That is, there is an element

Transcription

1 Section 3.1: Mathematical Induction Let N denote the set of natural numbers (positive integers). N = {1, 2, 3, 4, } Axiom: If S is a nonempty subset of N, then S has a least element. That is, there is an element m S such that m n for all n S.

2 THEOREM 1: (Mathematical Induction) Let S be a subset of N. If S has the following properties: (a) 1 S, and (b) k S implies k + 1 S, then S = N.

3 Examples: 1. Prove that n = n(n + 1) 2 for all n N. Let S be the set of positive integers for which the equation holds Step 1. Show that 1 S.

4 Step 2. Assume that k S.

5 Step 3. Prove that k + 1 S

6 2. Prove that n 1 = 2 n 1 for all n N.

7 Let R be the set of real numbers. Important subsets of R: N = {1, 2, 3,, } the natural numbers J = {0, 1, 1, 2, 2, 3, 3,, } the integers the ra- p Q = : p, q J, q 0 q tional numbers R Q the irrational numbers

8 There is a one-to-one correspondence between the set of real numbers and the set of points on the line This line is called the real line or the number line.

9 Section 3.2. Ordered Fields We consider R together with the arithmetic operations: addition (+) and multiplication ( )

10 Addition: A1. For all x, y R, x + y R (addition is a closed operation). A2. For all x, y R, x + y = y + x (addition is commutative) A3. For all x, y, z R, x + (y + z) = (x + y) + z addition is associative).

11 A4. There is a unique number 0 such that x + 0 = 0 + x = x for all x R. (0 is the additive identity.) A5. For each x R, there is a unique number x R such that x + ( x) = ( x) + x = 0. ( x is the additive inverse of x.)

12 Multiplication: M1. For all x, y R, x y R. (multiplication is a closed) M2. For all x, y R, x y = y x. (multiplication is commutative) M3. For all x, y, z R, x (y z) = (x y). (multiplication is associative).

13 M4. There is a unique number 1 such that x 1 = 1 x = x for all x R. (1 is the multiplicative identity.) M5. For each x R, x 0, there is a unique number 1/x = x 1 R such that x (1/x) = (1/x) x = 1. (1/x is the multiplicative inverse of x.)

14 Distributive Law: D. For all x, y, z R, x (y + z) = x y + x z.

15 A non-empty set S together with two operations, addition and multiplication which satisfies A1-A5, M1-M5, and D is called a field. The set R of real numbers with ordinary addition and multiplication is a field.

16 Other examples of fields 1. The set of rational numbers Q, together with ordinary addition and multiplication, is also a field, a subfield of R. 2. The set of complex numbers C is a field and R C. C = {(a, b) a, b R} C = {a + ib a, b R, i 2 = 1}

17 Theorem: A subset S of a field F is itself a field if it is closed with respect to addition and multiplication, if it contains the additive and multiplicative identities 0 and 1, and if it is closed with respect to additive and multiplicative inverses. 3. S = {a + b 2 a, b Q} is a field, a sub-field of R.

18 Order There is a subset P of R that has the following properties: (a) If x, y P, then x + y P. (b) If x, y P, then x y P. (c) For each x R exactly one of the following holds: x P, x = 0, x P. P is the set of positive numbers.

19 x P is written x > 0. Def. x is less than y (x < y) if and only if y x P i.e., y x > 0. less than or equal to x y greater than y > x greater than or equal to y x

20 Properties of < O1. For all x, y R, exactly one of the following holds: x < y, x = y, x > y. (Trichotomy Law) O2. For all x, y, z R, if x < y and y < z, then x < z.

21 O3. For all x, y, z R, if x < y, then x + z < y + z. O4. For all x, y, z R, if x < y and z > 0, then x z < y z. Note: These properties also hold for >,,.

22 {R, +,, <} is an ordered field. Any mathematical system {S, +,, <} satisfying these 15 axioms is an ordered field. In particular, the set of rational numbers Q, together with ordinary addition, multiplication and less than, is an ordered field, a subfield of R.

