Bartle Introduction to Real Analysis solutions

R. Bartle : Introduction To Real Analysis (4th edition) Solutions

1

Preliminaries

1.1

Sets and Functions

1. We have these sets : A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} B = {2, 5, 8, 11, 14, 17, 20, . . .} C = {3, 5, 7, 9, 11, 13, 15, 17, 19, . . .}

(a) A ∩ B ∩ C = {5, 11, 17} (b) (A ∩ B)\C = {2, 8, 14, 20} (c) (A ∩ C)\B = {3, 7, 9, 13, 15, 19} 2. (a) A\(B\A) A

B

A

B

(b) A\(A\B)

(c) A ∩ (B\A)

A

B

3. If A and B are sets, show that A ⊆ B if and only if A ∩ B = A. D´emonstration. By definition 1.1.1, if (A∩B) = A, A ⊆ (A∩B) ; so, for every x ∈ A, we have x ∈ (A ∩ B), then x ∈ B, therefore A ⊆ B. Conversely, if A ⊆ B, then for every x ∈ A, x ∈ (A ∩ B), therefore A ⊇ (A ∩ B) ; also, for every x ∈ (A ∩ B), x ∈ A, therefore (A ∩ B) ⊆ A ; hence, by definition 1.1.1, if A ⊆ (A ∩ B) and (A ∩ B) ⊆ A, A ∩ B = A. 4. Prove the second De Morgan Law [Theorem 1.1.4(b)] : A\(B ∩ C) = (A\B) ∪ (A\C). D´emonstration. For every x ∈ A\(B ∩ C), x ∈ A but x ∈ / (B ∩ C), which means that x ∈ (A\B) or x ∈ (A\C), therefore, by definition 1.1.3a, x ∈ (A\B) ∪ (A\C) ; we conclude that A\(B ∩ C) ⊆ (A\B) ∪ (A\C). Conversely, for every x ∈ (A\B) ∪ (A\C), x ∈ (A\B) or x ∈ (A\C), therefore x ∈ / (B ∩ C), so x ∈ A\(B ∩ C) ; we conclude that (A\B) ∪ (A\C) ⊆ A\(B ∩ C). By definition 1.1.1., we have A\(B ∩ C) = (A\B) ∪ (A\C). 5. Prove the distributive laws : (a) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) D´emonstration. For every x ∈ A ∩ (B ∪ C), x ∈ A and x ∈ (B ∪ C), which means that x ∈ B or x ∈ C, so x ∈ A ∩ B or x ∈ A ∩ C, therefore A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C). Conversely, for every x ∈ (A ∩ B) ∪ (A ∩ C), x ∈ A ∩ B or x ∈ A ∩ C, so x ∈ A and x ∈ B or x ∈ C, wich means that x ∈ A ∩ (B ∪ C), therefore (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C). By definition 1.1.1., we have A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). (b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) D´emonstration. For every x ∈ A ∪ (B ∩ C), x ∈ A or x ∈ B and C, which means that x ∈ A or B and x ∈ A or C, so x ∈ (A ∪ B) ∩ (A ∪ C), therefore A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C). Conversely, for every x ∈ (A ∪ B) ∩ (A ∪ C), x ∈ (A ∪ B) and x ∈ (A ∪ C), which means that x ∈ A ∪ (B ∩ C), therefore (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C). By definition 1.1.1., we have A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

6. The symmetric difference of two sets A and B is the set D of all elements that belong to either A or B but not both. Represent D with a diagram. A

