a question about diagonal prikry forcing Suppose <\kappa_n|n<\omega> is a strictly increasing sequence of measurable cardinals, 
\kappa is the limit of this sequence. For each n<\omega, U_n is a normal measure on 
\kappa_n. P is the diagonal Prikry forcing corresponding to \kappa_n's and U_n's. 
Suppose g is P-generic sequence over V. We have known that for each strictly increasing 
sequence x of length \omega such that each x(i)<\kappa_i and x\in{V}, x is eventually 
dominated by g. In V[g], suppose A is a subset of \kappa, A is not in V. Is there a strictly 
increasing sequence y of length \omega such that each y(i)<\kappa_i and y\in{V[A]}, y is not 
eventually dominated by g?
(g can eventually dominate all such sequences in V, V[A] is greater than V, I feel g can not 
eventually dominate all such sequences in V[A].)
 A: Yes.    The answer is obviously "yes" if $A$ is a subset of $g$, so it is sufficient to show for any subset $A\subset\kappa$ there is $A'\subset g$ such that $V[A']=V[A]$.
I assume that this is known, and it struck me at the start as obviously true, but I can't recall having seen it.  Here is an outline of a proof.
Write the conditions (following Gitik) as $x=\langle x_i\mid i \in\omega\rangle$, with
$x_i\in\kappa_i\cup U_i$.    Write $A_n=A\cap \kappa_n$.    There is
$x\leq^* 1^P$ which forces that $A_n$ is decided by conditions $y$
with $y_i\in U_i$ for $i>n$.   It follows in particular that $A_n\in V$.
Find $x'\leq^* x$ which decides, for each $n$, the sentence "there is
$z\in \dot G$  such that $z_i$ decides the value of $\dot A_n$ and
$z_i\in U_i$."   Let $b_n$  be the (finite) set of $i$ for which this
sentence is forced to be false.  Then there is $x''\leq^* x'$ and 
1--1 functions $h_n\colon \Pi_{i\in b_n} x_i''\to \kappa_n$ such that
$x''$ forces that $A_n=h_n(g\upharpoonright b_n)$.   
Set $A'=\bigcup_n b_n$.  Then $V[A] = V[A']$.