23 THEOREM 1: Let x, y R. If x y + ɛ for every positive number ɛ, then x y. Proof:

24 Absolute Value Def. Let x R. The absolute value of x, denoted x, is: x = x, if x 0 x, if x < 0

25 Properties of absolute value: Let x, y R and let a 0. Then (a) x 0. (b) x a iff a x a. (c) xy = x y. (d) x + y x + y. (Triangle inequality)

26 Section 3.3. Boundedness and the Completeness Axiom There is a one-to-one correspondence between the set R of real numbers and the points P on the number line. There are two basic types of real numbers: 1. The rational numbers Q, and 2. the irrational numbers R Q.

27 2 is a point on the number line, i.e., 2 is a real number. 2 is not a rational number.

28 In general, if r = p/q is a rational number which is not a perfect square, then p/q is an irrational number.

29 Def. Let S R, S Ø. S is bounded above if there exists a number m such that s m for all s S. m is called an upper bound for S. S is bounded below if there exists a number k such that s k for all s S. k is a lower bound for S. S is bounded if it is bounded above and below.

30 Examples: 1. S = (, 0] (1, 2] 2. S = ( 1, ] 3. S = {1, 1 2, 1 3,, 1 n, } 4. S = {sin(nπ/4)}, n = 1, 2, 3,

31 Thoerem 1: Let S R. If m is an upper bound for S, then any number p > m is also an upper bound for S. If k is a lower bound for S, then any number q < k is also a lower bound for S. Proof:

32 Def. Let S R, S Ø, be bounded above. A number α is the supremum or least upper bound for S if: 1. α is an upper bound for S, and 2. α m where m is any other upper bound for S. Notation: α = sup S or α = lub S

33 Def. Let S R, S Ø, be bounded below. A number β is the infimum (greatest lower bound) of S if: 1. β is a lower bound for S, and 2. β k where k is any other lower bound for S. Notation: β = inf S or β = glb S

34 Def. Let S R, S Ø. If there is an element p S such that s p for all s S, then p is the maximum (or largest element) of S. (p = max S) Def. Let S R, S Ø. If there is an element q S s q for all s S, then q is the minimum (or smallest element) of S. (q = min S)

35 THEOREM 2: Let S R, S Ø, be bounded above. If α 1 = sup S and α 2 = sup S, then α 1 = α 2. That is, the supremum of a set is unique. Proof:

36 The same result holds for inf, max, and min. Examples: 1. S = (0, 1] 2. S = { 1} [0, 1) 3. S = {1, 1 2, 1 3,, 1 n, }

37 4. S = N 5. S = (, 2] {5} 6. S = 1 + ( 1)n n

38 7. S = 1 + ( 1) n 2 8. S = {cos(nπ/3)} 9. S = { } 1 n + cos(nπ/3)

39 THEOREM 3: If α = sup S and m < α, then there exists an element x S such that m < x α. Proof:

40 Theorem 3 If α = sup S and ɛ is any positive number, then there exists an element x S such that α ɛ < x α.

41 THEOREM 4: If β = inf S and k > β, x S then there exists an element such that β x < k. THEOREM 4 : If β = inf S and ɛ is any positive number, then there exists an element x S such that β x < β + ɛ.

42 The Completeness Axiom: Let S R, S Ø. If S is bounded above, then S has a least upper bound. If S is bounded below, then S has a greatest lower bound. The real number system {R, +,, <} is a complete ordered field.

43 Notes: The rational number system {Q, +,, <} is not complete. S = {r Q r < 2} is bounded above. S does not have a (rational) least upper bound. The complex number system {C, +, } is complete; it is not ordered.

