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WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

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## 3.3 Continuity

Definition

Let E be a nonempty subset of R and f :ER.

(i)

f is said to be continuous at a point aE if and only if given ǫ >0 there is aδ >0 (which in general depends on ǫ, f, and a) such that

|xa| < δ and xE imply |f(x) −f(a)| < ǫ.

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## 3.3 Continuity

Definition

Let E be a nonempty subset of R and f :ER.

(i)

f is said to be continuous at a point aE if and only if given ǫ >0 there is aδ >0 (which in general depends on ǫ, f, and a) such that

|xa| < δ and xE imply |f(x) −f(a)| < ǫ.

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## 3.3 Continuity

Definition

Let E be a nonempty subset of R and f :ER.

(i)

f is said to be continuous at a point aE if and only if given ǫ >0 there is aδ >0 (which in general depends on ǫ, f, and a) such that

|xa| < δ and xE imply |f(x) −f(a)| < ǫ.

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## 3.3 Continuity

Definition

Let E be a nonempty subset of R and f :ER.

(i)

f is said to be continuous at a point aE if and only if given ǫ >0 there is aδ >0 (which in general depends on ǫ, f, and a) such that

|xa| < δ and xE imply |f(x) −f(a)| < ǫ.

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Definition (ii)

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

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Definition (ii)

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

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Definition (ii)

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

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### Remark:

Let I be an open interval that contains a point a and f :IR. Then f is continuous at aI if and only if

f(a) = lim

x→af(x).

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### Remark:

Let I be an open interval that contains a point a and f :IR. Then f is continuous at aI if and only if

f(a) = lim

x→af(x).

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### Example:

1. f(x) =x2, x ∈ [0,1].

2. f(x) =x2, xR.

3. f(x) = 1

x, x ∈ (0,1].

4. f(x) =√

x, xR+.

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Theorem

Suppose that E is a nonempty subset of R, aE, and f :ER. Then the following statements are equivalent:

(i) f is continuous at aE

(ii) If xn converges to a and xnE, then f(xn) →f(a)as n → ∞.

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Theorem

Suppose that E is a nonempty subset of R, aE, and f :ER. Then the following statements are equivalent:

(i) f is continuous at aE

(ii) If xn converges to a and xnE, then f(xn) →f(a)as n → ∞.

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Theorem

Suppose that E is a nonempty subset of R, aE, and f :ER. Then the following statements are equivalent:

(i) f is continuous at aE

(ii) If xn converges to a and xnE, then f(xn) →f(a)as n → ∞.

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Theorem

Suppose that E is a nonempty subset of R, aE, and f :ER. Then the following statements are equivalent:

(i) f is continuous at aE

(ii) If xn converges to a and xnE, then f(xn) →f(a)as n → ∞.

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Theorem

Suppose that A and B are subsets of R and that f :AR and g :BR with f(A) ⊆B.

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Theorem

Suppose that A and B are subsets of R and that f :AR and g :BR with f(A) ⊆B.

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Theorem (i)

If A:=I\ {a}, where I is a nondegenerate interval that either contains a or has a as one of its endpoint if

L:= lim

xa x∈I

f(x)

exists and belongs to B, and if g is continuous at LB, then

xlima x∈I

(g◦f)(x) =g lim

xa x∈I

f(x)

! .

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Theorem (i)

If A:=I\ {a}, where I is a nondegenerate interval that either contains a or has a as one of its endpoint if

L:= lim

xa x∈I

f(x)

exists and belongs to B, and if g is continuous at LB, then

xlima x∈I

(g◦f)(x) =g lim

xa x∈I

f(x)

! .

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Theorem (ii)

If f is continuous at aA and g is continuous at f(a) ∈B, then gf is continuous at aA.

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Theorem (ii)

If f is continuous at aA and g is continuous at f(a) ∈B, then gf is continuous at aA.

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B,f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B,f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB,it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB,it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17,gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17,gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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### Proof:

Suppose that xnI\ {a}and xna as n→ ∞. Since f(A) ⊆B, f(xn) ∈B.Also, by the Sequential

Characterization of Limits (Theorem 3.17), f(xn) →L as n → ∞. Since g is continuous at LB, it follows from Theorem 3.21 that gf(xn) :=g(f(xn)) →g(L)as n→ ∞. Hence by Theorem 3.17, gf(x) →g(L)as xa in I.

This prove (i). A similar proof establishes part (ii). 2

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Definition

Let E be a nonempty subset of R. A function f :ER is said to be bounded on E if and only if there is an MR such that|f(x)| ≤M for all xE.

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Definition

Let E be a nonempty subset of R. A function f :ER is said to be bounded on E if and only if there is an MR such that|f(x)| ≤M for all xE.

