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Advanced Calculus (II)

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Advanced Calculus (II)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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Ch8: Euclidean Spaces

8.3: Topology Of R

n

Definition (8.19) Leta ∈ Rn.

(i) For each r > 0, the open ball centered ata of radius r is the set of points

Br(a) := {x ∈ Rn : kx − ak < r }.

(ii) For each r ≥ 0, the closed ball centered ata of radius r is the set of points

{x ∈ Rn : kx − ak ≤ r }.

(3)

Definition (8.20) Let n ∈N.

(i) A set V inRn is said to be open if and only if for every a ∈ V there is an ε > 0 such that Bε(a) ⊆ V .

(ii) A set E inRnis said to be closed if and only if Ec :=Rn\E is open.

(4)

Remark (8.23)

For each n ∈N, the empty set ∅ and the whole space Rn are both open and closed.

(5)

Theorem (8.24) Let n ∈N.

(i) If {Vα}α∈A is any collection of open subsets ofRn, then [

α∈A

Vα

is open.

(ii) If {Vk :k = 1, 2, . . . , p} is a finite collection of open subsets ofRn, then

p

\

k =1

Vk := \

k ∈{1,2,...,p}

Vk

is open.

(6)

Theorem (8.24) Let n ∈N.

(i) If {Vα}α∈A is any collection of open subsets ofRn, then [

α∈A

Vα

is open.

(ii) If {Vk :k = 1, 2, . . . , p} is a finite collection of open subsets ofRn, then

p

\

k =1

Vk := \

k ∈{1,2,...,p}

Vk

is open.

(7)

Theorem (8.24)

(iii) If {Eα}α∈Ais any collection of closed subsets ofRn,

then \

α∈A

Eα is closed.

(iv) If {Ek :k = 1, 2, . . . , p} is a finite collection of closed subsets ofRn, then

p

[

k =1

Ek := [

k ∈{1,2,...,p}

Ek

is closed.

(v) If V is open and E is closed, then V \E is open and E \V is closed.

(8)

Theorem (8.24)

(iii) If {Eα}α∈Ais any collection of closed subsets ofRn,

then \

α∈A

Eα is closed.

(iv) If {Ek :k = 1, 2, . . . , p} is a finite collection of closed subsets ofRn, then

p

[

k =1

Ek := [

k ∈{1,2,...,p}

Ek

is closed.

(v) If V is open and E is closed, then V \E is open and E \V is closed.

(9)

Theorem (8.24)

(iii) If {Eα}α∈Ais any collection of closed subsets ofRn,

then \

α∈A

Eα is closed.

(iv) If {Ek :k = 1, 2, . . . , p} is a finite collection of closed subsets ofRn, then

p

[

k =1

Ek := [

k ∈{1,2,...,p}

Ek

is closed.

(v) If V is open and E is closed, then V \E is open and E \V is closed.

(10)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open,it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(11)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(12)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open,it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα;i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(13)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(14)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα;i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(15)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p.Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(16)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open,it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(17)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p.Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(18)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open,it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}.Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(19)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(20)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}.Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(21)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(22)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open,soT

α∈AEα is closed.

(23)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(24)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open,soT

α∈AEα is closed.

(25)

Proof.

(i)Letx ∈S

α∈AVα. Thenx ∈ Vα for some α ∈ A. Since Vα is open, it follows that there is an r > 0 such that Br(x) ⊆ Vα.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Letx ∈Tp

k =1Vk.Thenx ∈ Vk for k = 1, 2, . . . , p. Since each Vk is open, it follows that there are numbers rk >0 such that Brk(x) ⊆ Vk.Let r = min{r1, . . . ,rp}. Then r > 0 and Br(x) ⊆ Vk for all k = 1, 2, . . . , p;i.e., Br(x) ⊆Tp

k =1Vk. Hence this intersection is open.

(iii) By DeMorgan’s Law (Theorem 1.41) and part (i),



\

α∈A

Eα

c

= [

α∈A

Eαc

is open, soT

α∈AEα is closed.

(26)

Proof.

(iv) By DeMorgan’s Law and part (ii),

 p [

k =1

Ek

c

=

p

\

k =1

Ekc

is open, soSp

k =1Ek is closed.

(v) Since V \E = V ∩ Ec and E \V = E ∩ Vc,the former is open by part (ii), and the latter is closed by part (iii).

(27)

Proof.

(iv) By DeMorgan’s Law and part (ii),

 p [

k =1

Ek

c

=

p

\

k =1

Ekc

is open, soSp

k =1Ek is closed.

(v) Since V \E = V ∩ Ec and E \V = E ∩ Vc,the former is open by part (ii), and the latter is closed by part (iii).

(28)

Example (i)

\

k ∈N

(−1 k,1

k) = {0} is closed . (ii)

[

k ∈N

[ 1

k + 1, k

k + 1] = (0, 1) is open.

(29)

Definition (8.26) Let E ⊆Rn.

(i) A set U is said to be relatively open in E if and only if there is an open set A such that U = E ∩ A.

(ii) A set C is said to be relatively closed in E if and only if there is a closed set B such that C = E ∩ B.

(30)

Remark (8.27) Let U ⊆ E ⊆Rn.

(i) Then U is relatively open in E if and only if for each a ∈ U there is an r > 0 such that Br(a) ∩ E ⊂ U.

(ii) If E is open, then U is relatively open in E if and only if U is (plain old vanilla) open (in the usual sense).

(31)

Definition (8.28)

Let E be a subset ofRn.

(i) A pair of sets U, V is said to separate E if and only if U and V are nonempty, relatively open in E ,

E = U ∪ V , and U ∩ V = ∅.

(ii) E is said to be connected if and only if E cannot be separated by any pair of relatively open sets U, V .

(32)

Remark (8.29)

Let E ⊆Rn.If there exists a pair of open sets A, B such that E ∩ A 6= ∅, E ∩ B 6= ∅, E ⊆ A ∪ B, and A ∩ B = ∅, then E is not connected.

(33)

Theorem (8.30)

A subset E ofR is connected if and only if E is an interval.

(34)

Thank you.

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