44 Let S R, S Ø. Suppose that S is bounded above. sup S may or not be an element of S. Similarly for inf S. Examples:

45 THEOREM 5: Let S R, S Ø, be bounded above and let α = sup S. (a) If α S, then α = max S. Proof:

46 (b) If α / S, then S does not have a maximum element. Proof:

47 THEOREM 6: (Archimedean Property) N is not bounded above. Proof:

48 Corollaries re-statements of the theorem 1. For each z R there exists an n z N such that n z > z. 2. For each x > 0 and for each y R there exists an n N such that nx > y. 3. For each x > 0 there exists n N such that 0 < 1 n < x.

49 THEOREM 7: : Thm Thm 6 Proof:

50 THEOREM 8: Given any positive number y, there is a corresponding positive integer n such that n 1 y < n. Proof:

51 Density of the rationals THEOREM 9: Let x, y R, x < y. There exists a rational number r such that x < r < y. Proof:

52 Corollary: Let x, y R, x < y. There exists an irrational number t such that x < t < y. Proof:

53 Section 3.4: Topology of R Points and Neighborhoods Let x R. Def. An ɛ-neighborhood of x is the set N(x, ɛ) = {y R y x < ɛ} Note: N(x, ɛ) is the open interval (x ɛ, x + ɛ). centered at x with radius ɛ.

54 Neighborhoods and Deleted Neighborhoods Let x R. Def. An ɛ-neighborhood of x is the set N(x, ɛ) = {y R y x < ɛ} Note: N(x, ɛ) is the open interval (x ɛ, x + ɛ). centered at x with radius ɛ.

55 Def. A deleted ɛ-neighborhood of x is the set N (x, ɛ) = {y R 0 < y x < ɛ} Note: N (x, ɛ) = (x ɛ, x) (x, x + ɛ).

56 Let S R. Def. A point x is an interior point of S if there exists a neighborhood N(x) such that N(x) S. The interior of S is the set int S = {x S x is an interior point of S}

57 Negation: Let x S. x is not an interior point of S if there is no neighborhood N(x) which is contained in S. That is, for every neighborhood N(x) of x, N(x) S c.

58 1. S = (0, 3) 2. S = { 1} [0, 1) 3. S = {1, 1 2, 1 3,, 1 n, }

59 4. S = N 5. S = (, 2] {5} 6. S = { 0, 1 2, 2 3, 3 4,, n 1 } n, (2, 4] [5, )

60 Def. A point x is a boundary point of S if every neighborhood N(x) of x contains points of S and points of R/S = S c (the complement of S). That is, x is a boundary point of S if N(x) S and N(x) S c for every neighborhood N(x) of x.

61 The boundary of S is the set bd S = {x S x is a boundary point of S} Negation: A point x is not a boundary point of S if there is at least one neighborhood N(x) such that either N(x) S = or N(x) S c =

62 1. S = (0, 3) 2. S = { 1} [0, 1) 3. S = {1, 1 2, 1 3,, 1 n, }

63 4. S = N 5. S = (, 2] (2, 3) {5} 6. S = { 0, 1 2, 2 3, 3 4,, n 1 } n, (2, 4] (4, )

64 Open Sets/Closed Sets Let S R Def. S is open if every point x S is an interior point of S. Def. S is closed if S c is open. Examples:

65 1. S = (0, 3) S = R = (, ) S = the empty set. 2. S = { 1} (0, 2) 3. S = {1, 1 2, 1 3,, 1 n, }

66 4. S = N 5. S = [0, 1] [4, 5] 6. S = { 0, 1 2, 2 3, 3 4,, n 1 } n, (2, 4] [5, )

67 THEOREM 1: S is closed iff bd S S. Proof:

68 THEOREM 2: If F is a collection of open sets, then S = A F A is open. Proof:

69 THEOREM 3: If G = {A 1, A 2,, A n } is a finite collection of open sets, then S = A 1 A 2 A n is open. Proof:

70 Note: The intersection of an infinite collection of collection of open sets may not be open; the union of an infinite collection of closed sets may not be closed. Examples: 1. n=1 ( 1 n, ) n = [0, 1] 2. n=1 [ 1 n, 1 1 n ] = (0, 1)

71 DeMorgan s Laws: (a) (A B) c = A c B c (b) (A B) c = A c B c These laws extend to any collection of sets ( A α ) c = A c α ( A α ) c = A c α

72 Corollary: The intersection of any collection of closed sets is closed; the union of any finite collection of closed sets is closed.