B

(a) Show that D = (A\B) ∪ (B\A)). D´emonstration. By definition 1.1.3c, for every x ∈ A and x ∈ / B, we have A\B ; likewise, for every x ∈ B and x ∈ / A, we have B\A. Since every x ∈ D is the set of all elements that belong to either A or B but not both, by definition 1.1.3a, we conclude that D = (A\B) ∪ (B\A)). (b) Show that D = (A ∪ B)\(A ∩ B). D´emonstration. By definition 1.1.3a, for every x ∈ A or x ∈ B, we have A ∪ B. Since D is the set of all elements that belon to either A or B but not both, by definition 1.1.3c, if x ∈ (A∪B) but x ∈ / (A∩B), we have D = (A∪B)\(A∩B). 7. For each n ∈ N, let An = {(n + 1)k : k ∈ N}. (a) What is A1 ∩ A2 ? A1 = {2k : k ∈ N}, that consists of all natural numbers divisible by 2, and A2 = {3k : k ∈ N}, that consists of all natural numbers divisible by 3 ; then A1 ∩ A2 = {6k : k ∈ N}, that consists of all natural numbers divisible by 6. (b) Determine the sets

S

{An : n ∈ N} and

T

{An : n ∈ N}.

We have A = {An : n ∈ N} = N\1. For B = {An : n ∈ N}, a` d´eterminer 8. a` dessiner un jour S

T

9. We have C ⊆ A×B, a ∈ A and b ∈ B. By definition 1.1.6, if (a, b) ∈ f and (a, b0) ∈ f , with f ⊆ A × B, it implies that b = b0. But here, by the cartesian product (definition 1.1.5), we have (1, 1) and (1, −1) in C, hence, b 6= b0 ; in other words, C doesn’t pass the vertical line test. We conclude that C is not a function. 10. Let f (x) := 1/x2 , x 6= 0, x ∈ R. (a) Determine the direct image f (E) where E := {x ∈ R : 1 ≤ x ≤ 2}. We have f (E) := {x ∈ R : 1/4 ≤ x ≤ 1}. (b) Determine the inverse image f −1 (G) where G := {x ∈ R : 1 ≤ x ≤ 4} We have f −1 (G) := {x ∈ R : −1 ≤ x ≤ −1/2} ∪ {x ∈ R : 1/2 ≤ x ≤ 1}

11. Let g(x) := x2 and f (x) := x + 2 for x ∈ R, and let h be the composite function h := g ◦ f . (a) Find the direct image h(E) of E := {x ∈ R : 0 ≤ x ≤ 1}. h := g ◦ f = g(f (x)) = (x + 2)2 . We have h(E) = {x ∈ R : 4 ≤ x ≤ 9}. (b) Find the inverse image h−1 (G) of G := {x ∈ R : 0 ≤ x ≤ 4}. We have h−1 (G) := {x ∈ R : −4 ≤ x ≤ 0}. 12. Let f (x) := x2 for x ∈ R, and let E := {x ∈ R : −1 ≤ x ≤ 0} and F := {x ∈ R : 0 ≤ x ≤ 1}. Show that E ∩ F = {0} and f (E ∩ F ) = {0}, while f (E) = f (F ) = {y ∈ R : 0 ≤ y ≤ 1}. Hence f (E ∩ F ) is a proper subset of f (E) ∩ f (F ). What happens if 0 is deleted from the sets E and F ? Since 0 is the only common element between E and F , then E ∩ F = {0} ; and f (E ∩ F ) = f (0) = 02 = 0, then f (E ∩ F ) = {0}. If 0 is deleted from E and F , then E ∩ F := ∅ and f (E ∩ F ) is not defined. 13. Let f and E, F be as in Exercice 12. Find the sets E\F and f (E)\f (F ) and show that it is not true that f (E\F ) ⊆ f (E)\f (F ). We have E\F = {x ∈ R : −1 ≤ x < 0}, f (E) = f (F ) and f (E)\f (F ) = ∅, therefore f (E\F ) ⊆ f (E)\f (F ). 14. Show that if f : A −→ B and E, F are subsets of A, then f (E ∪ F ) = f (E) ∪ f (F ) and f (E ∩ F ) ⊆ f (E) ∩ f (F ). D´emonstration. If we suppose y ∈ f (E ∪ F ), by definition 1.1.7, it means that, for y = f (x), there is some x ∈ (E ∪ F ), which means that some x ∈ E or x ∈ F ; this impllies, by definition 1.1.7 again, that y ∈ f (E) of y ∈ f (F ), therefore f (E ∪ F ) ⊆ f (E) ∪ f (F ). Conversely, if we suppose y ∈ (f (E) ∪ f (F )), then y ∈ f (E) or y ∈ f (F ) ; it follows, by definition 1.1.7, that, for y = f (x), there is some x such that x ∈ E or x ∈ F , which means that x ∈ (E ∪ F ). This implies, by definition 1.1.7 again, that y ∈ f (E ∪ F ), then f (E) ∪ f (F ) ⊆ f (E ∪ F ). By definition 1.1.1, we conclude that f (E ∪ F ) = f (E) ∪ f (F ). D´emonstration. Now, if we suppose y ∈ f (E ∩ F ), then, by definition 1.1.7, it implies that for y = f (x), there is some x such that x ∈ (E ∩ F ), which means that x ∈ E and x ∈ F ; this implies, by definition 1.1.7 again, that y = f (E) and y = f (F ), then y ∈ (f (E) ∩ f (F )), therefore f (E ∩ F ) ⊆ f (E) ∩ f (F ). 15. Show that if f : A −→ B and G, H are subsets of B, then f −1 (G ∪ H) = f −1 (G) ∪ f −1 (H) and f −1 (G ∩ H) = f −1 (G) ∩ f −1 (H). D´emonstration. If we suppose x ∈ f −1 (G ∪ H), then, by definition 1.1.7, for x = f −1 (x), there is some f (x) such that f (x) ∈ f (G ∪ H), which means that some f (x) ∈

