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 $f$.
 A: 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.