73 Accumulation Points/Isolated Points Let S R Def. A point x R is an accumulation point of S if every deleted neighborhood N (x) contains a point s S. Equivalently, x is an accumulation point of S if every neighborhood N(x) contains a point s S, s x. S = {x R x is an accum. point of S}

74 Accumulation Points Let S R Def. A point x R is an accumulation point of S if every deleted neighborhood N (x, ɛ) contains a point s S. Equivalently, x is an accumulation point of S if for every ɛ > 0, the neighborhood N(x, ɛ) contains a point s S such that s x. S = {x R x is an accum. point of S}

75 Negation: x is not an accumulation point of S if there is at least one deleted neighborhood N (x) such that N (x) S =.

76 Examples: 1. S = {1, 1 2, 1 3,, 1 n, } 2. S = { 1} [0, 1) 3. S = {sin(nπ/6)} n = 1, 2, 3,... S = { 1 n + sin ( nπ 6 )}, n = 1, 2, 3,...

77 4. S = N 5. S = (, 2] (2, 3) {5} 6. S = { 0, 1 2, 2 3, 3 4,, n 1 } n, (2, 4] (4, ) Note: an accumulation point of S may or may not be an element of S.

78 Def. A point x S is an isolated point of S if there is a deleted neighborhood N (x) such that N (x) S = Ø. That is, x S but x / S. Note: an isolated point of S is a boundary point of S.

79 Negation: A point x S is not an isolated point of S if N (x) S for every deleted neighborhood N (x).

80 Examples: 1. S = (0, 3) 2. S = { 1} [0, 1) 3. S = {1, 1 2, 1 3,, 1 n, }

81 4. S = {r Q : 0 < r < 5} 5. S = (, 2] (2, 3) {5} 6. S = { 0, 1 2, 2 3, 3 4,, n 1 } n, (2, 4] (4, )

82 Def. The closure of S is the set cl S = S S Note: If x cl S and N(x) is any neighborhood of x, then N(x) S Ø.

83 Examples: 1. S = (0, 3) 2. S = { 1} [0, 1) 3. S = {1, 1 2, 1 3,, 1 n, }

84 4. S = N 5. S = (, 2] (2, 3) {5} 6. S = { 0, 1 2, 2 3, 3 4,, n 1 } n, (2, 4] (4, )

85 THEOREM 4: S is closed iff S S. Proof:

86 THEOREM 5: cl S is a closed set. Proof:

87 Let S be bounded above and let α = sup S. True/False: 1.If α / S, then α is an accumulation point of S. 2.If α / S, then S does not have a maximum. 3.If α S, then α is an accumulation point of S.

88 4.If α S, then α = max S. 5.If p is an accumulation point of S, then p α. Restate these in terms of bounded below and inf.

89 Section 3.5. Compact Sets Let S R. Def. A collection G of open sets such that S G G G is called an open cover of S. A subset F of G which also covers S is called a subcover.

90 Def. S is compact if for every open cover G of S there is a finite set G 1, G 2,, G n of elements of G such that S n i=1 G i That is, S is compact iff every open cover of S has a finite subcover.

91 Examples:

92 THEOREM 1: If S Ø is closed and bounded, then has a maximum m and a minimum k. Proof:

93 THEOREM 2: (Heine-Borel) S R is compact iff it is closed and bounded. Proof:

94 THEOREM 3: (Bolzano-Weierstrass) If S is a bounded infinite set, then S has at least one accumulation point. Proof: