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WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch9: Convergence in R

n

### 9.4: Compact Sets

Definition (9.35)

LetV = {Vα}α∈A be a collection of subsets of Rn, and suppose that ERn.

(i)V is said to cover E (or be a covering of E) if and only if E ⊆ [

α∈A

Vα.

(ii)V is said to be an open covering of E if and only ifV covers E and each Vα is open.

(iii) LetV be a covering of E. V is said to have a finite (respectively, countable) subcovering if and only if there is a finite (respectively, an at most countable) subset A0 of A such that{Vα}α∈A0 covers E.

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Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn.This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn, there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

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Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn. This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn, there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(5)

Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn.This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn, there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(6)

Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn. This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn, there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(7)

Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn. This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn,there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(8)

Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn. This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn, there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(9)

Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn. This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn,there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(10)

Theorem (9.39 Lindel ¨of)

Let nN and let E be a subst of Rn. If{Vα}α∈A is a collection of open sets and E ⊆ ∪α∈AVα, then there is an at most countable subset A0 of A such that

E ⊆ [

α∈A0

Vα.

Proof.

LetT be the collection of open balls with rational radii and rational centers, i.e., centers that belong to Qn. This

collection is countable. Moreover, by the proof of the Borel Covering Lemma, T ”approximates” the collection of open balls in the following sense: Given any open ball

Br(x) ⊆Rn, there is a ball Bρ(a) ∈ T such that xBρ(a) and Bρ(a) ⊆Br(x).

(11)

Proof.

To prove the theorem, let xE.By hypothesis, xVα for someα ∈A. Since Vα is open,there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(12)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A.Since Vα is open, there is a r >0 such that Br(x) ⊂Vα.SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(13)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open,there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(14)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα.SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα.The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(15)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(16)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα.The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(17)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(18)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(19)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction,

E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(20)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction, E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

(21)

Proof.

To prove the theorem, let xE. By hypothesis, xVα for someα ∈A. Since Vα is open, there is a r >0 such that Br(x) ⊂Vα. SinceT approximates open balls, we can choose a ball Bx∈ T such that xBxVα. The collectionT is countable, hence so is the subcollection

{U1,U2, . . .} := {Bx :xE}.

By the choice of the balls Bx, for each kN there is at least oneαkA such that UkVαk.

Hence, by construction, E ⊆ [

xE

Bx= [

kN

Uk ⊆ [

kN

Vαk.

Thus, set A0:= {αk :kN}.

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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

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

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