Analysis with an introduction to proof 5th by steven lay ch02

24 196 0
Analysis with an introduction to proof  5th by steven lay  ch02

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Thông tin tài liệu

Chapter Sets and Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-1 Section 2.3 Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-2 Definition 2.3.1 Let A and B be sets A function from A to B is a nonempty relation f ⊆ A × B that satisfies the following two conditions: Existence: For all a in A, there exists a b in B such that (a, b) ∈ f Uniqueness: If (a, b) ∈ f and (a, c) ∈ f , then b = c That is, given any element a in A, there is one and only one element b in B such that (a, b) ∈ f Set A is called the domain of f and is denoted by dom f Set B is referred to as the codomain of We write f : A → B to indicate f f has domain A and codomain B The range of f, denoted rng  f, is the set of all second elements of members of f That is, rng f = {b ∈ B : ∃ a ∈ A (a, b) ∈ f } If (x, y) ∈ f , we often say that f maps x onto y and use the notation y = f (x) Domain A x Codomain B • • f (x) rng  f f:A→B Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-3 *Similar to Practice 2.3.2 in the text Practice 2.3.2* Let A = {1, 2, 3} and B = {2, 4, 6, 8} Which of the following relations are functions from A to B? B A R = {(1, 6), (2, 4)} This is not a function It violates the existence condition The number in set A is not related to anything in set B B R = {(1, 2), (2, 8), (3, 6), (1, 4)} A This is not a function It violates the uniqueness condition The number in set A is related to two numbers in set B Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-4 *Similar to Practice 2.3.2 in the text Practice 2.3.2* Let A = {1, 2, 3} and B = {2, 4, 6, 8} Which of the following relations are functions from A to B? B A R = {(1, 6), (2, 8), (3, 2)} This is a function from A to B It is OK that nothing in set A is related to number in set B A B R = {(1, 4), (2, 6), (3, 4)} This is a function from A to B It is OK that both and in set A are related to in set B In both of these functions, there is at least one element in B that is not related to anything in A When there are no extra elements in B, the function is given a special name Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-5 Properties of Functions Definition 2.3.4 A function f : A → B is called surjective (or is said to map A onto B) if B = rng f A surjective function is also referred to as a surjection The question of whether or not a function is surjective depends on the choice of codomain A function can always be made surjective by restricting the codomain to being equal to the range, but sometimes this is not convenient If it happens that no member of the codomain appears more than once as a second element in one of the ordered pairs, then we have another important type of function Definition 2.3.5 A function f : A → B is called injective (or one-to-one) if, for all a and a′ in A, f (a) = f (a′) implies that a = a′ An injective function is also referred to as an injection If a function is both surjective and injective, then it is particularly well behaved Definition 2.3.6 A function f : A → B is called bijective or a bijection if it is both surjective and injective Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-6 Example 2.3.7 Consider the function given by the formula f (x) = x2 y y = h (x) If we take y = f (x) for both the domain and codomain so that f :    →  , then f is not surjective y = g (x) because there is no real number that maps onto –1 If we limit the codomain to be the set [0, ∞), then the function –2 g: → [0, ∞) such that g(x) = x2 is surjective x y = –1 Since g(−1) = g(1), we see that g is not injective