I was reading this lemma from Jech's tome and came across a "clearly, ..." so i tried to prove it on my own but i'm failing miserably. To give context:
Lemma 14.19. $V^B$ is full. Given a formula $\varphi(x, ...)$, there exists some $a \in V^B$ such that, $$ \|\varphi(a, ...)\| = \|\exists x\varphi(x, ...)\|$$ Proof. $\le$ holds for every $a$ because of the definition of $\|\exists x\varphi(x, ...)\|$, we wish to find $a \in V^B$ such that $\ge$ holds. Let $u_0 = \|\exists x\varphi(x, ...)\|$. Let $$ D = \{u \in B: \text{there is some } a_u \text{ such that } u \le \|\varphi(a_u, ...)\|\}.$$ It is clear that $D$ is open and dense below $u_0$. Let W be a maximal set of pairwise disjoint elements of $D$; clearly, $\sum\{u: u\in W\} \ge u_0$. By lemma 14.18 we are done.$\hspace{3cm}\Box$
Now if $u \in D$ we know that there exists some $a_u$ such that $u \le \|\varphi(a_u, ...)\|$, so $u \le \sum_{a \in V^B}\|\varphi(a, ...)\| = \|\exists x\varphi(x, ...)\| = u_0$ and again for each $u \in W$ we have $u \le u_0$ and in order to prove $\sum\{u: u\in W\} \ge u_0$ we have to prove $\sum\{u: u\in W\} = u_0$. And this is where i'm stuck. I have tried reaching a contradiction by assuming that equality doesn't hold but i got no results. Any hints or answers are appreciated.
Thanks for your patience.