## Main Proof (part 3)

We will now define certain increasing sequences ${\langle \alpha_n:n<\omega\rangle}$ and ${\langle i_n:n<\omega\rangle}$ as in the last post, along with another sequence ${\langle \epsilon_n:n<\omega\rangle}$.

Start by setting ${\alpha_0=\min({\rm nacc}(C_\delta))}$ and ${i_0=0}$. Given ${\alpha_n}$ and ${i_n}$, we let ${\alpha_{n+1}}$ be the least element of ${{\rm nacc}(C_\delta)}$ above ${\alpha_n}$ with the property that

$\displaystyle \gamma(\alpha_n)<\sup(C_{\alpha_{n+1}}), \ \ \ \ \ (1)$

and define

$\displaystyle \epsilon_n = \min(C_{\alpha_{n+1}}\setminus\gamma(\alpha_n). \ \ \ \ \ (2)$

What is the point? Note that this arrangement ensures that

$\displaystyle f_{\epsilon_n}\in \mathcal{N}(\alpha_{n+1}, i) \ \ \ \ \ (3)$

for any ${i<\kappa}$. Since ${\gamma(\alpha_n)\leq\epsilon}$, it follows that we can choose ${i_{n+1}\geq i_n}$ such that “${f_{\epsilon_n}}$ majorizes ${g(\alpha_n, i_n)}$ beyond ${\mu_{i_{n+1}}}$”, i.e.,

$\displaystyle \sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)

whenever ${\theta\in \mathfrak{a}_{\alpha_n, i_n}\setminus \mu_{i_{n+1}}}$.

Now define

$\displaystyle N=\bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n), \ \ \ \ \ (5)$

and

$\displaystyle N^+=\bigcup_{n<\omega}\mathcal{N}^+(\alpha_n, i_n). \ \ \ \ \ (6)$

We know the following facts from previous posts:

• ${N\in M^*\cap [\mu]^{<\mu}=\mathcal{P}}$,
• ${N\subseteq N^+}$, and
• ${a\subseteq N^+}$.

Thus, the theorem will be done provided we establish

$\displaystyle N\cap\mu = N^+\cap \mu. \ \ \ \ \ (7)$

Here we use the “Elementary submodel argument” from February 12, which tells us that we have what we want provided

$\displaystyle \theta\in N\cap\mu\cap{\sf Reg}\Longrightarrow \sup(N^+\cap\theta)=\sup(N\cap\theta). \ \ \ \ \ (8)$

The above is satisfied trivially for ${\theta\leq\mu_{i^*}}$ as ${\mu_{i^*}\subseteq N}$; thus, we need only verify things for ${\theta\in N\cap\mu\cap{\sf Reg}}$ which are greater than ${\mu_{i^*}}$. Furthermore, we know ${N\subseteq N^+}$, so

$\displaystyle \sup(N\cap\theta)\leq\sup(N^+\cap\theta) \ \ \ \ \ (9)$

in any case.

Assume now by way of contradiction that we have ${\theta}$ such that

• ${\theta\in N\cap\mu\cap{\sf Reg}}$,
• ${\mu_{i^*}<\theta}$, and
• ${\sup(N\cap\theta)<\sup(N^+\cap\theta)}$.

Choose ${n<\omega}$ so large that

• ${\sup(N\cap\theta)<\sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)}$, and
• ${\theta\in \mathcal{N}(\alpha_n, i_n)}$.

Since ${\mu_{i_{n+1}}<\mu_{i^*}}$ and ${\theta\in\mathfrak{a}_{\alpha_n, i_n}}$, it follows from the definition of ${\epsilon_n}$ that

$\displaystyle \sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)

But

$\displaystyle f_{\epsilon_n}(\theta)\leq \sup(\mathcal{N}(\alpha_{n+1}, i_{n+1})\cap\theta) \ \ \ \ \ (11)$

as both ${\mathfrak{a}_{\alpha_n, i_n}}$ and ${f_{\epsilon_n}}$ are in the latter model. Thus, on the one hand we have

$\displaystyle \sup(N\cap\theta)\leq \sup(\mathcal{N}^+(\alpha_n, i_n)\cap\theta)

and on the other,

$\displaystyle f_{\epsilon_n}(\theta)\leq \sup(\mathcal{N}(\alpha_{n+1}, i_{n+1})\cap\theta)\leq \sup(N\cap\theta). \ \ \ \ \ (13)$

Clearly, this is absurd.