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Advanced Calculus Lecture 9

Jeffrey M. Lee

Texas Tech University

Spring Semester 2013

(Institute) Calculus on Rn Spring Semester 2013 1 / 21

Euclidean Coordinate Space

Rn is the vector space of n-tuples of real numbers. We write them ascolumns only when doing matrix algebra.

x = (x1, x2, . . . , xn) ∈ Rn

r(a1, a2, . . . , an) + s(b1, b2, . . . , bn) =(ra1 + sb1, ra2 + sb2, , ...., ran + sbn

)e1, e2, . . . , en give the standard basis where ei = (0, 0, ..., 0, 1, 0, ..., 0)with the 1 in the i-th position.

Rn is a normed vector space:

|x | =(

∑i=1x2i

Euclidean Norm

x = (x1, x2, . . . , xn) ∈ Rn

|x | =(

∑i=1x2i

Euclidean Norm

x = (x1, x2, . . . , xn) ∈ Rn

e1, e2, . . . , en give the standard basis where ei = (0, 0, ..., 0, 1, 0, ..., 0)with the 1 in the i-th position.

|x | =(

∑i=1x2i

Euclidean Norm

x = (x1, x2, . . . , xn) ∈ Rn

|x | =(

∑i=1x2i

Euclidean Norm

x = (x1, x2, . . . , xn) ∈ Rn

|x | =(

∑i=1x2i

Euclidean Norm

A Norm on a vector space V are required to satisfy

1 ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 02 ‖rx‖ = |r | ‖x‖ for all x ∈ V3 ‖x + y‖ ≤ ‖x‖+ ‖y‖

Two norms ‖·‖1 and ‖·‖2 are equivalent if there exists positivenumbers c and C such that c ‖·‖1 ≤ ‖·‖2 ≤ C ‖·‖1 .

All norms on Rn are equivalent. Another good choice is|x | = max{|x1| , ..., |xn |}

For another example of a normed vector space take C ([a, b]) with‖f ‖∞ = sup |f |Yet another example of a normed vector space is C ([a, b]) with‖f ‖∞ = sup |f |

A Norm on a vector space V are required to satisfy1 ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 0

2 ‖rx‖ = |r | ‖x‖ for all x ∈ V3 ‖x + y‖ ≤ ‖x‖+ ‖y‖

A Norm on a vector space V are required to satisfy1 ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 02 ‖rx‖ = |r | ‖x‖ for all x ∈ V

3 ‖x + y‖ ≤ ‖x‖+ ‖y‖

A Norm on a vector space V are required to satisfy1 ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 02 ‖rx‖ = |r | ‖x‖ for all x ∈ V3 ‖x + y‖ ≤ ‖x‖+ ‖y‖

For another example of a normed vector space take C ([a, b]) with‖f ‖∞ = sup |f |

Yet another example of a normed vector space is C ([a, b]) with‖f ‖∞ = sup |f |

On Rn we have the standard inner product known as the “dotproduct"

〈x , y〉 :=n

∑i=1xiyi

An inner product on a real vector space is required to satisfy

1 〈x , x〉 ≥ 0 and 〈x , x〉 = 0 if and only if x = 0 (∀x)2 〈x , y〉 = 〈y , x〉3 〈ru + sv , y〉 = 〈ru, y〉+ 〈sv , y〉 for all vectors u, v , y and all scalars r , sand similarly for the second slot.

If we have an inner product then we automatically get a norm:‖x‖ = 〈x , x〉1/2

〈x , y〉 :=n

∑i=1xiyi

An inner product on a real vector space is required to satisfy

1 〈x , x〉 ≥ 0 and 〈x , x〉 = 0 if and only if x = 0 (∀x)2 〈x , y〉 = 〈y , x〉3 〈ru + sv , y〉 = 〈ru, y〉+ 〈sv , y〉 for all vectors u, v , y and all scalars r , sand similarly for the second slot.

〈x , y〉 :=n

∑i=1xiyi

An inner product on a real vector space is required to satisfy1 〈x , x〉 ≥ 0 and 〈x , x〉 = 0 if and only if x = 0 (∀x)

2 〈x , y〉 = 〈y , x〉3 〈ru + sv , y〉 = 〈ru, y〉+ 〈sv , y〉 for all vectors u, v , y and all scalars r , sand similarly for the second slot.

