Advanced Calculus (II)

(1)

Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

(2)

Ch10: Metric Spaces

10.4: Compact Sets

Definition (10.41)

Let V = {Vα}α∈A be a collection of subsets of a metric space X and suppose that E is a subset of X .

(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 if V covers E and each Vα is open.

(iii) Let V be a covering of E . V is said to have a finite (respectively, countable) subcovering if and only if there is

(3)

Definition (10.42)

A subset H of a metric space X is said to be compact if and only if every open covering of H has a finite subcover.

(4)

Remark (10.44)

A compact set is always closed.

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ . Since x does not belong to H, r (y ) > 0; hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact, we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(5)

Remark (10.44)

Proof.

Suppose that H is compact but not closed.Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ . Since x does not belong to H, r (y ) > 0; hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact, we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(6)

Remark (10.44)

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ .Since x does not belong to H, r (y ) > 0; hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact, we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(7)

Remark (10.44)

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ . Since x does not belong to H, r (y ) > 0;hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact, we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(8)

Remark (10.44)

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ .Since x does not belong to H, r (y ) > 0; hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact, we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(9)

Remark (10.44)

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ . Since x does not belong to H, r (y ) > 0;hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact, we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(10)

Remark (10.44)

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ . Since x does not belong to H, r (y ) > 0; hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact,we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(11)

Remark (10.44)

Proof.

(12)

Remark (10.44)

Proof.

Suppose that H is compact but not closed. Then H is nonempty and (by Theorem 10.16) there is a convergent sequence xk ∈ H whose limit x does not belong to H. For each y ∈ H, set r (y ) := ^{ρ(x ,y )}₂ . Since x does not belong to H, r (y ) > 0; hence, each Br (y )r (y ) is open and contains y ; i.e., {Br (y )r (y ) : y ∈ H} is an open covering of H. Since H is compact,we can choose points yj and radii is rj :=r (yj) such that {Br_j(yj) :j = 1, 2, . . . , N} covers H.

(13)

Remark (10.44)

Proof.

(14)

Proof.

Set r := min{r1, . . . ,rN}. (This is a finite set of positive numbers, so r is also positive.)Since xk → x as

k → ∞, xk ∈ Br(x ) for large k . But xk ∈ Br(x ) ∩ H implies xk ∈ B_r_j(yj)for some j ∈N.Therefore, it follows from the choices of rj and r , and from the triangle inequality, that

rj ≥ ρ(xk,yj) ≥ ρ(x , yj) − ρ(xk,x )

=2rj− ρ(x_k,x ) > 2rj− r ≥ 2r_j − r_j =rj

a contradiction.

(15)

Proof.

Set r := min{r1, . . . ,rN}. (This is a finite set of positive numbers, so r is also positive.) Since xk → x as

k → ∞, xk ∈ Br(x ) for large k .But xk ∈ Br(x ) ∩ H implies xk ∈ B_r_j(yj)for some j ∈N. Therefore, it follows from the choices of rj and r ,and from the triangle inequality, that

a contradiction.

(16)

Proof.

k → ∞, xk ∈ Br(x ) for large k . But xk ∈ Br(x ) ∩ H implies xk ∈ B_r_j(yj)for some j ∈N.Therefore, it follows from the choices of rj and r , and from the triangle inequality, that

rj ≥ ρ(xk,yj)≥ ρ(x, yj) − ρ(xk,x )

a contradiction.

(17)

Proof.

k → ∞, xk ∈ Br(x ) for large k . But xk ∈ Br(x ) ∩ H implies xk ∈ B_r_j(yj)for some j ∈N. Therefore, it follows from the choices of rj and r ,and from the triangle inequality, that

a contradiction.

(18)

Proof.

k → ∞, xk ∈ Br(x ) for large k . But xk ∈ Br(x ) ∩ H implies xk ∈ B_r_j(yj)for some j ∈N. Therefore, it follows from the choices of rj and r , and from the triangle inequality, that

rj ≥ ρ(xk,yj)≥ ρ(x, yj) − ρ(xk,x )

=2rj− ρ(x_k,x )>2rj− r ≥ 2r_j − r_j =rj

a contradiction.

(19)

Proof.

a contradiction.

(20)

Proof.

=2rj− ρ(x_k,x )>2rj− r ≥ 2r_j − r_j =rj

a contradiction.

(21)

Proof.

a contradiction.

(22)

Proof.

a contradiction.

(23)

Remark (10.45)

A closed subset of H of a compact set is compact.

(24)

Theorem (10.46)

Let H be a subset of metric space X . If H is compact, then H is closed and bounded.

Proof.

Suppose that H is compact. By Remark 10.44, H is closed. It is also bounded.Indeed, fix b ∈ X and observe that {Bn(b) : n ∈N} covers X . Since H is compact, it follows that

H ⊂

N

[

n=1

Bn(b)

for some N ∈N. Since these balls are nested, we conclude that H ⊂ BN(b); i.e., H is bounded.

(25)

Theorem (10.46)

Proof.

Suppose that H is compact.By Remark 10.44, H is closed. It is also bounded. Indeed, fix b ∈ X and observe that {Bn(b) : n ∈N} covers X .Since H is compact, it follows that

H ⊂

N

[

n=1

Bn(b)

(26)

Theorem (10.46)

Proof.

