1.4 Cauchy Sequence in R
A sequence xn ? R is said to converge to a limit x if Every convergent sequence is a Cauchy sequence. Proof. Assume xk ? x. Let ? > 0 be given.
Suggested Solution to Homework 1
If (xn) is Cauchy and has a convergent subsequence say
Aronszajns theorem
A function K : X ×X ? R is called a positive definite kernel on X iff it is This shows that for any x the sequence (fn(x))n?0 is Cauchy in R and has ...
Chapter 1. Metric spaces - Proofs covered in class
In a metric space every convergent sequence is a Cauchy sequence. Proof. Suppose that {xn} is a sequence which converges to x and let ? > 0 be given.
Week 3 Solutions Page 1 Exercise (2.4.1). Prove that ) is 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 =
Fixed point theorem and Cauchy-Lipschitz for linear ODEs
7 févr. 2018 of points in X. We say that x is a Cauchy sequence when ... Our goal here is to prove the Cauchy-Lipschitz theorem in the linear case.
ON I-CAUCHY SEQUENCES
In this section we prove a decomposition theorem for ? -convergent sequences. Theorem 1. Let (X p) be a linear metric space
Homework 4 1. Let X and Y be normed spaces T ? B(X
https://www.math.wustl.edu/~wick/teaching/math6338/math6338_hw4.pdf
The University of Sydney Pure Mathematics 3901 Metric Spaces
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
Basic Topology of R
Now assume that the limit of every Cauchy sequence (or convergent sequence) contained in F is also an element of F. We show F is closed. Let x be any limit
Lecture 2 - Vector Spaces Norms and Cauchy Sequences
Therefore we have the ability to determine if a sequenceis a Cauchy sequence Proposition 3 1If (X;k k)is a normed vector space then a sequence of pointsfXig1 i=1 is a Cauchy sequence i given any >0 there is anN2Nso thati; j > Nimplies kXi Xjk< : Proof Simple exercise in verifying the de nitions
How do I prove a sequence is Cauchy - Mathematics Stack Exchange
Theorem Cauchy sequences converge Homework problems 2 4 1: Show directly from the de nition that n21 n2 0
Is the sequence 1 n Cauchy?
Claim: The sequence { 1 n } is Cauchy. Proof: Let ? > 0 be given and let N > 2 ?. Then for any n, m > N, one has 0 < 1 n, 1 m < ? 2. Therefore, ? > 1 n + 1 m = | 1 n | + | 1 m | ? | 1 n ? 1 m |. Thus, the sequence is Cauchy as was to be shown. Everything you wrote is correct, but I think your point would be better illustrated by = ? 1.
Is x n Cauchy in R?
Examples: 1. (X;d) = Q, as a subspace of R with the usual metric. Take x 0= 2 and defne x n+1= xn 2 +1 xn The sequence continues 3=2, 17=12, 577=408;:::and indeed x n!xwhere x=x 2 +1 x , i.e., x2= 2. But this isn’t in Q. Thus (x n) is Cauchy in R, since it converges to p 2 when we think of it as a sequence in R.
How can we guarantee that (x n) will be Cauchy?
n 1) for n1. This gives a sequence (x n); if it is Cauchy and (X;d) is complete, then x= lim n!1x nexists and xshould solve x= ?(x). How can we guarantee that (x n) will be Cauchy? Note that d(x n;x n+1) = d(?(x n 1);?(x n)), so to get (x n) Cauchy we want ?to shrink distances.
How do you prove that a metric space is Cauchy?
Prove directly that it’s Cauchy, by showing how the nin the de nition depends upon . De nition: A metric space (X;d) is complete if every Cauchy sequence in Xconverges in X (i.e., to a limit that’s in X).
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] show that x is a markov chain
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