數學導論

數學導論

學數學

學 學 數學 學 數學 . 學數學

論 . 學 , .

(Logic), (Set) 數 (Function) . , 學

論. 論 學 數學 .

, ,

. , .

, . , 論

, . , .

v

Chapter 5

Function

function ( 數). Function 數學 學 數

學 . 數 . ,

, 數 . 數 ,

數 , 論 數, 數. ,

, 論 .

數 , , 論 數

學 數. , 數 , .

數, 數.

數 .

5.1. Basic Definition

nonempty sets X,Y . function X ,Y

relation. relation, X Y ,

X Y , “function”.

f ⊆ X ×Y X Y relation, ?

, x∈ X, y∈ Y

(x, y)∈ f . , ,

, ? , x∈ X, y∈ Y

(x, y)∈ f . 數學 (x, y)∈ f (x, y^′)∈ f , y = y^′.

function :

Definition 5.1.1. X ,Y nonempty sets f⊆ X ×Y, from X to Y relation.

f , f from X to Y function ( 數). function

mapping map ( ).

(1) x∈ X, y∈ Y (x, y)∈ f .

(2) x∈ X, y,y^′∈ Y (x, y)∈ f (x, y^′)∈ f , y = y^′.

69

relation function “ ” ,

f : X → Y, f X Y function. x∈ X,

f (x) = y x f “ ” y . f (x) = y

(x, y)∈ f . f function f

function, (1), (2) ( Example 5.1.2).

f : X→ Y . f : X→ Y, f

, .

X ,Y . X f domain ( ),

. Y f codomain ( ), “ ”

. “ ” , function

Y X .

function “well-defined function”,

function ( ). , “well-defined”

, function well-defined.

Example 5.1.2. relations, well-defined function.

(A) X ={x ∈ R : x ≥ 0}, Y = R relation f ⊆ X ×Y f ={(x,y) ∈ X ×Y :

y²= x}. relation f function (1), x∈ X, x≥ 0,

y =√

x, y∈ Y = R y²=√

x²= x. x∈ X, y∈ Y (x, y)∈ f . f (2). 1²= (−1)²= 1, (1, 1)∈ f (1,−1) ∈ f .

f function.

(B) X ={x ∈ R : x ≥ 0}, Y = {y ∈ R : y ≤ 0} relation f ⊆ X × Y f ={(x,y) ∈ X ×Y : y²= x}. relation f function (1) ,

x∈ X, x≥ 0, y =−√

x, y∈ R y≤ 0, y∈ Y. y²=√

x²= x,

x∈ X, y∈ Y (x, y)∈ f . f (2). x∈ X, y,y^′∈ Y

(x, y)∈ f (x, y^′)∈ f , y²= x = y^′2. (y− y^′)(y + y^′) = 0, y = y^′ y =−y^′. x = 0, y = y^′= 0. x̸= 0, y̸= 0 y^′̸= 0, y, y^′∈ Y,

y < 0 y^′< 0. y =−y^′ , y = y^′. f : X→ Y function.

(C) (B) X X =R, f ={(x,y) ∈ X ×Y : y²= x} function.

−1 ∈ X, y∈ Y ⊆ R y²=−1. y∈ Y (−1,y) ∈ f .

f (1), f function. , (B) Y

Y ={y ∈ R : y < 0}, f ={(x,y) ∈ X ×Y : y²= x} function. 0∈ X,

y∈ Y y²= 0, y∈ Y (0, y)∈ f . f function.

(D) X =R, Y = P(R) function f : X → Y x∈ X,

f (x) ={y ∈ R : y²= x}.

f well-defined function. x∈ X = R, x

x > 0, x = 0, x < 0 . x > 0 , f (x) ={√x,−√x} R subset,

Y =P(R) . x = 0 , f (0) ={0}, Y =P(R) .

5.1. Basic Definition 71

x < 0 , f (x) = /0, Y =P(R) . x∈ X,

R subset A∈ Y (x, A)∈ f . , x < 0 , f (x) = /0,

, /0∈ Y (x, /0)∈ f . y∈ Y, (x, y)∈ f , f

(1). f (2). , x∈ X,

R subset A f (x) = A. , x > 0, f (x) {√

x,−√x}

Y . (x,{√

x,−√

x}) ∈ f ; (x,√

x)∈ f (x,−√ x)∈ f .

f (2) .

Example 5.1.2 , ,

, 數. , 數 f : X→ Y f^′: X^′→ Y^′

X = X^′, Y = Y^′ x∈ X, f (x) = f^′(x) , f f^′

數. Example 5.1.2 (D) , f

. 數 . 數 ,

數 .

數 . 論 數 , 論 .

數 , 數, identity function. ,

數. :

Definition 5.1.3. X nonempty set. id_X : X → X, id_X(x) = x, ∀x ∈ X. idX

function, the identity function on X.

Question 5.1. f : X → X function. f relation on X.

f : X→ X identity function? f identity

function, X ={1,2} , f identity function.

(1) f is reflexive.

(2) f is symmetric.

(3) f is transitive.

數, 數 . 數 f : X → Y,

X Y , “ ” 數. ,

數 . , X nonempty subset X^′,

f|X^′ : X^′→ Y, 數. f|X^′ : x∈ X^′, f|X^′(x) = f (x).

f|X^′ f X^′ subset, f .

f X Y function f|X^′ X^′ Y function. f|X^′ the restriction of f to X^′. Example 5.1.2 (B) , X^′={x ∈ R : x > 1},

f|X^′ : X^′→ Y, function. function . ,

. , 數

. ,

( Example 5.1.2 (C) ). Example 5.1.2 (D) , Y^′={A ∈ P(R) : #(A) ≤ 2}, f : X→ Y^′ function.

f : X→ Y g : Y → Z functions, f , g X Z function, g◦ f : X → Z. g ◦ f : x∈ X, g ◦ f (x) = g( f (x)). g◦ f (x) Z

x f f (x) Y , f (x) g Z

g( f (x)). x7−→ f (x)^f 7−→ g( f (x)).^g g◦ f : X → Z function.

