By Charles Hayes

ISBN-10: 047136472X

ISBN-13: 9780471364726

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IfA C = BuC,thenA =B. =B. 41. 42. A A c B if and A B = 48. (A — B) only if A Ac u B= B (A u A u B. B are disjoint. A (B — A) = 49. 50. (A —B)u(B—C)=A —C. 51. (A —B)u(A —A) =A uB. 53. IfA —B=C,thenA =BuC. 54. If A c: B u C then Ac: Bor A c: C. 55. Show that x (x) for all Hint: If x = (x), then show that x must be a member of; which is impossible. 56. If A, then (x) c: A. 57. If A c C, B c C, then C — (A u B) = (C — A) (C — B) and C — (A n B) (C — A) u (C — B). 58. If A C and B c C then A = 0 if and only if B = (A n (C — B)) u 59.

5 In the following problems you may assume and use elementary facts of algebra. 23. Prove that 0 c A, where A is an arbitrary set, by assuming the contrary and showing that a contradiction arises. 24. If A = {x x is the square of some positive integer) and B = {x I x is the cube of some positive integer), prove that A 4: B and B 4: A. Show that 64 E A and 64 e B. Find a few other numbers belonging to both sets. 25. Given A = {x x is the square of some positive integer}, B = {x x is the f fourth power of some positive integer), prove that A (that is, B is a proper subset of A).

R e S. Any number z satisfying this condition is called a lower bound of S. In case S is bounded both above and below, we say simply that S is bounded. Now we let T= is an upper bound of S}, V = {z z is a lower bound of S}. — Now T 0 if and only ifS is bounded above; and V 0 if and only if Sis bounded below. 4). If Tdoes have a smallest member t, then clearly I is an upper bound of S, and so it is the smallest of all upper bounds of S. b. S or sup 5, respectively. Similarly, if V has a largest member v, then v is a lower bound of S, and it is evidently the biggest of all lower bounds of S.

### Concepts of Real Analysis by Charles Hayes

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