δ ⇒ ρ(f (x), f (x0)) < ε

⇒ f (x) ∈ Bε(y0)⇒ f (x) ∈ A⇒ x ∈ f −1(A)

So Bδ(x0) ⊂ f −1(A), and therefore f −1(A) is open.

The inverse image of any open set in the range is an open set in the domain ⇒ fis continuous.

Proof.Suppose

f −1(A) is open in X for each A ⊂ Y such that A is open in Y

We need to show that f is continuous.

Let x0 ∈ X , with ε > 0, and let A = Bε(f (x0)).

A is an open ball, hence an open set, so f −1(A) is open in X .

x0 ∈ f −1(A), so there exists δ > 0 such that Bδ(x0) ⊂ f −1(A).

d(x, x0) < δ ⇒ x ∈ Bδ(x0)⇒ x ∈ f −1(A)

⇒ f (x) ∈ A which is nothing but Bε(f (x0))

⇒ ρ(f (x), f (x0)) < ε

Thus, we have shown that f is continuous at x0.Since x0 is an arbitrary point in X , f is continuous.

Algebra of ContinuityThe composition of continuous functions is continuous:

TheoremLet (X , dX ), (Y , dY ) and (Z , dZ ) be metric spaces. If f : X → Y and g : Y → Zare continuous, then g ◦ f : X → Z is continuous.

Proof.Suppose A ⊂ Z is open.

g is continuous, thus g−1(A) is open in Y ; f is continuous, thus f −1(g−1(A))is open in X .

I claim that f −1(g−1(A)) = (g ◦ f )−1(A)

Observex ∈ f −1(g−1(A)) ⇔ f (x) ∈ g−1(A)

⇔ g (f (x)) ∈ A⇔ (g ◦ f )(x) ∈ A⇔ x ∈ (g ◦ f )−1(A)

which establishes the claim.

This shows that (g ◦ f )−1(A) is open in X , so g ◦ f is continuous.

Uniform Continuity

The definition of continuity at a point allows for δ to depend on that point.

When it does not, the function is, for lack of a better expression, morecontinuous.

Definition (Uniform Continuity)Let (X , d) and (Y , ρ) be metric spaces and f : X → Y . We say f is uniformlycontinuous if

∀ε > 0 ∃δ(ε) > 0 such that ∀x0 ∈ X , d(x, x0) < δ(ε)⇒ ρ(f (x), f (x0)) < ε

Contrast with the earlier definition of continuity at a point x0:∀ε > 0 ∃δ(x0, ε) > 0 such that d(x, x0) < δ(x0, ε)⇒ ρ(f (x), f (x0)) < ε. Thisallows δ to depend on x0 and ε.Uniform continuity requires δ to depend only on ε. It is more restrictive thancontinunity.

Uniform Continuity: Example

A continuous function that is not uniformly continuous

Consider f : (0, 1]→ R defined as f (x) = 1x .

f is continuous (check this) but not uniformly continuous.

Fix ε > 0 and x0 ∈ (0, 1]. Let x = x01+εx0

. Clearly, x < x0. Then

1x− 1x0

> 0

|f (x)− f (x0)| =

∣∣∣∣1x − 1x0

∣∣∣∣ =1x− 1x0

=1+ εx0x0

− 1x0

=εx0x0

= ε

Thus, in the definition of continuity δ(x0, ε) must be chosen small enough sothat

δ(x0, ε) ≤ x0 −x0

1+ εx0=

ε(x0)2

1+ εx0< ε(x0)2

which converges to zero as x0 → 0.

Hence there is no δ(ε) > 0 that will work for all x0 ∈ (0, 1].

Uniform Continuity: ExampleAn f : [a, b]→ R that has a bounded derivative is uniformly continuous on [a, b].However, a function with an unbounded derivative can also be uniformlycontinuous.

A function with an unbounded derivative that is uniformly continuousConsider f : [0, 1]→ R defined as f (x) =

√x .

f is continuous (verify this) and also uniformly continuous.

Given ε > 0, let δ = ε2 (notice this does not depend on x).

Then given any x0 ∈ [0, 1], |x − x0| < δ implies (by the Fundamental Theoremof Calculus)

|f (x)− f (x0)| =

∣∣∣∣∫ x

x0

1

2√tdt

∣∣∣∣≤

∫ |x−x0|0

1

2√tdt =

√|x − x0|

<√δ =√ε2

= ε

Thus, f is uniformly continuous on [0, 1], even though f ′(x)→∞ as x → 0.

Lipschitz Continuity

DefinitionLet X ,Y be normed vector spaces. A function f : X → Y is Lipschitz on E ⊂ X if

∃K > 0 such that ‖f (x)− f (z)||Y ≤ K‖x− z‖X ∀x, z ∈ Ef is locally Lipschitz on E ⊂ X if

∀x0 ∈ E ∃ε > 0 such that f is Lipschitz on Bε(x0) ∩ E

The function is locally Lipschitz if every point in its domain has aneighborhood on which the function is Lipschitz.

RemarkLipschitz continuity is stronger than either continuity or uniform continuity:

locally Lipschitz ⇒ continuous

Lipschitz ⇒ uniformly continuous

Every C 1 function is locally Lipschitz (a function f : Rm → Rn is said to be C 1 ifall its first partial derivatives exist and are continuous).

HomeomorphismsDefinitionLet (X , d) and (Y , ρ) be metric spaces. A function f : X → Y is called ahomeomorphism if it is one-to-one, onto, continuous, and its inverse function iscontinuous.

NoteSuppose that f is a homeomorphism and U ⊂ X , and let g = f −1 : Y → X .

Theny ∈ g−1(U)⇔ g(y) ∈ U ⇔ y ∈ f (U)

and

U open in X ⇒ g−1(U) is open in (f (X ), ρ)⇒ f (U) is open in (f (X ), ρ)

Therefore (X , d) and(f (X ), ρ|f (X )

)are identical in terms of properties that

can be characterized solely in terms of open sets.

Such properties are called “topological properties.”

RemarkTopological properties are invariant under homeomorphisms.

Tomorrow

We study more properties of functions from R to R.1 Boudedness and Extreme Value Theorem2 Intermediate Value Theorem and Fixed Points3 Monotonicity4 Complete Spaces and Cauchy Sequences5 Contraction Mappings