Prove that if $F$ is an ordered field in which every non empty set which has an upper bound also has a supremum , then $F$ is archimedean
Attempt: If $F$ is an ordered field, then it possesses a positive class $P$ such that :
$(i) ~a+b \in P~~\forall ~a,~b \in P $
$(ii)~ ab \in P~~\forall~~a,b \in P$
$(iii) ~$ If $a \in F$, then precisely one of the following conditions can occur :
$a \in P, -a \in P , a =\{0\}$
We need to prove that $F$ is archimidean means, we need to prove that $~\forall~x \in F,~~\exists~~n \in \mathbb N$ such that $x < n$
Let $S$ be a non empty in $F$ such that $S$ has an upper bound, then $S$ has a supremum
Let $u$ be the supremum of $S$ and $v$ be an upper bound of $S$ such that $v \geq u$
I am not able to proceed from here in proving that $~\exists~~n \in \mathbb N$ such that $n>x ~\forall~x \in F$.
How can a condition of existence of upper bounds and supremum on $S$ ensure that $F$ as a whole is archimedian?
Thank you for your help
A start: Suppose to the contrary that our field is not Archimedean. Let $W$ be the set of all field elements of the form $1+1+\cdots+1$. Since $F$ is not Archimedean, the set $W$ is bounded above, so it has a least upper bound. Now derive a contradiction from this.