TITLE: Doubt on a Kunen's proof ($V=L \rightarrow \diamondsuit^+$)
QUESTION [5 upvotes]: I'm studying set theory using Kunen's "Set Theory: An introduction to Independence Proofs" and I'm stuck in a line of the proof of theorem 5.2 of chapter 6 ($V=L \rightarrow \diamondsuit^+$) on page 179. I can't figure out why $C\cap\alpha=\{\beta<\alpha:\beta=cl^1(\beta\cup\{A\cap\alpha\})\cap\alpha\}$. I think $cl^1(G(\beta \cup \{A\cap\alpha\}))=cl(\beta \cup \{A\cap\alpha\})$, but I'm not sure. Can someone help me?
Ps: I understood the whole proof but this sentence.

REPLY [2 votes]: First, some context:
$K_{nm}:L(\omega_2)^n\rightarrow L(\omega_2)$ is a Skolem function for the $m$th formula in the Gödel numbering. We have that $M=L(\delta)$ for some $\delta<\omega_1$.
Fix some countable $A\subseteq\omega_1$. Let $\alpha<\omega_1$ be such that $\alpha=cl(\alpha\cup \{A\});$ under the Skolem functions above. Then $(Y,\in)\preceq (L(\omega_2),\in)$, so in particular $\omega_1\in Y$, and $\in$ is extensional on $Y$. Then if $G$ is the Mostowski collapse from $Y$ onto some transitive set $M$, $G$ is an isomorphism; notice that $G(A)=A\cap\alpha$. Thus $M\vDash\mathsf{ZF-P+V=L},$  in consequence $M=L(\delta)$ for some $\delta<\omega_1$. Define $K_{nm}^1:L(\delta)^n\rightarrow L(\delta)$ by $$K_{nm}(G(y_1),\ldots,G(y_n))=G(K_{nm}(y_1,\ldots,y_n)).$$
We have indeed that $K_{nm}^1[L(\delta)^n]\subseteq L(\delta);$ as $Y$ is closed under the $K_{nm}$ and for all $x\in Y$, $(x,y)\in K_{nm}$ if and only if $(G(x),G(y))\in K_{nm}^1$, because of the definition of $K_{nm}^1$ and since $G$ is an isomorphism.
Let us see $G(cl(\beta\cup\{A\}))=cl^1(\beta\cup\{A\cap\alpha\})$ for any $\beta<\alpha$. Let $H\in Y$, then $G(K_{nm}[H])=K_{nm}^1[G(H)],$ this implies $G(cl(H))=cl^1(G(H)).$ But $G(\beta)=\beta$ for $\beta<\alpha$, $G(A)=A\cap\alpha$, and $G(\omega_1)=\alpha$, thus $$C\cap\alpha=\{\beta<\alpha:\beta=cl^1(\beta\cup\{A\cap\alpha\})\cap\alpha\},$$ where $$C=\{\beta<\omega_1:\beta=cl(\beta\cup\{A\})\cap\omega_1\}.$$