Show that the subset of a real ordered field defined by a ring formula has a least upper bound. 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!!
 A: 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.
