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We promise that $F(g_1, \ldots, g_k)$ denotes the composition for each $F \colon \mathbb{R}^k \to \mathbb{R}$ and $g_1,\ldots ,g_k \colon \omega \to \mathbb{R}$. Let $M$ be a subset of $\mathbb{R}^\omega$ and $M^\wedge$ denotes the set of all $F(g_1, \ldots, g_k)$ where $k \in \omega$, $g_1,\ldots, g_k \in M$, and $\ F \colon \mathbb{R} ^k \to \mathbb{R}$ where $F$ is definable in $\langle\mathbb{R},1,0, +,\cdot,<\rangle $ without parameters. Notice that, considering the case $k=0$, $\emptyset^\wedge =$ the set of all definable elements in $\mathbb{R}$. We say that a filter $\mathcal{F}$ on $\mathcal{P}(\omega)$ is an $M$-ultrafilter if either $\{i \in\omega \mid g(i) < g'(i) \}\in \mathcal{F}$, $\{i \in\omega \mid g(i) = g'(i) \}\in \mathcal{F}$, or $\{i \in\omega \mid g(i) > g'(i) \}\in \mathcal{F}$ for each $g,g' \in M$. Let $f \in \mathbb{R}^\omega$ and $\mathcal{F}$ be an $M^\wedge \cup \{f\} $-ultrafilter.

My question is: Is $\mathcal{F}$ an $(M\cup \{ f\} )^\wedge$-ultrafilter?

I know that $M^\wedge/\mathcal{F}$ is a real closed field. Thus, by the $o$-minimality of the theory of real closed fields, for each definable functions $F_1, F_2 \colon \mathbb{R}^{k+1} \to \mathbb{R}$ and $g,g_1,\ldots , g_k \in M$, there are $h_1^-,\ldots, h^-_m,h_1^+,\ldots, h_m^+, h_1,\ldots,h_n \in M^\wedge$ where maybe $h_1^-$ and $h_m^+$ are $-\infty$ and $+\infty$, respectively, such that \begin{align*} F_1(g_1,\ldots, g_k,g) \leq_{\mathcal{F}} F_2(g_1,\ldots,g_k,g) \iff \bigvee_{1\leq l \leq m}h_l^- <_\mathcal{F} g <_\mathcal{F} h_l^+ \lor \bigvee_{1\leq l \leq n} h_l =_{\mathcal{F}} g. \end{align*} But I don't know whether this is valid if we replace $g$ by $f$.

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    $\begingroup$ I still have to read the second paragraph but I don't know anything about $o$-minimality. I just want to ask you this: your statement implies that an $\{g\}^{∧} ∪ \{f\}$-ultratilfter is an $\{f,g\}^{∧}$-ultrafilter. But this is false, right? Take for instance $g=(0,0,0,…)$ and $f = (1,1/2,1/3,…)$. The filter of all the $\{n∈𝜔\,|\,n≥K\}$ for $K∈𝜔$ is then an $\{g\}^{∧} ∪ \{f\}$-ultratilfter but is not an $\{g,f\}^{∧}$-ultratilfter, right? $\endgroup$ – Idéophage Jan 17 at 19:47
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    $\begingroup$ I have added a remark. In your case, $\{g\}^\wedge = \emptyset^\wedge$. By the way, you are not right. If $F\colon \mathbb{R} \to \mathbb{R}$ is definable, then $F| [0, \epsilon)$ is either strictly increasing, strictly decreasing, or constant for some $\epsilon >0$. Thus, the sequence $F(f)$ is strictly increasing, strictly decreasing, or constant, eventually. So your filter is strong enough to compare $F(f)$ and $0$. $\endgroup$ – Yushiro Aoki Jan 17 at 20:11
  • $\begingroup$ Ah, you're right, sorry. $\endgroup$ – Idéophage Jan 17 at 20:13
  • $\begingroup$ Don't be. Thank you so much. $\endgroup$ – Yushiro Aoki Jan 17 at 20:35
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I have solved it.

Suppose that \begin{align*} A := \{ i \in \omega \mid F(g_1(i),\ldots, g_k(i),f(i)) \leq 0 \} \notin \mathcal{F}. \end{align*} Since $M^\wedge/ \mathcal{F} \models \mathrm{RCF}$ and $\mathrm{RCF}$ is an $o$-minimal theory, there are $h^-_1,\ldots, h^-_m,h_1^+,\ldots, h_m^+, h_1,\ldots,h_n \in M^\wedge$ where maybe $h_1^-$ and $h_m^+$ are $-\infty$ and $+\infty$, respectively, such that \begin{align*} M^\wedge / \mathcal{F} \models \forall v \left( F_0([g_1]_\mathcal{F} ,\ldots, [g_k]_\mathcal{F} ,v) \leq 0 \iff \left( \bigvee_{1\leq l \leq m}[h_l^-]_\mathcal{F} < v < [h_l^+]_\mathcal{F} \lor \bigvee_{1\leq l \leq n} [h_l]_\mathcal{F} = v. \right) \right) \end{align*} Define \begin{align*} X := \left\{ i \in \omega \mid \mathbb{R} \models \forall v \left( F(g_1(i),\ldots, g_k(i) ,v) \leq 0 \iff \left( \bigvee_{1\leq l \leq m}h_l^-(i) < v < h_l^+(i) \lor \bigvee_{1\leq l \leq n} h_l(i)= v \right) \right) \right\} \end{align*} and \begin{align*} Y := \left\{ i \in \omega \mid \mathbb{R} \models \left( \bigvee_{1\leq l \leq m}h_l^-(i) < f(i) < h_l^+(i) \lor \bigvee_{1\leq l \leq n} h_l(i)= f(i) \right) \right\}. \end{align*} Since $\mathcal{F}$ is an $M^\wedge \cup \{f\}$ -ultrafilter, either $Y \in \mathcal{F}$ or $\omega \setminus Y \in \mathcal{F}$. If $Y \in \mathcal{F}$, then $X \cap Y \in \mathcal{F}$ and $\mathbb{R} \models F(g_1(i), \ldots , g_k(i) , f(i) ) \leq 0$ for each $i \in X\cap Y$, it is a contradiction. Thus $\omega \setminus Y \in \mathcal{F}$. For each $i \in X \setminus Y$, $\mathbb{R} \models F(g_1(i),\ldots , g_k(i), f(i) ) > 0 $ so that $[F(g_1,\ldots g_k , f)] _\mathcal{F} > _\mathcal{F} 0$. Therefore, $(M \cup \{ f\} ) ^\wedge / \mathcal{F}$ is totally ordered.

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