# Analysis 2

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 Author: shm224 ID: 138773 Filename: Analysis 2 Updated: 2012-03-01 11:43:06 Tags: real analysis algebraic Structure Folders: Description: Algebraic Structure, Chapter 8 from An Introduction to Analysis by William Wade, 4th edition, Show Answers:

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1. Algebraic Structure
Let x = ( x, ... , x ), y = ( y, ... , y ) ∈ Rn and α ∈ R.

1) sum x + y := ( x1 + y1, x2+ y2, ... , xn + yn )

2) difference x - y := ( x1 - y1, x2 - y2, ... , xn - yn )

3) product αx := ( αx1, αx2, ... , αx3 )

4) dot product x · y : = x1y1 + x2y2 + ... + xnyn
2. Algebraic Definition 1

i) Euclidean Norm
ii) L-one-norm
iii) sup-norm
iv) distance
x ∈ Rn.

• i) Euclidean Norm of x the scalar:

• ii) L-one-norm of x is the scalar:

• iii) sup-norm of x is the scalar:

• iv) distance between two points a, b ∈ Rn is the scalar
3. Algebraic Definition

i) Euclidean Norm :
ii)
iii)

4. Algebraic Definition:

For a, b,
i) Orthogonality
ii) Parallel
i) a and b are said to be parallel if and only if there is a scalar t ∈ R s.t. a = tb

ii) a and b are said to be orthogonal if and only if a b = 0.
5. Algebraic Structure : Inequality

• Proof:
• i)

• ii)

6. Cauchy-Schwartz Inequality

prove | x ⋅ y | = || x || || y ||
• Recall:
• 1)
• 2) adding a scalar, t, to get a better estimate to (1) :

• Proof: when y = 0, it's trivial.
• If , substitute

7. Topology of Rn: Definition

i) open ball
ii) closed ball
i) For each r > 0, the open ball centered at a of radius r is the set of points

ii) For each , the closed ballat centered at a of radius r is the set of points

8. Topology of Rn: Definition

i) Open set
ii) Closed set
i) A subset V of Rn is said to be open (in Rn)

• ii) A subset V of Rn is said to be closed (in Rn)
9. Topology: Remark (8.21)

Prove:

Let . Set . If , then by the Triangle Inequality and the choice of e,

Thus, by definition, . In particular,
10. Topology: Remark (8.22)

Prove:

Let and set . Then, by definition, , so . Therefore, Ec is open.
11. Topology: Remark (8.22)

Prove:

For each n ∈ N, the empty set ∅ and the whole space Rn are both open and closed.
Since Rn = ∅c and ∅ = (Rn)c, suffices to prove that ∅ and Rn are both open.

Since the empty set contains no points, "every" point x ∈ ∅ satisfies Be(x) ⊆ ∅ (vacuously). Therefore, ∅ is open.

On the other hand, since Be(x) ⊆ Rn, ∀ x ∈ Rn and e > 0, Rn is open
12. Topology : Theorem

If
i) {Vα}α∈A any collection of open subsets of Rn
ii) {Vk : k=1, 2, ..., p} a finite collection of open subsets of Rn
iii) {Eα}α∈Aany collection of closed subsets of Rn
iv) {Ek : k = 1, 2, ... , p} a finite collection of closed subsets of Rn,
v) V is open and E is closed,

then
i) is open

• ii) is open
• #### iii)

13. Topology : Theorem

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