(1): x∈ X, f : X → Y function y∈ Y f (x) = y.

y, g : Y→ Z function, z∈ Z, g(y) = z. z∈ Z,

g◦ f (x) = g( f (x)) = g(y) = z. (2), x∈ X z∈ Z

g◦ f (x) = z. f : X→ Y function, x∈ X, y∈ Y f (x) = y.

z, z^′∈ Z g◦ f (x) = z g◦ f (x) = z^′, g◦ f (x) = g( f (x)) = g(y),

z, z^′∈ Z g(y) = z g(y) = z^′. g : Y → Z function ,

z = z^′. g◦ f : X → Z 數, 數 f g composite function

( 數). 數 composition.

, 數 , 數 數

. 數 數 ,

. , 數 , ,

數 數 . 數 .

, 數 . x f f (x),

f (x) g g( f (x)). 數 數 ,

, .

Example 5.1.4. X ={1,2,3}, Y = {a,b,c,d}, Z = {α,β,γ}. f : X → Y : f (1) = a, f (2) = a, f (3) = c g : Y → Z : g(a) =γ,g(b) = β,g(c) = γ,g(d) = α, g◦ f : X → Z : g◦ f (1) = g( f (1)) = g(a) =γ, g ◦ f (2) = g( f (2)) = g(a) = γ, g◦ f (3) = g( f (3)) = g(c) =γ.

identity function 數, 數 ,

“ ”, 數 . .

Lemma 5.1.5. f : X→ Y function. X identity function id_X : X→ X

Y identity function idY: Y → Y, :

f◦ idX = f , idY◦ f = f .

Proof. f◦ idX f . idX : X → X

f : X → Y, 數 , f◦ idX : X → Y. x∈ X,

f◦ idX(x) = f (id_X(x)) = f (x). f◦ idX = f .

id_Y◦ f f . f : X→ Y id_Y: Y → Y,

數 , idY◦ f : X → Y. x∈ X, idY◦ f (x) = idY( f (x)).

f (x)∈ Y, id_Y( f (x)) = f (x). id_Y◦ f = f . 

5.2. Image and Inverse Image 73

composition . f : X→ Y g : Y → Z functions,

g◦ f f◦ g. , Z̸= X , f◦ g ( ),

. Z = X g◦ f f◦ g .

Question 5.2. X ={1,2}, f : X→ X, g : X → X g◦ f ̸= f ◦ g.

composition , composition .

.

Proposition 5.1.6. X ,Y, Z,W nonempty sets. f : X → Y, g : Y → Z h : Z→ W functions,

h◦ (g ◦ f ) = (h ◦ g) ◦ f .

Proof. h◦ (g ◦ f ) (h◦ g) ◦ f .

g◦ f : X → Z, h◦ (g ◦ f ) : X → W. h◦ (g ◦ f ) X , W .

h◦ g : Y → W, (h◦ g) ◦ f : X → W. (h◦ g) ◦ f X , W .

, x∈ X h◦ (g ◦ f )(x) = (h ◦ g) ◦ f (x). h◦ (g ◦ f )(x) g◦ f (x) h h((g◦ f )(x)). g◦ f (x) = g( f (x)),

h◦ (g ◦ f )(x) = h((g ◦ f )(x)) = h(g( f (x))).

h◦ (g ◦ f )(x) x f f (x), g g( f (x)),

h h(g( f (x))). (h◦g)◦ f (x) f (x) h◦g (h◦g)( f (x)).

(h◦ g)( f (x)) f (x) g g( f (x)) h,

(h◦ g) ◦ f (x) = (h ◦ g)( f (x)) = h(g( f (x))).

h◦ (g ◦ f ) (h◦ g) ◦ f 數. 

數 , 數 , .

5.2. Image and Inverse Image

前 數 數 ,

數 . 數

, image . , ,

, inverse image .

數學 , image inverse image 數 論 .

, function f : X→ Y X subset A, A f

image A f . .

Definition 5.2.1. f : X → Y function A⊆ X. f (A) ={ f (a) : a ∈ A}, f (A) the image of A under f . , the image of X under f , f (X ) f range ( ).

f (A) , f (A) Y subset. ,

. , (

Example 5.2.2). , 學 f (a)∈ f (A) a∈ A.

, b̸∈ A f (b)∈ f (A). ,

f (A) Y . y∈ f (A), a∈ A y = f (a).

, y∈ A a∈ A y = f (a) y∈ f (A). f (A)

f (A) ={y ∈ Y : ∃a ∈ A,y = f (a)}.

, f (A) . .

Example 5.2.2. X =R \ {3}, f : X → R f (x) = (x + 1)/(x− 3), ∀x ∈ X.

, f well-defined function. f range, f (X ). ,

f (X ) ={(x + 1)/(x − 3) : x ∈ X}, f (X ) .

, y∈ f (X), y∈ R x∈ X y = f (x) = (x + 1)/(x− 3).

y 數, y = (x + 1)/(x− 3) X . y

數, x 數, y(x− 3) = x + 1 (y− 1)x = 3y + 1, x = (3y + 1)/(y− 1).

, , y∈ R x∈ X y = (x + 1)/(x− 3),

x = (3y + 1)/(y− 1). x , x .

x f (x) = y.

x = (3y + 1)/(y− 1), y̸= 1. y = 1, x∈ X

1 = (x + 1)/(x− 3), x + 1 = x− 3, 1 =−3 . y̸= 1, x = (3y + 1)/(y− 1),

x + 1 x− 3=

3y+1y−1 + 1

3y+1 y−1 − 3=

y−14y y−14

=4y 4 = y.

, y̸= 1 , x = (3y + 1)/(y− 1) ∈ R (x + 1)/(x− 3) = y.

x̸= 3, x∈ X. x = (3y + 1)/(y− 1) = 3, 3y + 1 = 3y− 3,

1 =−3 , x∈ X. , y̸= 1 , x∈ X y = f (x).

y = 1 x∈ X y = f (x). f (X ) =R \ {1}.

, image .

Lemma 5.2.3. f : X→Y function A, B X subsets. A⊆ B, f (A)⊆ f (B).

Proof. , y∈ f (A), a∈ A, y = f (a). A⊆ B, a∈ B.

a∈ B y = f (a), y∈ f (B). f (A)⊆ f (B). 

X subsets A, B, A⊆ A∪B B⊆ A∪B. Lemma 5.2.3,

f (A)⊆ f (A ∪ B) f (B)⊆ f (A ∪ B). Corollary 3.2.4, f (A)∪ f (B) ⊆ f (A ∪ B).

