I'm kinda stuck on this problem. Please help me. Thank you much.
Since x is the supremum of {$\displaystyle x_{n}:n\in n$},
GIVEN ε>0 there exists kεN AND SUCH that:
$\displaystyle x-\epsilon< x_{k}\leq x<x+\epsilon$
But since {$\displaystyle x_{n}$} is increasing and bounded by x:
for all n $\displaystyle x_{n}\leq x$ and
for $\displaystyle n\geq k$ $\displaystyle x_{n}\geq x_{k}$
hence :
$\displaystyle x-\epsilon<x_{n}\leq x<x+\epsilon\Longleftrightarrow|x_{n}-x|<\epsilon$
THUS $\displaystyle \lim_{n\rightarrow\infty}{x_{n}} = x$