Skolem Functions on Kunen I'm studying Kunen's "Introduction to Set Theory" and I can't figure out something.
On the proof of the lemma 5.1 on Chapter 6, the last line states that the set $\{\rho<\omega_1:L(\rho) \text{ is closed under all the } H_{nm}\}$ is c.u.b. and states that this follows from lemma 6.13 from chapter 2.
Lemma 6.13 from chapter 2 states that whenever $\kappa>\omega$ and $\mathscr A$ is a set of less than $\kappa$ finitary functions, then the set $\{\rho < \kappa: \rho \text{ is closed under } \mathscr A \}$ is c.u.b.
Well, it's not exactly the same thing. Can someone tell me how to apply lemma 6.13 in this case?
Edit:
$H_{nm}$ are the Skolem functions defined using the functions $En$ in the same book, but I don't think that their definition is relevant to prove that the set above is c.u.b. The important thing is that they are all finitary.
 A: Let's prove the following lemma:

Let $\kappa>\omega$ be regular and suppose $\mathscr A$ is a set of less than $\kappa$ finitary functions, then the set $X=\{\rho < \kappa: L(\rho) \text{ is closed under } \mathscr A \}$ is c.u.b.

It's easy to see that $L(\rho)$ is closed: Suppose $\gamma<\kappa$ is a limit ordinal and suppose $\gamma\cap X$ is unbounded in $\gamma$. We will show that $\gamma \in X$.
Let $f \in \mathscr A$ and $x \in L(\gamma)$. Then $x \in L(\beta)$ for some $\beta < \gamma$. Choose $\alpha < X \cap \gamma$ such that $\alpha>\beta$. Since $x\in L(\beta) \subset L(\alpha)$ and $\alpha$ is closed under $f$, we see that $f(\alpha)\in L(\beta) \subset L(\gamma)$. Therefore, $L(\gamma)$ is closed under $\mathscr A$.
Now we will show that $X$ is unbounded in $\kappa$. Let $\omega<\xi<\kappa$. We will see that $\exists \beta \in \kappa\cap X(\xi < \beta)$, which completes the proof. We define $f^n(\xi)\,(n \in \omega)$ by recursion, as follows: we let $f^0(\xi)=\xi$. Suppose we have chosen $f^l(\xi)<\kappa$ for all $l < n+1$ such that $f^i(\xi)<f^j(\xi)$ whenever $i<j$. Let $Cl(L(f^n(\xi)))$ be the closure of $L(f^n(\xi))$ under $\mathscr A$. We know that $|Cl(L(f^n(\xi)))|=|L(f^n(\xi))|=|f^n(\xi)|<\kappa$. Since $\kappa$ is regular, we may pick $f^n(\xi)<\alpha<\kappa$ such that $Cl(L(f^n(\xi)))\subset L(\alpha)$ (looking at the ranks). We let $f^{n+1}(\xi)=\alpha$.
Now we let $f^\omega(\xi)=\bigcup_{n<\omega}f^n(\xi)$. Notice that $f^\omega(\xi)<\kappa$ is a limit ordinal bigger than $\xi$. We now show that $f^\omega(\xi)\in X$.
Let $x \in L(f^\omega(\xi))$ and let $g \in \mathscr A$. Since $f^\omega(\xi)$ is limit, there exists $n<\omega$ such that $x \in L(f^n(\xi))$. By construction, $g(x)\in L(f^{n+1}(\xi))\subset L(f^\omega(\xi))$. $\Box$
