Sorry for asking but I'm really stuck, I have absolutely no idea of where start solving exercices 3. Can someone give to me some hints? Down I say what is my idea.
Let $ L = \{ 0,1,<,+, \cdot \} $ be a language of first order containing the symbol $=$. Let $ R = \left< \mathbb{R} ,0,1,<,+, \cdot \right> $ and let $ Q =\left< \mathbb{Q} ,0,1,<,+, \cdot \right> $ be the $L$-structures corresponding to the real and rational number with the usual interpretation.
- Existe a $L$-structure that is countable elementary equivalent to $R$ ? (Easy with downward SK theorem)
- Exsite a $L$-structure that is equipotent to $\mathbb{R}$ that is elementary equivalent to $Q$? (Easy with upward SK theorem)
- Prove that there exsits a $\mathcal{M}, \mathcal{M}' $ structure such that
i) The domain of $\mathcal{M}, \mathcal{M}' $ is countable
ii) $\mathcal{M}, \mathcal{M}' $ are elementary equivalent to $R$
iii) There exsits a dense embendding of $Q$ into $\mathcal{M} $ and there is not a dense embendding of $Q$ into $\mathcal{M}'$
where a for dense embendding i mean that $ f: Q \rightarrow \mathcal{M} $ is an embendding such that for all $a,b, \in \mathcal{M} $ if $a < b$ exsits $q \in Q$ such that $a < f(q) < b $.
My ideas for 3. Since any formula parametrized in $Q$ is of finite height (the height of its tree of decomposition) moreover for a given height there are only countable many formula since $Q$ is countable. Then we have only countable many $ \varphi $ parametrized in $Q$ such that $ R \models \varphi $ and $ Q \not\models \varphi $. Let's define some enumeration of these formule, and consider the set $ T = \{ \varphi_n : n \in \mathbb{N}, \varphi_n \text{ parametrized in } Q \} $, i.e. $ R \models \varphi $ and $ Q \not\models \varphi $ if and only if $ \varphi \in T $ or $ \varphi $ is not parametrized in $Q$. Then consider the theory of $ Q$, i.e. $ Th(Q)= \{ \varphi : \varphi \text{ closed formula such that } Q \models \varphi \} $. Clearly $ R \models Th(Q) \cup T $, by upward SK theorem we have that exist a model $ \mathcal{M} $ that is countable. Similary we have that $Th(R) $ possesses by upward SK-theorem a model $ \mathcal{M} ' $ that is countable. My idea is to shows that these models $ \mathcal{M} $ , $ \mathcal{M} ' $ satisfies property iii) but I don't know how and also I don't know if my intuition is correct.