Lecture 17Section 10.1 Least Upper Bound Axiom

Size: px

Start display at page:

Download "Lecture 17Section 10.1 Least Upper Bound Axiom"

Dortha Paula Cannon
5 years ago
Views:

1 Lecture 7Sectio 0. Least Upper Boud Axiom Sectio 0.2 Sequeces of Real Numbers Jiwe He Real Numbers. Review Basic Properties of R: R beig Ordered Classificatio N = {0,, 2,...} = {atural umbers} Z = {..., 2,, 0,, 2,..., } = {itegers} Q = { p q : p, q Z, q 0} = {ratioal umbers} R = {real umbers} = Q {irratioal umbers (π, 2,...)} R is A Ordered Field x y ad y z x z. x y ad y x x = y. x, y R x y or y x. x y ad z R x + z y + z. x 0 ad y 0 xy 0. Archimedea Property ad Dedekid Cut Axiom Archimedea Property x > 0, y > 0, N such that x > y. Dedekid Cut Axiom Let E ad F be two oempty subsets of R such that E F = R;

2 E F = ; x E, y F, we have x y. The, z R such that x z, x E ad z y, y F. Least Upper Boud Theorem Every oempty subset S of R with a upper boud has a least upper boud (also called supremum)..2 Least Upper Boud Basic Properties of R: Least Upper Boud Property Defiitio. Let S be a oempty subset S of R. S is bouded above if M R such that x M for all x S; M is called a upper boud for S. S is bouded below if m R such that x m for all x S; m is called a lower boud for S. S is bouded if it is bouded above ad below. Least Upper Boud Theorem Every oempty subset S of R with a upper boud has a least upper boud (also called supremum). Proof. Let F = {upper bouds for S} ad E = R \ E (E, F ) is a Dedekid cut b R such that x b, x E ad b y, y F ; b is also a upper boud of S b is the lub of S. Supremum or Ifimum of a Set S Defiitio 2. Let S be a oempty subset of R with a upper boud. We deote by sup(s) or lub(s) the supremum or least upper boud of S. Theorem 3. Let M = sup(s). The x M, x S; 2

3 ɛ > 0, (M ɛ, M] S Defiitio 4. Let S be a oempty subset of R with a lower boud. We deote by if(s) or glb(s) the ifimum or greatest lower boud of S. Theorem 5. Let m = if(s). The x m, x S; ɛ > 0, [m, m + ɛ] S s: Supremum or Ifimum of a Set S s 6. Every fiite subset of R has both upper ad lower bouds: sup{, 2, 3} = 3, if{, 2, 3} =. If a < b, the b = sup[a, b] = sup[a, b) ad a = if[a, b] = if(a, b]. If S = {q Q : e < q < π}, the if S = e, sup S = π. If S = {x R : x 2 < π}, the if S = 3, sup S = 3. If S = {x Q : x 2 < π}, the if S = 3, sup S = 3. Theorem 7. The otios of ifimum ad supremum are dual i the sese that where S = { s s S}. if(s) = sup( S) 2 Sequeces of Real Numbers Sequeces: Defiitio Defiitio 8. A sequece of real umbers is a real-valued fuctio defied o the set of positive itegers N : N = {, 2,...} a = f() R. where the th term f() is deoted by a. [2ex] The sequece a, a 2,..., is deoted by (a ) = or (a ). s 9. a =, N, is the sequece, 2, 3, 4,...; a = +, N, is the sequece 2, 2 3, 3 4, 4 5,...; a = 2, N, is the sequece, 4, 9, 6,...; a = cos π = ( ), N, is the sequece,,,.... 3

4 Limit of a Sequece Defiitio 0. Let (a ) = be a sequece of real umbers. A real umber L is a limit of (a ) =, deoted by L = lim a, s if. ɛ > 0, If a N= N such that a L < ɛ, > N. +, N, the lim a = 0. For ay ɛ > 0 give, choose N > 0 such that ɛn >, i.e., N < ɛ. The, if > N, we have 0 < + < N+ < ɛ. If a = +, N, the lim a =. For ay ɛ > 0 give, choose N > 0 such that ɛn >, i.e., N < ɛ. The, if > N, we have 0 < + = < N < ɛ. Coverget Sequece Defiitio 2. A sequece that has a limit is said to be coverget. A sequece that has o limit is said to be diverget. Uiqueess of Limit If lim a = L ad lim a = M, the L = M. Proof. ɛ > 0, N > 0 such that a L < ɛ 2 ad a M < ɛ 2, > N. L M a L + a M < ɛ L = M. 3. If a = cos π = ( ), N, the sequece (a ) is diverget. If ɛ = 3, the the iterval (x 3, x + 3 ) has a legth 2 3 that is < ; x R, it ca ot cotai ad at the same time. Therefore it is ot possible to fid N such that a x < 3 if > N. Boudedess of a Sequece Defiitio 4. A sequece (a ) = is bouded above or bouded below or bouded if the set S = {a, a 2,...} is bouded above or bouded below or bouded. s 5. If a = +, N, the the sequece (a ) is bouded above by M ad bouded below by m 0. If a = cos π = ( ), N, the M is a upper boud for the sequece (a ) ad m is a lower boud for the sequece (a ). Theorem 6. Every coverget sequece is bouded. lim a = L ɛ > 0, N N s.t. a L < ɛ, > N a a L + L < ɛ + L, > N doe! Theorem 7. Every ubouded sequece is diverget. 4

5 Mootoic Sequece Defiitio 8. A sequece (a ) = is icreasig if a a + for all N. A sequece (a ) = is decreasig if a a + for all N. Theorem 9. A bouded, icreasig sequece coverges to its lub; a bouded, decreasig sequece coverges to its glb. s 20. If a = +, N, the (a ) is decreasig, bouded, ad lim a = if(a ) = 0. a+ a = + +2 = + +2 < If a = 2, N, the (a ) is icreasig, but ubouded above, therefore is diverget. 2. Let a = +, N. (a ) is icreasig a+ a = = > The sequece displays as 2, 2 if(a ) = 2 3, 3 4,..., 99 00,.... sup(a ) = ad lim a = sup(a ) =. 5

6 22. Let a = 2 with = ( ). (a ) is decreasig a + = 2+ a ( + ) 2 = 2 + < sup(a ) = 2 ad if(a ) = 0 lim a = if(a ) = 0. 6

7 23. Let a = e. (a ) is decreasig[4ex] Let f(x) = x e x. [ex] f (x) = ex xe x e 2x sup(a ) = e ad if(a ) = 0. = x e x < 0 lim a = if(a ) = Let a =, =, 2,... (a ) is decreasig for 3 [ex] Let f(x) = x /x = e (/x) l x. [ex] f (x) = e (/x) l x( (/x) l x ) < 0 if(a ) = lim a = if(a ) =. Outlie Cotets 7

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca