In this question, I consider the language $L_{or} = \{=,<,+,\cdot,1,0\}$ of ordered rings.
My question: Is there an extension of real closed fields $F \subset G$ such that:
- every $a \in G \setminus F$ defines a countable gap (i.e., a cut of countable cofinality on both sides) on $F$ but,
- there exists $\bar{b} \in (G \setminus F)^{< \omega}$ such that $type(\bar{b}/F)$ is uncountably generated ?
My observation: Let $(x_\alpha \mid \alpha < \omega_1)$ be a decreasing sequence such that: \begin{align*} 0 < \epsilon < x_\alpha < n^{-1} \end{align*} for each $\alpha < \omega_1$ and each $1\leq n \in \mathbb{N}$. Let $\mathbb{F}$ be a minimal extension of $\mathbb{Q}$ such that,
- $\mathbb{F}$ is a real closed field and
- each $x_\alpha \in \mathbb{F}$.
Then, $y_1= \pi - \epsilon$ and $y_2 = \pi + \epsilon$ defines a countable gap on $\mathbb{F}$. However, $type(y_1,y_2/\mathbb{F})$ is uncountably generated.
In this case, an extension $\mathbb{G}$ must have $\epsilon$, so some single element of $\mathbb{G}$ must define an uncountable gap on $\mathbb{F}$. So this case is not an answer.