Advanced Calculus (I)

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Advanced Calculus (I)

WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN Advanced Calculus (I)

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2.3 Bolzano-Weierstrass Theorem

Definition

Let{xn}n∈N be a sequence of real numbers.

(i){xn}is said to be increasing (respectively, strictly increasing) if and only if x1≤x2 ≤ · · · (respectively, x1 <x2 <· · ·)

(ii){xn}is said to be decreasing (respectively, strictly decreasing) if and only if x1≥x2≥ · · · (respectively, x1 >x2 >· · ·)

(iii){xn}is said to be monotone if and only if it is either increasing or decreasing.

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2.3 Bolzano-Weierstrass Theorem

Definition

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2.3 Bolzano-Weierstrass Theorem

Definition

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2.3 Bolzano-Weierstrass Theorem

Definition

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2.3 Bolzano-Weierstrass Theorem

Definition

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Theorem (Monotone Convergence Theorem) If{xn}is increasing and bounded above, or if it is decreasing and bounded below, then{xn}has a finite limit.

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Theorem (Monotone Convergence Theorem) If{xn}is increasing and bounded above, or if it is decreasing and bounded below, then{xn}has a finite limit.

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Proof:

We shall actually prove that an increasing sequence converges to its supremum, and a decreasing sequence converges to its infimum.

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Suppose that{xn}is increasing and bounded above. By the Completeness Axiom, the supremum

a:=sup{xn:n ∈N}exists and is finite. Letǫ >0. By the Approximation Property for Supremum, choose N ∈N such that

a− ǫ <xN ≤a.

Since xN ≤xnfor n ≥N. In particular, xn↑a as n→ ∞.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a:=sup{xn:n ∈N}exists and is finite. Letǫ >0. By the Approximation Property for Supremum, choose N ∈N such that

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a:=sup{xn:n ∈N}exists and is finite. Letǫ >0. By the Approximation Property for Supremum, choose N ∈N such that

a− ǫ <xN ≤a.

Since xN ≤xnfor n ≥N. In particular, xn↑a as n→ ∞.

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Proof:

(i)

a− ǫ <xN ≤a.

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Proof:

(i)

a− ǫ <xN ≤a.

Since xN ≤xnfor n ≥N. In particular, xn↑a as n→ ∞.

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Proof:

(i)

a− ǫ <xN ≤a.

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(ii)

If{xn}is decreasing with infimum b:=inf{xn :n∈N}, then{−xn}is increasing with supremum−b (see Theorem 1.28). Hence, by part(i) and Theorem 2.12(ii)

b = −(−b) = − lim

n→∞(−xn) = lim

n→∞xn2

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(ii)

b = −(−b) = − lim

n→∞(−xn) = lim

n→∞xn2

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(ii)

b = −(−b) = − lim

n→∞(−xn) = lim

n→∞xn2

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(ii)

b = −(−b) = − lim

n→∞(−xn) = lim

n→∞xn2

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(ii)

b = −(−b) = − lim

n→∞(−xn) = lim

n→∞xn2

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Definition

A sequence of sets{In}_n∈Nis said to be nested if and only if

I1⊇I2 ⊇ · · · .

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Definition

A sequence of sets{In}_n∈Nis said to be nested if and only if

I1⊇I2 ⊇ · · · .

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Theorem (Nested Interval Property)

If{In}_n∈N is a nested sequence of nonempty closed bounded intervals, then

E = \

n∈N

In := {x :x ∈In for all n∈N}

contains at least one number. Moreover, if the lengths of these intervals satisfy|In| →0 as n → ∞, then E contains exactly one number.

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Theorem (Nested Interval Property)

If{In}_n∈N is a nested sequence of nonempty closed bounded intervals, then

E = \

n∈N

In := {x :x ∈In for all n∈N}

contains at least one number. Moreover, if the lengths of these intervals satisfy|In| →0 as n → ∞, then E contains exactly one number.

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Proof:

Let In = [an,bn]. Since{In}is nested, the real sequence {an}is increasing and bounded above by b1, and{bn}is decreasing and bounded below by a1. Thus by M.C.T, there exist a,b ∈R such that an ↑a and bn ↓b as n→ ∞.

Since an ≤bnfor all n ∈N, it also follows from the Comparison Theorem that an ≤a≤b ≤bn. Hence, a number x belongs to Infor all n ∈N if and only if a ≤x ≤b. This proves that E = [a,b].

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Proof:

Let In = [an,bn]. Since{In}is nested, the real sequence {an}is increasing and bounded above by b1,and{bn}is decreasing and bounded below by a1. Thus by M.C.T, there exist a,b ∈R such that an ↑a and bn ↓b as n→ ∞.

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Proof:

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Proof:

Let In = [an,bn]. Since{In}is nested, the real sequence {an}is increasing and bounded above by b1,and{bn}is decreasing and bounded below by a1. Thus by M.C.T, there exist a,b ∈R such that an ↑a and bn ↓b as n→ ∞.

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Proof:

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Proof:

Since an ≤bnfor all n ∈N,it also follows from the Comparison Theorem that an ≤a≤b ≤bn. Hence, a number x belongs to Infor all n ∈N if and only if a ≤x ≤b. This proves that E = [a,b].

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Proof:

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Proof:

Since an ≤bnfor all n ∈N,it also follows from the Comparison Theorem that an ≤a≤b ≤bn. Hence,a number x belongs to Infor all n ∈N if and only if a ≤x ≤b. This proves that E = [a,b].

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Proof:

Since an ≤bnfor all n ∈N, it also follows from the Comparison Theorem that an ≤a≤b ≤bn. Hence, a number x belongs to Infor all n ∈N if and only if a ≤x ≤b. This proves that E = [a,b].

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Proof:

Since an ≤bnfor all n ∈N, it also follows from the Comparison Theorem that an ≤a≤b ≤bn. Hence,a number x belongs to Infor all n ∈N if and only if a ≤x ≤b. This proves that E = [a,b].

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Proof:

Since an ≤bnfor all n ∈N, it also follows from the Comparison Theorem that an ≤a≤b ≤bn. Hence, a number x belongs to Infor all n ∈N if and only if a ≤x ≤b. This proves that E = [a,b].

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Proof:

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Suppose now that |In| →0 as n→ ∞. Then bn−an→0 as n → ∞,and we have by Theorem2.12 that b−a=0.

In particular, E = [a,a] = {a}contains exactly one number. 2

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Suppose now that |In| →0 as n→ ∞. Then bn−an→0 as n → ∞, and we have by Theorem2.12 that b−a=0.

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Suppose now that |In| →0 as n→ ∞. Then bn−an→0 as n → ∞,and we have by Theorem2.12 that b−a=0.

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Suppose now that |In| →0 as n→ ∞. Then bn−an→0 as n → ∞, and we have by Theorem2.12 that b−a=0.

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Remark:

The Nested Interval Property might not hold if“closed^′′ is omitted.

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Remark:

The Nested Interval Property might not hold if“closed^′′ is omitted.

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Remark:

The Nested Interval Property might not hold if“bounded^′′

is omitted.

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Remark:

The Nested Interval Property might not hold if“bounded^′′

is omitted.

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Theorem (Bolzano-Weierstrass Theorem)

Every bounded sequence of real numbers has a convergent subsequence.

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Theorem (Bolzano-Weierstrass Theorem)

Every bounded sequence of real numbers has a convergent subsequence.

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Advanced Calculus (I)