# Subfields of an ultrapower $\mathbb{R}^\omega/U$ obtained by reducing $U$.

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 $$f$$.

• 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? – Idéophage Jan 17 at 19:47
• 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 $0$. – Yushiro Aoki Jan 17 at 20:11
• Ah, you're right, sorry. – Idéophage Jan 17 at 20:13
• Don't be. Thank you so much. – Yushiro Aoki Jan 17 at 20:35

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.