# Least upper bound property implies that the orderered field in question is Archimedean

Let $(F,+_F,*_F, \le_F)$ be a totally ordered field with zero $0_F$ and unity $1_F$. Let $(F,+_F,*_F, \le_F)$ have the least upper bound property.

I know that there is a proof saying that if the field is non-Archimedean, then it doesn't have this property.

I have been wondering if there is another proof of the theorem in title, which, rather then starting with the assumption that a field isn't Archimedean, starts with the assumption that it has L.U.B. property and from that proves that the field is Archimedean.

Does such a proof exist?

• So you're asking for a direct proof, rather than one that uses contradiction? – mweiss Jan 17 '17 at 18:26

I'm not really sure if this is what you want -- in my opinion, the distinction between "prove $A \implies B$" and "prove $\neg B \implies \neg A$" is more or less a matter of phrasing.
Let $F$ be an ordered field with the LUB property. We want to prove that $F$ is Archimedean. So, let $x>0$ be any positive element of $F$, and let $y>x$ be a second positive element. We need to prove that for some $N \in \mathbb{N}$, $N\cdot x > y$.
Suppose no such $N$ exists. Then the set $S=\{n\dot x | n\in \mathbb{N}\}$ is bounded above by $y$. Since the field has the least-upper-bound property, $S$ must have a least upper bound $b$.
We now have that for all $n$, $$(n+1)x < b$$ and hence for all $n$ $$nx < b - x$$ whence $b-x$ is also an upper bound of $S$, contradicting the leastness of $b$.
Therefore no such $b$ exists, which means that $S$ is unbounded -- and in particular that $y$ is not an upper bound for $S$, which is what we needed to prove.