, y∈ f (A∪B), x∈ A∪B, y = f (x). , x∈ A, y = f (x)∈ f (A),

5.2. Image and Inverse Image 75

x∈ B, y = f (x)∈ f (B). y∈ f (A) y∈ f (B). y∈ f (A) ∪ f (B),

f (A∪ B) ⊆ f (A) ∪ f (B). .

Proposition 5.2.4. f : X→ Y function A, B X subsets.

f (A)∪ f (B) = f (A ∪ B).

, A∩ B ⊆ A A∩ B ⊆ B, Lemma 5.2.3, f (A∩ B) ⊆ f (A) f (A∩B) ⊆ f (B). Corollary 3.2.4, f (A∩B) ⊆ f (A)∩ f (B). f (A)∩ f (B)

f (A∩B). y∈ f (A)∩ f (B), y∈ f (A) y∈ f (B), a∈ A

b∈ B y = f (a) y = f (b). a = b, a∈ A∩B.

數 f :{1,2} → {0} f (1) = f (2) = 0. A ={1}, B = {2}, A∩ B = /0, f (A∩ B) = /0. f (A) = f (B) ={0} f (A)∩ f (B) = {0}. f (A)∩ f (B)

f (A∩ B). f (A∩ B) ⊆ f (A) ∩ f (B) .

, f (A\ B) f (A)\ f (B) . y∈ f (A) \ f (B),

a∈ A y = f (a) y̸∈ f (B). a∈ B, y = f (a)∈ f (B) . a∈ A \ B, y = f (a)∈ f (A \ B). f (A)\ f (B) ⊆ f (A \ B). ,

y∈ f (A \ B), a∈ A \ B. (A\ B) ⊆ A, f (a)∈ f (A).

a̸∈ B, y = f (a)̸∈ f (B), b∈ B f (a) = f (b). 前

f :{1,2} → {0} f (1) = f (2) = 0 . A ={1},B = {2}, A\ B = A, f (A\ B) = f (A) = {0}. f (A) = f (B) ={0}, f (A)\ f (B) = /0. f (A\ B)

f (A)\ f (B). f (A)\ f (B) ⊆ f (A \ B) .

Question 5.3. X , A⊆ X f : X → X function. f (A^c)⊆ f (A)^c

? f (A)^c⊆ f (A^c) ?

, inverse image. , function f : X→ Y

Y subset C, C f inverse image f C

. .

Definition 5.2.5. f : X→ Y function C⊆ Y. f⁻¹(C) ={x ∈ X : f (x) ∈ C}, f⁻¹(C) the inverse image of C under f .

f⁻¹(C) , f⁻¹(C) X subset. inverse image

, inverse image .

, , inverse image image .

Proposition 5.2.6. f : X→ Y function C, D Y subsets.

(1) C⊆ D, f⁻¹(C)⊆ f⁻¹(D).

(2) f⁻¹(C∪ D) = f⁻¹(C)∪ f⁻¹(D).

(3) f⁻¹(C∩ D) = f⁻¹(C)∩ f⁻¹(D).

(4) f⁻¹(C\ D) = f⁻¹(C)\ f⁻¹(D).

Proof. (1) x∈ f⁻¹(C), f (x)∈ C. C⊆ D, f (x)∈ D, x∈ f⁻¹(D).

f⁻¹(C)⊆ f⁻¹(D).

(2) C⊆C∪D D⊆C∪D, (1) f⁻¹(C)⊆ f⁻¹(C∪D) f⁻¹(D)⊆ f⁻¹(C∪D).

Corollary 3.2.4 f⁻¹(C)∪ f⁻¹(D)⊆ f⁻¹(C∪ D). , x∈ f⁻¹(C∪ D), f (x)∈ C ∪ D, f (x)∈ C f (x)∈ D. x∈ f⁻¹(C) x∈ f⁻¹(D),

x∈ f⁻¹(C)∪ f⁻¹(D). f⁻¹(C∪ D) ⊆ f⁻¹(C)∪ f⁻¹(D), f⁻¹(C∪ D) = f⁻¹(C)∪ f⁻¹(D).

(3) C∩D ⊆C C∩D ⊆ D, (1) f⁻¹(C∩D) ⊆ f⁻¹(C) f⁻¹(C∩D) ⊆ f⁻¹(D).

Corollary 3.2.4 f⁻¹(C∩ D) ⊆ f⁻¹(C)∩ f⁻¹(D). , x∈ f⁻¹(C)∩ f⁻¹(D), x∈ f⁻¹(C) x∈ f⁻¹(D), f (x)∈ C f (x)∈ D. f (x)∈ C ∩ D,

x∈ f⁻¹(C∩ D). f⁻¹(C)∩ f⁻¹(D)⊆ f⁻¹(C∩ D), f⁻¹(C∩ D) = f⁻¹(C)∩ f⁻¹(D).

(4) x∈ f⁻¹(C\D), f (x)∈ C\D, f (x)∈ C f (x)̸∈ D. x∈ f⁻¹(C).

x∈ f⁻¹(D), f (x)∈ D, 前 f (x)̸∈ D , x̸∈ f⁻¹(D). x∈ f⁻¹(C) x̸∈ f⁻¹(D), x∈ f⁻¹(C)\ f⁻¹(D). f⁻¹(C\ D) ⊆ f⁻¹(C)\ f⁻¹(D). , x∈ f⁻¹(C)\ f⁻¹(D), x∈ f⁻¹(C) x̸∈ f⁻¹(D). f (x)∈ C. f (x)∈ D,

x∈ f⁻¹(D), 前 x̸∈ f⁻¹(D) , f (x)̸∈ D. f (x)∈ C f (x)̸∈ D, f (x)∈ C \ D, x∈ f⁻¹(C\ D). f⁻¹(C)\ f⁻¹(D)⊆ f⁻¹(C\ D),

f⁻¹(C\ D) = f⁻¹(C)\ f⁻¹(D). 

