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) Conversely. which means that x ∈ A or B and x ∈ A or C. so x ∈ A\(B ∩ C) ..1. then for every x ∈ A. therefore. so. therefore (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C). therefore A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C).1. hence. . For every x ∈ A ∪ (B ∩ C). show that A ⊆ B if and only if A ∩ B = A. we have x ∈ (A ∩ B). so x ∈ (A ∪ B) ∩ (A ∪ C). Prove the distributive laws : (a) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) D´emonstration. Conversely. x ∈ A ∩ B or x ∈ A ∩ C. For every x ∈ A\(B ∩ C).1. therefore A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C). therefore x ∈ / (B ∩ C). by definition 1. also. 4. For every x ∈ A ∩ (B ∪ C). 5. x ∈ A or x ∈ B and C. x ∈ (A ∩ B). Conversely. therefore (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C). if A ⊆ (A ∩ B) and (A ∩ B) ⊆ A.1.1. D´emonstration.1. which means that x ∈ B or x ∈ C.1. we conclude that A\(B ∩ C) ⊆ (A\B) ∪ (A\C).1. for every x ∈ (A ∩ B) ∪ (A ∩ C).4(b)] : A\(B ∩ C) = (A\B) ∪ (A\C). (b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) D´emonstration.. for every x ∈ A. x ∈ (A\B) or x ∈ (A\C). if (A∩B) = A.1. for every x ∈ (A\B) ∪ (A\C). therefore (A ∩ B) ⊆ A . we have A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). x ∈ (A ∪ B) and x ∈ (A ∪ C). so x ∈ A ∩ B or x ∈ A ∩ C. If A and B are sets. x ∈ (A\B) ∪ (A\C) . then x ∈ B. By definition 1. therefore A ⊆ B. x ∈ A but x ∈ / (B ∩ C). By definition 1. we have A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). A ⊆ (A∩B) . which means that x ∈ A ∪ (B ∩ C). which means that x ∈ (A\B) or x ∈ (A\C). we have A\(B ∩ C) = (A\B) ∪ (A\C)..3a. by definition 1.1. if A ⊆ B. x ∈ A and x ∈ (B ∪ C). x ∈ A. we conclude that (A\B) ∪ (A\C) ⊆ A\(B ∩ C). wich means that x ∈ A ∩ (B ∪ C). therefore A ⊇ (A ∩ B) .1. A ∩ B = A. for every x ∈ (A ∩ B). By definition 1.A B 3. By definition 1. Conversely. Prove the second De Morgan Law [Theorem 1. for every x ∈ (A ∪ B) ∩ (A ∪ C).1. D´emonstration. so x ∈ A and x ∈ B or x ∈ C. 1. if x ∈ (A∪B) but x ∈ / (A∩B). Since D is the set of all elements that belon to either A or B but not both. 1) and (1. −1) in C. that consists of all natural numbers divisible by 6. for every x ∈ A and x ∈ / B. C doesn’t pass the vertical line test.3c. We have A = {An : n ∈ N} = N\1. For each n ∈ N. let An = {(n + 1)k : k ∈ N}. By definition 1. But here. D´emonstration. 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. (a) What is A1 ∩ A2 ? A1 = {2k : k ∈ N}.5).1. we have A ∪ B. and A2 = {3k : k ∈ N}.1. We conclude that C is not a function.6.6. By definition 1. for every x ∈ B and x ∈ / A. D´emonstration. A B (a) Show that D = (A\B) ∪ (B\A)). We have C ⊆ A×B.1. with f ⊆ A × B. By definition 1. 7.1.1.3a. Since every x ∈ D is the set of all elements that belong to either A or B but not both. x 6= 0. a` d´eterminer 8. For B = {An : n ∈ N}. (a) Determine the direct image f (E) where E := {x ∈ R : 1 ≤ x ≤ 2}. by definition 1. we have D = (A∪B)\(A∩B). Let f (x) := 1/x2 . then A1 ∩ A2 = {6k : k ∈ N}. that consists of all natural numbers divisible by 3 . b0) ∈ f . we have A\B . We have f (E) := {x ∈ R : 1/4 ≤ x ≤ 1}. that consists of all natural numbers divisible by 2. hence. it implies that b = b0. b) ∈ f and (a. (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} . for every x ∈ A or x ∈ B. if (a. likewise. we conclude that D = (A\B) ∪ (B\A)). a` dessiner un jour S T 9. by the cartesian product (definition 