# Is the order arithmetic of the positive reals o-minimal?

Consider the structure of the positive real numbers $$(0, \infty)$$ with its unit $$1$$, its addition $$+$$, its multiplication $$\times$$, and its strict ordering $$>$$. Is this structure $$( (0, \infty), 1, +, \times, >)$$ o-minimal?

• Note that $>$ is definable from $+$ here: $a<b$ iff $\exists c(a+c=b)$. – Noah Schweber Feb 27 '19 at 16:36

Another way to see o-minimality is to note that $$log$$ induces an isomorphism $$log : ((0,\infty), 1, +, \times, >) \cong (\mathbb R, 0, \oplus, +, >),$$ where $$\oplus$$ is the binary operation defined by $$\oplus(x,y) = \log(e^x + e^y)$$. But $$\oplus$$ is definable in $$\mathbb R_{exp}$$. Hence the latter structure is a reduct of $$\mathbb R_{exp}$$ and therefore is o-minimal.

Yes. To see this, first note that we can interpret the positive reals in $$(\mathbb{R},0,1,+,-,\times)$$ as squares of non-zero real numbers. We can by induction on complexity of formulas define a translation map from formulas in the language of the positive reals to formulas in the language of the reals as follows:

• For an atomic formula $$\phi(x_1,\dots x_k)$$ by its translation is $$T(\phi)(y_1, \dots, y_k) = \phi(y_1^2, \dots , y_k^2)$$

• $$T(\phi \wedge \psi) = T(\phi) \wedge T(\psi)$$

• $$T(\neg \phi) = \neg T(\phi)$$

• $$T(\exists x \phi(x)) = \exists y (y \neq 0 \wedge T(\phi)(y^2))$$

Then the postive reals model $$\phi(c)$$ for $$c > 0$$ if and only the reals models $$T(\phi)(\sqrt{c})$$, for formulas with a single free variable of arbitrary complexity.

But then using o-minimality of the reals, we know $$\{y : T(\phi)(y)\}$$ is a finite union of points and intervals, so the same is true of $$\{c : \phi(c)\} = \{c : T(\phi)(\sqrt{c})\}$$

• So by the same argument, also the nonnegative reals $[0, \infty)$ is o-minimal, right? Thank you, James! – Colin Tan Feb 27 '19 at 14:59
• @ColinTan More simply, your structure is an expansions-by-definitions of a definable substructure of $(\mathbb{R}; +,\times)$, and so o-minimality immediately follows. – Noah Schweber Feb 27 '19 at 16:35