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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch10: Metric Spaces

### 10.2: Limit Of Functions

Definition (10.22)

A point a is said to be a cluster point (of X ) if and only if Bδ(a) contains infinitely many points for each δ > 0.

Notation:

E is a subspace of X .

BrE(a) := {x ∈ E : ρ(x , a) < r }.(relative balls in E )

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## Ch10: Metric Spaces

### 10.2: Limit Of Functions

Definition (10.22)

A point a is said to be a cluster point (of X ) if and only if Bδ(a) contains infinitely many points for each δ > 0.

Notation:

E is a subspace of X .

BrE(a) := {x ∈ E : ρ(x , a) < r }.(relative balls in E )

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Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0. Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b). But (a, b) contains infinitely many points, e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

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Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0.Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b).But (a, b) contains infinitely many points, e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

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Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0. Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b). But (a, b) contains infinitely many points,e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

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Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0. Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b).But (a, b) contains infinitely many points, e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

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Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0. Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b). But (a, b) contains infinitely many points,e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

(9)

Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0. Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b). But (a, b) contains infinitely many points, e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

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Example (10.24)

Show that every point in the interval [0,1] is a cluster point of the open interval (0,1).

Proof.

Let x0 ∈ [0, 1] and δ > 0. Then x0+ δ >0 and x0− δ < 1.

In particular, (x0− δ, x0+ δ) ∩ (0, 1) is itself a

nondegenerate interval, say (a, b). But (a, b) contains infinitely many points, e.g., (a + b)/2, (2a + b)/3, (3a + b)/4, . . .. Therefore, x0 is a cluster point of (0, 1).

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Definition (10.25)

Given (X , ρ), (Y , τ ), let a be a cluster point of X and f : X \{a} → Y . Then f (x ) is said to

converge to L, as x approaches a, if and only if for every ε >0 there is a δ > 0 such that

(1) 0 < ρ(x , a) < δ implies τ (f (x ), L) < ε.

In this case we write

L = lim

x →af (x )

and call L the limit of f (x ) as x approaches a.

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Theorem (10.26)

Let a be a cluster point of X and f , g : X \{a} → Y . (i) If f (x ) = g(x ) for all x ∈ X \{a} and f (x ) has a limit as x → a, then g(x ) also has a limit as x → a and

x →alimg(x ) = lim

x →af (x ).

(ii) [Sequential Characterizations of Limits]. The limit L := lim

x →af (x )

exists if and only if f (xn) →L as n → ∞ for every sequence xn ∈ X \{a} that converges to a as n → ∞.

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Theorem (10.26)

Let a be a cluster point of X and f , g : X \{a} → Y . (i) If f (x ) = g(x ) for all x ∈ X \{a} and f (x ) has a limit as x → a, then g(x ) also has a limit as x → a and

x →alimg(x ) = lim

x →af (x ).

(ii) [Sequential Characterizations of Limits]. The limit L := lim

x →af (x )

exists if and only if f (xn) →L as n → ∞ for every sequence xn ∈ X \{a} that converges to a as n → ∞.

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Theorem (10.26)

Let a be a cluster point of X and f , g : X \{a} → Y . (i) If f (x ) = g(x ) for all x ∈ X \{a} and f (x ) has a limit as x → a, then g(x ) also has a limit as x → a and

x →alimg(x ) = lim

x →af (x ).

(ii) [Sequential Characterizations of Limits]. The limit L := lim

x →af (x )

exists if and only if f (xn) →L as n → ∞ for every sequence xn ∈ X \{a} that converges to a as n → ∞.

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Theorem (10.26)

(iii) Suppose that Y =Rn. If f (x ) and g(x ) have a limit as x approaches a, then so do (f + g)(x ), (f · g)(x ), (αf )(x ), and (f /g)(x ) (when Y =R and the limit of g(x ) is

nonzero). In fact,

x →alim(f + g)(x ) = lim

x →af (x ) + lim

x →ag(x ),

x →alim(αf )(x ) = α lim

x →af (x ),

x →alim(f · g)(x ) = lim

x →af (x ) · lim

x →ag(x ), and (when Y =R and the limit of g(x ) is nonzero)

x →alim

 f g



(x ) = limx →af (x ) limx →ag(x ).

