MA2108S Tutorial 4 Answer Sheet

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1 MA208S Tutorial 4 Aswer Sheet Qia Yujie, Shi Yagag, Xu Jigwei 2 Februar 20 Sectio Perform the computatios i (a) ad (b) of the precedig eercise for the fuctio h : X Y R defied b { 0 if <, h(, ) := if. Proof. (a) Note that h(, ) =, hece f() := sup{h(, ) : Y } = for all X. Thus if{f() : X} =. (b) For each X, we take = +, the we have h(, + ) = 0,the g() = 2 2 if{h(, ) : X} = 0 for all Y. Thus sup{g() : Y } = Let X ad Y be oempt sets ad let h : X Y R have bouded rage i R. Let f : X R ad g : Y R be defied b f() := sup{h(, ) : Y }, g() := if{h(, ) : X}. Prove that sup{g() : Y } if{f() : X}. Proof. Sice f( 0 ) h( 0, ) Y, ad g( 0 ) h(, 0 ) X. Take a 0 X, the f( 0 ) h( 0, ) g() Y, amel f() g() X Y. Hece f() is a upper boud of {g() : Y }, sup{g() : Y } f() X, Therefore sup{g() : Y } is a lower boud of {f() : X}. Hece sup{g() : Y } if{f() : X}.. Let X ad Y be oempt sets ad let h : X Y R have bouded rage i R. Let F : X R ad G : Y R defied b F () := sup{h(, ) : Y }, G() := sup{h(, ) : X}.

2 Establish the Priciple of the Iterated Suprema sup h(, ) = sup sup h(, ) = sup sup h(, ). Proof. 0 X, Y, h( 0, ) F ( 0 ). Hece sup h(, ) sup{f () : X}. h(, ) sup{f () : O the other had, sup X}. Hece sup B smmetr, sup Hece sup h(, ) = sup h(, ) h( 0, ) 0 X, Y, hece sup h(, ). h(, ) = sup{f () : X} = sup h(, ) = sup sup sup h(, ). h(, ) = sup sup h(, ). sup 2. Give a R, show that there eists uique Z such that <. Proof. () If = 0, the let :=, = 0 < =. (2) If > 0, Let S := {N : > }. B the Archimedea Propert, N, such that >, S is oempt. B the well orderig propert, a smallest elemet 0 ca be picked i S. Sice 0 < 0, therefore 0 / S. Namel 0. Hece 0 < 0. (3) If < 0, the > 0. B the Archimedea Propert, m N with m >. Therefore m + > 0. B the result of part (2), N m + <. Hece ( m) < ( m) with ( m) Z. Take := m, the <. Uiqueess: suppose to the cotrar, for a R, m, Z ad m, such that < ad m < m. Without loss of geeralit, assume m >, because m, Z, therefore m +. The, m >, which cotradicts with the fact that m, therefore m =. Hece give a R, there is a uique Z such that <. 3. If > 0, show that there eists N such that 2 <. Proof. Sice > 0, b The Archimedea Propert, N, such that ( ) 2 = ( + ) + > for N, 2 < <. Hece if > 0, there eists N such that 2 <. <. Sice 2

3 8. If u > 0 is a real umber ad <, show that there eists a ratioal umber r such that < ru <. (Hece the set {ru : r Q is dese i R.) Proof. Sice u > 0, ad <, therefore u < u. Sice u > 0, b the Archimedea u Propert, N, such that < u u. Hece + <. Appl Corollar u u (or result of Eercise 2.4.2), m N, such that m < m. Therefore, u m u + < u, whece < m < u u. Thus the ratioal umber r := m satisfies < ru <. EtraA Let A ad B be oempt subsets of R. Let A + B = {a + b : a A, b B}. ad AB = {ab : a A, b B}. Show that if(a+b) = if A+if B. If A ad B are both subsets of positive real umber, show also that if(ab) = if A if B. Proof. () a A, b B, if A + if B a + b. Thus if A + if B is a lower boud of A + B. For a ɛ > 0, a ɛ A, such that a ɛ < if A+ ɛ ; ad b ɛ B, such that b ɛ < if B + ɛ Therefore, for a ɛ > 0, a ɛ A, b ɛ B, such that a ɛ + b ɛ < if A + if B + ɛ. B Lemma if A + if B = if(a + B). (2) a A a if A 0. b B, b if B 0. Hece a A, b B, if A if B a b. Hece if A if B is a lower boud of AB. Let be a lower boud of AB. The a A, b B, ab u. a, b > 0, thereforea u b, u is a lowerboud of A. b Heceif A u which implies u if A b. Sice if A 0, b Case : if if A > 0 the b B b u. Therefore if B u the if A if B u; if A if A Case 2: if if A = 0, > 0, therefore u = 0, thus if B if A = 0 = u. u is a lower boud of AB, hece if A if B = if(ab) Hece if A if B = if(ab). 3

