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

(b) A\(A\B)

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.

Demonstration. By definition 1.1.1, if (AB) = A, A (AB) ; so, for every x A,we have x (A B), then x B, therefore A B. Conversely, if A B, then forevery 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).

Demonstration. For every x A\(B C), x A but x / (B C), which means thatx (A\B) or x (A\C), therefore, by definition 1.1.3a, x (A\B) (A\C) ; weconclude 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 concludethat (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)

Demonstration. For every x A (B C), x A and x (B C), which meansthat 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 orx 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 haveA (B C) = (A B) (A C).

(b) A (B C) = (A B) (A C)

Demonstration. For every x A (B C), x A or x B and C, whichmeans that x A or B and x A or C, so x (A B) (A C), thereforeA (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 belongto either A or B but not both. Represent D with a diagram.

A B

(a) Show that D = (A\B) (B\A)).

Demonstration. 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 setof 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).

Demonstration. 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, bydefinition 1.1.3c, if x (AB) but x / (AB), we haveD = (AB)\(AB).

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, andA2 = {3k : k N}, that consists of all natural numbers divisible by 3 ; thenA1 A2 = {6k : k N}, that consists of all natural numbers divisible by 6.

(b) Determine the sets{An : n N} and {An : n N}.

We have A = {An : n N} = N\1. For B = {An : n N}, a` determiner8. a` dessiner un jour

9. We have C AB, a A and b B. By definition 1.1.6, if (a, b) f and (a, b) f ,with f AB, it implies that b = b. But here, by the cartesian product (definition1.1.5), we have (1, 1) and (1,1) in C, hence, b 6= b ; in other words, C doesnt passthe 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 f1(G) where G := {x R : 1 x 4}

We have f1(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 functionh := 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 h1(G) of G := {x R : 0 x 4}.

We have h1(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 isdeleted from the sets E and F ?

Since 0 is the only common element between E and F , then E F = {0} ; andf(E F ) = f(0) = 02 = 0, then f(E F ) = {0}. If 0 is deleted from E and F , thenE 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 thatit 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 ) = , thereforef(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 ).

Demonstration. If we suppose y f(E F ), by definition 1.1.7, it means that, fory = f(x), there is some x (E F ), which means that some x E or x F ; thisimpllies, 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 orx F , which means that x (E F ). This implies, by definition 1.1.7 again, thaty f(E F ), then f(E) f(F ) f(E F ). By definition 1.1.1, we conclude thatf(E F ) = f(E) f(F ).

Demonstration. Now, if we suppose y f(EF ), then, by definition 1.1.7, it impliesthat for y = f(x), there is some x such that x (E F ), which means that x Eand x F ; this implies, by definition 1.1.7 again, that y = f(E) and y = f(F ), theny (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 f1(G H) = f1(G) f1(H) and f1(G H) = f1(G) f1(H).

Demonstration. If we suppose x f1(G H), then, by definition 1.1.7, for x =f1(x), there is some f(x) such that f(x) f(GH), which means that some f(x)

f(G) or f(x) f(H) ; this implies, by definition 1.1.7 again, that x f1(G) orx f1(H), then x f1(H) f1(G), therefore f1(G H) f1(G) f1(H).Conversely, if we suppose x f1(G) f1(H), then x f1(G) or x f1(H) ;it follows, by definition 1.1.7, that for x = f1(x), there is some f(x) such thatf(x) f(G) or f(x) f(H), which means that f(x) f(G H) ; this implies, bydefinition 1.1.7 again, that x f1(GH), therefore f1(G)f1(H) f1(GH).By definition 1.1.1, we conclude that f1(G H) = f1(G) f1(H).

Demonstration. Now, if we suppose we have x f1(GH), then, by definition 1.1.7,for x = f1(x), there is some f(x) such that f(x) f(GH), which means that somef(x) f(G) and f(x) f(H) ; this implies, by definition 1.1.7 again, that x f1(G)and x f1(H), then x f1(G)f1(H), therefore f1(GH) f1(G)f1(H).Conversely, if we suppose we have x f1(G) f1(H), then, by definition 1.1.7,for x = f1(x), there is some f(x) such that f(x) f(G) f(H), which means thatf(x) f(G) and x f(H) ; this implies, by definition 1.1.7 again, that x f1(G)and x f1(H), which means that x f1(G H), therefore f1(G) f1(H) f1(GH). By definition 1.1.1, we conclude that f1(GH) = f1(G)f1(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}.

Demonstration. We have the domain of f , which is A := {x R}. By, definition1.1.9, to determine if f is a bijection, we first assume that f is injective and verifythat, for all x1, x2 in the domain of f , if f(x1) = f(x2), then x1 = x2. :

f(x1) = f(x2) =x1x21 + 1

= x2x22 + 1

x21x21 + 1

= x22

x22 + 1x21(x22 + 1) = x22(x21 + 1)x21x

22 + x21 = x21x22 + 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 = xx2 + 1

y2(x2 + 1) = x2

y2x2 + y2 = x2

y2x2 + x2 = y2x2(1 y2) = y2

x = y1 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} ontoB := {y : 0 < y < 1}.

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18. (a) Lets 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) Lets take f(x) =x, g(x) = x and h(x) = x2, for x > 0 R. We have

[f (g + h)](x) = x+ x2 6= (f g)(x) + (f h)(x) = x+ x2.19. (a) Show that if f : A B is injective and E A, then f1(f(E)) = E. Give an

example to show that equality need not hold if f is not injective.

Demonstration. In general, by definition 1.1.7, if E is a subset of A, then thedirect 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) isf1(f(E)) := {x A : f(x) f(E)} ; thus, for x E, we have x f1(f(E)) A ; therefore, E f1(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 f1(f(E)), we have x E, therefore f1(f(E)) E. By definition1.1.1, we conclude that f1(f(E)) = E.

In general, it is true that E f1(f(E)), but many times f1(f(E)) 6 E. Forexample, let f(x) = sin x ; for x = 0 we have 0 E, and f(0) = sin 0 = 0, then0 f(E), but f1(f(E)) = {npi : n Z} 6= E.

(b) Show that if f : A B is surjective and H B, then f(f1(H)) = H. Give anexample to show that equality need not hold if f is not surjective.

Demonstration. By definition 1.1.7, if H is a subset of B, the inverse image ofH under f is f1(H) := {x A : f(x) H} ; thus, if f(x) f(f1(H)) for somex f1(H), then f(x) H, therefore f(f1(H)) H. Now, suppose we havef(x) H, then, by definition 1.1.7, we have x f1(H) A, but f is surjective,

so, by definition 1.1.9b, for every x A, f(A) = B, then f(f1(H)) = B ; thus,if f(x) H, we have f(x) f(f1(H)), therefore H f(f1(H)). By definition1.1.1, we conclude that f(f1(H)) = H.

In general, it is true that f(f1(H)) H, but many times H 6 f(f1(H)). Forexample, let f(x) = x2 ; if we suppose there is y = 1 H, we have f1(H) = and f(f1(H)) = 6= H.

20.