In Kunen's book (introduction to independence proofs, ) the following lemma is proved (chapter 8, lemma 5.14):


Assume that in M, $\alpha$ is a limit ordinal, 
$( ( \mathbb{P}\_\xi : \xi \leq \alpha) , (\pi\_\xi : \xi < \alpha) )$
is an $\alpha$-stage iterated forcing construction with supports in $\mathcal{I}$ , and each element of $\mathcal{I}$ is bounded in $\alpha$. suppose G is $\mathbb{P}\_\alpha$-generic over $M \:,\: S \in M \:,\: X\subseteq S \:,\: X \in M[G] \:,\: and \: (|S| < cf(\alpha))^{M[G]}.$ then for some $\eta<\alpha \:,\: X\in M[i\_{\eta\alpha}^{-1}(G)]. $


the proof seemed very simple at first glance. however: 


kunen shows that "for $s \in S \:,\: s \in X $ if and only if there is a $ \xi = \xi\_s < \alpha$ such that $\exists p \in G\_\xi (i\_{\xi\alpha}(p)\Vdash\_{\mathbb{P}\_\alpha}\check{s} \in \sigma )$". 


pretty obvious. but then kunen uses this to form, **in M[G]** , the set $\{\xi\_s : s \in X\}$. why is this set in M[G]? the "$\exists p \in G\_\xi $" part is ok, since the sequence $(G\_\xi : \xi < \alpha)$ is in M[G]. but the "$(i\_{\xi\alpha}(p)\Vdash\_{\mathbb{P}\_\alpha}\check{s} \in \sigma )$" , although it is indeed definable in M by definability of forcing,
it is not necessarily definable in M[G]. 


then he does it again by showing that X = $\{s \in S : \exists p \in G\_\eta (i\_{\eta\alpha}(p)\Vdash\_{\mathbb{P}\_\alpha}\check{s} \in \sigma ) \}$ and concluding that $X \in M[G\_\eta]$, "since $\Vdash\_{\mathbb{P}\_\alpha}$ is defined in M " , although not necessarily in M[G].


why are these two arguments valid?



proof that "small" sets in an extension by iterated forcing already appear in an earlier stage
Another way to argue is simply that the forcing relation is definable in the ground model $M$, and so the part of it to which you want to appeal forms a set in $M$, and this set still exists in $M[G]$. So you can use this relation in $M[G]$ when constructing other sets. 


