Elementary Submodels in Set Theory I was reading the following summary on elementary submodels:
http://boolesrings.org/mpawliuk/2012/01/26/a-practical-guide-to-using-countable-elementary-submodels/
Say $M\prec N$. The link above indicates (Super Useful Fact) that any set definable by parameters in $M$ is an element of $M$. In particular, let $\phi(v_1,v_2)$ be a first order formula of set theory. My understanding of this is that, if $x\in M$ and $\{t: t\in N\wedge \phi(x,t)\}\in N$, then $\{t: t\in M\wedge \phi(x,t)\}\in M$. What I am wondering is whether we can guarantee that $\{t: t\in M\wedge \phi(x,t)\}\subset M$. As a first order sentence, we have
($\exists z) \ (\forall t)t\in z\leftrightarrow(\phi(x,t))$
Assuming this is satisfied by $N$, it must be satisfied by $M$ since $M\prec N$. In particular, there exists $z'\in M$ such that $(\forall t\in M)t\in z'\leftrightarrow(\phi(x,t))$. Certainly, $z'$ is a subset of the class we want but perhaps it contains elements which are not in $M$. Can we guarantee that $z'\subset M$?
 A: This all comes from the definition of elementary, and is a little more delicate than you might think.
Model theoretically, a model $M$ is an elementary submodel of a model $N$ if $M$ is a submodel of $M$ (the universe of $M$ is a subset of that of $N$, and all interpretations of function/predicate/constant symbols are just the restrictions of those of $N$), and every statement about the elements of $M$ is true in $N$ iff it is true in $M$; meaning that given a formula $\varphi ( x_1 , \ldots , x_n )$ and $a_1 , \ldots , a_n \in M$, $$M \models \varphi ( a_1 , \ldots , a_n ) \Leftrightarrow N \models \varphi ( a_1 , \ldots , a_n ).$$
Now if $M$ is a (countable) elementary submodel of $H$ (for some model of a suitable fragment of $\mathsf{ZFC}$) and there is a formula $\varphi ( x, x_1 , \ldots , x_n )$ and elements $a_1 , \ldots , a_n \in M$ such that $$H \models ( \exists z ) ( \forall u ) ( u \in z \leftrightarrow \varphi ( u , a_1 , \ldots , a_n ) ), \tag{1}$$ then this statement must also be true in $M$, namely $$M \models ( \exists z ) ( \forall u ) ( u \in z \leftrightarrow \varphi ( u , a_1 , \ldots , a_n ) ). \tag{2}$$  Letting $b \in H$ be a witness for (1) and $c \in M$ a witness for (2), note that $c \in H$.  Note, also, that $$M \models ( \forall u ) ( u \in c \leftrightarrow \varphi ( u , a_1 , \ldots , a_n ) ), \tag{3}$$ and so by elementarity $$H \models ( \forall u ) ( u \in c \leftrightarrow \varphi ( u , a_1 , \ldots , a_n ) ).\tag{4}$$  But now logic intervenes and tells us that $$H \models ( \forall  u ) ( u \in b \leftrightarrow u \in c )$$ and if the Axiom of Extensionality is true in $H$ (which in practice it always will), then $$H \models c = b.$$  And so that set $b$ is an element of $M$.
We cannot guarantee that any set in $M$ is a subset of $M$ (in particular, the set $\mathbb{R}$ of real numbers will be an element of any countable elementary submodel $M$ of a suitable $H$, but since $M$ is countable it cannot be that $\mathbb{R} \subseteq M$).
