So currently trying to get well practiced in model theory, and i have come across the following question which i need some help with.
Esentially let $S \subseteq \mathbb{R}$ be a non empty set, bounded above defined by an $L_{oring}$ formula $\phi(x)$. Suppose also that $<F;+,.,0,1,< >$ is another real closed field. Show that the subset of F that is defined by $\phi(x)$ also has a least upper bound.
Literally no idea where to begin. Heeeeelp!!
Note that the theory of real closed fields is complete. This means that any two real closed fields are elementarily equivalent. You can use these facts to solve your problem: Since the subset defined by $\phi(x)$ in $\mathbb{R}$ is bounded from above, it must have a least upper bound in $\mathbb{R}$ (this follows from the ordering completeness of the ordered field $\mathbb{R}$). Now, you have to verify that the statement
$\phantom{aaaaaaaaaaaaaaaaa}$"the set defined by $\phi(x)$ has a least upper bound"
can be formalized by a sentence in the language of ordered rings. Assume $\psi$ is the corresponding sentence. Then, because of my previous remarks, $\mathbb{R} \vDash \psi$ and since two real closed fields are elementarily equivalent, it follows that $F \vDash \psi$.
$\textbf{Edit}:$ Here is a hint concerning the formalization of the above statement. Start with $$\exists x[\forall y(\phi(y) \to (y<x \vee y=x)) \wedge \phantom{a}...\phantom{a}]$$ This formula expresses that there is an upper bound $x$ for the set defined by $\phi$. Now, try to complete this formula such that it expresses that $x$ is also a least upper bound.