Answer to Question #170537 in Real Analysis for Prathibha Rose

Question

Show that a sequence in R² is convergent if and only if it is bounded and all it's convergent subsequences have the same limit

Answer 1

Solution:

Every convergent sequence is bounded.

Proof. Let $\left(x_{n}\right)$ be a sequence which converges to $\ell$ , and let $\epsilon=1$ and note that $\epsilon>0$ . By the definition of convergence of a sequence, there exists some $N \in \mathbb{N}$ such that if $n \geq N$ , then $\left|x_{n}-\ell\right|<\epsilon$ . We compute

$\begin{aligned} &\left|x_{n}-\ell\right|<\epsilon \\ & \Longrightarrow-\epsilon<x_{n}-\ell<\epsilon \\ \Longrightarrow &-\epsilon+\ell<x_{n}<\epsilon+\ell \\ \Longrightarrow & \ell-1<x_{n}<\ell+1 \end{aligned}$

We make the following two calculations.

$\begin{aligned} \ell & \leq|\ell| \\ \Longrightarrow \ell+1 & \leq|\ell|+1 \end{aligned}$

and

$\begin{aligned} -\ell & \leq|\ell| \\ \Longrightarrow-|\ell| & \leq \ell \\ \Longrightarrow-|\ell|-1 & \leq \ell-1 \end{aligned}$

Therefore, we can compute

$\begin{aligned} -|\ell|-1 \leq \ell-1<x_{n}<\ell+1 \leq \mid \ell+1 \\ \Longrightarrow-|\ell|-1<x_{n}<\mid \ell+1 \\ \Longrightarrow\left|x_{n}\right|<|\ell|+1 \end{aligned}$

Let $M=\max \left\{\left|x_{1}\right|,\left|x_{2}\right|, \ldots,\left|x_{N-1}\right|,|\ell|+1\right\}$ . Then given any $x_{n}$ , if $n<N,\left|x_{n}\right| \leq M$ , and if $n \geq N$ , then

$\left|x_{n}\right|<|\ell|+1 \leq M$.

Hence the sequence $\left(x_{n}\right)$ is bounded by M.

Next: If a sequence converges then all subsequences converge and all convergent subsequences converge to the same limit. 

Proof: Let $\left\{a_{n}\right\}_{n \in \mathbb{N}}$ be any convergent sequence. Denote the limit by $\ell$ . Let $\left\{b_{n}\right\}_{n \in \mathbb{N}}$ be any subsequence. Let $\varepsilon>0$ be given. By the definition of convergence for $\left\{a_{n}\right\}_{n \in \mathbb{N}}$ there exists $N \in \mathbb{N}$ such that $\left|a_{n}-\ell\right|<\varepsilon$ for all $n \geq N$ . But this value N will also work for $\left\{b_{n}\right\}_{n \in \mathbb{N}}$ . This is because if $n \geq N,\ then\ b_{n}=a_{m}$ for some $m \geq n \geq N$ and so $\left|b_{n}-\ell\right|=\left|a_{m}-\ell\right|<\varepsilon$ . Thus $\left|b_{n}-\ell\right|<\varepsilon$ for all $n \geq N$

as required.