〈x , y〉 :=n

∑i=1xiyi

An inner product on a real vector space is required to satisfy1 〈x , x〉 ≥ 0 and 〈x , x〉 = 0 if and only if x = 0 (∀x)2 〈x , y〉 = 〈y , x〉

3 〈ru + sv , y〉 = 〈ru, y〉+ 〈sv , y〉 for all vectors u, v , y and all scalars r , sand similarly for the second slot.

〈x , y〉 :=n

∑i=1xiyi

An inner product on a real vector space is required to satisfy1 〈x , x〉 ≥ 0 and 〈x , x〉 = 0 if and only if x = 0 (∀x)2 〈x , y〉 = 〈y , x〉3 〈ru + sv , y〉 = 〈ru, y〉+ 〈sv , y〉 for all vectors u, v , y and all scalars r , sand similarly for the second slot.

Schwartz Inequality

TheoremFor any inner product we have

|〈x , y〉| ≤ ‖x‖ ‖y‖

with equality if and only if x is a multiple of y or visa versa.

Proof.

We may assume that y is not zero. If we let z = x − 〈x , y〉y/ ‖y‖2 then〈y , z〉 = 0 and we can write x = 〈x ,y 〉

〈y ,y 〉y + z .

‖x‖2 =

∥∥∥∥ 〈x , y〉〈y , y〉y + z∥∥∥∥2 = |〈x , y〉|2|〈y , y〉|2

‖y‖2 + ‖z‖2 ≥ |〈x , y〉|2

|〈y , y〉|2‖y‖2

= |〈x , y〉|2 / ‖y‖2

and the result follows by multiplying by ‖y‖2.(Institute) Calculus on Rn Spring Semester 2013 5 / 21

Triangle inequality

Also, if equality holds it can only be because z is zero and that would forcex = 〈x , y〉y/ ‖y‖2

TheoremFor any inner product we have

‖x + y‖ ≤ ‖x‖+ ‖y‖

Proof.

‖x + y‖2 = ‖x‖2 + 2〈x , y〉+ ‖y‖2 ≤ ‖x‖2 + 2 ‖x‖ ‖y‖+ ‖y‖2 =(‖x‖+ ‖y‖)2

Distance and Topology

If we use one of our two norms on Rn to define distance then weobtain a metric space.

For d(x , y) := ‖x − y‖ we have

1 d(x , y) ≥ 0 for all x and d(x , y) = 0 iff x = y ,2 d(x , y) = d(y , x)3 d(x , z) = d(x , y) + d(y , z)

Set of the form B(x0, r) := {x ∈ Rn : d(x0, x) < r} are called openballs(radius r and center x0).

For d(x , y) := ‖x − y‖ we have

1 d(x , y) ≥ 0 for all x and d(x , y) = 0 iff x = y ,2 d(x , y) = d(y , x)3 d(x , z) = d(x , y) + d(y , z)

For d(x , y) := ‖x − y‖ we have1 d(x , y) ≥ 0 for all x and d(x , y) = 0 iff x = y ,

2 d(x , y) = d(y , x)3 d(x , z) = d(x , y) + d(y , z)

For d(x , y) := ‖x − y‖ we have1 d(x , y) ≥ 0 for all x and d(x , y) = 0 iff x = y ,2 d(x , y) = d(y , x)

3 d(x , z) = d(x , y) + d(y , z)

For d(x , y) := ‖x − y‖ we have1 d(x , y) ≥ 0 for all x and d(x , y) = 0 iff x = y ,2 d(x , y) = d(y , x)3 d(x , z) = d(x , y) + d(y , z)

Interior points and Open Sets

DefinitionLet us be given a subset A ⊆ Rn. A point x ∈ A is called an interiorpoint of A if there exists an open ball with center x that is contained in A.

