Answer to Question #170323 in Real Analysis for Prathibha Rose

Question

Given a sequence ((x_n,y_n)) is R² .prove that the following,

1. ((x_n,y_n)) is convergent implies ((x_n,y_n)) is bounded.

2. If ((x_n,y_n)) is a bounded sequence the ((x_n,y_n)) has a convergent subsequence

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

4. ((x_n,y_n)) is caught if and only if ((x_n,y_n)) is convergent

Answer 1

Solution:

1.

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

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

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

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

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

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

2.

Statement: Every bounded sequence contains a convergent subsequence. 

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

- 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 $\left(a_{n}\right)$ . Let $I_{1}$ be that interval, and let $a_{n_{1}}$ be any point in $I_{1}$ .

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

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

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

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

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

3.

Statement: Let $\left\{x_{n}\right\}$ be a bounded sequence all of whose convergent proper subsequences converge to $\ell$ . Prove that $\left\{x_{n}\right\}$ 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 $\left(x_{n_{k}}\right)_{k=1}^{\infty}$ . So for every $\varepsilon>0$ , all except finitely many terms of $\left(x_{n_{k}}\right)_{k=1}^{\infty}$ are in the interval with endpoints $\ell \pm \varepsilon$ . 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 $\left(x_{n}\right)$ is a convergent sequence, and $\lim \left(x_{n}\right)=x$ . Let $\epsilon>0$ We can find $N \in \mathbb{N}$ such that for all $n \geq N,\left|x_{n}-x\right|<\epsilon / 2$ . Therefore, by the triangle inequality, for all $m, n \geq \mathbb{N},\left|x_{m}-x_{n}\right| \leq\left|x_{m}-x\right|+\left|x-x_{n}\right|< \epsilon / 2+\epsilon / 2=\epsilon$ . So $(x_n​)$ is Cauchy.

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

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