Generic filters in the ground model (forcing) In his book Set Theory (second edition), page 203, Thomas Jech writes

Lemma 14.4 Let $\mathfrak{M}$ be a countable model of ZFC and $P$ be a partially ordered set. If $p\in P$, there exists a generic filter on $P$.

Then, the following exercise is given:

Exercise Let $\mathfrak{M}$ as above. Let $P\in M$ be a partially ordered set with no atom (for every $p\in P$, then there exist $q,r\in P$ such that $q,r\le p$ and
  $q,r$ are incompatible). If $G\subset P$ is generic over $\mathfrak{M}$,
  then $G\not\in \mathfrak{M}$.

The short proof of Lemma 14.4 is given here:

According to the exercise, the filter $G$ constructed in this proof must fail to belong to $\mathfrak{M}$ in some cases. But why ? Since $P\in \mathfrak{M}$ it seems that $G\in \mathfrak{M}$ using comprehension, doesn't it ?
 A: If $G$ belongs to $\mathfrak{M}$, then so does $P \setminus G$.  Since $P$ has no atoms, it is easy to show that $P \setminus G$ is dense in $P$, and therefore $G \cap ( P \setminus G ) \neq \emptyset$ by genericity, which is absurd!
The point is that in the proof of Lemma 14.4 you enumerated the dense subsets of $P$ which belong to $\mathfrak{M}$, and this enumeration will generally take place outside of $\mathfrak{M}$, and in the "real world".  From this there is no reason to expect that the sequence of conditions of $P$ you base your generic set $G$ on belongs to $\mathfrak{M}$, and therefore there is little reason to expect that the generic set itself belongs to $\mathfrak{M}$.  In fact, the exercise makes it clear that under fairly mild (and very common) assumptions the generic set will provably not belong to $\mathfrak{M}$.
A: Comprehension says (eliding some details) that $\{x\in A | \varphi(x)\}$ exists in $\mathfrak{M}$ for any formula $\phi$ (with parameters); presumably you intend $A$ here to be $P$.  But there is not generally going to be any such formula $\varphi$ to define a generic filter $G$, so comprehension won't tell you that such a $G \in \mathfrak{M}$.