4 Sectio 2.5 Questio If I := [a, b] ad I := [a, b ] are closed itervals i R, show that I I if ad ol if a a ad b b, which leads to cotradictio. Proof. If a, b I a, b I, the a a, b b Coversel, if a a, b b, the a b wheever a b. Thus I I. Questio 2 If S R is oempt, show that S is bouded if ad ol if there eists a closed bouded iterval I such that S I. Proof. () Claim: The set S is bouded if there eists a closed bouded iterval I such that S I. Take I = [a, b], the for a S, a b. Hece for a S, a S is bouded below. For a S, b S is bouded above. Hece S is bouded. The set S is bouded if there eists a closed bouded iterval I such that S I. (2) Claim: The set S is bouded ol if there eists a closed bouded iterval I such that S I. As S is bouded above, there eists a umber u R such that s u for all s S. Also, S is bouded below, there eists a umber w R such that w s for all s S. Hece w s u. Hece there eists closed bouded iterval I = [w, u] such that S I. Hece S is bouded ol if there eists a closed bouded iterval I such that S I. Hece S is bouded if ad ol if there eists a closed bouded iterval I such that S I. Questio 3 If S R is a oempt bouded set, ad I s := [ifs, sups], show that S I s. Moreover, if J is a closed bouded iterval cotaiig S, show that I s J. Proof. () Claim: The set S I s. Take a s S, the if S s sup S s I s. Hece S I s. (2) Claim: The set I s J if S J. Suppose J = [a, b]. The for a s S, sice S J, we have a s b. Hece a is a lower boud of S ad b is a upper boud of S. Therefore a if S. For a s S,s J, the b s. Therefore b is a upper boud of S. Hece sup S b a b. Hece I s J b Questio. Hece if J is a closed bouded iterval cotaiig S, I s J. 4

5 Questio 7 Let I := [0, /] for N. Prove that I = {0}. Proof. Clearl 0 I for a N. Hece 0 I. Claim: The set {0} = I. Suppose {0} = I. Sice 0 I, there eists a 0, such that I.Hece I for all N. As I = [0, /], > 0. Hece b the Archimedea Propert, there eists a K N such that /K <. Hece / I K. Hece / I. This is a cotradictio! Hece I = {0}. Questio 8 Let J := (0, /) for N. Prove that J =. Proof. Suppose there eists a, such that J. Hece J for all N. As J = (0, /), we have > 0. Hece b the Archimedea Propert, there eists a K N such that /K <. Hece / J K. Hece / J. This is a cotradictio! Hece J =. Questio 9 Let K := (, ) for N. Prove that K =. 5