Interior points and Open Sets

We can also base things on cubes which just means we are using ball asseen by our max norm |x | = max{|x1| , ..., |xn |}:

Interior Point

Boundary Point

Exterior Point

Open Sets, Closed Sets

DefinitionWith respect to the norm on Rn, a set is called open if and only if each ofits point are interior points. A set is called closed if it is the complementof an open set.

Let’s go to the board and explore some examples and further concepts

Open ball, closed ball

Open rectangle, closed rectangle

[a, b]× [c, d ]× · · · (notation)Interior, exterior, boundary, cluster

Set theoretic behavior of open sets of

The family T of all open set as defined above has the followingproperties:

1 Both ∅ and Rn are in T (That is,they are both open)2 If U and V are in T then U ∩ V ∈ T(The intersection of two open sets is an open set)

3 The union of any family of members of T is in T(A union of open sets is open)

Number 2 above generalizes to any finite family of open sets.

Notice carefully the⋂n>0(1/n, 1) is not open in R.

Can we make an abstraction? (Topological Space)

Activity. Discussion Possible Examples

1 Both ∅ and Rn are in T (That is,they are both open)

2 If U and V are in T then U ∩ V ∈ T(The intersection of two open sets is an open set)

DefinitionA topological space is then a set X together with a collection T ofsubsets of X , called open sets and satisfying the following axioms:

1.The empty set and X itself are open.2. Any union of open sets is open.3. The intersection of any finite number of open sets is open.

The collection T of open sets is then also called a topology on X, or,if more precision is needed, an open set topology. The sets in T arecalled the open sets, and their complements in X are called closedsets.

A subset of X may be neither closed nor open, either closed or open,or both. A set that is both closed and open is called a clopen set.

DefinitionA topological space is then a set X together with a collection T ofsubsets of X , called open sets and satisfying the following axioms:

1.The empty set and X itself are open.2. Any union of open sets is open.3. The intersection of any finite number of open sets is open.

The collection T of open sets is then also called a topology on X, or,if more precision is needed, an open set topology. The sets in T arecalled the open sets, and their complements in X are called closedsets.

A subset of X may be neither closed nor open, either closed or open,or both. A set that is both closed and open is called a clopen set.

Open covers and Compactness

DefinitionsLet A be a subset of Rn. A family of open sets {Uα} is called an opencover of A if A ⊆ ⋃Uα.

Example

The family {Un} where Un = (1/n, 1) is an open cover of the openinterval (0, 1). Notice that no finite subfamily will still cover the set.

A subfamily of a cover of a set is called a subcover if the union stillcontains the set.

DefinitionA subset A of Rn is called compact if ever open cover of A has a finitesubcover.

Let’s get in groups and see if we can come up with a proof that thewhole real line is not compact.

Example

Theorem (Heine-Borel)A closed and bounded interval is compact.

Compactness of Products

Suppose that B ⊆ Rm is compact. Let x ∈ Rn and consider the set{x} × B ⊆ Rn ×Rm .

Is {x} × B compact in Rn+m?Note that if W is a open in Rn ×Rm , and (x , y) ∈ W then there areopen balls U x ⊆ Rn and Vy ⊆ Rm with x ∈ U x , y ∈ Vy andU x × Vy ⊆ W

Is {x} × B compact in Rn+m?

Note that if W is a open in Rn ×Rm , and (x , y) ∈ W then there areopen balls U x ⊆ Rn and Vy ⊆ Rm with x ∈ U x , y ∈ Vy andU x × Vy ⊆ W

Is {x} × B compact in Rn+m?Note that if W is a open in Rn ×Rm , and (x , y) ∈ W then there areopen balls U x ⊆ Rn and Vy ⊆ Rm with x ∈ U x , y ∈ Vy andU x × Vy ⊆ W

TheoremIf B ⊆ Rm is compact and {Wα} is an open cover of {x} × B ⊆ Rm+n

then there is an open set U ⊆ Rn so that U × B is covered by an a finitesubcover of {Wα}

Proof.Since {x} × B is compact we can immediately assume that {Wα} is afinite family. Our task is simply to find U so that U × B is covered by{Wα}. For each y ∈ B the point (x , y) is in some W from the cover andsince W is open there is an open set Uy × Vy ⊆ W with (x , y) ∈ Uy × Vyand Uy and Vy open. The family {Vy } covers B and so there is a finitenumber Vy1 ,Vy2 , ...,Vyk that also cover B. Each Uyi × Vyi is in some Wwhich we rename Wi . In any case, we let U = Uy1 ∩ · · ·Uyk . Now let’slook at the picture so far....

