• 沒有找到結果。

# Advanced Calculus (II)

N/A
N/A
Protected

Share "Advanced Calculus (II)"

Copied!
48
0
0

(1)

## Advanced Calculus (II)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

WEN-CHINGLIEN Advanced Calculus (II)

(2)

## Ch10: Metric Spaces

### 10.6: Continuous Functions

Recall:

Given (X , ρ), (Y , τ ), a function f : X → Y is continuous.

⇔ Given a ∈ X , and ε > 0, there is a δ > 0 such that ρ(x , a) < δ implies

τ (f (x ), f (a)) < ε.

(i.e., Bδ(a) ⊂ f−1(Bε(f (a))). )

WEN-CHINGLIEN Advanced Calculus (II)

(3)

Theorem (10.58)

Let X and Y be metric spaces, and let f : X → Y . Then f is continuous if and only if f−1(V ) is open in X for every open V in Y .

WEN-CHINGLIEN Advanced Calculus (II)

(4)

Corollary (10.59)

Let E ⊆ X and f : E → Y . Then f is continuous on E if and only if f−1(V ) ∩ E is relatively open in E for all open sets V in Y .

WEN-CHINGLIEN Advanced Calculus (II)

(5)

Theorem (10.61)

If H is compact in X and f : H → Y is continuous on H, then f (H) is compact in Y .

Proof.

Suppose that {Vα}α∈Ais an open covering of f (H). By Theorem 1.43,

H ⊆ f−1(f (H))⊆ f−1

 [

α∈A

Vα



= [

α∈A

f−1(Vα).

WEN-CHINGLIEN Advanced Calculus (II)

(6)

Theorem (10.61)

If H is compact in X and f : H → Y is continuous on H, then f (H) is compact in Y .

Proof.

Suppose that {Vα}α∈Ais an open covering of f (H). By Theorem 1.43,

H ⊆ f−1(f (H)) ⊆ f−1

 [

α∈A

Vα



= [

α∈A

f−1(Vα).

WEN-CHINGLIEN Advanced Calculus (II)

(7)

Theorem (10.61)

If H is compact in X and f : H → Y is continuous on H, then f (H) is compact in Y .

Proof.

Suppose that {Vα}α∈Ais an open covering of f (H). By Theorem 1.43,

H ⊆ f−1(f (H))⊆ f−1

 [

α∈A

Vα



= [

α∈A

f−1(Vα).

WEN-CHINGLIEN Advanced Calculus (II)

(8)

Theorem (10.61)

If H is compact in X and f : H → Y is continuous on H, then f (H) is compact in Y .

Proof.

Suppose that {Vα}α∈Ais an open covering of f (H). By Theorem 1.43,

H ⊆ f−1(f (H)) ⊆ f−1

 [

α∈A

Vα



= [

α∈A

f−1(Vα).

WEN-CHINGLIEN Advanced Calculus (II)

(9)

Theorem (10.61)

If H is compact in X and f : H → Y is continuous on H, then f (H) is compact in Y .

Proof.

Suppose that {Vα}α∈Ais an open covering of f (H). By Theorem 1.43,

H ⊆ f−1(f (H)) ⊆ f−1

 [

α∈A

Vα



= [

α∈A

f−1(Vα).

WEN-CHINGLIEN Advanced Calculus (II)

(10)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H.since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316). It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj) =

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(11)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact,there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316).It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj) =

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(12)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316). It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj) =

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(13)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316).It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj)=

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(14)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316). It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj) =

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(15)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316). It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj)=

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(16)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316). It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj) =

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(17)

Proof.

Hence, by Corollary 10.59, {f−1(Vα)}α∈Ais a covering of H whose sets are all relatively open in H. since H is compact, there are indices α1, α2, . . . , αN such that

H ⊆

N

[

j=1

f−1(Vαj)

(see Exercise 7, p.316). It follows from Theorem 1.43 that

f (H) ⊆ f

 N [

j=1

f−1(Vαj)



=

N

[

j=1

(f ◦ f−1)(Vαj) =

N

[

j=1

(Vαj).

