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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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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., Bδ(a) ⊂ f−1(Bε(f (a))). )

(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 .

(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 .

(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).

(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).

(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).

(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).

(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).

(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.

(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.

(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.

(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.

(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.

(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.

(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.

(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.

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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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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).

(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

(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

(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

(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

(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

(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

(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

(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

(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).

(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).

(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).

(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).

(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).

(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).

(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).

(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).

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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)

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)

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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## Thank you.

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