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Let $\bar{\mathbb{R}}$ be the structure of the real field, that is $(\mathbb{R},0,1,+,-,*,<)$ . We say that a function $f$ is of growth higher than exponential if for all $N\in \mathbb{N}$ there $f(x)$ is ultimately greater than $\exp^N(x)$. That is there is we can find $r\in \mathbb{R}$ such that for all $x>k$ we have $\exp^N(x) < f(x)$ where $\exp^N(x)$ stands for $\exp(\exp(\cdots \exp(x)\cdots))$ $N$ times.

Is it possible to find an o-minimal expansion (in the model theoretic sense) of $\bar{\mathbb{R}}$ where an $f$ as above is definable?

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  • $\begingroup$ Is there such a high-growth function that is Pfaffian? $\endgroup$
    – Joel David Hamkins
    Commented Feb 7, 2013 at 0:22

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This is a well-known open problem, see e.g. http://www1.maths.leeds.ac.uk/maloa/lecturenotes/lyon.pdf . All currently known o-minimal expansions of the reals (such as the pfaffian closure of $\mathbb R$) are exponentially bounded.

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    $\begingroup$ By contrast, one might also ask if there are o-minimal expansions where you have functions that grow faster than polynomials but no functions of exponential growth. A beautiful and surprising theorem of Chris Miller showed that you can't. Indeed exponentiation is definable in any o-minimal expansion of the reals containing any function of super-polynomial growth. $\endgroup$
    – Dave Marker
    Commented Feb 8, 2013 at 17:11

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