1
$\begingroup$

Let $\left<F,+,\cdot,\leq\right>$ be a real closed field. Consider the following schema:

$(*)$ Let $\phi(\cdot)$ be a first-order predicate. Let there exist $x \neq y$ such that $\phi(x),\phi(y)$. For all $z$, let $\phi(z)$ imply $\phi(2z-x)$ and $\phi(2x-z)$. Then, for all $t$, there are $x',y':\phi(x'),\phi(y'),x'\leq t\leq y'$.

This schema seems to be a first-order one. This also seems to imply Archimedean property:

We have $(2z-x)-x=2(z-x)$ and $(2x-z)-x=(x-z)$. So if $\phi(y)$ and $|x-z|=2^n|x-y|, n \in \mathbb N$, then $\phi(z)$.

Take any $x', y': x' \leq y'$. By $(*)$, we can find $u,z,t,w: u \leq x' \leq z, t \leq y' \leq w, |x-u|=2^n|x-y|,|x-z|=2^m|x-y|$,$|x-t|=2^k|x-y|,|x-w|=2^l|x-y|$. $u \leq x' \leq y' \leq w$. $|x'-y'| \leq |u - w| \leq (2^l+2^n)|x-y|$.

But I know that, at least under GCH, $\mathbb R$ is the only real closed Archimedean field; I also know that real closed fields are indistinguishable by first order properties.

So, at least one of my assertions has to be false. Which one?

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.