## Advanced Calculus (II)

W^{EN}-C^{HING}L^{IEN}

Department of Mathematics National Cheng Kung University

2009

## Ch8: Euclidean Spaces

### 8.3: Topology Of **R**

^{n}

Definition (8.19)
Let**a ∈ R**^{n}.

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

Br(a) := {x ∈ R^{n} : kx − ak < r }.

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

**{x ∈ R**^{n} : kx − ak ≤ r }.

Definition (8.20)
Let n ∈**N.**

(i) A set V in**R**^{n} is said to be open if and only if for every
**a ∈ V there is an ε > 0 such that B**ε(a) ⊆ V .

(ii) A set E in**R**^{n}is said to be closed if and only if
E^{c} :=**R**^{n}\E is open.

Remark (8.23)

For each n ∈**N, the empty set ∅ and the whole space R**^{n}
are both open and closed.

Theorem (8.24)
Let n ∈**N.**

(i) If {V_{α}}α∈A is any collection of open subsets of**R**^{n}, then
[

α∈A

Vα

is open.

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

p

\

k =1

Vk := \

k ∈{1,2,...,p}

Vk

is open.

Theorem (8.24)
Let n ∈**N.**

(i) If {V_{α}}α∈A is any collection of open subsets of**R**^{n}, then
[

α∈A

Vα

is open.

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

p

\

k =1

Vk := \

k ∈{1,2,...,p}

Vk

is open.

Theorem (8.24)

(iii) If {Eα}_{α∈A}is any collection of closed subsets of**R**^{n},

then \

α∈A

E_{α}
is closed.

(iv) If {Ek :k = 1, 2, . . . , p} is a finite collection of closed
subsets of**R**^{n}, 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.

Theorem (8.24)

(iii) If {Eα}_{α∈A}is any collection of closed subsets of**R**^{n},

then \

α∈A

E_{α}
is closed.

(iv) If {Ek :k = 1, 2, . . . , p} is a finite collection of closed
subsets of**R**^{n}, 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.

Theorem (8.24)

(iii) If {Eα}_{α∈A}is any collection of closed subsets of**R**^{n},

then \

α∈A

E_{α}
is closed.

(iv) If {Ek :k = 1, 2, . . . , p} is a finite collection of closed
subsets of**R**^{n}, 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.

Proof.

(i)Let**x ∈**S

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

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

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

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

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

α∈AVα;i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

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

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

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

α∈AVα;i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

k =1Vk.Then**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open,soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open,soT

α∈AEα is closed.

Proof.

(i)Let**x ∈**S

**x ∈ V**α for some α ∈ A. Since
Vα is open, it follows that there is an r > 0 such that
Br(x) ⊆ V_{α}.Thus Br(x) ⊆S

α∈AVα; i.e., this union is open.

(ii) Let**x ∈**Tp

**x ∈ V**k 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

E_{α}^{c}

is open, soT

α∈AEα is closed.

Proof.

(iv) By DeMorgan’s Law and part (ii),

^{p}
[

k =1

Ek

c

=

p

\

k =1

E_{k}^{c}

is open, soSp

k =1Ek is closed.

(v) Since V \E = V ∩ E^{c} and E \V = E ∩ V^{c},the former is
open by part (ii), and the latter is closed by part (iii).

Proof.

(iv) By DeMorgan’s Law and part (ii),

^{p}
[

k =1

Ek

c

=

p

\

k =1

E_{k}^{c}

is open, soSp

k =1Ek is closed.

(v) Since V \E = V ∩ E^{c} and E \V = E ∩ V^{c},the former is
open by part (ii), and the latter is closed by part (iii).

Example (i)

\

k ∈N

(−1 k,1

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

[

k ∈N

[ 1

k + 1, k

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

Definition (8.26)
Let E ⊆**R**^{n}.

(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.

Remark (8.27)
Let U ⊆ E ⊆**R**^{n}.

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

Definition (8.28)

Let E be a subset of**R**^{n}.

(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 .

Remark (8.29)

Let E ⊆**R**^{n}.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.

Theorem (8.30)

A subset E of**R is connected if and only if E is an interval.**