Analysis 2

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