Advanced Calculus (II)
WEN-CHINGLIEN
Department of Mathematics National Cheng Kung University
2009
Ch8: Euclidean Spaces
8.3: Topology Of R
nDefinition (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 }.
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.
Remark (8.23)
For each n ∈N, the empty set ∅ and the whole space Rn are both open and closed.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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).
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).
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 ⊆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.
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).
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 .
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.
Theorem (8.30)
A subset E ofR is connected if and only if E is an interval.