Aufgaben 03 - Vollständigkeitsaxiom, Supremum, Infimum, .. PDF

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Vollständigkeitsaxiom, Supremum, Infimum, .....

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I := [a, b) ⊆ R

sup I = b

b

I

b

I

I

b

I

sup I = b k ∈ N

Ak ⊆ R sup Ak := sk

k∈N

n∈N

A :=

n S

Ak

k=1

A A Ak , k = 1, . . . , n ∞ S

Ak

k=1

A, B

A+B

R

A + B := {c ∈ R | ∃a ∈ A ∧ ∃b ∈ B : c = a + b} = {a + b | a ∈ A ∧ b ∈ B} −A := {−a | a ∈ A} sup(A + B) = sup A + sup B sup(−A) = − inf A inf(A + B) = inf A + inf B R

A, B a ≤ b a≤s≤b

a∈A

a ∈ A

b∈B (A1)

(A13)

b ∈ B (A12)

s ∈ R a)

0 0

H(a, b) ≤ G(a, b) ≤ A(a, b), a, b > 0 In = [an , bn ], n ∈ N,

I1 , I2 , . . . In+1 ⊆ In

n ∈ N,

ε>0

n∈N

|In | = bn − an < ε s∈R

I1 , I2 , . . . (A13) an := 1+ 1n n∈N In := [an , bn ]

bn := 2+ n1

In ⊆

In In In In Mk M1 := {−2, −1, 0, 1, 2, . . . , n, . . . } M2 := {n ∈ N | n

}

M3 := {1} ∪ {x ∈ R | x ≥ 2} M4 := {1} ∪ {x ∈ R | sin(πx) = 1} M5 := {x ∈ R | sin(πx) = 0} an a0 := 1, an+1 := a · an .

a

n ∈ N ∪ {0} =: N0

a 6= 0 a−n := (a−1 )n , n ∈ N. ∀a ∈ R \ {0} ∀m, n ∈ Z : an am = an+m m, n ∈ N

k ∈ N

Ak ⊆ R sup Ak := sk n T

a=0

k ∈ N

n ∈ N

B :=

Ak

k=1

B B

sup B = min{s1 , . . . , sn } Ak , k =

1, . . . , n ∞ T

Ak

k=1

A⊆R inf A−1 = (sup A)−1

sup A < 0

A−1 :=

a, b ∈ R \ {0} (an )m = an·m = (am )n , a=0∨b=0

an · bn = (a · b)n . n, m ∈ N

1

a

|a ∈ A



m, n ∈ Z...