# Are archimedean subextensions of ordered fields dense?

Let $$E$$ be an ordered field and let $$F$$ be a real closed subfield. We say that $$E$$ is $$F$$-archimedean if for each $$e\in E$$ there is $$x\in F$$ such that $$-x\le e\le x$$.

Is it true that if $$E$$ is $$F$$-archimedean then every interval in $$E$$ contains an element in $$F$$? That is, is it true that for every $$e in $$E$$ there is an element $$x\in F$$ such that $$e?

This is known if $$F$$ is the field of real algebraic numbers (in which case $$E$$ is an ordered subfield of $$\mathbb{R}$$), and it seems to me that it should have an easy proof in the general case. However I cannot find neither an easy proof nor a counterexample.

• Have you tried $E = \mathbb{Q}((\infty)), F = \mathbb{Q}((\infty^2))$? – user44191 Dec 17 '18 at 19:49
• What is $\mathbb{Q}((\infty))$? Do you mean $\mathbb{Q}(t)$ with the order making $t$ an infinite positive element (I'm not sure this order can be extended to the Laurent series)? – Denis Nardin Dec 17 '18 at 19:50
• Yes, sorry, my idiocy. – user44191 Dec 17 '18 at 19:51
• No need to be self-deprecating it's a good counterexample... Uhm there might in fact be no number between $t$ and $t+1$. Can you do a counterexample where $F$ is real closed? Sorry for changing the goal posts, I should have put it from the beginning since I was really thinking about that case. – Denis Nardin Dec 17 '18 at 19:52
• Try $E$ the real closure of $\mathbb{Q}(x, y)$ with $x > \mathbb{Q}, y > \mathbb{Q}(x)$, and $F$ the real closure of $\mathbb{Q}(y)$? – user44191 Dec 17 '18 at 20:00

Let $$F$$ be any real-closed field of uncountable cofinality. That is, every countable subset of $$F$$ is bounded. One can make such a field in a process of $$\omega_1$$-many field extensions; alternatively, the ultrapower of $$\mathbb{R}$$ by nonprincipal ultrafilter on $$\omega$$ also has uncountable cofinality.

Let $$E=F^\omega/\mu$$ be an ultrapower of $$F$$ by a nonprincipal ultrafilter $$\mu$$ on $$\omega$$. We may identify $$F$$ with its canonical copy in $$E$$ using equivalence classes of constant functions $$x\mapsto [c_x]_\mu$$.

Since every function from $$\omega$$ to $$F$$ is bounded by a constant function, it follows that $$E$$ is $$F$$-archimedean. But $$F$$ is not dense in $$E$$, since there are no constant functions between $$[\text{id}]_\mu$$ and $$[\text{id}+1]_\mu$$, where $$\text{id}:n\mapsto n$$, viewing $$\omega\subset F$$.

• Oh well... I guess that was too good to be true. Thanks for the answer, I'll wait till tomorrow to see if other interesting answers pop up and then accept it. – Denis Nardin Dec 17 '18 at 20:08
• Sure thing, no problem. I liked your question. The argument I give shows that no first-order property of the ordered field (like being real-closed) can have the consequence you want, since we can always find such fields of uncountable cofinality and then take an ultrapower. – Joel David Hamkins Dec 17 '18 at 20:22

Let $$E$$ be the real closure of $$\mathbb{Q}(x, y) = (\mathbb{Q}(x))(y)$$, with order given by $$x > \mathbb{Q}$$ and$$y > \mathbb{Q}(x)$$. In other words, positivity on $$\mathbb{Q}(x, y)$$ is determined first by degree in $$y$$, and then by degree in $$x$$. Then let $$F$$ be the real closure of $$\mathbb{Q}(y)$$.

First, we prove that $$E$$ is $$F$$-archimedean. Let $$e \in E$$. There is some $$e' \in \mathbb{Q}(x, y)$$ with $$e' > e > -e'$$. Then $$e'$$ has degree $$n$$ in $$y$$ for some $$y$$; let $$f = y^{n + 1}$$. By the order on $$\mathbb{Q}(x, y)$$, we have $$f > e' > e > -e' > -f$$. Therefore $$E$$ is $$F$$-archimedean.

On the other hand, clearly, there is no element of $$F$$ between $$x$$ and $$x + 1$$.

• Thank you for your answer! Now I'm going to have a hard time deciding which one to accept :) – Denis Nardin Dec 17 '18 at 20:35
• I ended up accepting JDH's answer because it provides me with a "machine" to generate counterexamples to similar statements, but your simple answer was very appreciated (and possibly shows that I hadn't thought through the situation as much as I thought...) – Denis Nardin Dec 18 '18 at 22:37

This is similar to user44191's answer but I want to put the emphasis on the fact that there is no reason that the cofinality of $$F$$ in $$E$$ (which is what ou call [$$E$$ is $$F$$-archimedean]) imply the density of $$F$$ in $$E$$.

Indeed, if $$F$$ is any non-archimedean ordered field, then the field $$F(t)$$ can be equipped with an order where $$t$$ is positive infinite but smaller than any positive infinite element of $$F$$. Thus $$F$$ is cofinal in $$F(t)$$ but not dense in it since no element of $$F$$ is close to $$t$$. To do so, say that a fraction $$\frac{P(t)}{Q(t)}$$ is positive if $$P(t)$$ and $$Q(t)$$ have the same sign, where the sign of a polynomial $$R(t)$$ is positive if $$R(s)$$ is positive for sufficiently large finite element $$s$$ of $$F$$.

This is a special case of filling a cut in an ordered field using a simple extension.