Definition of a definable set of a substructure Let $M$ be a structure. Let $N$ be a substructure of $M$ (not necessarily an elementary substructure). Is it the case that every definable set $Y\subseteq$ $N^n$ in $N$ can be written as $Y=X\cap N^n$ where $X\subseteq M^n$ is a definable set in $M$?
 A: No.
It turns out that $\Bbb Z$ is definable in $\Bbb Q$ when consider it as an ordered field. However, a definable subset of $\Bbb R$ (when considered as an ordered field) is a finite union of intervals, therefore $\Bbb Z$ is not definable there.
In particular, there is no set of real numbers $A$ which is definable over $\Bbb R$ in the language of ordered fields, and $A\cap\Bbb Q=\Bbb Z$.

Under the assumption that the substructure is an elementary submodel, we actually do get this. In fact, here is a very nice observation I realized now:

The following are equivalent:
  
  
*
  
*$N\prec M$.
  
*If $A\subseteq M^n$ is definable by $\varphi$ in $M$, then $A\cap N^n$ is definable by $\varphi$ in $N$.
  

Proof.
Assume $N\prec M$, and $A=\{\bar m\mid M\models\varphi(\bar m)\}$. Then given any $k$-tuple from $N$, $\bar n$, $\bar n\in A\cap N^k$ if and only if $M\models\varphi(\bar n)$ if and only if $N\models\varphi(\bar n)$. So we get the wanted equality.
On the other hand, if we have relativization, suppose that $\bar n$ is a $k$-tuple from $N$, then for all $\varphi$, let $A_\varphi^M\subseteq M^k$ be $\{\bar m\mid M\models\varphi(\bar m)\}$, then $A_\varphi^N=A_\varphi^M\cap N^k$ and therefore $M\models\varphi(\bar n)$ if and only if $N\models\varphi(\bar n)$.
So your requested property is almost stating that $N$ is an elementary submodel of $M$. But it certainly holds for an elementary submodel.
