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WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch9: Convergence in R

### 9.3: Continuous functions

Definition (9.22)

Let E be a nonempty subset of Rn and let f : ERm. (i) f is said to be continuous at aE if and only if for every ε >0 there is aδ > 0 (which in general depends on ε, f , and a) such that

(3) kx−ak < δand xE imply kf(x) −f(a)k < ε.

(ii) f is said to be continuous on E (notation: f :ERm is continuous) if and only if f is continuous at every xE.

(3)

Let E be a nonempty subset of Rn and let f : ERm. Then f is said to be uniformly continuous on E (notation:

f :ERm is uniformly continuous) if and only if for every ε >0 there is aδ >0 such that

kx−ak < δ and x,aE imply kf(x) −f(a)k < ε.

(4)

Let E be a nonempty compact subset of Rn.If f is continuous on E, then f is uniformly continuous on E.

(5)

Suppose that f is continuous on E.Given ε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(6)

Suppose that f is continuous on E.Givenε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(7)

Suppose that f is continuous on E.Given ε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E,we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(8)

Suppose that f is continuous on E.Given ε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(9)

Suppose that f is continuous on E.Given ε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E,we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(10)

Suppose that f is continuous on E.Given ε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(11)

Suppose that f is continuous on E.Given ε >0 and aE, chooseδ(a) >0 such that

xBδ(a)(a)and xE imply kf(x) −f(a)k < ε 2. Sinceδ(a)/2 is positive for all a ∈E, we can choose finitely many points ajE and numbersδj := δ(aj)/2 such that

(4) E

[N

j=1

Bδj(aj).

Setδ :=min{δ1, . . . , δN}.

(12)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(13)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(14)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(15)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(16)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj),i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(17)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(18)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj),i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(19)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(20)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(21)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(22)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(23)

Suppose that x,aE and kx−ak < δ.By (4), x belongs to Bδj(aj)for some 1≤jN.Hence,

ka−ajk ≤ ka−xk + kxajk< δj+ δj =2δj = δ(aj), i.e., a also belongs to Bδ(aj)(aj).It follows, therefore, from the choice ofδ(aj)that

(5)

kf(x) −f(a)k ≤ kf(x) −f(aj)k+kf(aj) −f(a)k < ε 2+ε

2 = ε.

This proves that f is uniformly continuous on E.

(24)

Let n,mN and f :RnRm.Then the following three conditions are equivalent.

(i) f is continuous on Rn.

(ii) f−1(V)is open in Rnfor every open subset V of Rm. (iii) f−1(E)is closed in Rn for every closed subset E of Rm.

(25)

Let n,mN and f :RnRm.Then the following three conditions are equivalent.

(i) f is continuous on Rn.

(ii) f−1(V)is open in Rnfor every open subset V of Rm. (iii) f−1(E)is closed in Rn for every closed subset E of Rm.

(26)

Let n,mN and f :RnRm.Then the following three conditions are equivalent.

(i) f is continuous on Rn.

(ii) f−1(V)is open in Rnfor every open subset V of Rm. (iii) f−1(E)is closed in Rn for every closed subset E of Rm.

(27)

Let n,mN and f :RnRm.Then the following three conditions are equivalent.

(i) f is continuous on Rn.

(ii) f−1(V)is open in Rnfor every open subset V of Rm. (iii) f−1(E)is closed in Rn for every closed subset E of Rm.

(28)

Let n,mN, let E be open in Rn, and suppose that f :ERm. Then f is continuous on E if and only if f−1(V)is open in E for every open set V in Rm.

(29)

(i) If f(x) = x21+1 and E = (0,1], then f is continuous on R and E is bounded, but f−1(E) = (−∞, ∞)is not bounded.

(ii) If f(x) = x2 and E = (1,4), then f is continuous on R and E is connected, but f−1(E) = (−2, −1) ∪ (1,2)is not connected.

(30)

Let n,mN. If H is compact in Rnand f :HRm is continuous on H, then f(H)is compact in Rm.

(31)

Let n,mN. If E is connected in Rn and f :ERm is continuous on E, then f(E)is connected in Rm.

(32)

The graph y =f(x)of a continuous real function f on an interval[a,b]is compact and connected.

(33)

Suppose that H is a nonempty subset of Rnand

f :HR. If H is compact, and f is continuous on H, then M :=sup{f(x) :xH} and m :=inf{f(x) :xH}

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

(34)

Let n,mN. If H is a compact subset of Rnand

f :HRm is 1-1 and continuous, then f1 is continuous on 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