Advanced Calculus (II)

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Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

WEN-CHINGLIEN Advanced Calculus (II)

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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., B_δ(a) ⊂ f⁻¹(B_ε(f (a))). )

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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 .

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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 .

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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).

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Theorem (10.61)

Proof.

H ⊆ f⁻¹(f (H)) ⊆ f⁻¹

[

α∈A

Vα

= [

α∈A

f⁻¹(Vα).

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Theorem (10.61)

Proof.

H ⊆ f⁻¹(f (H))⊆ f⁻¹

[

α∈A

Vα

= [

α∈A

f⁻¹(Vα).

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Theorem (10.61)

Proof.

H ⊆ f⁻¹(f (H)) ⊆ f⁻¹

[

α∈A

Vα

= [

α∈A

f⁻¹(Vα).

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Theorem (10.61)

Proof.

H ⊆ f⁻¹(f (H)) ⊆ f⁻¹

[

α∈A

Vα

= [

α∈A

f⁻¹(Vα).

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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, . . . , α_N 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(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j) =

N

[

j=1

(Vα_j).

Therefore, f (H) is compact.

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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, . . . , α_N 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(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j) =

N

[

j=1

(Vα_j).

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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, . . . , α_N such that

H ⊆

N

[

j=1

f⁻¹(Vαj)

f (H) ⊆ f

^N [

j=1

f⁻¹(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j) =

N

[

j=1

(Vα_j).

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Proof.

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(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j)=

N

[

j=1

(Vα_j).

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Proof.

H ⊆

N

[

j=1

f⁻¹(Vαj)

f (H) ⊆ f

^N [

j=1

f⁻¹(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j) =

N

[

j=1

(Vα_j).

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Proof.

H ⊆

N

[

j=1

f⁻¹(Vαj)

f (H) ⊆ f

^N [

j=1

f⁻¹(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j)=

N

[

j=1

(Vα_j).

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Proof.

H ⊆

N

[

j=1

f⁻¹(Vαj)

f (H) ⊆ f

^N [

j=1

f⁻¹(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j) =

N

[

j=1

(Vα_j).

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Proof.

H ⊆

N

[

j=1

f⁻¹(Vαj)

f (H) ⊆ f

^N [

j=1

f⁻¹(Vα_j)

=

N

[

j=1

(f ◦ f⁻¹)(Vα_j) =

N

[

j=1

(Vα_j).

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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

connected, a contradiction.

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Theorem (10.62)

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 ).

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Theorem (10.62)

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

E = (f⁻¹(U) ∩ E ) ∪ (f⁻¹(V ) ∩ E ).

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Theorem (10.62)

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 ).

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Theorem (10.62)

Proof.

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

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Theorem (10.62)

Proof.

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

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Theorem (10.62)

Proof.

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

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Theorem (10.62)

Proof.

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

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Theorem (10.62)

Proof.

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

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Theorem (10.62)

Proof.

E = (f⁻¹(U) ∩ E ) ∪ (f⁻¹(V ) ∩ E ).

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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).

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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 (x_M). A similar argument shows that m is finite and attained on H

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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).

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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).

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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).

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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 (x_M).A similar argument shows that m is finite and attained on H

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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).

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Proof.

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

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Proof.

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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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Theorem (10.64)

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).

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Advanced Calculus (II)