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This question is closely related to this one. Belatedly, it occurred to me that I'd probably picked the wrong test question. I'm leaving that question up because I don't think it's bad per se, but I think this one is what I should have asked:

Suppose $\mathfrak{A}=(\mathbb{R};<,\ldots)$ is an o-minimal structure in a countable language $\Sigma$. Letting $U$ be a unary predicate symbol not in $\Sigma$, say that an $\mathfrak{A}$-ordinator is a function $F:\omega_1\rightarrow\omega_1$ such that there is some $\Sigma[U]$-formula $\varphi(x)$ such that for club-many $\alpha<\omega_1$, if $A\subseteq\mathbb{R}$ has ordertype $\alpha$, then $$\operatorname{otp}(\varphi^{\mathfrak{A},U\mapsto A})=F(\alpha).$$

(Note that this differs from the definition in the above-linked post; that post worked "club-many" into the specific question, rather than the definition.)

It's easy to show that there are $(\mathbb{R};<,+,\times)$-ordinators which are not $(\mathbb{R};<,+)$-ordinators (basically, the only non-constant example of the latter is the identity function). I'm curious whether we can keep improving; in particular, whether adding the function $e^x$ gives us any power in this sense.

Question: Are there $(\mathbb{R};<,+,\times,\exp)$-ordinators which are not $(\mathbb{R};<,+,\times)$-ordinators?

I'm not quite sure where to look for a positive example; each map $\alpha\mapsto\alpha^n$ is an $(\mathbb{R};<,+,\times)$-ordinator, but I don't see that $\alpha\mapsto\alpha^\omega$ is an $(\mathbb{R};<,+,\times,\exp)$-ordinator. So the first natural function which might be a candidate for escaping one structure isn't obviously accessible by the other.

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    $\begingroup$ What can you say about the constant ordinators? $\endgroup$ Commented Aug 24, 2023 at 12:04
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    $\begingroup$ If $M=(\mathbb{R}; {<}, {+}, {\times}, \mathrm{exp})$ is o-minimal, then how could $\alpha\mapsto\alpha^\omega$ be an $M$-ordinator? Wouldn't that give that there is an $M$-definable subset of $\mathbb{R}$ of ordertype $\omega=2^\omega$? (Since we can define over $(M,A)$ where $A=\{0,1\}$, which has ordertype 2.) $\endgroup$
    – Farmer S
    Commented Aug 24, 2023 at 13:42
  • $\begingroup$ @FarmerS Argh, I messed up the definition. Fixing ... $\endgroup$ Commented Aug 24, 2023 at 16:43
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    $\begingroup$ @MichaelHardy The "o" in "o-minimal" stands for "order". It makes no sense to put it in math italics. $\endgroup$ Commented Aug 24, 2023 at 18:06
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    $\begingroup$ @JoelDavidHamkins Well, due to the "equality-on-a-club" allowance, maps like $\alpha\mapsto\omega$ are ordinators (and this is true even for $(\mathbb{R};<,+)$, contra a claim I made in my previous question now fixed). I don't immediately see the sup of (the output values of) the constant ordinators, but I suspect it's $\omega^\omega$. $\endgroup$ Commented Aug 25, 2023 at 3:58

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