I am referring to the Jech's "Set Theory" book. He wants to show that, if $(P,<)$ is the notion of forcing defined as:
$p \in P$ iff $p$ is a finite sequence of ordinals less than $\omega_{1}^{L}$ and $p<q$ iff $p \supset q$
then $L[G]\models \omega_{1}^{L}$ is countable.
So the author wants to show that is consistent that $P$ has a generic filter in $L$. Let $B$ be the completion of $P$ in $L$. I'm working in the boolean value model $L^{B}$. Let's call $D=\mathcal{P}^{L}(B)$ and let G be the canonical ultrafilter.
I already know that $\lVert$ G is a $\check{D}$-generic ultrafilter $\rVert$=1
I can't understand how to show $\lVert$ $\check{D}$ is the set of all constructible subset of $\check{B}$ $\rVert$ =1
But let's say I demonstrate that last line above. Why isn't this enought for what I want to proof? Why Jech also say that we use that construibility is absolute in boolean universe and so is the definition of $P$? Where am I wrong when i just say:
1=$\lVert$ G is a $\check{D}$-generic ultrafilter $\rVert$ $\lVert$ $\check{D}=\mathcal{P}^{L}(\check{B})$ $\rVert$ $\le$ $\lVert$ G is a $\mathcal{P}^{L}(\check{B})$-generic ultrafilter $\rVert$