In Proper and Improper forcing, VI.5; Claim 5.1 part 1 is the following:


If $F$ is a P-point in $V$, $P$ is a proper forcing notion and 
$\Vdash\_P `` F$ generates an ultrafilter"
Then the ultrafilter generated by $F$ is a P-point in $V^P$. 


The proof starts like this: let $p\Vdash \{\tilde A\_n : n<\omega\}$ be such that $p\Vdash \tilde A\_n\in F$. 


**By properness, for some $q\geq p$ and $A\_{n,m}\in F$ for $n,m<\omega$ we have $q\Vdash \tilde A\_n\in\{A\_{n,m} : m <\omega\}$**


To finish the proof, he uses that $F$ is a P-point in V, and finds an almost-subset of all the $A\_{n,m}$. 


My question is how is the bolded statement true?  It appears (since each $A\_{n,m}\in V$) that it would follow that P cannot add any subsets of $\omega$, which is clearly not true. What am I missing?


**EDIT**: Thanks Mohammad Golshani for the comment. I now believe that the bolded statement can be true. But how do we get $q$ (and the collection of $A\_{n,m}$)? Another question is once we have $q$, why is $q$ in $G$? Is the collection of such $q$ dense below $p$?



Shelah's proof that proper forcing preserves P-points
In more details: Let $N$ be a countable elementary submodel of some sufficiently large $H(\chi)$ containing $p$, $P$, $F$, and the sequence of names $\{\dot A\_n:n<\omega\}$. Then the countable set $N\cap F$ works for the collection of $A\_{n, m}$, and any $(N,P)$-generic extension of $p$ will work for $q$. 


