[PDF] Let xn be a monotone increasing sequence bounded above and con PDF MATH3150-fa14-hmw2-solutions.pdf

every monotone increasing bounded sequence bounded above converges that for every ε > 0 there exists some n > 1/ε such that xn < ε 1 (c) Using the Archimedean property, argue that yk cannot be bounded above by M, hence

(n2−1 n2 ) is Cauchy using directly the definition of Cauchy sequences Proof Given ϵ > 0 Let {xn} be a sequence such that there exists a 0

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Let X = (xn) be a sequence of positive real numbers such that lim (xn+1 xn ) is Cauchy 4 Let (fn) ∈ C[0,1] be such that there exists M > 0 such that fn ∞ ≤

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MATH 3150 | HOMEWORK 2

Problem 1(p. 97, #5).Letxnbe a monotone increasing sequence bounded above and con- sider the setS=fx1;x2;:::g. Show thatxnconverges to sup(S):Make a similar statement for decreasing sequences. Remark.This shows that theleast upper bound property| that every nonempty set with an upper bound has a least upper bound | implies themonotone sequence property| that every monotone increasing bounded sequence bounded above converges. Combined with the reverse implication proved in class, it follows that the least upper bound property is equivalent to completeness. Solution.Sis a set with an upper bound, so it has a supremum x= sup(S): Let" >0. By our characterization of the supremum, there is somexN2Ssuch that x" < xNand sincexnis increasing it follows that x" < xn;8nN =) jxnxj< "8nN

Thus lim

n!1xn=x= sup(S): Ifxnis a decreasing sequence bounded below, thenxnconverges to inf(fxng) by a similar proof. Problem 2(p. 97, #7).For nonempty setsA;BR, letA+B=x+yx2Aandy2B:

Show that sup(A+B) = sup(A) + sup(B):

Solution.Leta= sup(A) andb= sup(B):Then sinceaxfor allx2Aandbyfor all y2B, it follows thata+bis an upper bound forA+B, i.e. a+bx+y;8x+y2A+B: Let" >0. Then there is somex2Aandy2Bsuch thata"=2< xandb"=2< y, which means that a+b" < x+y2A+B; and it follows thata+b= sup(A+B):

Problem 3(p. 52, #4).

(a) Let xnbe a Cauchy sequence. Suppose that for every" >0 there is somen >1="such thatjxnj< ":Prove thatxn!0: (b) Sho wthat the h ypothesisthat xnbe Cauchy in (a) is necessary, by coming up with an example of a sequencexnwhich does not converge, but which has the other property: that for every" >0 there exists somen >1="such thatjxnj< ": 1 Solution.(a)Let " >0 be given. Sincexnis Cauchy, there exists anNsuch thatjxnxmj< "=2 for allm;nN:If we now let

1= min("=2;1=N)

then it follows from the other assumption that there is ak >1="1Nsuch that jxkj< "1"=2:

Thus, forxnN, we have

jxn0j=jxnxk+xkj jxnxkj+jxkj< "=2 +"=2 ="; so thatxn!x: (b)

Con siderthe sequence

x n=( 1nodd

1=n neven.

This clearly does not converge, and yet for any" >0 we can choose an evenn >1="for whichjxnj< ": Problem 4(p. 99 #15).Letxnbe a sequence inRsuch thatjxnxn+1j 12 jxn1xnj:

Show thatxnis a Cauchy sequence.

Solution.To show thatxnis Cauchy, we must comparexnandxmfor alln;mNfor variousN, not just subsequent elements. To do this we rst note that for arbitraryk >0, jxnxn+kj=jxnxn+1+xn+1 xn+k1+xn+k1xn+kj jxnxn+1j+jxn+1xn+2j++jxn+k1xn+kj jxnxn+1j+12 jxnxn+1j+14 jxnxn+1j++12 k1jxnxn+1j = (1 + 12 ++12 k1)jxnxn+1j <2jxnxn+1j 22
njx0x1j=12 n1jx0x1j: LetM=jx0x1j 2R:Then for an arbitrary" >0, we may chooseNsuciently large that 12 N1<"M :(This uses the fact that 1=2n!0.) Thus for anyn;mN, supposing thatmn, we can writem=n+kfor somek0 and then jxnxmj=jxnxn+kjN1< "; soxnis Cauchy. Problem 5.Prove that an Archimedean ordered eld in which every Cauchy sequence converges is complete (i.e. has the monotone sequence property). Here are some suggested steps: (a) Denote the eld b yF, and supposexnis a monotone increasing sequence bounded above by someM2F: 2 (b)Pro ceedingb ycon tradiction,supp osexnis not Cauchy. Deduce the existence of a sub- sequenceyk=xnkwith the property that y kyk1+";8k(1) for some xed positive number" >0 which does not depend onk: (c) Using the Arc himedeanprop erty,argue that ykcannot be bounded above byM, hence obtaining a contradiction. (d)

Con cludethat xnconverges.

Proof.SupposeFis Archimedean and has the property that every Cauchy sequence inF converges. Letxnbe a monotone sequence inF, with an upper boundM, and suppose that x nis not Cauchy. Then there exists an" >0 such that, for allN2N, there is a pair n;mNfor which jxnxmj ": (This is just the negation of the statement thatxnis Cauchy.) We construct a subsequece as suggested by the hint. Choosen1= 1 (really it doesn't matter where you start), and by induction suppose that we haven1< n2<< nksuch thatxnkxnk1+". SetN=nk; then by assumption there is a pairnk+1;mk+1nk(and without loss of generality we can suppose thatnk+1> mk+1) such thatxnk+1xmk+1"; =)xnk+1xmk+1+" xnk+" since the sequence is increasing. This completes the induction step and gives a subsequence y k=xnksatisfying (1), where" >0 is a xed positive number, per our assumption thatxn is not Cauchy. Letd=My1be the distance from the rst element of the subsequence to the upper bound forxn. By the Archimedean property ofF, there exists someN2Nsuch that

N > d=";()"N > d:

By the property (1) on the subsequenceyk, it follows that y

Ny1+N" > y1+d=M;

SinceyN=xnNis an element of the original sequence, this contradicts the assumption that x nis bounded. Since we reached this conclusion by assuming that our bounded increasing sequencexn was not Cauchy, it follows thatxnmust be Cauchy, hence convergent by the assumption onF. Sincexnwas an arbitrary increasing bounded sequence, it follows thatFhas the monotone sequence property. 3quotesdbs_dbs6.pdfusesText_12

[PDF] [PDF] Let xn be a monotone increasing sequence bounded above and con

[PDF] Week 3 Solutions Page 1 Exercise (241) Prove that ) is Cauchy