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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch9: Convergence in R

n

### 9.1: Limits Of Sequences

Definition (9.1)

Let {xk} be a sequence points in Rn.

(i) {xk} is said to converge to some point a ∈ Rn(call the limit ofxk)if and only if for every ε > 0 there is an N ∈N such that

k ≥ N implies kxk − ak < ε.

(ii) {xk} is said to be bounded if and only if there is an M > 0 such that kxkk ≤ M for all k ∈ N.

(iii)xk ∈ Rn is said to be Cauchy if and only if for every ε >0 there is an N ∈N such that

k , m ≥ N imply kxk − xmk < ε.

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Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem, xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Theorem (9.2)

Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤√

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem, xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem, xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Theorem (9.2)

Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤√

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem,xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem, xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Theorem (9.2)

Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem,xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem, xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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Theorem (9.2)

Leta := (a(1), . . . , a(n)) and xk := (xk(1), . . . , xk(n)) belong toRn for k ∈N. Then xk → a, as k → ∞, if and only if the component sequences xk(j) → a(j), as k → ∞, for all j = 1, 2, . . . , n.

Proof.

Fix j ∈ {1, . . . , n}. By Remark 8.7, kxk(j) − a(j)k ≤ kxk − ak ≤

n max

1≤`≤n|xk(`) −a(`)|.

Hence, by the Squeeze Theorem, xk(j) → a(j) as k → ∞ for all 1 ≤ j ≤ n if and only if the real sequence

kxk − ak → 0 as k → ∞. Since kxk − ak → 0 if and only if xk → a, as k → ∞, the proof of the theorem is complete.

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For eacha ∈ Rnthere is a sequence xk ∈ Qn such that xk → a as k → ∞.

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Theorem (9.4) Let n ∈N.

(i) A sequence inRn can have at most one limit.

(ii) If {xk}k ∈Nis a sequence inRnthat converges toa and {xkj}k ∈N is any subsequence of {xk}k ∈N, thenxkj

converges toa as j → ∞.

(iii) Every convergent sequence inRnis bounded, but not conversely.

(iv) Every convergent sequence inRnis Cauchy.

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Let n ∈N.

(i) A sequence inRn can have at most one limit.

(ii) If {xk}k ∈Nis a sequence inRnthat converges toa and {xkj}k ∈N is any subsequence of {xk}k ∈N, thenxkj

converges toa as j → ∞.

(iii) Every convergent sequence inRnis bounded, but not conversely.

(iv) Every convergent sequence inRnis Cauchy.

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Theorem (9.4) Let n ∈N.

(i) A sequence inRn can have at most one limit.

(ii) If {xk}k ∈Nis a sequence inRnthat converges toa and {xkj}k ∈N is any subsequence of {xk}k ∈N, thenxkj

converges toa as j → ∞.

(iii) Every convergent sequence inRnis bounded, but not conversely.

(iv) Every convergent sequence inRnis Cauchy.

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Let n ∈N.

(i) A sequence inRn can have at most one limit.

(ii) If {xk}k ∈Nis a sequence inRnthat converges toa and {xkj}k ∈N is any subsequence of {xk}k ∈N, thenxkj

converges toa as j → ∞.

(iii) Every convergent sequence inRnis bounded, but not conversely.

(iv) Every convergent sequence inRnis Cauchy.

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Theorem (9.4)

(v) If {xk} and {yk} are convergent sequences in Rn and α ∈R, then

k →∞lim(xk +yk) = lim

k →∞xk+ lim

k →∞yk,

k →∞lim(αxk) = α lim

k →∞(xk), and

k →∞lim(xk · yk) =



k →∞lim(xk)



·



k →∞lim(yk)

 . Moreover, when n = 3,

k →∞lim(xk × yk) =



k →∞lim(xk)



×



k →∞lim(yk)

 .

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Theorem (9.5 Bolzano-Weierstrass Theorem forR ) Every bounded sequence inRnhas a convergent subsequence.

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Theorem (9.6)

A sequence {xk} in Rnis Cauchy if and only if it converges.

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Letxk ∈ Rn.Thenxk → a as k → ∞ if and only if for every open set V that containsa there is an N ∈ N such that k ≥ N impliesxk ∈ V .

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Theorem (9.8)

Let E ⊆Rn.Then E is closed if and only if E contains all its limit points; i.e.,xk ∈ E and xk → x imply x ∈ E.

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Let E be a subset ofRn.

(i) An open covering of E is a collection of sets {Vα}α∈A

such that each Vα is open and E ⊆ [

α∈A

Vα.

(ii) The set E is said to be compact if and only if every open covering of E has a finite subcovering; i.e., if {Vα}α∈A is an open covering of E , then there is a finite subset A0 of A such that

E ⊆ [

α∈A0

Vα.

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Lemma (9.10 Borel Covering Lemma)

Let E be a closed, bounded subset ofRn.If r is any function from E into (0, ∞), then there exist finitely many pointsy1, . . . ,yN ∈ E such that

E ⊆

N

[

j=1

Br (yj)(yj).

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Let E be a subset ofRn.Then E is compact if and only if E is closed and bounded.

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## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung