# Real Closed Fields with Predicate for a Dense Subfield

Consider $M = (\mathbb{R};+,<, \times, 0, 1, K)$ where $K$ is a unary predicate which holds on $\mathbb{Q}$ (or any dense subfield of $\mathbb{R}$).

Question: Is it true that the parametrically definable sets in 1-free variable is the finite union of points and 'locally dense' sets?

(I'm not sure if locally dense is a real term, so I will definite it as such: $U$ is locally dense if there exists some interval $I \subset M \cup \{\pm \infty\}$ such that for every $a,b \in I$ there exists a $q \in U$ such that $a<q<b$).

Motivation: It is a corollary of the Tarski–Seidenberg theorem that $(\mathbb{R};<, +, \times, 0, 1)$ is O-minimal (i.e. every parametrically definable subset is the finite union of points and intervals).

If $\varphi(x)$ is a quantifier free formula in the language (with parameters), then $\varphi(x)$ is equivalent to a boolean combination of $p(x) = 0$, $q(x) >0$, $K(x)$ and $\neg K(x)$. This case is clear. When we begin to add quantifiers, things become much more tricky to pin down. For instance, $(\exists y)(K(y) \wedge y^2 = x)$ - there does not seem to be a way to eliminate quantifiers.

• If you want to talk about quantifier elimination for $\mathbb{R}$, you have to add the symbol $<$ and interpret it as the strict order. – nombre Jul 26 '15 at 22:23
• @nombre: I can't believe I forgot to add $<$ in the question! Right, you can't even talk about O-minimality without linear order. – Kyle Gannon Jul 26 '15 at 22:24

In Definability and decision problems in arithmetic Julia Robinson has proven (among other interesting things) that $\mathbb{Z}$ was definable in $(\mathbb{Q},+,.,0,1)$.
So if $K$ is interpreted as $\mathbb{Q}$, definable sets in $\mathcal{M}$ can be more complicated than that, and there is no quantifier elimination in $\left\langle +,.,0,1,<,K \right\rangle$.
Now, if $K$ is interpreted as another dense subset of $\mathbb{R}$, then it probably depends on the subset. I guess the result holds for some real closed subfields of $\mathbb{R}$ and fails for some other subsets, among which there is $\mathbb{Q}$.
• Yes, but it seems that you have to pay to download the pdf; didn't it use to be free to download JSTOR articles? i can't find a note expressing any change in their policy. You can read the article online apparently, but still, it's too bad. By the way, the problem of knowing if your statement is true if $K$ is a real closed subfield of $\mathbb{R}$ seems pretty interesting, and it might not be too difficult. – nombre Jul 26 '15 at 22:40