Suppose that H is compact. By Remark 10.44, H is closed. It is also bounded.Indeed, fix b ∈ X and observe that {Bn(b) : n ∈N} covers X . Since H is compact,it follows that

H ⊂

N

[

n=1

Bn(b)

(27)

Theorem (10.46)

Proof.

Suppose that H is compact. By Remark 10.44, H is closed. It is also bounded. Indeed, fix b ∈ X and observe that {Bn(b) : n ∈N} covers X .Since H is compact, it follows that

H ⊂

N

[

n=1

Bn(b)

for some N ∈N.Since these balls are nested, we conclude that H ⊂ BN(b); i.e., H is bounded.

(28)

Theorem (10.46)

Proof.

Suppose that H is compact. By Remark 10.44, H is closed. It is also bounded. Indeed, fix b ∈ X and observe that {Bn(b) : n ∈N} covers X . Since H is compact,it follows that

H ⊂

N

[

n=1

Bn(b)

(29)

Theorem (10.46)

Proof.

Suppose that H is compact. By Remark 10.44, H is closed. It is also bounded. Indeed, fix b ∈ X and observe that {Bn(b) : n ∈N} covers X . Since H is compact, it follows that

H ⊂

N

[

n=1

Bn(b)

for some N ∈N.Since these balls are nested, we conclude that H ⊂ BN(b); i.e., H is bounded.

(30)

Theorem (10.46)

Proof.

H ⊂

N

[

n=1

Bn(b)

(31)

Theorem (10.46)

Proof.

H ⊂

N

[

n=1

Bn(b)

(32)

Remark (10.47)

The converse of Theorem 10.46 is false for arbitrary metric spaces.

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R,every subset of X is bounded. Since xk → x in X implies x_k → x for large k , every subset of X is closed. Thus [0, 1] is a closed bounded subset of X . Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover, we conclude that [0, 1] is closed and bounded, but not compact.

(33)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3.Since σ(0, x ) ≤ 1 for all x ∈R, every subset of X is bounded.Since xk → x in X implies x_k → x for large k , every subset of X is closed. Thus [0, 1] is a closed bounded subset of X . Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover, we conclude that [0, 1] is closed and bounded, but not compact.

(34)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R,every subset of X is bounded. Since xk → x in X implies x_k → x for large k , every subset of X is closed.Thus [0, 1] is a closed bounded subset of X . Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover, we conclude that [0, 1] is closed and bounded, but not compact.

(35)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R, every subset of X is bounded.Since xk → x in X implies x_k → x for large k , every subset of X is closed. Thus [0, 1] is a closed bounded subset of X .Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover, we conclude that [0, 1] is closed and bounded, but not compact.

(36)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R, every subset of X is bounded. Since xk → x in X implies x_k → x for large k , every subset of X is closed.Thus [0, 1] is a closed bounded subset of X . Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover,we conclude that [0, 1] is closed and bounded, but not compact.

(37)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R, every subset of X is bounded. Since xk → x in X implies x_k → x for large k , every subset of X is closed. Thus [0, 1] is a closed bounded subset of X .Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover, we conclude that [0, 1] is closed and bounded, but not compact.

(38)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R, every subset of X is bounded. Since xk → x in X implies x_k → x for large k , every subset of X is closed. Thus [0, 1] is a closed bounded subset of X . Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover,we conclude that [0, 1] is closed and bounded, but not compact.

(39)

Remark (10.47)

Proof.

Let X =R be the discrete metric space introduced in Example 10.3. Since σ(0, x ) ≤ 1 for all x ∈R, every subset of X is bounded. Since xk → x in X implies x_k → x for large k , every subset of X is closed. Thus [0, 1] is a closed bounded subset of X . Since {x }x ∈[0,1] is an uncountable open covering of [0, 1] that has no proper finite subcover, we conclude that [0, 1] is closed and bounded, but not compact.

(40)

Definition (10.48)

A metric space X is said to be separable if and only if it contains a countable dense subset; i.e., there is a

countable set Z of X such that for every point a ∈ X there is a sequence xk ∈ Z such that xk → a as x → ∞.

(41)

Theorem (10.49 Lindel ¨of)

Let E be a subset of separable metric space X . If {Vα}_α is a collection of open sets and E ⊆ ∪α∈AVα, then there is a countable subset A0of A such that

E ⊆ [

α∈A₀

Vα.

(42)

Theorem (10.50 Heine-Borel)

Let X be a separable metric space that satisfies the Bolzano-Weierstrass Property and H be a subset of X . Then H is compact if and only if it is closed and bounded.

(43)

Definition (10.51)

Let X be a metric space, E be a nonempty subset of X , and f : E → Y . Then f is said to be uniformly continuous on E (notation: f : E → Y is uniformly continuous) if and only if given ε > 0 there is a δ > 0 such that

ρ(x , a) < δ and x , a ∈ E imply τ (f (x ), f (a)) < ε.

(44)

Theorem (10.52)

Suppose that E is a compact subset of X and f : X → Y . Then F is uniformly continuous on E if and only if f is continuous on E .

(45)

Advanced Calculus (II)