when defined on all of But restricting g to be defined on only [0, ∞), it becomes injective Thus h: [0, ∞) → [0, ∞) such that h(x) = x2 is bijective Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-7 Practice 2.3.12 Consider the four functions pictured below For each function, the domain and codomain are sets consisting of two or three points as indicated Classify each function as being surjective, injective, bijective, or none of these • • • • • • • Copyright © 2013, 2005, 2001 Pearson Education, Inc This function is neither • • • • • This function is surjective • • • • • • • • This function is injective This function is bijective Section 2.3, Slide 1-8 Functions Acting on Sets When thinking of a function as transforming its domain into its range, we may wish to Or we may wish to identify the consider what happens to certain subsets of the domain set of all points in the domain that are mapped into a particular subset of the range To this we use the following notation: Notation 2.3.13 Suppose that f : A → B If C ⊆ A, we let f (C ) represent the subset { f (x): x ∈ C} of B The set f (C ) is called the image of C in B f A B f –1 f (D) f (A) –1 D f (C) C f If D ⊆ B, we let f – 1(D) represent the subset {x ∈ A: f (x) ∈ D} of A called the pre-image of D Note: At this time, f –1 in A or “ inverse of D  f The set – f   1(D) is  ” is only applied to sets, not points Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 1-9 Example 2.3.13 Let f :   →  be given by y f (x) = x2 Then the following hold y=x If C = [0, 2], then f (C1) f (C1) = [0, 4] –2 C x –2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 10 1-10 Example 2.3.13 Let f :   →  y be given by f (x) = x2 Then the following hold y=x If C = [0, 2], then If C = [−1, 2], then f (C2) f (C1) = [0, 4] f (C2) = [0, 4] –2 C x –2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 11 1-11 Example 2.3.13 Let f :   →  y be given by f (x) = x2 Then the following hold y=x If C = [0, 2], then If C = [−1, 2], then f (C3) f (C1) = [0, 4] f (C2) = [0, 4] If C = [−2, −1] ∪ [0, 1], then –2 C f (C3) = [0, 4] x –2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 12 1-12 Example 2.3.13 Let f :   →  y be given by f (x) = x2 Then the following hold y=x f (C1) = [0, 4] If C = [0, 2], then If C = [−1, 2], then D f (C2) = [0, 4] If C = [−2, −1] ∪ [0, 1], then If D = [0, 4], then –2 f –1 f  –  1 f (C3) = [0, 4] (D ) = [−2, 2] x (D ) –2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 13 1-13 Example 2.3.13 Let f :   →  y be given by f (x) = x2 Then the following hold y=x f (C1) = [0, 4] If C = [0, 2], then If C = [−1, 2], then D f (C2) = [0, 4] 2 If C = [−2, −1] ∪ [0, 1], then If D = [0, 4], then –2 x If D = [−1, 4], then f –1 f –  1 f  –  1 f (C3) = [0, 4] (D ) = [−2, 2] (D ) = [−2, 2] (D ) –2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 14 1-14 Example 2.3.13 Let f :   →  be given by y f (x) = x2 Then the following hold y=x f (C1) = [0, 4] If C = [0, 2], then If C = [−1, 2], then D f (C2) = [0, 4] If C = [−2, −1] ∪ [0, 1], then If D = [0, 4], then –2 x If D = [−1, 4], then f –1 (D ) If D = [1, 4], then f –  1 f  – f –  1 f (C3) = [0, 4] (D ) = [−2, 2] (D ) = [−2, 2]  1 (D ) = [−2, −1] ∪ [1, 2] –2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 15 1-15 Given a function f : A → B, there are many relationships that hold between the images and pre-images of subsets of A and B Several of these are included in the next theorem The proofs are left to the practice problems and the exercises Theorem 2.3.16 Suppose that f : A → B Let C, C , and C be subsets of A and let D, D , and D be 2 subsets of B Then the following hold: Copyright © 2013, 2005, 2001 Pearson Education, Inc – 1 (a) C (b) f [ f  – 1(D)] ⊆ D (c) f (C ∩ C ) (d) f (C ∪ C ) = f (C ) ∪ f (C ) 2 (e)  f  – 1(D1 ∩ D2) =  f  – 1(D1) ∩  f – 1(D2) (f )  f  (g)  f  ⊆ f  – 1 – 1 [ f  (C)] ⊆ f (C1) ∩ (D ∪ D ) =  f (B \D) = A\ f  – 1 – 1 (D ) ∪  f – 1 f (C ) (D ) (D) Section 2.3, Slide 16 1-16 To see that equality may not hold in Theorem 2.3.16 (a) and (c), consider the following situation: Given f : A → B, suppose C1 and C2 are nonempty subsets of A such that C1 ∩ C2 = ∅ and f (C1) = f (C2) A B f C1 f (C1) C2 f We see that f  – f (C2)  (C1)] = C1 ∪ C2, and this is larger than C1  1 [ f This shows that equality may not hold in part (a): C Also, since C1 ∩ C2 = ∅, we have f (C1 ∩ C2 ) = ∅ ⊆ f  – 1 [ f  (C)] But f (C1) ∩ f (C2) = f (C1) ≠ ∅, This shows that equality may not hold in part (c): f (C1 ∩ C2) ⊆ f (C1) ∩ Copyright © 2013, 2005, 2001 Pearson Education, Inc f (C ) Section 2.3, Slide 17 1-17 While Theorem 2.3.16 states the strongest results that hold in general, if we apply certain restrictions on the functions involved, then the containment symbols in parts (a), (b), and (c) may be replaced by equality Theorem 2.3.18 Suppose that f : A → B Let C, C1, and C2 be subsets of A and let D be a subset of B Then the following hold: (a) If f is injective, then f  (b) If f is surjective, then f [ f (c) If f is injective, then f (C   ∩ C2 ) –  1 [ f (C)] = C –  1 = (D)] = D f Proof of (c): We only have to show that f (C1) ∩  (C1) ∩ f  (C2) f (C ) ⊆ f (C ∩ C ), since the 2 converse inclusion is Theorem 2.3.16(c) Then y ∈ f (C1) and To this end, let y ∈ f (C ) ∩ f (C ) y ∈  f (C2) It follows that there exists a point x in C such that f (x ) = y 1 Similarly, there exists a point x in C such that f (x ) = y 2 Since f is injective and f (x ) = y = f (x ), we must have x = x 2 That is, x ∈ C  ∩ C 1 Copyright © 2013, 2005, 2001 Pearson Education, Inc But then y = f (x1) ∈ f (C ∩ C ) ♦ Section 2.3, Slide 18 1-18 Composition of Functions then for any a ∈ A, f (a) ∈ B If f and g are functions with f : A → B and g : B → C, But B is the domain of g, so g can be applied to f (a) This yields g  f (a)), an element of C  ( B A g f C g  f (a))  ( f (a) • • a • ( g ○ f )(a)   g ○ f  This establishes a correspondence between a in A and g ( f (a)) in C This correspondence is called the composition function of by g   (read “g of f ”) It defines a function g ○f ( g ○ f )(a) Copyright © 2013, 2005, 2001 Pearson Education, Inc f and g and is denoted   ○f : A → C given by = g ( f (a)) for all a ∈ A   Section 2.3, Slide 19 1-19 Our next theorem tells us that composition of functions preserves the properties of being surjective or injective The proof is in the book and the exercises Theorem 2.3.20 Let f : A → B and g : B → C Then (a) If f and g are surjective, then g ○ f  is (b) If f and g are injective, then injective g ○ f  is surjective (c) If f and g are bijective, then g ○ f  is bijective Inverse Functions Given a function f : A → B, we have seen how f determines a relationship between That is, given D ⊆ B, we have the pre-image subsets of B and subsets of A We would like to be able to extend this idea so that to obtain a point in A That is, suppose There are two things that can prevent It may be that f  D = { y}, – 1 can be applied to a point in B where y ∈ B – 1 (D) from being a point in A: – f   1(D) is empty, and it may be that Copyright © 2013, 2005, 2001 Pearson Education, Inc f  – f   1(D) in A f  – 1 (D) contains several points instead of just one Section 2.3, Slide 20 1-20 Practice 2.3.21 Given f : A → B and y ∈ B, under what conditions on assert that there exists an x in A such that Practice 2.3.22 f can we Ans: f must be surjective f (x) = y? Given f : A → B and y ∈ B, under what conditions on f can we Ans: f must be bijective f (x) = y? assert that there exists a unique x