Let $R$ be a real closed field and $f: R \to R$ a map. Then let $\textrm{F}(f)$ be the set of semialgebraic subsets of $R^2$, which contain $(t,f(t))$ for all $0< t< \epsilon$ for some $\epsilon >0$. Clearly $\textrm{F}(f)$ is a filter on the set of semialgebraic subsets subsets of $R^2$ and if $f$ is defineable, then $\textrm{F}(f)$ is an ultrafilter. An other example would be the exponential function. Does anyone know for which functions $f$ the set $\textrm{F}(f)$ is also an ultrafilter?
1 Answer
Whenever $(R,f)$ is an o-minimal structure, $F(f)$ will be an ultrafilter.
This includes your examples: $f$ definable and the exponential. Also, since the property defining $F(f)$ depends only locally on $f$, using the fact that $(R,f\upharpoonright _{[0,1]})$ is o-minimal for any analytic function $f$, one gets that $F(f)$ will be an ultrafilter for any analytic function $f$.
I suspect some kind of converse should be true, which would give you a characterization (perhaps something along the lines "$F(f)$ is ultrafilter iff $(R,f\upharpoonright _{(0,\epsilon)})$ is o-minimal for some $\epsilon>0$"), but I'm not so sure about that.