TheoremIf B ⊆ Rm is compact and {Wα} is an open cover of {x} × B ⊆ Rm+n

then there is an open set U ⊆ Rn so that U × B is covered by an a finitesubcover of {Wα}

Proof.Since {x} × B is compact we can immediately assume that {Wα} is afinite family. Our task is simply to find U so that U × B is covered by{Wα}. For each y ∈ B the point (x , y) is in some W from the cover andsince W is open there is an open set Uy × Vy ⊆ W with (x , y) ∈ Uy × Vyand Uy and Vy open. The family {Vy } covers B and so there is a finitenumber Vy1 ,Vy2 , ...,Vyk that also cover B. Each Uyi × Vyi is in some Wwhich we rename Wi . In any case, we let U = Uy1 ∩ · · ·Uyk . Now let’slook at the picture so far....

If (a, b) ∈ U × B then b ∈ Vyi for some yi . Also, a ∈ Uyi since a ∈ U.Thus (a, b) ∈ Uyi × Vyi ⊆ Wi . Done! Look at next picture......

(a, b) ∈ Uyi × Vyi ⊆ Wi . We ended up with U × B being covered byW1, ....,Wk .

Main Theorem on compactness of products of compacta

TheoremIf A is compact in Rn and B is compact in Rm then A× B is compact inRn+m

Proof: If {Wα} is an open cover of A× B then for eaxh x it is also acover of {x} ×B. But from above we can find an open Ux containingx such that Ux × B is covered by a finite number of the Wα’s. Butsince A itself is compact, a finite number of Ux cover A . So let’s saythat Ux1 , ...,Uxl covers A. Then Ux1 × B, ...,Uxl × B covers A× Band since each of these is covered by a finite number of W ′s we seethat A× B itself is also covered by a finite number of the W ′s.It follows by a simple induction that any closed rectangle in RN iscompact.

Proof: If {Wα} is an open cover of A× B then for eaxh x it is also acover of {x} ×B. But from above we can find an open Ux containingx such that Ux × B is covered by a finite number of the Wα’s. Butsince A itself is compact, a finite number of Ux cover A . So let’s saythat Ux1 , ...,Uxl covers A. Then Ux1 × B, ...,Uxl × B covers A× Band since each of these is covered by a finite number of W ′s we seethat A× B itself is also covered by a finite number of the W ′s.

It follows by a simple induction that any closed rectangle in RN iscompact.

Proof: If {Wα} is an open cover of A× B then for eaxh x it is also acover of {x} ×B. But from above we can find an open Ux containingx such that Ux × B is covered by a finite number of the Wα’s. Butsince A itself is compact, a finite number of Ux cover A . So let’s saythat Ux1 , ...,Uxl covers A. Then Ux1 × B, ...,Uxl × B covers A× Band since each of these is covered by a finite number of W ′s we seethat A× B itself is also covered by a finite number of the W ′s.It follows by a simple induction that any closed rectangle in RN iscompact.

TheoremAny closed and bounded subset of Rn is compact.

Proof. Let {Wα} be an open cover of the closed and bounded set K .Since K is bounded it is inside some closed rectangle E . Then thebigger family {Wα} ∪ {Rn\E} is an open cover of E and so there isa finite subfamily that covers E and hence K . But Rn\K is clearlynot needed so actually a finite subfamily of the W ′s still coveres K .

TheoremAny closed and bounded subset of Rn is compact.

Proof. Let {Wα} be an open cover of the closed and bounded set K .Since K is bounded it is inside some closed rectangle E . Then thebigger family {Wα} ∪ {Rn\E} is an open cover of E and so there isa finite subfamily that covers E and hence K . But Rn\K is clearlynot needed so actually a finite subfamily of the W ′s still coveres K .

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