1. x ∈ R. we have B\A. (b) Determine the sets S {An : n ∈ N} and T {An : n ∈ N}. a ∈ A and b ∈ B. in other words. (b) Show that D = (A ∪ B)\(A ∩ B).3a. we have (1.3c. 10. b 6= b0 . by definition 1. Represent D with a diagram. If 0 is deleted from E and F . then f (E ∩ F ) = {0}. it follows. it implies that for y = f (x). then f (E ∪ F ) = f (E) ∪ f (F ) and f (E ∩ F ) ⊆ f (E) ∩ f (F ). D´emonstration. therefore f (E\F ) ⊆ f (E)\f (F ). Let f and E. f (E) = f (F ) and f (E)\f (F ) = ∅.7.1. this impllies. it means that.11. 14.1.1. which means that some x ∈ E or x ∈ F . We have E\F = {x ∈ R : −1 ≤ x < 0}. while f (E) = f (F ) = {y ∈ R : 0 ≤ y ≤ 1}. by definition 1. By definition 1. then y ∈ f (E) or y ∈ f (F ) . then E ∩ F := ∅ and f (E ∩ F ) is not defined. which means that x ∈ (E ∪ F ). 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 ).1. h := g ◦ f = g(f (x)) = (x + 2)2 . 13. Show that if f : A −→ B and G.7. This implies. Conversely. that y ∈ f (E) of y ∈ f (F ).1. then E ∩ F = {0} . Let g(x) := x2 and f (x) := x + 2 for x ∈ R. for y = f (x). for y = f (x). then f −1 (G ∪ H) = f −1 (G) ∪ f −1 (H) and f −1 (G ∩ H) = f −1 (G) ∩ f −1 (H). If we suppose x ∈ f −1 (G ∪ H). therefore f (E ∩ F ) ⊆ f (E) ∩ f (F ). by definition 1. D´emonstration. (a) Find the direct image h(E) of E := {x ∈ R : 0 ≤ x ≤ 1}. then y ∈ (f (E) ∩ f (F )). therefore f (E ∪ F ) ⊆ f (E) ∪ f (F ). H are subsets of B. What happens if 0 is deleted from the sets E and F ? Since 0 is the only common element between E and F . if we suppose y ∈ f (E ∩ F ). by definition 1. 12. Hence f (E ∩ F ) is a proper subset of f (E) ∩ f (F ). there is some f (x) such that f (x) ∈ f (G ∪ H). and f (E ∩ F ) = f (0) = 02 = 0. by definition 1. and let h be the composite function h := g ◦ f .1. Show that E ∩ F = {0} and f (E ∩ F ) = {0}. We have h−1 (G) := {x ∈ R : −4 ≤ x ≤ 0}. by definition 1. We have h(E) = {x ∈ R : 4 ≤ x ≤ 9}. 15. that. which means that x ∈ E and x ∈ F .7 again. D´emonstration. this implies.1. for x = f −1 (x). there is some x such that x ∈ (E ∩ F ).1.7. if we suppose y ∈ (f (E) ∪ f (F )). which means that some f (x) ∈ . there is some x such that x ∈ E or x ∈ F . that y ∈ f (E ∪ F ). Now.7 again. F be as in Exercice 12. then.7. then f (E) ∪ f (F ) ⊆ f (E ∪ F ). by definition 1. If we suppose y ∈ f (E ∪ F ). that y = f (E) and y = f (F ). Let f (x) := x2 for x ∈ R. Show that if f : A −→ B and E. and let E := {x ∈ R : −1 ≤ x ≤ 0} and F := {x ∈ R : 0 ≤ x ≤ 1}. (b) Find the inverse image h−1 (G) of G := {x ∈ R : 0 ≤ x ≤ 4}.7 again. we conclude that f (E ∪ F ) = f (E) ∪ f (F ). by definition 1. F are subsets of A. there is some x ∈ (E ∪ F ). then.1. which means that f (x) ∈ f (G) and x ∈ f (H) .7. by definition 1. this implies. therefore f −1 (G) ∩ f −1 (H) ⊆ f −1 (G∩H). if we suppose we have x ∈ f −1 (G∩H). for x = f −1 (x). Conversely. by definition 1. there is some f (x) such that f (x) ∈ f (G ∩ H). then x ∈ f −1 (G) ∩ f −1 (H). it follows.1. √ 16. D´emonstration. if we suppose we have x ∈ f −1 (G) ∩ f −1 (H).7 again.f (G) or f (x) ∈ f (H) . there is some f (x) such that f (x) ∈ f (G) or f (x) ∈ f (H). We then verify if f is surjective . which is A := {x ∈ R}. if we suppose x ∈ f −1 (G) ∪ f −1 (H).1. this implies.7 again. which means that some f (x) ∈ f (G) and f (x) ∈ f (H) . for x = f −1 (x).1. by definition 1.1.1.1. that x ∈ f −1 (G) and x ∈ f −1 (H). then. x2 in the domain of f .1. therefore f −1 (G ∩ H) ⊆ f −1 (G) ∩ f −1 (H). that x ∈ f −1 (G∪H). by definition 1.1. by