Doubt on a Kunen's proof ($V=L \rightarrow \diamondsuit^+$) 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.
 A: 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 $$C=\{\beta<\omega_1:\beta=cl(\beta\cup\{A\})\cap\omega_1\}.$$
A: First, we show that $\forall y \in Y( y \in cl(\beta \cap\{A\}) \Leftrightarrow G(y) \in cl^1(\beta \cup \{A \cap\alpha\})$.
$\Rightarrow$: We know that $$cl(\beta \cap\{A\})=\bigcup_{n \in \omega}C_n$$
Where $C_0=\beta \cap {A}$ and $C_{k+1}=C_k\cup\{K_{nm}*C_k:n,m \in \omega \}$. By induction, we show that $y \in C_n \Rightarrow G(y) \in cl^1(\beta \cup \{A \cap\alpha\})$.
Step 0: Since $G(\xi)=\xi$ for all $\xi \in \beta$ and $G(A)=A\cap\alpha$, the thesis holds.
Successor step: Suppose the thesis holds for $C_k$. Suppose $\xi_1, ..., \xi_n \in C_k$. Since $K_{nm}^1(G(\xi_1), ..., G(\xi_n))=G(K_{nm}(\xi_1, ..., \xi_n))$, the thesis holds for $C_{k+1}$.
$\Leftarrow$ is proven by an analogous argument.
$\\$
Now we want to show that $\beta \in C\cap \alpha \Leftrightarrow \beta=cl^1(\beta\cup\{A \cap \alpha\})\cap\alpha$. We will show only the $\Rightarrow$ direction, since the other is analogous.
Suppose $\beta=cl(\beta\cup\{A\})\cap\omega_1$ and $\beta\in\alpha$. It's easy to see that $\beta\subset cl^1(\beta\cup\{A \cap \alpha\})$, therefore all we need to do is to show that $ cl^1(\beta\cup\{A \cap \alpha\})\subset \beta$. Since before, we write $$cl^1(\beta\cup\{A \cap \alpha\})=\bigcup_{n<\omega}C_n$$
Where $C_0=\beta\cup\{A\cap\alpha\}$ and $C_{k+1}=C_k\cup\{K_{nm}^1*C_k: n, m \in \omega\}$. Again, by induction, we show that $C_k \cap \alpha \subset \beta$, which completes the proof.
Step 0: Since $\beta\cap\alpha=\beta\subset \beta$, all we need to do here is to show that if $A\cap\alpha\in\alpha$ then $A \cap \alpha \in \beta$. This holds, because $A\cap \alpha \in \alpha \Leftrightarrow G(A)\in G(\omega_1) \Leftrightarrow A \in\omega_1$ imples that $A \in \beta$, by the hypothesis, therefore $G(A)\in G(\beta) \Rightarrow A\cap\alpha \in \beta$.
Successor step: Suppose $C_k \cap \alpha \subset \beta$. We need to show that $\{K_{nm}^1*C_k: n, m \in \omega\}\cap \alpha \subset \beta$, which holds since, fixing $G(\xi_1), ..., G(\xi_n) \in C_k$ such that $K_{nm}^1(G(\xi_1), ..., G(\xi_n))=G(K_{nm}(\xi_1, ..., \xi_n))\in \alpha$ implies  $K_{nm}(\xi_1, ..., \xi_n) \in \omega_1$, and by the precceding lemma, since $G(\xi_1), ..., G(\xi_n) \in cl^1(\beta \cup \{A \cap\alpha\})$, then $\xi_1, ..., \xi_n \in cl(\beta \cap\{A\})$, and, therefore, $K_{nm}(\xi_1, ..., \xi_n) \in cl(\beta \cap\{A\}) \cap \omega_1=\beta$, and now it follows from the isomorphism that $K_{nm}^1(G(\xi_1), ..., G(\xi_n))=G(K_{nm}(\xi_1, ..., \xi_n))\in \beta$.
