I'm reading Kunen's Set Theory and the last line of the proof of the Reflection theorem (page 131) is a bit puzzling to me. To those not in possession of Kunen at the moment, the book states verbatim:
Theorem II.5.3 (Reflection Theorem) Let $\varphi_0, \varphi_1, \ldots, \varphi_{n-1}$ be any list of formulas of $\mathcal{L}=\{\epsilon\}$. Assume that $B$ is a non-empty class and $A(\xi)$ is a set for each $\xi \in O N$, and assume that: I. $\xi<\eta \rightarrow A(\xi) \subseteq A(\eta)$. II. $A(\eta)=\bigcup_{\xi<\eta} A(\xi)$ for limit $\eta$. III. $B=\bigcup_{\xi \in O N} A(\xi)$.
Then $\forall \xi \exists \eta>\xi\left[A(\eta) \neq \emptyset \wedge \bigwedge_{i<n}\left(A(\eta) \preccurlyeq \varphi_i B\right) \wedge \eta\right.$ is a limit ordinal $]$.
Proof. We may assume that our list is subformula-closed; if not, replace each $\varphi_i$ by a logically equivalent formula not using $\forall$ (replace the $\forall$ by $\neg \exists \neg$ ), and then add to the list all subformulas of formulas appearing on the list.
For each existential $\varphi_i(\vec{x})$ (of form $\exists y \varphi_j(\vec{x}, y)$, where $\vec{x}$ denotes an $r$-tuple; $\left.r=r_i\right)$, define $F_i: B^r \rightarrow O N$ as follows: If $\varphi_i^B(\vec{a})$, then $F_i(\vec{a})$ is the least $\zeta$ such that $\exists b \in A(\zeta) \varphi_j^B(\vec{a}, b)$. If $\neg \varphi_i^B(\vec{a})$, then $F_i(\vec{a})=0$. These $F_i$ are analogous to the Skolem functions in the proof of Theorem I.15.10, but we are not assuming $A C$, so our functions don't choose some $b$; rather, we use the well-order on $O N$ to define a $\zeta$ with $A(\zeta)$ containing at least one such $b$.
Next, define $G_i: O N \rightarrow O N$ by: $G_i(\xi)=\sup \left\{F_i\left(a_1, \ldots, a_r\right): a_1, \ldots, a_r \in\right.$ $A(\xi)\}$ whenever $\varphi_i$ is existential, with $r=r_i$. When $\varphi_i$ is not existential, let $G_i(\xi)=0$. Finally, let $K(\xi)$ be the larger of $\xi+1$ and $\max \left\{G_i(\xi): i<n\right\}$.
Now, fix $\xi$; it is sufficient to produce an $\eta>\xi$ such that $A(\eta) \neq \emptyset$ and (2) of Lemma II.5.2 holds for $A(\eta), B$. So, let $\zeta_0$ be the least $\zeta>\xi$ such that $A(\zeta) \neq \emptyset$, and let $\zeta_{n+1}=K\left(\zeta_n\right)$. Then $\xi<\zeta_0<\zeta_1<\cdots$. Let $\eta=\sup \left\{\zeta_k: k \in \omega\right\}$. □
My question is the following, why does $A(\eta)$ contain all the witnesses when the parameters are in $A(\eta)$? I feel that it is intuitively true, since it is the limiting value, but I am having a hard time formally arguing this.
Any help would be greatly appreciated!
Given a finite collection of parameters $\vec a\in A(\eta),$ since $A(\eta ) = \bigcup_{\xi < \eta} A(\xi)$ and the $A(\xi)$ are increasing, there is a $\xi < \eta$ such that $\vec a \in A(\xi).$ So since $\eta = \lim_k \zeta_k,$ there is a $k$ such that $\vec a\in A(\zeta_k).$ Then, if $B\models \varphi(b,\vec a)$ for some $b\in B,$ per the construction, there is a $b'\in A(\zeta_{k+1})\subseteq A(\eta)$ such that $B\models \varphi(b',\vec a).$