Therefore, f (H) is compact.

WEN-CHINGLIEN Advanced Calculus (II)

(18)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ).By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(19)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected.By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E .Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(20)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ).By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(21)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E .Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(22)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅,we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(23)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅.Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(24)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅,we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E .Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(25)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅.Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(26)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E .Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(27)

Theorem (10.62)

If E is connected in X and f : E → Y is continuous on E , then f (E ) is connected in Y .

Proof.

Suppose that f (E ) is not connected. By Definition 10.53, there exist a pair U, V ⊂ Y of relatively open sets in f (E ) that separate f (E ). By Exercise 4, f−1(U) ∩ E and

f−1(V ) ∩ E are relatively open in E . Since f (E ) = U ∪ V , we have

E = (f−1(U) ∩ E ) ∪ (f−1(V ) ∩ E ).

Since U ∩ V = ∅, we also have f−1(U) ∩ f−1(V ) = ∅. Thus f−1(U) ∩ E , f−1(V ) ∩ E is a pair of relatively open sets that separate E . Hence, by Definition 10.53, E is not

WEN-CHINGLIEN Advanced Calculus (II)

(28)

Theorem (10.63 Extreme Value Theorem)

Let H be a nonempty, compact set in a metric space X and suppose that f : H →R is continuous. Then

M := sup{f (x ) : x ∈ H} m := inf{f (x ) : x ∈ H}

are finite real numbers and there exist points xM,xm ∈ H such that M = f (xM)and m = f (xm).

WEN-CHINGLIEN Advanced Calculus (II)

(29)

Proof.

By symmetry, it suffices to prove the result for M.Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded.Since f (H) is bounded, M is finite. By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞. Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM). A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(30)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact.Hence, by Theorem 10.46, f (H) is closed and bounded. Since f (H) is bounded, M is finite.By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞. Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM). A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(31)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded.Since f (H) is bounded, M is finite. By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞.Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM). A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(32)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded. Since f (H) is bounded, M is finite.By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞. Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM). A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(33)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded. Since f (H) is bounded, M is finite. By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞.Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM).A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(34)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded. Since f (H) is bounded, M is finite. By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞. Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM). A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(35)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded. Since f (H) is bounded, M is finite. By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞. Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM).A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(36)

Proof.

By symmetry, it suffices to prove the result for M. Since H is compact, f (H) is compact. Hence, by Theorem 10.46, f (H) is closed and bounded. Since f (H) is bounded, M is finite. By the Approximation Property, choose xk ∈ H such that f (xk) →M as k → ∞. Since f (H) is closed, M ∈ f (H).

Therefore, there is an xM ∈ H such that M = f (xM). A similar argument shows that m is finite and attained on H

WEN-CHINGLIEN Advanced Calculus (II)

(37)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X .Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(38)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H).Let E be closed in X . Then E ∩ H is a closed subset of H,so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(39)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X .Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact.Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(40)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H,so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed.Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(41)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact.Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33).Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(42)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed.Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed,it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(43)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33).Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H).Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(44)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed,it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(45)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H).Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(46)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(47)

Theorem (10.64)

Let X and Y be metric spaces. If H is a compact subset of X and f : H → Y is 1-1 and continuous, then f−1is continuous on f (H).

Proof.

By Exercise 4a, it suffices to show that (f−1)−1takes closed sets in X to relatively closed sets in f (H). Let E be closed in X . Then E ∩ H is a closed subset of H, so by Remark 10.45, E ∩ H is compact. Hence, by Theorem 10.61, f (E ∩ H) is compact, in particular closed. Since f is 1-1, f (E ∩ H) = f (E ) ∩ f (H) (see Exercise 6, p.33). Since f (E ∩ H) and f (H) are closed, it follows that f (E ) ∩ f (H) is relatively closed in f (H). Since (f−1)−1=f , we conclude that (f−1)−1(E ) ∩ f (H) is relatively closed in f (H).

WEN-CHINGLIEN Advanced Calculus (II)

(48)

## Thank you.

WEN-CHINGLIEN Advanced Calculus (II)

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