Real sequences/Convergence and boundedness/Section

Definition

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

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

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

Lemma

A convergent real sequence is

bounded.

Proof

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

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

So in particular

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

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

${}B:=\max _{n<n_{0}}\{\vert {x_{n}}\vert ,\,\vert {x}\vert +\epsilon \}\,$

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

$\Box$

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

Example

The alternating sequence

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

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

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

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