Question 5.4. X , A⊆ X f : X→ X function. f⁻¹(A^c)⊆ ( f⁻¹(A))^c

? ( f⁻¹(A))^c⊆ f⁻¹(A^c) ?

f : X → Y function A X subset , f (A) Y subset,

f⁻¹( f (A)). a∈ A, f (a)∈ f (A), inverse image a∈ f⁻¹( f (A)), A⊆ f⁻¹( f (A)). , x∈ f⁻¹( f (A)), f (x)∈ f (A),

x∈ A. 前 f :{1,2} → {0} f (1) = f (2) = 0 . A ={1},

f (A) ={0}, f⁻¹( f (A)) = f⁻¹({0}) = {1,2} ̸= A. f⁻¹( f (A)) A.

A⊆ f⁻¹( f (A)) .

Question 5.5. f : X→ Y function. f⁻¹( f (X )) = X .

C Y subset , f⁻¹(C) X subset, f ( f⁻¹(C)).

y∈ f ( f⁻¹(C)), x∈ f⁻¹(C) y = f (x). inverse image x∈ f⁻¹(C) f (x)∈ C, y = f (x)∈ C. f ( f⁻¹(C))⊆ C. , y∈ C,

y∈ f ( f⁻¹(C)), x∈ X y = f (x). 數

f :{1,2} → {3,4} f (1) = f (2) = 3. C ={3,4}, f⁻¹(C) ={1,2}, f ( f⁻¹(C)) = f ({1,2}) = {3} ̸= C. C f ( f⁻¹(C)). y∈ C

x∈ X y = f (x), . .

5.3. Onto, One-to-One and Inverse 77

Proposition 5.2.7. f : X→ Y function C Y subset, f ( f⁻¹(C)) = C∩ f (X).

Proof. 前 f ( f⁻¹(C))⊆ C, f⁻¹(C)⊆ X, f ( f⁻¹(C))⊆ f (X), f ( f⁻¹(C))⊆ C ∩ f (X). y∈ C ∩ f (X), y∈ C x∈ X y = f (x).

, x f (x) = y∈ C, x∈ f⁻¹(C). y = f (x)∈ f ( f⁻¹(C)), C∩ f (X) ⊆

f ( f⁻¹(C)). f ( f⁻¹(C)) = C∩ f (X). 

Proposition 5.2.7, . 數 f : X → Y X subset A.

f (A) Y subset, f (A)⊆ f (X). Proposition 5.2.7 (C = f (A) ), f ( f⁻¹( f (A))) = f (A)∩ f (X) = f (A).

Question 5.6. f : X→ Y function C Y subset. Proposition 5.2.7, Proposition 5.2.6 Question 5.5

f⁻¹( f ( f⁻¹(C))) = f⁻¹(C).

5.3. Onto, One-to-One and Inverse

Onto one-to-one 數 . 數

數. 數學 . 學 onto

one-to-one 數, .

onto ( ) 數, ,

. 數 range ( ) codomain ( ) onto 數.

:

Definition 5.3.1. f : X → Y function. f (X ) = Y , f onto.

y∈ Y x∈ X f (x) = y. onto 數 surjective function.

inverse image f : X→ Y onto y∈ Y, f⁻¹({y}) ̸= /0.

數 onto, 前 image .

.

Example 5.3.2. (A) Example 5.2.2 數 f :R \ {3} → R f (x) =

(x + 1)/(x− 3), ∀x ∈ X. f range R \ {1}. f onto. “ ”

數 g :R \ {3} → R \ {1} g(x) = (x + 1)/(x− 3), ∀x ∈ X, g(x) onto.

(B) 數 f :Z → N ∪ {0}

f (n) =

{ 2n, if n≥ 0;

−2n − 1, if n < 0.

f onto. f f

數 數. k∈ N ∪ {0} 數, k/2∈ Z k/2≥ 0. n = k/2,

f (n) = 2n = k. k∈ N ∪ {0} 數, k + 1∈ Z 數 k + 1 > 0.

n =−(k+1)/2, n∈ Z n < 0, f (n) =−2n−1 = (−2(−(k+1)/2)−1 = k.

f onto.

數 ( 數 ) , onto .

數 onto .

Theorem 5.3.3. f : X→ Y function. f onto g : Y → X function f◦ g = idY.

Proof. (⇒) f : X→ Y onto , f 數 g : Y → X f◦g = idY.

Axiom of Choice , ,

. . f

onto, y∈ Y, f⁻¹({y}) ̸= /0. y∈ Y, g(y)

f⁻¹({y}) . Y X 數 g.

f◦ g : Y → Y y∈ Y, g(y) = x, x∈ f⁻¹({y}), f (x) = y.

f◦ g(y) = f (g(y)) = f (x) = y. f◦ g = idY.

(⇐) g : Y → X function f◦ g = idY, f : X→ Y onto,

y∈ Y, x∈ X y = f (x). y∈ Y, g(y)∈ X.

x = g(y)∈ X, f (x) = f (g(y)) = f◦g(y) = idY(y) = y. x∈ X y = f (x),

f : X→ Y onto. 

Example 5.3.4. X ={1,2,3}, Y = {a,b} f : X → Y, f (1) = f (2) = a,

f (3) = b. f : X → Y onto. g : Y → X f◦ g = idY.

Y X 數, Y . f⁻¹({a}) = {1,2},

f⁻¹({a}) , 2, g(a) = 2. f⁻¹({b}) = {3}

, g(b) = 3. g : Y→ X function

f◦ g(a) = f (g(a)) = f (2) = a f◦ g(b) = f (g(b)) = f (3) = b. f◦ g = idY.

Theorem 5.3.3 onto onto .

.

Proposition 5.3.5. f1: X → Y, f2: Y → Z onto function, f2◦ f1: X → Z onto.

Proof. ( ) onto , z∈ Z, x∈ X f₂◦ f1(x) =

z. f2: Y→ Z onto, z∈ Z, y∈Y f2(y) = z. f1: X→Y onto, y∈ Y, x∈ X f₁(x) = y. x, f₂◦ f1(x) = f₂( f₁(x)) = f₂(y) = z.

f2◦ f1: X → Z onto.

