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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.

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