If $G$ is a generic ultrafilter, why $(\exists a\in A)(a\in G)\leftrightarrow \Sigma A\in G$? Let $B$ be a complete Boolean Algebra. Let $G$ be a generic ultrafilter of $B$, that is, such that for any dense $D\subset B$ we have $D\cap G\neq \emptyset$.
Why for all $A\subseteq B,$ $\Sigma A\in G$ if and only if $(\exists a\in A)(a\in G)$?
I have no idea why this is true, and this fact is used many times in the section of Forcing of Jech's Set Theory.
Thanks

Let $D=(\downarrow\!\!A)\setminus\{0\}\cup\left\{x\in B:x\land\bigvee A=0\right\}$; $D$ is dense in $B$. To see that $D$ is dense, note that if $x\land\bigvee A\ne 0$, then $x\land a\ne 0$ for some $a\in A$, and $x\land a\in(\downarrow\!\!A)\setminus\{0\}$. Let $g\in G\cap D$. If $\bigvee A\in G$, then clearly $g\land\bigvee A\ne 0$, so $g\in(\downarrow\!\!A)\setminus\{0\}$. Thus, there is an $a\in A$ such that $g\le a$, and therefore $a\in G\cap A$.
The other direction is clear.
