## Full text

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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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