## Main Proof (part 2)

Proposition 1 Suppose ${\langle\alpha_n:n<\omega\rangle}$ is increasing in ${{\rm nacc}(C_\delta)}$ and ${\langle i_n:n<\omega\rangle}$ is increasing in ${\kappa}$. Then

$\displaystyle \bigcup_{n<\omega}\mathcal{N}(\alpha_n,i_n)\in M^*. \ \ \ \ \ (1)$

Proof: Define

• ${\alpha^*:=\sup\{\alpha_n:n<\omega\}}$,
• ${C^*=\bigcup_{n<\omega}C_{\alpha_n}}$, and
• ${i^*:=\sup\{i_n:n<\omega\}}$.

Here is where the choice of ${\langle C_\alpha:\alpha<\mu^+\rangle}$ comes to our rescue. Since ${\alpha_n\in{\rm nacc}(C_\delta)}$ for each ${n}$, it follows that

$\displaystyle C_{\alpha_n}=C_\delta\cap\alpha_n \ \ \ \ \ (2)$

for each ${n}$, and therefore

$\displaystyle C_{\alpha_n} = C_{\alpha_{n+1}}\cap\alpha_n\subseteq C_{\alpha_{n+1}}, \ \ \ \ \ (3)$

and

$\displaystyle C^*=C_\delta\cap\alpha^*. \ \ \ \ \ (4)$

Now it follows easily that

$\displaystyle \bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n)={\rm Sk}_{\mathfrak{A}}(\mu_{i^*}\cup\{f_\xi:\xi\in C^*\}). \ \ \ \ \ (5)$

Next, we define

$\displaystyle \epsilon:=\min({\rm nacc}(C_\delta)\setminus\alpha^*). \ \ \ \ \ (6)$

Given our definition of ${\langle M_\alpha:\alpha<\mu^+\rangle}$, it is not hard to see

• ${\epsilon\cup\mu\subseteq M_\epsilon}$, and
• every proper initial segment of ${\langle f_\xi:\xi<\epsilon\rangle}$ is in ${M_\epsilon}$.

In particular,

$\displaystyle \mu_{i^*}\cup\{f_\xi:\xi\in C^*\}\subseteq M_\epsilon. \ \ \ \ \ (7)$

Note that ${C^*\in M_{\epsilon+1}}$, as it is definable from ${\bar{C}}$, ${\epsilon}$, and ${\alpha^*}$:

$\displaystyle C^*= C_\epsilon\setminus \{\alpha^*\}, \ \ \ \ \ (8)$

and so

$\displaystyle \bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n)\in M_{\epsilon+1}\subseteq M^*, \ \ \ \ \ (9)$

as it can be computed in ${M_{\epsilon+1}}$ by taking the Skolem hull of ${\mu_{i^*}\cup\{f_\xi:\xi\in C^*\}}$ in the model ${M_\epsilon}$.

$\Box$