( ) Theorem 5.3.3, f₂◦ f1: X → Z onto, g : Z→ X

( f2◦ f1)◦ g = idZ . f1: X → Y, f2: Y → Z onto, Theorem 5.3.3 g₁: Y→ X, g2: Z→ Y f₁◦ g1= id_Y f₂◦ g2= id_Z. g = g₁◦ g2: Z→ X,

5.3. Onto, One-to-One and Inverse 79

( f₂◦ f1)◦ g = ( f2◦ f1)◦ (g1◦ g2). 數 (Proposition 5.1.6) Lemma 5.1.5, ( f₂◦ f1)◦ (g1◦ g2) = f₂◦ ( f1◦ g1)◦ g2= f₂◦ (idY◦ g2) = f₂◦ g2= id_Z.

( f2◦ f1)◦ g = idZ. 

Proposition 5.3.5 , f2◦ f1 onto f1, f2

onto. Example 5.3.4 g :{a,b} → {1,2,3} g(a) = 2, g(b) = 3, onto.

f◦ g = id_{a,b} onto. .

Corollary 5.3.6. f₁: X → Y, f2: Y→ Z function f₂◦ f1: X → Z onto, f₂ onto.

Proof. f₂◦ f1: X→ Z onto, Theorem 5.3.3 g : Z→ X ( f₂◦ f1)◦g = idZ. 數 f2◦ ( f1◦ g) = idZ. g2= f1◦ g, g2: Z→ Y

f₂◦ g2= f₂◦ ( f1◦ g) = idZ. Theorem 5.3.3 f₂: Y → Z onto.  Question 5.7. onto Corollary 5.3.6.

Corollary 5.3.6 , f2 onto

f₂◦ f1 onto.

Question 5.8. X ={a,b}, Y = {1,2,3}, f1: X→ Y, f2: Y→ X functions f₂ onto, f₂◦ f1 onto.

one-to-one ( ) 數,

. :

Definition 5.3.7. f : X → Y function. X x₁̸= x2, f (x1)̸= f (x2), f one-to-one. one-to-one 數 injective function.

inverse image f : X → Y one-to-one y ∈ Y,

#( f⁻¹({y})) ≤ 1 ( f⁻¹({y})) = /0). ,

one-to-one , Definition 5.3.7 contrapositive . x1, x2∈ X f (x1) = f (x2), x1= x2. .

Example 5.3.8. Example 5.3.2 數 one-to-one.

(A) 數 f :R\{3} → R f (x) = (x + 1)/(x−3), ∀x ∈ X. x₁, x₂∈ R\{3}

f (x1) = f (x2), (x1+ 1)/(x1−3) = (x2+ 1)/(x2−3), (x1+ 1)(x2−3) = (x2+ 1)(x1−3).

x₂− 3x1= x₁− 3x2, x₁= x₂. f one-to-one.

(B) 數 f :Z → N ∪ {0}

f (n) =

{ 2n, if n≥ 0;

−2n − 1, if n < 0.

n1, n2∈ Z f (n1) = f (n2). n1, n2 0

0, f f (n₁) f (n₂) , f (n₁) = f (n₂) .

n₁≥ 0,n2≥ 0 n₁< 0, n₂< 0. n₁≥ 0,n2≥ 0, f (n₁) = 2n₁, f (n₂) = 2n₂, f (n1) = f (n₂) n1= n₂. , n1< 0, n₂< 0, f (n1) =−2n1− 1, f (n2) =

−2n2− 1, f (n1) = f (n2) n1= n2. f one-to-one.

onto , 數 one-to-one .

Theorem 5.3.9. f : X→ Y function. f one-to-one h : Y → X function h◦ f = idX.

Proof. (⇒) f : X → Y one-to-one , f 數 h : Y → X h◦ f = idY. f one-to-one, y∈ Y, #( f⁻¹({y})) ≤ 1.

y∈ Y, f⁻¹({y}) = /0 h(y) X . f⁻¹({y}) ̸= /0,

f⁻¹({y}) . f⁻¹({y}) = {x} h(y) = x.

Y X 數 h. h◦ f : X → X x∈ X, f (x) = y,

x∈ f⁻¹({y}), h(y) = x. h◦ f (x) = h( f (x)) = h(y) = x. h◦ f = idX.

(⇐) h : Y → X function h◦ f = idX, f : X → Y one-

to-one, x1, x2∈ X f (x1) = f (x2), x1= x2. x1∈ X, x₁= id_X(x₁) = h◦ f (x1) = h( f (x₁)). x₂∈ X, x₂= h( f (x₂)).

f (x1) = f (x₂)∈ Y h : Y → X function h( f (x1)) = h( f (x₂)).

x₁= h( f (x₁)) = h( f (x₂)) = x₂.

 Example 5.3.10. X ={a,b}, Y = {1,2,3} f : X→ Y, f (a) = 3, f (b) = 1.

f : X → Y one-to-one. h : Y → X h◦ f = idX.

Y X 數, Y . f⁻¹({2}) = /0,

X , a, h(2) = a. f⁻¹({1}) = {b}

h(1) = b. f⁻¹({3}) = {a} h(3) = a h : Y → X

function h◦ f (a) = h( f (a)) = h(3) = a h◦ f (b) = h( f (b)) = h(1) = b.

h◦ f = idX.

Theorem 5.3.9 one-to-one one-to-one .

.

Proposition 5.3.11. f1: X→ Y, f2: Y → Z one-to-one function, f2◦ f1: X→ Z one-to-one.

Proof. ( ) one-to-one . x₁, x₂∈ X f₂◦ f1(x₁) = f2◦ f1(x₂). f2( f₁(x₁)) = f₂( f₁(x₂)), f2: Y → Z one-to-one, f1(x₁) = f1(x2). f1: X → Y one-to-one, x1= x2

( ) Theorem 5.3.9, f2◦ f1: X → Z one-to-one,

h : Z→ X h◦ ( f2◦ f1) = id_X . f₁: X → Y, f2: Y → Z one-to-one,

5.3. Onto, One-to-One and Inverse 81

Theorem 5.3.9 h₁: Y → X, h2: Z → Y h₁◦ f1= id_X h₂◦ f2= id_Y. h = h1◦ h2 : Z→ X, h◦ ( f2◦ f1) = (h₁◦ h2)◦ ( f2◦ f1). 數

(Proposition 5.1.6) Lemma 5.1.5, (h1◦ h2)◦ ( f2◦ f1) = h1◦ (h2◦ f2)◦ f1= h₁◦ (idY◦ f1) = h₁◦ f1= id_X. h◦ ( f2◦ f1) = id_X. 

Proposition 5.3.11 , f2◦ f1 one-to-one

f₁, f₂ one-to-one. Example 5.3.10 h :{1,2,3} → {a,b} h(1) = b, h(2) = a, h(3) = a, one-to-one. h◦ f = id_{a,b} one-to-one.

.

Corollary 5.3.12. f1: X→ Y, f2: Y → Z function f2◦ f1: X→ Z one-to-one, f₁ one-to-one.

Proof. f2◦ f1: X → Z one-to-one, Theorem 5.3.9 h : Z→ X h◦ ( f2◦ f1) = idX. 數 (h◦ f2)◦ f1= idX. h1 = h◦ f2,

h₁: Y → X h₁◦ f1= (h◦ f2)◦ f1= id_X. Theorem 5.3.9 f₁: X→ Y

one-to-one. 

Question 5.9. one-to-one Corollary 5.3.12.

Corollary 5.3.12 , f₁ one-to-one

f2◦ f1 one-to-one.

Question 5.10. X ={a,b}, Y = {1,2,3}, f1 : X → Y, f2 : Y → X functions f₁ one-to-one, f₂◦ f1 one-to-one.

one-to-one and onto 數. 數 bijective

function bijection. f : X → Y bijective, f onto g : Y → X

f◦ g = idY (Theorem 5.3.3). f one-to-one h : Y → X h◦ f = idX

(Theorem 5.3.9). Lemma 5.1.5,

h = h◦ idY = h◦ ( f ◦ g) = (h ◦ f ) ◦ g = idX◦ g = g.

f : X → Y bijective , g : Y → X, f◦ g = idY

g◦ f = idX. 數 g . g : Y → X g^′: Y → X

f◦ g = f ◦ g^′= id_Y g◦ f = g^′◦ f = idX,

g^′= g^′◦ idY = g^′◦ ( f ◦ g) = (g^′◦ f ) ◦ g = idX◦ g = g.

數 g f , f⁻¹, f

inverse. , bijective function invertible function.

Question 5.11. f : X → Y injective. g : Y → X f◦ g = idY, g = f⁻¹. h : Y → X h◦ f = idX, h = f⁻¹.

, f⁻¹ inverse image . 數 inverse image, f : X→ Y bijective, Y subset C, inverse image f⁻¹(C)

. Y y, f bijective f⁻¹(y) . ,

y∈ Y, f⁻¹({y}) , f⁻¹(y) f bijective .

f : X→ Y bijective , inverse image f⁻¹: Y → X . ,

y∈ Y, f onto one-to-one, #( f⁻¹({y})) = 1. f⁻¹({y})

. f⁻¹({y}) = {x}, f⁻¹(y) = x. , f (x) = y

f⁻¹(y) = x, f◦ f⁻¹= id_Y f⁻¹◦ f = idX.

Example 5.3.13. Example 5.3.2 bijective function inverse . (A) 數 g :R \ {3} → R \ {1} g(x) = (x + 1)/(x− 3), ∀x ∈ X, g(x) onto. Example 5.2.2 y∈ R \ {1}, g⁻¹({y}) = {(3y + 1)/(y − 1)}.

g⁻¹:R \ {1} → R \ {3} g⁻¹(x) = (3x + 1)/(x− 1), ∀x ∈ R \ {1}.

(B) 數 f :Z → N ∪ {0}

f (n) =

{ 2n, if n≥ 0;

−2n − 1, if n < 0.

Example 5.3.2 k∈ N ∪{0} 數, f⁻¹({k}) = {k/2}. k∈ N ∪{0}

數, f⁻¹({k}) = {−(k + 1)/2}. f⁻¹:N ∪ {0} → Z f⁻¹(n) =

{ n/2, if n is even;

−(n + 1)/2, if n is odd.

f : X→ Y bijective , f inverse . , f inverse ,

f⁻¹: Y → X f◦ f⁻¹= id_Y f⁻¹◦ f = idX, Theorem 5.3.3 Theorem 5.3.9

f bijective. .

Theorem 5.3.14. f : X→ Y function. f bijection f⁻¹: Y → X f◦ f⁻¹= id_Y f⁻¹◦ f = idX.

Question 5.12. f : X→ Y bijective function. f⁻¹: Y → X bijective ( f⁻¹)⁻¹= f .

Proposition 5.3.5 Proposition 5.3.11 :

Proposition 5.3.15. f₁: X→ Y, f2: Y→ Z bijective function, f₂◦ f1: X→ Z bijective function.

( f₂◦ f1)⁻¹= f₁⁻¹◦ f₂⁻¹.

Proof. ( f₂◦ f1)◦ ( f₁⁻¹◦ f₂⁻¹) = id_Z ( f₁⁻¹◦ f₂⁻¹)◦ ( f2◦ f1) = id_X. Theorem 5.3.14 f₂◦ f1: X→ Z bijective. inverse function ,

( f2◦ f1)⁻¹= f₁⁻¹◦ f₂⁻¹.

( f₂◦ f1)◦ ( f₁⁻¹◦ f₂⁻¹) = f₂◦ ( f1◦ f₁⁻¹)◦ f₂⁻¹= ( f₂◦ idY)◦ f₂⁻¹= f₂◦ f₂⁻¹= id_Z,

5.4. Equivalent Sets and Cardinal Number 83

( f₁⁻¹◦ f₂⁻¹)◦ ( f2◦ f1) = f₁⁻¹◦ ( f₂⁻¹◦ f2)◦ f1= f₁⁻¹◦ (idY◦ f1) = f₁⁻¹◦ f1= id_X.

. 

Question 5.13. f₁: X→ Y, f2: Y → Z function f₂◦ f1: X→ Z bijective.

f1: X→ Y, f2: Y→ Z bijective? f1, f2 bijective, bijective?

5.4. Equivalent Sets and Cardinal Number

數 , 數

數 . A n , 數 A

, {1,...,n} A bijective function.

.

Definition 5.4.1. A, B set, bijection f : A→ B, A is equivalent to

B, |A| = |B| .

A finite set, |A| A 數 #(A).

A finite set , |A| , A cardinal number.

A is equivalent to B A B cardinal number.

Equivalent set equivalence relation.

Proposition 5.4.2. sets A, B,C, .