in A such that Given a bijection f : A → B, we see that each y in B corresponds to exactly one x in A, the unique x such that f (x) = y This correspondence defines a function from B into A called the inverse of f and denoted f  –   Thus x = f  –  1 A  1 ( y) B f  x = f  –1   y = f  ( x) ( y)   • • f  –1   Definition 2.3.23 Let f : A → B be bijective The inverse function of f is the function given by f  –   = {(  1 Copyright © 2013, 2005, 2001 Pearson Education, Inc  y, x) ∈ B × A : (x, y) ∈ f } Section 2.3, Slide 21 1-21 If f : A → B is bijective, then it follows that f  – 1  : B → A is also bijective – – Indeed, since dom f = A and rng f = B, we have dom f   = B and rng f    = A  1 Thus f – 1  1 is a mapping from B onto A Since f is a function, a given x in A can correspond to only one y in B This means that f  –  1 is injective, and hence bijective When f is followed by f  That is, ( f  –  1  ○ –  1, the effect is to map x in A onto f (x) in B and then back to x in A  f )(x) = x, for every x ∈ A A function defined on a set A that maps each element in A onto itself is called the identity function on A, and is denoted by iA Furthermore, if f (x) = Thus f ○ f  – 1 = iB y, then x = f So, f  – 1 –  1  ○  f  = iA – f   1 )( y) = f ( f – 1 ( y)) = f (x) = ( y), so that ( f ○ y We summarize these results in the following theorem Theorem 2.3.24 Let f : A → B be bijective Then (a) f  – 1  : B → A is bijective Copyright © 2013, 2005, 2001 Pearson Education, Inc (b) f  – 1 ○ A and f f =i ○ f  – 1 =i B Section 2.3, Slide 22 1-22 Our final theorem relates inverse functions and composition Theorem 2.3.28 Let f : A → B and and ( g ○ f ) − 1 = f  – 1 ○ g  − 1 g : B → C be bijective Then the composition g ○ f : A → C is bijective Proof: We know from Theorem 2.3.20 that g ○ f is bijective, so g ○ f has an inverse denoted by ( g ○ f ) − 1, and this inverse maps C onto A g  f ○ f g A B f  –1 C –1 g  ( g  f ) ○   –1 –1 –1 = f    g  ○ We are asked to verify the equality of the two functions (   − 1 – 1 − 1  g ○ f )  and f  ○ g  , as sets of ordered pairs Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 23 1-23 To show (  g ○ f )  − 1   – 1 ⊆ f  ○ g − 1, we suppose (c, a) ∈ ( g ○ f ) − 1 By the definition of an inverse function, this means (a, c) ∈ g  ○ f The definition of composition implies that ∃ b ∈ B such that (a, b) ∈ Since f and g are bijective, this means that (b, a) ∈ f  That is, f  – f and (b, c) ∈ g – 1 and (c, b) ∈ g  But then, – 1  1 (b) = a and g  (c) = b ( f  so that (c, a) ∈ ( f  – 1 – 1 ○ –  1 ○ g  ( − 1 – 1 − 1 – 1 )(c) = f  g  (c)) = f  (b) = a, − 1 g  − 1 ) and ( g  ○  f ) − 1 ⊆ ( f – 1 ○ g  ) The reverse set inclusion is Exercise 31 ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.3, Slide 24 1-24 ... transforming its domain into its range, we may wish to Or we may wish to identify the consider what happens to certain subsets of the domain set of all points in the domain that are mapped into... Functions then for any a ∈ A, f (a) ∈ B If f and g are functions with f : A → B and g : B → C, But B is the domain of g, so g can be applied to f (a) This yields g  f (a)), an element of C  (... there are many relationships that hold between the images and pre-images of subsets of A and B Several of these are included in the next theorem The proofs are left to the practice problems and the

Ngày đăng: 14/08/2017, 16:16

Từ khóa liên quan

Mục lục

  • Slide 1

  • Slide 2

  • Slide 3

  • Slide 4

  • Slide 5

  • Slide 6

  • Slide 7

  • Slide 8

  • Slide 9

  • Slide 10

  • Slide 11

  • Slide 12

  • Slide 13

  • Slide 14

  • Slide 15

  • Slide 16

  • Slide 17

  • Slide 18

  • Slide 19

  • Slide 20

Tài liệu cùng người dùng

  • Đang cập nhật ...

Tài liệu liên quan