Question #170323

Given a sequence ((xn,yn)) is R2 .prove that the following,

1. ((xn,yn)) is convergent implies ((xn,yn)) is bounded.

2. If ((xn,yn)) is a bounded sequence the ((xn,yn)) has a convergent subsequence

3. ((xn,yn)) is convergent if and only if ((xn,yn)) is bounded and every convergent subsequence of ((xn,yn)) has the same limit.

4. ((xn,yn)) is caught if and only if ((xn,yn)) is convergent


1
Expert's answer
2021-04-14T14:19:12-0400

Solution:

1.

Statement: Every convergent sequence is a bounded sequence, that is the set {xn:nN}\left\{x_{n}: n \in \mathbb{N}\right\} is bounded.

Proof : Suppose a sequence (xn)\left(x_{n}\right) converges to x. Then, for ϵ=1\epsilon=1 , there exist N such that

xnx1 for all nN\left|x_{n}-x\right| \leq 1 \text { for all } n \geq N

This implies xnx+1\left|x_{n}\right| \leq|x|+1 for all nNn \geq N . If we let

M=max{x1,x2,,xN1}M=\max \left\{\left|x_{1}\right|,\left|x_{2}\right|, \ldots,\left|x_{N-1}\right|\right\}

then xnM+x+1\left|x_{n}\right| \leq M+|x|+1 for all n. Hence (xn)\left(x_{n}\right) is a bounded sequence.

2.

Statement: Every bounded sequence contains a convergent subsequence. 

Proof. Let (an)\left(a_{n}\right) be a bounded sequence. Then there exists some M>0 such that anM\left|a_{n}\right| \leq M for all nNn \in \mathbb{N} . This implies that an[M,M]a_{n} \in[-M, M] for all nNn \in \mathbb{N} .

- We bisect the interval [-M, M] into two closed intervals of equal length. One of these intervals must contain infinitely many terms of the sequence (an)\left(a_{n}\right) . Let I1I_{1} be that interval, and let an1a_{n_{1}} be any point in I1I_{1} .

- Next we bisect I1I_{1} into two closed intervals, and note that one of these intervals must contain infinitely many terms of the sequence (an)\left(a_{n}\right). Let I2I_{2} be that interval, and choose an2a_{n_{2}} inside this interval which satisfies n2n1n_{2} \geq n_{1}

- In general, we bisect Ik1I_{k-1} into two closed intervals, one which must contain infinitely many terms of (an)\left(a_{n}\right) . Let IkI_{k} be this closed interval, and choose ankIka_{n_{k}} \in I_{k} such that nk>nk1n_{k}>n_{k-1} . Therefore we have obtained a subsequence (an1,an2,)\left(a_{n_{1}}, a_{n_{2}}, \ldots\right) of (an)\left(a_{n}\right) and a sequence of nested intervals

I1I2I3I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \ldots

By the Nested Interval Property (i), the intersection n=1In\bigcap_{n=1}^{\infty} I_{n} is nonempty, and therefore contains some element

We claim that (ank)\left(a_{n_{k}}\right) converges to xx . Let ϵ>0\epsilon>0 be arbitrary. Then the length of the interval IkI_{k} is M(12)k1M\left(\frac{1}{2}\right)^{k-1} . By (i), the sequence (12k1)\left(\frac{1}{2}^{k-1}\right) converges to 0. Therefore, by the Algebraic Limit Theorem, the sequence (M(12)k1)\left(M\left(\frac{1}{2}\right)^{k-1}\right) converges to 0 as well. Hence, there exists some NNN \in \mathbb{N} such that if kNk \geq N , then M(12)k1<ϵ\left|M\left(\frac{1}{2}\right)^{k-1}\right|<\epsilon But then ank,xIka_{n_{k}}, x \in I_{k} , hence ankx<ϵ\left|a_{n_{k}}-x\right|<\epsilon . Therefore (ank)\left(a_{n_{k}}\right) converges to xx .

3.

Statement: Let {xn}\left\{x_{n}\right\} be a bounded sequence all of whose convergent proper subsequences converge to \ell . Prove that {xn}\left\{x_{n}\right\} converges to \ell .

Proof: A bounded sequence of real numbers must have at least one convergent subsequence. All of those converge to \ell , and there is at least one of those; call it (xnk)k=1\left(x_{n_{k}}\right)_{k=1}^{\infty} . So for every ε>0\varepsilon>0 , all except finitely many terms of (xnk)k=1\left(x_{n_{k}}\right)_{k=1}^{\infty} are in the interval with endpoints ±ε\ell \pm \varepsilon . If infinitely many terms of the original sequence lie outside that small interval, then that is a bounded sequence, which therefore has at least one convergent subsequence. That subsequence converges to some point outside that interval small interval. But that contradicts the hypothesis.

4.

Statement: Cauchy Convergence Criterion: A sequence \left(x_{n}\right) is Cauchy if and only if it is convergent.

Proof: Suppose (xn)\left(x_{n}\right) is a convergent sequence, and lim(xn)=x\lim \left(x_{n}\right)=x . Let ϵ>0\epsilon>0 We can find NNN \in \mathbb{N} such that for all nN,xnx<ϵ/2n \geq N,\left|x_{n}-x\right|<\epsilon / 2 . Therefore, by the triangle inequality, for all m,nN,xmxnxmx+xxn<ϵ/2+ϵ/2=ϵm, n \geq \mathbb{N},\left|x_{m}-x_{n}\right| \leq\left|x_{m}-x\right|+\left|x-x_{n}\right|< \epsilon / 2+\epsilon / 2=\epsilon . So (xn)(x_n​) is Cauchy.

Conversely, suppose (xn)\left(x_{n}\right) is Cauchy. Let ϵ>0\epsilon>0 . By a result proved in class, (xn)\left(x_{n}\right) is bounded. By Bolzano-Weierstrass, it has a convergent subsequence (xnk)\left(x_{n_{k}}\right) with lim(xnk)=x\lim \left(x_{n_{k}}\right)=x for some xx . We can find KNK \in \mathbb{N} such that for all kK,xnkx<ϵ/2k \geq K,\left|x_{n_{k}}-x\right|<\epsilon / 2 . We can also find M such that for all m,nM,xmxn<ϵ/2m, n \geq M, \left|x_{m}-x_{n}\right|<\epsilon / 2 . Let N=sup{K,M}N=\sup \{K, M\} . Then since nkkn_{k} \geq k for all k, if kNk \geq N , we have that k, nkM and nkKn_{k} \geq M\ and\ n_{k} \geq K . Therefore, for all kNxnxxnxnk+xnkx<ϵ/2+ϵ/2=ϵk \geq N \left|x_{n}-x\right| \leq\left|x_{n}-x_{n_{k}}\right|+\left|x_{n_{k}}-x\right|<\epsilon / 2+\epsilon / 2=\epsilon by the Triangle inequality.

Therefore, (xn)\left(x_{n}\right) is Cauchy.


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