6 Proof. Suppose there eists a, such that K. Hece K for all N. Also, K = (, ), we have > 0. Hece b the Archimedea Propert, there eists a K N such that < K. Hece / K K. Hece / K. This is a cotradictio! Hece K =. Questio 0 With the otatio i the proofs of Theorems ad 2.5.3, show that we have η I. Also show that [ξ, η] = I. Proof. Recall that η = if{b : N} ad the fact that for each N, a is a lower boud of the set {b : N}, it is clear that η a for each. Thus η ξ = sup{a : N}. Net, it is clear that if [ξ, η], the a ξ η b for a. Thus [a, b ] = I for each. It is the clear that [ξ, η] N I. Coversel, if N I, the a b for all ad hece is a upper boud of {a : N} ad a lower boud of {b : N}. Thus ξ η ad hece [ξ, η] N I. Thus [ξ, η] = N I. Questio 4 4.Show that if a k, b k {0,,..., 9} ad if a 0 + a a 2 0 = b 0 + b b m 2 0 0, m the a k = b k for k =,,. Proof. Suppose ot. The there is a least umber i such that a k 0 = b k whe k < i ad k 0k a i 0 b i i 0. i Without loss of geeralit, we ma assume a i 0 > b i i 0. Sice a i, b i i {0,,..., 9}, we have a i b i +. Hece a i 0 b i i 0 +. Note that i 0i b i+ 0 + b i+2 i b m i m i i+2 0 m = 9 [ ( 0 i+ 0 )m i ] 0 = 0 i [ 0 m i ] < 0 i. 6

7 Hece The we have a i 0 i b i 0 i + 0 i > b i 0 i + b i+ 0 i+ + b i+2 0 i b m 0 m. a 0 + a a 0 a 0 + a a i 0 i > b 0 + b b 0. This is a cotradictio. Hece a k = b k for all k. Questio 6 Epress 7 ad 2 9 as periodic decimals. Solutio: B computatio, we have 7 = ad 2 9 = Questio 7 What ratioals are represeted b the periodic decimals ad ? Solutio: Take = The we have ( ) = Hece = 3253/ Hece = 3253/ Take = The we have ( ) = Hece = 3539/ Hece = 3539/ Etra Questio (C): Let z > 0. Show that for a N, there eists > 0 such that = z. Proof. Case : z > Let S = {s R : s 0, s < z}. Sice S, the S is ot a empt set. Also, S is bouded above b z, as z > z ad s < z s < z (recall that s, z 0). Therefore, the Completeess Propert implies that the set S has a supremum i R. Let = sups. Note that >. First assume that < z. (+δ) = (+δ )((+δ) +(+δ) ) < δ(+δ) < δ(+), for 2, N ad 0 < δ <. Sice z > 0, b the Archimedea Propert, there eists N N such that z >. N Let δ = N( + ). The ( + δ) < δ( + ) = N < z. 7

8 Hece ( + δ) < z. It is a cotradictio to the fact := sups. Hece z. Net assume that > z. The ( δ) = ( + δ)( + 2 ( δ) + + ( δ) ) < δ, for 2, N ad 0 < δ <. Sice z > 0, b the Archimedea Propert, there eists N N such that z > N. Let δ = N. The ( δ) < δ = N < z. Hece ( δ) > z. It is a cotradictio to the fact := sups. Hece z. Combiig these two cases, we coclude that = z. Case 2: z = Clearl, if =, the =. Case 3: 0 < z < The /z >. B Case (), there eists > such that = /z. Observe that (/) = z. Sectio3. Questio: 5(a) Use the defiitio of the limit of a sequece to establish the limit of lim( 2 + ) = 0. Proof. Observe that 2 + < = 2. Give a ɛ > 0, b the Archimedea Propert, there eists k N, such that k < ɛ. Therefore if k, the k < ɛ. Hece, < ɛ, for all k. Hece lim( ) = 0. Questio: 5(d) Use the defiitio of limit of a sequece to establish the limit of lim( ) = 2. Proof. Observe that = = < 5 4 < Give ɛ > 0, b the Archimedea Propert, there eists k N, such that < ɛ 5. Therefore take k, the 5 5 k < ɛ. Hece, < 5 < ɛ, for all k. ( ) 2 Hece, lim =

9 Questio: 6(c) Show that lim( ) = Proof. Observe that + 7 < =. Give a ɛ > 0, b the Archimedea Propert, there eists k N, such that < k ɛ2. Therefore if k, the k ɛ2, thus < ɛ. Hece, < ɛ, for all k. + Questio: 6(d) Show that lim( ( ) 2 + ) = 0. Proof. Observe that ( ) < = 2. Give a ɛ > 0, b the Archimedea Propert, there eists k N, such that k < ɛ 5. Therefore if k, the k < ɛ. Hece, ( ) < ɛ, for all k