# 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

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?

• Overkill, but an alternative proof would be to use the fact that every completely ordered field is isomorphic as an ordered field to $\mathbb{R}$. (Though the proof of this does use the fact that it is Archimedean by using the fact that a copy of the rationals is dense in any completely ordered field) – Hayden Aug 29 '14 at 22:56
• Interesting result. I haven't come across this result as of yet in Bartle : Elements of real analysis though. I will anyways, try to prove this result – MathMan Aug 29 '14 at 23:07
• I recommend you look at the construction of the reals using Dedekind cuts; the usual proof shows that every completely ordered field is isomorphic (as an ordered field) to this specific construction. – Hayden Aug 29 '14 at 23:10
• I will. Thanks! – MathMan Aug 29 '14 at 23:14
• Wow... I did not initially realize that 'completeness' necessitated 'Archimedean'! I remember learning both properties at the same time a long time ago and I guess that led to me thinking they might be inequivalent. Not so :) thanks for asking the question. – rschwieb Mar 3 '15 at 11:08

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.
• okay .. Here's an attempt :Suppose that $F$ is not archimedean. Let $W = \{n~~|~~n \in \mathbb N \}$. . Suppose that sup $W = y \implies y \geq n ~\forall~ n \in N.$ . – MathMan Aug 29 '14 at 23:06
• But, then, won't our strategy be confined to only subsets of the form $W = \{n~|~n \in \mathbb N \}$ ? – MathMan Aug 29 '14 at 23:08
• Informal continuation: Since $y$ is the sup, $y-1$ is not an upper bound. So there is a $w\gt y-1$. But then $w+1\gt y$. – André Nicolas Aug 29 '14 at 23:10
• Remember, given an $x$ we have to produce an "integer" greater than $x$, If we cannot, then $W$ is bounded above. It is from the $W$ bounded above that we derive a contradiction. – André Nicolas Aug 29 '14 at 23:13
• okay .. so we brought a contradiction first assuming that $F$ is not archimedean. But, then we proved that $~\exists~w+1 >y$ . Hence, $w+1 > x~~\forall~~x \in W$ . which means there is an integer greater than all members of $W$. My query here is that we have obtained this contradiction only on sub sets of the form $W=\{n~:~n \in \mathbb N\}$ . How can we be sure that elements belonging to other subsets of $F$ shall also follow the same contradiction? – MathMan Aug 29 '14 at 23:25