definition 1. Show that the function f defined by f (x) := x/ x2 + 1. We have the domain of f . then x ∈ f −1 (G) or x ∈ f −1 (H) . then x1 = x2 . we first assume that f is injective and verify that.1. if f (x1 ) = f (x2 ). x ∈ R is a bijection of R onto {y : −1 < y < 1}.1. there is some f (x) such that f (x) ∈ f (G) ∩ f (H). : 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. which means that x ∈ f −1 (G ∩ H). f is an injection. this implies. then. that for x = f −1 (x). definition 1. that x ∈ f −1 (G) or x ∈ f −1 (H). by definition 1. by definition 1. that x ∈ f −1 (G) and x ∈ f −1 (H). By definition 1.7.7 again. therefore f −1 (G)∪f −1 (H) ⊆ f −1 (G∪H). Now.7.7 again.9. By.1. By definition 1. we determine the range . Conversely. D´emonstration. for all x1 . to determine if f is a bijection. therefore f −1 (G ∪ H) ⊆ f −1 (G) ∪ f −1 (H). this implies. we conclude that f −1 (G ∪ H) = f −1 (G) ∪ f −1 (H).1. then x ∈ f −1 (H) ∪ f −1 (G). we conclude that f −1 (G∩H) = f −1 (G)∩f −1 (H). which means that f (x) ∈ f (G ∪ H) . if f (x) ∈ f (f −1 (H)) for some x ∈ f −1 (H). we have f (x) ∈ f (E). but f −1 (f (E)) = {nπ : n ∈ Z} = 6 E. thus. we conclude that f −1 (f (E)) = E.7. thus. a` faire 18. and f (0) = sin 0 = 0.7. By definition 1. then f (f −1 (H)) = H. Give an example to show that equality need not hold if f is not injective. but many times f −1 (f (E)) 6⊆ E. √ (b) Let’s take f (x) =√ x. (a) Let’s take f (x) = x + a and g(x) = x + b. suppose we have f (x) ∈ H.1. 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}. 19. For example. D´emonstration. We have 2 [f ◦ (g + h)](x) = x + x 6= (f ◦ g)(x) + (f ◦ h)(x) = x + x . the inverse image of H under f is f −1 (H) := {x ∈ A : f (x) ∈ H} . (a) Show that if f : A −→ B is injective and E ⊆ A. then the direct image of E under f is the subset f (E) := {f (x) : x ∈ E} .1. for x ∈ f −1 (f (E)). We have (f ◦ g)(x) = x + a + b = (g ◦ f )(x). (b) Show that if f : A −→ B is surjective and H ⊆ B. but f (x) 6= g(x). the inverse image of f (E) is f −1 (f (E)) := {x ∈ A : f (x) ∈ f (E)} .7. by definition 1. and by definition 1. Thus. For a. if H is a subset of B. but f is surjective. for each x ∈ E. . E ⊆ f −1 (f (E)).of f by solving the equation for y. if E is a subset of A. However. for a 6= b ∈ R. 17.9a. we have x ∈ f −1 (f (E)) ∈ A . for x ∈ E. it is true that E ⊆ f −1 (f (E)). D´emonstration. therefore f (f −1 (H)) ⊆ H.7. Give an example to show that equality need not hold if f is not surjective. we have x ∈ f −1 (H) ∈ A. for x ∈ E.1.1. by definition 1.1. In general. find an explicit bijection of A := {x : a < x < b} onto B := {y : 0 < y < 1}. there is. In general. for x = 0 we have 0 ∈ E. b ∈ R with a < b.1. g(x) = x and h(x) = x2 . by definition 1. because f is injective. for √ x > 2 0 ∈ R. then. thus. then 0 ∈ f (E). therefore. then f −1 (f (E)) = E. Now. let f (x) = sin x . an unique f (x) such that f (x) ∈ f (E) . therefore.1. therefore f −1 (f (E)) ⊆ E. we have x ∈ E. then f (x) ∈ H. By definition 1. f is a bijection of A onto B. 1. thus.9b. if f (x) ∈ H. by definition 1. for every x ∈ A. but many times H 6⊆ f (f −1 (H)). we have f −1 (H) = ∅ and f (f −1 (H)) = ∅ = 6 H. we have f (x) ∈ f (f −1 (H)). we conclude that f (f −1 (H)) = H.1. f (A) = B. let f (x) = x2 . if we suppose there is y = −1 ∈ H. therefore H ⊆ f (f −1 (H)).1. In general. By definition 1.so. For example. it is true that f (f −1 (H)) ⊆ H. . 20. then f (f −1 (H)) = B .
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