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Theorem (10.26)

(iv) [Squeeze Theorem for Functions] Suppose that Y =R. If h : X \{a} → R satisfies g(x ) ≤ h(x ) ≤ f (x ) for all x ∈ X \{a}, and

x →alimf (x ) = lim

x →ag(x ) = L, then the limit of h exists, as x → a, and

x →alimh(x ) = L.

(v) [Comparison Theorem for Functions] Suppose that Y =R. If f (x ) ≤ g(x ) for all x ∈ X \{a}, and f and g have a limit as x approaches a, then

x →alimf (x ) ≤ lim

x →ag(x ).

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Theorem (10.26)

(iv) [Squeeze Theorem for Functions] Suppose that Y =R. If h : X \{a} → R satisfies g(x ) ≤ h(x ) ≤ f (x ) for all x ∈ X \{a}, and

x →alimf (x ) = lim

x →ag(x ) = L, then the limit of h exists, as x → a, and

x →alimh(x ) = L.

(v) [Comparison Theorem for Functions] Suppose that Y =R. If f (x ) ≤ g(x ) for all x ∈ X \{a}, and f and g have a limit as x approaches a, then

x →alimf (x ) ≤ lim

x →ag(x ).

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Definition (10.27)

Let E be a nonempty subset of X and f : E → Y .

(i) f is said to be continuous at a point a ∈ E if and only if given ε there is a δ > 0 such that

ρ(x , a) < δ and x ∈ E imply τ (f (x ), f (a)) < ε.

(ii) f is said to be continuous on E (Notation: f : E → Y is continuous) if and only if f is continuous at every x ∈ E .

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Theorem (10.28)

Let E be a nonempty subset of X and f , g : E → Y . (i) f is continuous at a ∈ E if and only if f (xn) →f (a), as n → ∞, for all sequences xn∈ E that converge to a.

(ii) Suppose that Y =Rn. If f , g are continuous at a point a ∈ E (respectively, continuous on a set E), then so are f + g, f · g, and αf (for any α ∈R). Moreover, in the case Y =R, f /g is continuous at a ∈ E when g(a) 6= 0

(respectively, on E when g(x ) 6= 0 for all a ∈ E .)

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Theorem (10.28)

Let E be a nonempty subset of X and f , g : E → Y . (i) f is continuous at a ∈ E if and only if f (xn) →f (a), as n → ∞, for all sequences xn∈ E that converge to a.

(ii) Suppose that Y =Rn. If f , g are continuous at a point a ∈ E (respectively, continuous on a set E), then so are f + g, f · g, and αf (for any α ∈R). Moreover, in the case Y =R, f /g is continuous at a ∈ E when g(a) 6= 0

(respectively, on E when g(x ) 6= 0 for all a ∈ E .)

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Theorem (10.28)

Let E be a nonempty subset of X and f , g : E → Y . (i) f is continuous at a ∈ E if and only if f (xn) →f (a), as n → ∞, for all sequences xn∈ E that converge to a.

(ii) Suppose that Y =Rn. If f , g are continuous at a point a ∈ E (respectively, continuous on a set E), then so are f + g, f · g, and αf (for any α ∈R). Moreover, in the case Y =R, f /g is continuous at a ∈ E when g(a) 6= 0

(respectively, on E when g(x ) 6= 0 for all a ∈ E .)

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Theorem (10.29)

Suppose that X , Y , and Z are metric spaces, a is a cluster point of X , f : X → Y , and g : f (X ) → Z . If f (x ) → L as x → a and g is continuous at L, then

x →alim(g ◦ f )(x ) = g



x →alim(f )(x )

 .

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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