f (G) or f (x) ∈ f (H) ; this implies, by definition 1.1.7 again, that x ∈ f −1 (G) or x ∈ f −1 (H), then x ∈ f −1 (H) ∪ f −1 (G), therefore f −1 (G ∪ H) ⊆ f −1 (G) ∪ f −1 (H). Conversely, if we suppose x ∈ f −1 (G) ∪ f −1 (H), then x ∈ f −1 (G) or x ∈ f −1 (H) ; it follows, by definition 1.1.7, that for x = f −1 (x), there is some f (x) such that f (x) ∈ f (G) or f (x) ∈ f (H), which means that f (x) ∈ f (G ∪ H) ; this implies, by definition 1.1.7 again, that x ∈ f −1 (G∪H), therefore f −1 (G)∪f −1 (H) ⊆ f −1 (G∪H). By definition 1.1.1, we conclude that f −1 (G ∪ H) = f −1 (G) ∪ f −1 (H). D´emonstration. Now, if we suppose we have x ∈ f −1 (G∩H), then, by definition 1.1.7, for x = f −1 (x), there is some f (x) such that f (x) ∈ f (G ∩ H), which means that some f (x) ∈ f (G) and f (x) ∈ f (H) ; this implies, by definition 1.1.7 again, that x ∈ f −1 (G) and x ∈ f −1 (H), then x ∈ f −1 (G) ∩ f −1 (H), therefore f −1 (G ∩ H) ⊆ f −1 (G) ∩ f −1 (H). Conversely, if we suppose we have x ∈ f −1 (G) ∩ f −1 (H), then, by definition 1.1.7, for x = f −1 (x), there is some f (x) such that f (x) ∈ f (G) ∩ f (H), which means that f (x) ∈ f (G) and x ∈ f (H) ; this implies, by definition 1.1.7 again, that x ∈ f −1 (G) and x ∈ f −1 (H), which means that x ∈ f −1 (G ∩ H), therefore f −1 (G) ∩ f −1 (H) ⊆ f −1 (G∩H). By definition 1.1.1, we conclude that f −1 (G∩H) = f −1 (G)∩f −1 (H). √ 16. Show that the function f defined by f (x) := x/ x2 + 1, x ∈ R is a bijection of R onto {y : −1 < y < 1}. D´emonstration. We have the domain of f , which is A := {x ∈ R}. By, definition 1.1.9, to determine if f is a bijection, we first assume that f is injective and verify that, for all x1 , x2 in the domain of f , if f (x1 ) = f (x2 ), then x1 = x2 . : f (x1 ) = f (x2 ) = q

x1

x2 =q x21 + 1 x22 + 1

x21 x22 = x21 + 1 x22 + 1 x21 (x22 + 1) = x22 (x21 + 1) x21 x22 + x21 = x21 x22 + x22 x21 = x22 |x1 | = |x2 | x1 = x2 (square root > 0 ⇒ numerator signs must agree)

