# Real sequences/Convergence and boundedness/Section

## Definition

A subset ${\displaystyle {}M\subseteq \mathbb {R} }$ of the real numbers is called bounded, if there exist real numbers ${\displaystyle {}s\leq S}$ such that

${\displaystyle {}M\subseteq [s,S]}$.

In this situation, ${\displaystyle {}S}$ is also called an upper bound for ${\displaystyle {}M}$ and ${\displaystyle {}s}$ is called a lower bound for ${\displaystyle {}M}$. These concepts are also used for sequences, namely for the image set, the set of all members ${\displaystyle {}{\left\{x_{n}\mid n\in \mathbb {N} \right\}}}$. For the sequence ${\displaystyle {}1/n}$, ${\displaystyle {}n\in \mathbb {N} _{+}}$, ${\displaystyle {}1}$ is an upper bound and ${\displaystyle {}0}$ is a lower bound.

## Lemma

### Proof

Let ${\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}$ be the convergent sequence with ${\displaystyle {}x\in \mathbb {R} }$ as its limit. Choose some ${\displaystyle {}\epsilon >0}$. Due to convergence there exists some ${\displaystyle {}n_{0}}$ such that

${\displaystyle \vert {x_{n}-x}\vert \leq \epsilon {\text{ for all }}n\geq n_{0}.}$

So in particular

${\displaystyle \vert {x_{n}}\vert \leq \vert {x}\vert +\vert {x-x_{n}}\vert \leq \vert {x}\vert +\epsilon {\text{ for all }}n\geq n_{0}.}$

Below ${\displaystyle {}n_{0}}$ there are ony finitely many members, hence the maximum

${\displaystyle {}B:=\max _{n

is welldefined. Therefore ${\displaystyle {}B}$ is an upper bound and ${\displaystyle {}-B}$ is a lower bound for ${\displaystyle {}{\left\{x_{n}\mid n\in \mathbb {N} \right\}}}$.

${\displaystyle \Box }$

It is easy to give a bounded but not convergent sequence.

## Example

The alternating sequence

${\displaystyle {}x_{n}:=(-1)^{n}\,}$

is bounded, but not convergent. The boundedness follows directly from ${\displaystyle {}x_{n}\in [-1,1]}$ for all ${\displaystyle {}n}$. However, there is no convergence. For if ${\displaystyle {}x\geq 0}$ were the limit, then for positive ${\displaystyle {}\epsilon <1}$ and every odd ${\displaystyle {}n}$ the relation

${\displaystyle {}\vert {x_{n}-x}\vert =\vert {-1-x}\vert =1+x\geq 1>\epsilon \,}$

holds, so these members are outside of this ${\displaystyle {}\epsilon }$-neighbourhood. In the same way we can argue against some negative limit.