(1) |A| = |A|.

(2) |A| = |B| |B| = |A|.

(3) |A| = |B| |B| = |C|, |A| = |C|.

Proof. (1) A, idA: A→ A. idA bijective, |A| = |A|.

(2) |A| = |B|, f : A→ B bijective. f⁻¹: B→ A, bijective ( Question 5.12), |B| = |A|.

(3) |A| = |B| |B| = |C|, f : A→ B, g : B →C bijective, Proposition

5.3.15 g◦ f : A → C bijective. |A| = |C|. 

, , 數 n, I_n 1 n 數

, I1={1}, I2={1,2},…, In={1,...,n}. A n finite set,

|A| = |In|. A B n , |A| = |In| |B| = |In|,

Proposition 5.4.2 |A| = |B|.

A, B finite set A 數 n B 數 m,

|A| = |B| ? |In| |Im| . n̸= m, ,

m > n. |Im| = |In|, bijection f : I_m→ In. Theorem 2.2.3 (

I_m m , I_n n ), f : I_m→ In one-to-one

( 數 數, 1 ), f bijective

, |In| ̸= |Im|. |A| = |B|, |A| = |In|, |B| = |Im| Proposition 5.4.2

|In| = |Im| , |A| ̸= |B|.

前 , finite set , cardinal number

數 數 . cardinal number infinite set. ,

, 數 , nonempty set A, n∈ N,

|A| ̸= |In|, A infinite set.

cardinal number 數 , A∩ B = /0, C ∩ D = /0,

|A| = |C|, |B| = |D|, 數 , |A ∪ B| = |C ∪ D|. ,

.

Lemma 5.4.3. I index set, {Ai, i∈ I}, {Bi, i∈ I} A, B partition.

i∈ I, |Ai| = |Bi|, |A| = |B|.

Proof. , {Ai, i∈ I} A partition A =^∪

i∈I

A_i i, j∈ I, i̸= j, A_i∩ Aj= /0. , i∈ I, |Ai| = |Bi|, f_i: A_i→ Bi bijective

function. fi, bijective function f : A→ B. |A| = |B|.

f : A→ B : a∈ A, {Ai, i∈ I} A partition,

i∈ I a∈ Ai, f (a) = fi(a)∈ Bi. {Ai, i∈ I} A partition, f

A B_i⊆ B. A B

function. f : A→ B one-to-one and onto.

b∈ B, {Bi, i∈ I} B partition, i∈ I, b∈ Bi.

f_i: A_i→ Bi onto, a∈ Ai f_i(a) = b. f , b,

a∈ A, a∈ Ai, f f (a) = fi(a) = b. f : A→ B onto.

a, a^′∈ A f (a) = f (a^′). f (a)∈ B, i∈ I f (a) = f (a^′)∈ Bi. f , a∈ Aj, f (a) = fj(a)∈ Bj. f (a)∈ Bi∩ Bj i = j, a∈ Ai. a^′∈ Ai. f f (a) = f_i(a) f (a^′) = f_i(a^′).

f (a) = f (a^′) fi(a) = f_i(a^′), fi one-to-one a = a^′. f

one-to-one. 

cardinal number 數 , . finite

set , one-to-one .

.

Definition 5.4.4. A, B set, |A| ≤ |B| one-to-one function f : A→ B.

. A⊆ B, f : A→ B, f (a) = a,

∀a ∈ A. f one-to-one function, |A| ≤ |B|. ,

5.4. Equivalent Sets and Cardinal Number 85

m, n∈ N m > n, I_n⊆ Im, |In| ≤ |Im|. 前

one-to-one function f : Im→ In, |Im| ≤ |In| .

數 ,

cardinal number . .

Proposition 5.4.5. A, B set. |A| ≤ |B| onto function h : B→ A.

Proof. (⇒) |A| ≤ |B|, one-to-one function : A→ B. Theorem 5.3.9 h : B→ A h◦ f = idA. id_A: A→ A onto function, Corollary 5.3.6 h : B→ A onto.

(⇐) h : B→ A onto , g : A→ B h◦ g = idA (Theorem 5.3.3). idA

one-to-one g : A→ B one-to-one (Corollary 5.3.12). |A| ≤ |B|.  Definition 5.4.4 cardinal number partial order. ( cardinal number total order, Axiom of Choice

, .) reflexive , A,

idA: A→ A, idA one-to-one, |A| ≤ |A|. transitive ,

|A| ≤ |B| |B| ≤ |C|, f : A→ B g : B→ C one-to-one, g◦ f : A → C one-to-one (Proposition 5.3.11). |A| ≤ |C|. anti-symmetric ,

, Cantor–Schröder–Bernstein Theorem.

Theorem 5.4.6 (Cantor–Schröder–Bernstein). A, B sets |A| ≤ |B| |B| ≤ |A|,

|A| = |B|.

Proof. |A| ≤ |B| f : A→ B one-to-one function. |B| ≤ |A|,

g : B→ A one-to-one function. f , g A, B partition, Lemma

5.4.3 |A| = |B|.

a∈ A, A∪ B 數 . :

x1= a, inverse image g⁻¹({a}). g one-to-one, g⁻¹({a})

. g⁻¹({a}) = /0, 數 x1 . g⁻¹({a}) = {b},

x₂= b. b∈ B, f⁻¹({b}). , f one-to-one, f⁻¹({b})

. f⁻¹({b}) = /0, 數 x1, x2 . f⁻¹({b}) = {a^′},

x₃= a^′. a^′∈ A, g⁻¹({a^′}), 前 . ⟨a⟩

a 數 ( 數 ,

Example 5.4.7). a∈ A 數 ⟨a⟩ = x1, x₂, . . . , .

數 數 . a∈ A g⁻¹({a}) = /0, ⟨a⟩ , 數

. 數 數 . a∈ A g⁻¹({a}) = {b} f⁻¹({b}) = /0, ⟨a⟩

a, b , 數 . 數 , a∈ A

數 inverse image .

A_o={a ∈ A : ⟨a⟩ 數 }, Ae={a ∈ A : ⟨a⟩ 數 }, A_∞={a ∈ A : ⟨a⟩ }.

a∈ A, 數 ⟨a⟩, a Ao, A_e, A_∞

, . Ao, Ae, A_∞ A partition.

