Assume that (xn)n∈N is a bounded sequence in R and that there exists x ∈ R such that any convergent subsequence (xni )i∈N converges to x Then limn→∞ xn
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[PDF] Prove that ) is Cauchy using directly the definition of Cauchy
Assume that (xn)n∈N is a bounded sequence in R and that there exists x ∈ R such that any convergent subsequence (xni )i∈N converges to x Then limn→∞ xn
[PDF] 14 Cauchy Sequence in R
A sequence xn ∈ R is said to converge to a limit x if • ∀ϵ > 0, ∃N s t n > N ⇒ xn − x < ϵ A sequence xn ∈ R is called Cauchy sequence if • ∀ϵ, ∃N s t n > N m > N ⇒ xn − xm < ϵ Every convergent sequence is a Cauchy sequence Proof
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Show that xn → x as well; i e to prove that a Cauchy sequence is convergent, we only need i e any bounded sequence has a convergent subsequence Show
[PDF] 1 Cauchy sequences - ntc see result
A sequence {an}is called a Cauchy sequence if for any given ϵ > 0, there exists N ∈ N Proof Since {an}forms a Cauchy sequence, for ϵ = 1 there exists N ∈ N such that (i) lima1/n = 1, if a > 0 (ii) limnαxn = 0, if x < 1 and α ∈ IR Solution:
ON U-CAUCHY SEQUENCES - Project Euclid
If a sequence of points in X is U-convergent in X then it fulfills U-Cauchy condition Proof Let U − lim n→∞ xn = x and ε > 0 Thus A(ε/2) = {n ∈ N : ρ(xn, x) ≥
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Definition A sequence (an) is said to be a Cauchy sequence iff for any ϵ > 0 there exists N such prove (over the course of 2 + ϵ lectures) the following theorem: If a subsequence of a Cauchy sequence converges to x, then the sequence
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Let X = (X, d) be a metric space Let (xn) and (yn) be two sequences in X such that (yn) is a Cauchy sequence and d(xn,yn) → 0 as n → ∞ Prove that (i) (xn) is a
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Every complex Cauchy sequence is convergent Proof Put zn = x + iy Then xn is Cauchy: xx − xm ⩽ zn − zm (as
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(c) Show that the sequence xn is bounded below by 1 and above by 2 (d) Use (e) Use (d) in a proof to show that Sn is Cauchy and thus converges In the first case, if k = 0, then we wish to prove that k · f(x) = 0, the zero function, has limit 0
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If (X, d) is a complete metric space and Y is a closed subspace of X, then (Y,d) is complete Proof Let (xn) be a Cauchy sequence of points in Y Then (xn) also
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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.