I'm working through Theorem 16.13 in Jech, i.e., the consistency of MA with $\neg CH$. I understand the broad idea of the proof, but there are some details that I'm missing. I'm more familiar with the poset approach to forcing, so these missing details might be something about Boolean-valued models that I'm overlooking.
My first question is in the context of inductively defining the $\langle \dot Q_\alpha\mid \alpha<\kappa\rangle$ (4th paragraph into the proof). In the construction, we let $\alpha\in\kappa$ and assume $\langle\dot Q_\alpha\mid \beta<\alpha\rangle$ has been defined. We've fixed in advance a bijection $\pi:\kappa\leftrightarrow\kappa\times\kappa$, satisfying $\pi(\alpha)=(\beta,\gamma)$ implies $\beta\leq\alpha$. Let $\dot Q$ be the $\gamma$th name in $V^{P_{\beta}}$ for a partial order of size $<\kappa$ (we previously concluded that there can be at most $\kappa$ such names).
My question concerns the next sentence: "Let $b=||\dot Q$ satisfies the countable chain condition$||_{P_\alpha}$ and let $\dot Q_\alpha\in V^{P_\alpha}$ be such that $||\dot Q_\alpha=\dot Q||_{P_\alpha}=b$ and $||\dot Q_\alpha=\{1\}||_{P_\alpha}=-b$." Question: Why can we pick such $\dot Q_\alpha$ (i.e., why is $\dot Q_\alpha$ well-defined)?
My second question has to do the proof of Lemma 16.14 in the next paragraph, which states: If $\lambda<\kappa$ and $X\subseteq \lambda$ is in $V[G]$, then $X\in V[G_\alpha]$ for some $\alpha<\kappa$.
The proof starts with picking an arbitrary name $\dot X$ for $X$, and then asserts "Every Boolean value $||\xi \in \dot X||$ (where $\xi < \lambda$)is determined by a countable antichain in $P$ and hence $\dot X$ is determined by at most $λ$ conditions in $P$." And we then use the finite support condition to show that certain bounds exists for how far up these conditions can be.
Question: why is the quoted sentence true? ("Every Boolean value $||\xi \in \dot X||$ (where $\xi < \lambda$) is determined by a countable antichain in $P$ and hence $\dot X$ is determined by at most $λ$ conditions in $P$.")
Much thanks in advance!
Disclaimer: This is a partial answer, as I don't quite see how to get equality in your first question.
You can apply the Mixing Lemma (Lemma 14.18 in Jech) with the antichain $\{b,-b\}$ to find $\dot Q_\alpha$ such that $\|\dot Q=\dot Q_\alpha\|\ge b$ and $\|\dot Q_\alpha=\{1\}\|\ge-b$. Note that this is sufficient to conclude that $\|\dot Q_\alpha \text{ has the ccc}\|=1$, which is what is needed for the proof to go through. For more information on the Mixing Lemma, see Bell's Set Theory: Boolean-Valued Models and Independence Proofs.
For your second question, let $B\supset P$ be the completion of $P$. For each $b\in B$, let $A_b\subset \{p\in P: p\le b\}$ be a maximal antichain (in the poset $\{p\in P: p\le b\})$. This antichain will be maximal in the poset $\{c\in B^+:c\le b\}$ by the density of $P$. Because $A_b\subset P$ and $A_b$ is an antichain in $P$ (because $P$ is dense in $B$), we have that $A_b$ is countable. The claim is that $A_b=A_c\to b=c$.
Suppose $A_b=A_c$ but $b\not\le c$. Put $d=b\cdot (-c)$, so that $d\le b$. By the maximality of $A_b$, it follows that $d\cdot a\neq 0$ for some $a\in A_b$. As $A_b=A_c$, we get that $a\in A_c$, so $a\le c$, so $d\cdot a\le d\cdot c=0$, contradiction. Thus, $b\ge c$. By symmetry, $b=c$.
Applying the above to each $\|\xi\in\dot X\|$, we get a countable antichain $A(\xi)$ for each $\xi<\lambda$. Thus, $A:=\bigcup_{\xi<\lambda}A(\xi)$ is a set of size at most $\lambda$. Since $\kappa>\lambda$ is regular, the set $$ \{\max \text{spt}(p):p\in A\} $$ is bounded in $\kappa$, and we're done.
I personally find the proof without Boolean algebras to be easier to follow. In particular, Kunen's presentation is very readable and pays close attention to the details, as it's the first iterated forcing proof done in the book (Theorem V.4.1 in the new edition). Another option for the poset-based approach is Section 3 of Baumgartner's Iterated Forcing.
If you want to see an alternate exposition using Boolean algebras, you can find it in chapter 6 of Bell's text, although I haven't read that one (I just know it's there).