數 , 數 A , 數 B

. ⟨a⟩ = x1, x₂, . . . , i 數 x_i∈ A. x_i∈ f⁻¹({xi−1}), f (xi) = xi−1. i 數 xi∈ B. xi∈ g⁻¹({xi−1}), g(xi) = xi−1.

b∈ B, b 數 , y1= b,

f⁻¹({b}) , . ⟨b⟩ b 數 .

, B partition B₀, B_e, B_∞,

B_o={b ∈ B : ⟨b⟩ 數 }, Be={b ∈ B : ⟨b⟩ 數 }, B∞={b ∈ B : ⟨b⟩ }.

restriction map f|Ao : A_o→ B, a∈ Ao, f|Ao(a) = f (a). f one-to-one, f|Ao one-to-one. f|Ao range f|Ao(Ao) Be.

a∈ Ao, f|Ao(a) = f (a)∈ B. f (a) sequence, y₁= f (a), f⁻¹({ f (a)}) = {a} ( f (a)∈ { f (a)}), ⟨ f (a)⟩ y2= a. 言 ,

數 ⟨ f (a)⟩ 前 y1= f (a), y2= a, ( ) y3 g⁻¹({a})

. a 數 ⟨a⟩ x₂ . 言 , 數 ⟨ f (a)⟩ 數

⟨a⟩ 前 f (a) . a∈ Ao, ⟨a⟩ 數 , ⟨ f (a)⟩

數 , f (a)∈ B f (a)∈ Be. f|Ao(A_o)⊆ Be. , b∈ Be, b 數

⟨b⟩ 數 . ⟨b⟩ ( , ). ⟨b⟩

f⁻¹({b}) , a∈ A f (a) = b. a ⟨b⟩

, 前 , ⟨a⟩ ⟨b⟩ = ⟨ f (a)⟩ , ⟨a⟩ 數 ,

a∈ Ao. b∈ Be, a∈ Ao f (a) = f|Ao(a) = b.

B_e⊆ f |Ao(A_o), f|Ao range f|Ao(A_o) B_e. 言 , f|Ao A_o Be one-to-one and onto function. |Ao| = |Be|.

, g : B→ A Bo restriction g|Bo : Bo→ A, |Bo| = |Ae| (

f , g ). g B_∞ restriction g|B_∞ ( f

A_∞ restriction). g|B_∞ one-to-one. b∈ B_∞, g(b)

數 ⟨g(b)⟩. g(b)∈ A, g⁻¹({g(b)}) = {b}, ⟨g(b)⟩ g(b),

b, ⟨b⟩ . b∈ B∞ ⟨b⟩ ⟨g(b)⟩ .

g(b)∈ A_∞, g|B_∞ range g|B_∞(B_∞) A_∞. , a∈ A_∞, a

數 ⟨a⟩ . ⟨a⟩ . ⟨a⟩ g⁻¹({a})

, b∈ B g(b) = a. b ⟨a⟩ , 前

,⟨b⟩ ⟨a⟩ , ⟨b⟩ , b∈ B_∞.

a∈ A∞, b∈ B∞ g(b) = g|B_∞(b) = a. A_∞⊆ g|B_∞(B_∞),

g|B_∞ range g|B_∞(B_∞) A_∞. 言 , g|B_∞ B_∞ A_∞ one-to-one and onto function. |B∞| = |A∞|.

Ao, Ae, A_∞ A partition Bo, Be, B_∞ B partition, |Ao| = |Be|,

|Ae| = |Bo| |A_∞| = |B_∞|, Lemma 5.4.3 |A| = |B|. 

5.4. Equivalent Sets and Cardinal Number 87

Question 5.14. Theorem 5.4.6 , |Ae| = |Bo| g B_o restriction f Ae restriction? f Ae restriction f|Ae: Ae→ B,

range ? |A∞| = |B∞| f A_∞ restriction f|A_∞ : A_∞→ B ? Example 5.4.7. A ={1,2,...} 數 , B ={−1,−2,...} 數

. f : A→ B f (a) =−1 − a, ∀a ∈ A g : B→ A g(b) = 1− b,

∀b ∈ B. Theorem 5.4.6 數 .

數 :

..

1 .

2 .

3 .

4 .

···

−1.. . −2 . −3 . −4

···

f , g.

3∈ A, g⁻¹({3}) = {−2}, f⁻¹({−2}) = {1} g⁻¹({1}) = /0,

3 數 ⟨3⟩ 3,−2,1. 數 3 , 3∈ Ao. ,

4 數 ⟨4⟩ 4,−3,2,−1, 4∈ Ae. , a∈ A

數 a∈ Ao, a∈ A 數 a∈ Ae. A_∞= /0. A_o, A_e A

partition ( 數 數 partition).

−3 ∈ B, f⁻¹({−3}) = {2}, g⁻¹({2}) = {−1} f⁻¹({−1}) = /0,

−3 數 ⟨−3⟩ −3,2,−1. 數 3 , −3 ∈ Bo. ,

−4 數 ⟨−4⟩ −4,3,−2,1, −4 ∈ Be. ,

b∈ B 數 b∈ Bo, b∈ B 數 b∈ Be. B_∞= /0. Bo, B_e

B partition.

f Ao Be ( one-to-one correspondence).

a∈ Ao a 數, f (a) =−(1+a) f (a) 數, f (a)∈ Be. f

Ao Be. b∈ Be, b 數. a =−b − 1,

a > 0 ( b≤ −2) a 數, a∈ Ao. a =−b−1 ∈ Ao f f (a) =−(1+a) = b.

f A_o B_e. f A_e B_o. B_o

inverse image . −1 ∈ Bo f⁻¹({−1}) = /0.

g B_o A_e one-to-one correspondence. b∈ Bo, b

數, g(b) = 1−b 數, g(b)∈ Ae. , a∈ Ae, b = 1−a < 0

( a≥ 2) 數. b = 1− a ∈ Bo g, g(b) = 1− b = 1 − (1 − a) = a,

g B_o A_e.

Question 5.15. Example 5.4.7 f g A B bijective function h : A→ B h|Ao = f|Ao.

, cardinal number “strict order”. A, B sets, |A| ≤ |B|

|A| ̸= |B| , |A| < |B| . 前 m, n 數 m > n ,