Is o-minimality preserved in expansions by unary functions? It is a well-known result that the ordered field $(\mathbb R,+,\cdot,0,1,<)$ is o-minimal. That means that in the language $(+,\cdot,0,1,<)$ all definable subsets of $\mathbb R$ are finite unions of points and open intervals. For example, a subset of $\mathbb R$ definable in this language is $\{x \mid \exists y \ y \cdot y = x\}$, which is just the open interval $(0,\infty)$ together with the singleton $\{0\}$. Now we can add unary function symbols to the language like $\exp$, which on $\mathbb R$ we interpret as the standard exponential. As a corollary of Wilkie's Theorem, also $(\mathbb R,+,\cdot,0,1,<,\exp)$ is o-minimal. This just serves as an example that we can expand the ordered field of real numbers by unary functions and still keep this new structure o-minimal.
My question is now the following: Suppose that $f$ and $g$ are unary functions on $\mathbb R$ such that both $(\mathbb R,+,\cdot,0,1,<,f)$ and $(\mathbb R,+,\cdot,0,1,<,g)$ are o-minimal. Is then also $(\mathbb R,+,\cdot,0,1,<,f,g)$ o-minimal?
All the examples I can come up with seem to confirm that this is indeed the case. But in the general case I do not know how $f$ and $g$ interact with each other and can thus not make a statement about their definable sets. Is there an easy counterexample? Or maybe there is even a (simple) proof which generalises to arbitrary o-minimal structures.
 A: No. A class of counterexamples is given in the paper Quasianalytic Denjoy-Carleman classes and o-minimality, by Rolin, Speissegger, and Wilkie.
Theorem 2(1) implies that for any $C^\infty$ function $f\colon \mathbb{R}\to \mathbb{R}$, there exist functions $g,h\colon [-1,1]\to \mathbb{R}$ such that $(\mathbb{R},\leq,+,\times,-,0,1,g)$ and $(\mathbb{R},\leq,+,\times,-,0,1,h)$ are both o-minimal, and $g(x) + h(x) = f(x)$ for all $x\in [-1,1]$. (In fact, you can even expand the real field by a whole "Denjoy-Carleman class" of functions containing $g$ or $h$, while retaining o-minimality.) On the other hand, for any closed set $A\subseteq \mathbb{R}$, there is a $C^\infty$ function $f\colon \mathbb{R}\to \mathbb{R}$ with zero set exactly $A$ (see here, for example). 
Taking $f$ to be such a function with zero set $A = \{0\}\cup \{1/n\mid n\in \mathbb{N}\}$ and picking $g$ and $h$ for $f$ as in the Rolin-Speissegger-Wilkie theorem, the structure $(\mathbb{R},\leq,+,\times,-,0,1,g,h)$ is not o-minimal, since we can define $A$ by the formula $g(x) + h(x) = 0$.