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Theorem (Extreme Value Theorem)

If I is a closed, bounded interval and f :IR is continuous on I, then f is bounded on I. Moreover, if

M =sup

x∈I

f(x) and m=inf

x∈If(x), then there exist points xm,xMI such that (6)

f(xM) =M and f(xm) =m.

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Theorem (Extreme Value Theorem)

If I is a closed, bounded interval and f :IR is continuous on I, then f is bounded on I. Moreover, if

M =sup

x∈I

f(x) and m=inf

x∈If(x), then there exist points xm,xMI such that (6)

f(xM) =M and f(xm) =m.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I.Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence,say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence,say xnka as k → ∞. Since I is closed,we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed,we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞,we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞,a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞,we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞,a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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### Proof:

Suppose first that f is not bounded on I. Then there exist xnI such that

(7) |f(xn)| > n, nN.

Since I is bounded, we know (by the Bolzano-Weierstrass Theorem) that{xn}has a convergent subsequence, say xnka as k → ∞. Since I is closed, we also know (by the Comparison Theorem) that aI. In particular, f(a) ∈R.

On the other hand, substituting nk for n in (7) and taking the limit of this inequility as k → ∞, we have|f(a)| = ∞, a contradiction. Hence, the function f is bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M,suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1

Mf(x) is continuous, hence, bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M, suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1

Mf(x) is continuous, hence, bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M,suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1 Mf(x) is continuous, hence, bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M, suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1 Mf(x) is continuous, hence,bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M, suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1 Mf(x) is continuous, hence, bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M, suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1 Mf(x) is continuous, hence,bounded on I.

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We have proved that both M and m are finite real numbers. To show that there is an xMI such that f(xM) =M, suppose to the contrary that f(x) <M for all xI. Then the function

g(x) = 1 Mf(x) is continuous, hence, bounded on I.

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI,we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI,we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence,there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence,there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

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In particular, there is a C >0 such that|g(x)| =g(x) ≤C.

It follows that

(8) f(x) ≤M− 1

c

for all xI. Taking the supremum of (8) over all xI, we obtain MM−1/C <M, a contradiction. Hence, there is an xMI, such that f(xM) =M.A similar argument proves that there is an xmI such that f(xm) = m. 2

(70)

### Lemma:

[Sign-Preserving Property].

Let f :IR where I is an open, nondegenerate interval.

If f is continuous at a point x0I and f(x0) >0, then there are positive numberǫandδ such that|xx0| < δ implies f(x) > ǫ.

(71)

### Lemma:

[Sign-Preserving Property].

Let f :IR where I is an open, nondegenerate interval.

If f is continuous at a point x0I and f(x0) >0, then there are positive numberǫandδ such that|xx0| < δ implies f(x) > ǫ.

(72)

Theorem (Intermediate Value Theorem)

Let I be a nondegenerate interval and f :IR be continuous. If a,bI with a<b, and if y0lies between f(a)and f(b), then there is an x0∈ (a,b)such that f(x0) =y0.

(73)

Theorem (Intermediate Value Theorem)

Let I be a nondegenerate interval and f :IR be continuous. If a,bI with a<b, and if y0lies between f(a)and f(b), then there is an x0∈ (a,b)such that f(x0) =y0.

(74)

### Example:

Assuming that sin x is continuous on R, prove that

f(x) = (

sin 1

x x 6=0

1 x =0

is continuous on (−∞,0)and(0, ∞), discontinuous at 0, and neither f(0+)nor f(0−)exists. (see Figure 3.1 on p.61.)

(75)

### Example:

Assuming that sin x is continuous on R, prove that

f(x) = (

sin 1

x x 6=0

1 x =0

is continuous on (−∞,0)and(0, ∞), discontinuous at 0, and neither f(0+)nor f(0−)exists. (see Figure 3.1 on p.61.)

(76)

### Example:

The Dirchlet function is defined on R by

f(x) :=

 1 xQ 0 xQ

Prove that every point xR is a point of discontinuity of f.

(Such functions are called nowhere continuous.)

(77)

### Example:

The Dirchlet function is defined on R by f(x) :=

 1 xQ 0 xQ

Prove that every point xR is a point of discontinuity of f.

(Such functions are called nowhere continuous.)

(78)

### Example:

Prove that the function

f(x) =

 1

q x = p

qQ (in reduced form) 0 xQ.

is continuous at every irrational in the interval (0,1)but discontinuous at every rational in (0,1).

(79)

### Example:

Prove that the function

f(x) =

 1

q x = p

qQ (in reduced form) 0 xQ.

is continuous at every irrational in the interval (0,1)but discontinuous at every rational in (0,1).

(80)

## Thank you.

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