Ordered field/Convergent sequence/Bounded/Fact/Proof

Proof

Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ be the convergent sequence with ${}x\in K$ 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 ${}{\left\{x_{n}\mid n\in \mathbb {N} \right\}}$.

To fact