Therefore, f is an injection. We then verify if f is surjective ; we determine the range

of f by solving the equation for y. We have : y=√

x +1

x2

y 2 (x2 + 1) = x2 y 2 x2 + y 2 = x2 −y 2 x2 + x2 = y 2 x2 (1 − y 2 ) = y 2 x= √

y 1 − y2

The range of f is B := {y ∈ R : −1 < y < 1}. Thus, f is a bijection of A onto B. 17. For a, b ∈ R with a < b, find an explicit bijection of A := {x : a < x < b} onto B := {y : 0 < y < 1}. a` faire 18. (a) Let’s take f (x) = x + a and g(x) = x + b, for a 6= b ∈ R. We have (f ◦ g)(x) = x + a + b = (g ◦ f )(x), but f (x) 6= g(x). √ (b) Let’s take f (x) =√ x, g(x) = x and h(x) = x2 , for √ x > 2 0 ∈ R. We have 2 [f ◦ (g + h)](x) = x + x 6= (f ◦ g)(x) + (f ◦ h)(x) = x + x . 19. (a) Show that if f : A −→ B is injective and E ⊆ A, then f −1 (f (E)) = E. Give an example to show that equality need not hold if f is not injective. D´emonstration. In general, by definition 1.1.7, if E is a subset of A, then the direct image of E under f is the subset f (E) := {f (x) : x ∈ E} ; thus, for x ∈ E, we have f (x) ∈ f (E), and by definition 1.1.7, the inverse image of f (E) is f −1 (f (E)) := {x ∈ A : f (x) ∈ f (E)} ; thus, for x ∈ E, we have x ∈ f −1 (f (E)) ∈ A ; therefore, E ⊆ f −1 (f (E)). However, because f is injective, for each x ∈ E, there is, by definition 1.1.9a, an unique f (x) such that f (x) ∈ f (E) ; therefore, for x ∈ f −1 (f (E)), we have x ∈ E, therefore f −1 (f (E)) ⊆ E. By definition 1.1.1, we conclude that f −1 (f (E)) = E. In general, it is true that E ⊆ f −1 (f (E)), but many times f −1 (f (E)) 6⊆ E. For example, let f (x) = sin x ; for x = 0 we have 0 ∈ E, and f (0) = sin 0 = 0, then 0 ∈ f (E), but f −1 (f (E)) = {nπ : n ∈ Z} = 6 E. (b) Show that if f : A −→ B is surjective and H ⊆ B, then f (f −1 (H)) = H. Give an example to show that equality need not hold if f is not surjective. D´emonstration. By definition 1.1.7, if H is a subset of B, the inverse image of H under f is f −1 (H) := {x ∈ A : f (x) ∈ H} ; thus, if f (x) ∈ f (f −1 (H)) for some x ∈ f −1 (H), then f (x) ∈ H, therefore f (f −1 (H)) ⊆ H. Now, suppose we have f (x) ∈ H, then, by definition 1.1.7, we have x ∈ f −1 (H) ∈ A, but f is surjective,

so, by definition 1.1.9b, for every x ∈ A, f (A) = B, then f (f −1 (H)) = B ; thus, if f (x) ∈ H, we have f (x) ∈ f (f −1 (H)), therefore H ⊆ f (f −1 (H)). By definition 1.1.1, we conclude that f (f −1 (H)) = H. In general, it is true that f (f −1 (H)) ⊆ H, but many times H 6⊆ f (f −1 (H)). For example, let f (x) = x2 ; if we suppose there is y = −1 ∈ H, we have f −1 (H) = ∅ and f (f −1 (H)) = ∅ = 6 H. 20.