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Let $\mathcal{R}_0=\{\mathbb{R},+,<\}$ be the order divisible abelian group of reals and $\mathcal{R}=\{\mathbb{R},+,\cdot,<\}$ be the real closed field of reals. Both $\mathcal{R}_0$ and $\mathcal{R}$ are o-minimal structures.

Let $f:\mathbb{R}\setminus\{0\}\rightarrow \mathbb{R}$ be the map given by $f(x)=1/x$. Consider $\mathcal{R}_f=\{\mathbb{R},+,f,<\}$, the structure generated after expanding $\mathcal{R}_0$ by adding function $f$ as a definable set.

Is $\mathcal{R}_f = \mathcal{R}$, i.e. is the field product in $\mathbb{R}$ definable in $\mathbb{R}_f$?

The question arises from the more general doubt of whether there exists an o-minimal group containing a definable homeomorphism between a bounded and an unbounded set that doesn't end up defining also a field structure.

Anguepa
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1 Answers1

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Multiplication is definable in $\mathcal R_f$.

Step 0. All the additive group operations ($+, -, 0$) are definable.

The operation $+$ is given in the language, while $0$ is the unique identity for $+$ and $-x$ is the unique additive inverse of $x$.

Step 1. The singleton $\{1\}$ is definable.

$1$ is the unique element $x$ satisfying the formula $(0<x)\wedge (f(x)=x)$.

Step 2. The squaring function $y = x^2$ is definable.

$x^2=y$ holds exactly when $x=\pm 1$ and $y=1$, or else $x\neq \pm 1$ and

$$ y = \frac{1}{\frac{1}{x-1} - \frac{1}{x+1}} + \frac{1}{\frac{1}{x-1} - \frac{1}{x+1}} + 1, $$

which is expressible using only the additive structure, $1$, and $f(x) = \frac{1}{x}$.

Step 3. The product function $z = xy$ is definable.

$xy = z$ holds exactly when $(x+y)^2 = x^2 + y^2 + z+z$, which is expressible using only the additive structure and squaring.

Keith Kearnes
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