Week 3 Solutions Page 1 Exercise (2.4.1). Prove that ) is Cauchy
(n2?1 n2. ) is Cauchy using directly the definition of. Cauchy sequences. Let {xn} be a sequence such that there exists a 0 <C< 1 such that.
Chapter 2. Sequences §1. Limits of Sequences Let A be a nonempty
A sequence (an)n=12
Midterm Solutions
Show that X is not a bounded sequence and hence is not convergent. Solution. Since lim. (xn+1 xn. ) = L given ? > 0
1.4 Cauchy Sequence in R
Suppose xn is a bounded sequence in R. ?M such that but there are no such pt. ... Let V = C([01])=all continuous functions on the interval.
Chapter 6: Limits of Functions
Suppose that (xn) is any sequence in A with xn = c that converges to c and let ? > 0 be given. From. Definition 6.1
Sequences
Proposition 3.19. A convergent sequence is bounded. Proof. Let (xn) be a convergent sequence with limit x. There exists N ? N such that.
Solutions to Assignment-3
(b) Let E ? R be a subset such that there exists a sequence {xn} in E with Solution: For any x = 0 there exists an N such that
MA 101 (Mathematics I) Hints/Solutions for Practice Problem Set - 2
converges to 0 then the sequence (xn n) must converge to 0. Solution: The given statement is TRUE. If xn ? 0
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continuous at c if for every ? > 0 there exists a ? > 0 such that In particular f is discontinuous at c ? A if there is sequence (xn) in the domain.
Lecture 2 : Convergence of a Sequence Monotone sequences
Let us now state the formal definition of convergence. Definition : We say that a sequence (xn) converges if there exists x0 ? IR such that for every.
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(c) If {xn} is a sequence of real (or complex) numbers that converges to Now suppose {xn} converges to x i e for all ? > 0 there exists N ? N such
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Proposition 3 19 A convergent sequence is bounded Proof Let (xn) be a convergent sequence with limit x There exists N ? N such
[PDF] Chapter 2 Sequences §1 Limits of Sequences Let A be a nonempty
xn = s Proof Let ? > 0 be given Since limn?? an = s there exists a positive integer N1 such that
[PDF] 14 Cauchy Sequence in R
Suppose xn is a bounded sequence in R ?M such that but there are no such pt Let V = C([01])=all continuous functions on the interval
[PDF] MA 101 (Mathematics I) Hints/Solutions for Practice Problem Set - 2
converges to 0 then the sequence (xn n) must converge to 0 Solution: The given statement is TRUE If xn ? 0 then there exists n0 ? N such that xn < 1
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Let us now state the formal definition of convergence Definition : We say that a sequence (xn) converges if there exists x0 ? IR such that for every
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(b) Let E ? R be a subset such that there exists a sequence {xn} in E with the property that xn ? x0 /? E Show that there is an unbounded continuous
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Let X = (xn) be a sequence of positive real numbers such that lim (xn+1 xn ) Let (fn) ? C[01] be such that there exists M > 0 such that fn ? ?
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Let A be a bounded subset of R Show that there exists a sequence (an) of elements of A such that lim(an) = sup(A)
[PDF] 174 Let {an} be a sequence with positive terms such that limn
Let {an} be a bounded sequence such that every convergent subsequence of {an} has a limit L Prove that limn?? an = L Solution Method 1: Note that La = {L}
Week 3 Solutions Page 1
Exercise(2.4.1).Prove that
n21n 2 is Cauchy using directly the denition ofCauchy sequences.
Proof.Given >0, letM2Nbe such thatr2
< M.Then, for anym;nM,
jxmxnj=m21m 2n21n 2 1n 21m2 1n 2+1m 2 1M 2+1M 2 2M 2
Therefore,
n21n 2 is a Cauchy sequence.Exercise(2.4.2).Letfxngbe a sequence such that there exists a0< C <1 such that jxn+1xnj Cjxnxn1j: Prove thatfxngis Cauchy. Hint: You can freely use the formula (forC6= 1)1 +C+C2++Cn=1Cn+11C:
Proof.Let >0 be given. Note that
jx3x2j Cjx2x1j jx4x3j Cjx3x2j CCjx2x1j=C2jx2x1j and in general, one could prove that jxn+1xnj Cjxnxn1j C2jxn1xn2j Cn1jx2x1j:Week 3 Solutions Page 2
Now, form > n, we can evaluate the quantity
jxmxnj jxmxm1j+jxm1xm2j++jxn+1xnjCm2jx2x1j+Cm3jx2x1j++Cn1jx2x1j
= (Cm2++Cn1)jx2x1j =Cn1(1 +C++Cmn1)jx2x1j =Cn11Cmn1C jx2x1jCn111C
jx2x1j Now, since 0< C <1,Cn1!0 asn! 1. Therefore, there existsN2N such that whenevernN, jCn10j< 11C jx2x1j:For this sameN, wheneverm > nN
jxmxnj Cn111C jx2x1j 11C jx2x1j11C jx2x1jTherefore,fxngis a Cauchy sequence.Exercise.Prove the following statement using Bolzano-Weierstrass theorem.
Assume that(xn)n2Nis a bounded sequence inRand that there existsx2Rsuch that any convergent subsequence(xni)i2Nconverges tox. Thenlimn!1xn=x. Proof.Assume for contradiction thatxn6!x. Then9 >0 such that jxnxj for innitely manyn. From this, we can create a subsequencefxnjg such thatjxnjxj for allj2N. Since our original sequence is bounded, this subsequence is bounded, and so, by Bolzano-Weierstrass, there is a convergent subsequence of this subsequence, fxnjkg. By assumption,fxnjkgconverges tox. However, this is a contradiction sincejxnjkxj for allk2N.Hence, we must havexn!xasn! 1.
Week 3 Solutions Page 3
Exercise.Show that
a) the setZ=f:::;1;0;1;:::ghas no cluster points. b) every point inRis a cluster point ofQ. Proof.a) Ifx2Z, then (x1=2;x+1=2)\Znfxg=;and soxis not a cluster point ofZ. Ifx62Z, then9k2Zsuch thatx2(k;k+ 1). Choose= minfjxkj;jx (k+1)jg. Then (x;x+)\Znfxg=;, and soxis not a cluster point of Z.ThereforeZhas no cluster points.
b) Letx2R. Let >0. By the density ofQ,9r2Qsuch thatx < r < x+. Then (x;x+)\Qnfxg 6=;. Sincewas arbitrary, this shows thatxis a cluster point ofQ. Sincexwas arbitrary, every point inRis a cluster point ofQ.Exercise.In the lecture we have shown any Cauchy sequence(xn)n2NRhas a limit inR, i.e. there existsx2Rwithlimn!1xn=x.(1) The same statement is false inQ, the following is false: any Cauchy sequence(xn)n2NQhas a limit inQ, i.e. there existsx2Qwithlimn!1xn=x.(2) a) Give a counterexample to (2). b) Which part of the proof of (1) (from the lecture) fails when we attempt to prove (2)? Proof.a) Dene a sequencefxngas follows. For alln2N, choosexn2Qsuch thatp21=n < xn[PDF] letter and sound assessment form
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