0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ So you're asking for a direct proof, rather than one that uses contradiction? $\endgroup$ – mweiss Jan 17 '17 at 18:26
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.