Monday, July 28, 2014

Towards simultaneous resolutions Part II

\def\REG{\sf {REG}}

Suppose now that $A$ is $N$-admissible. Since

\(N\subseteq\bigcup_{\alpha<\kappa} M_\alpha\),

we know there is an $\alpha\in C$ with $A\in M_\alpha$.

The closed unbounded set $C[A]$ from our Technical Lemma is therefore also a member of $M_\alpha$
(it is definable from parameters available in $M_\alpha$), and since $\alpha = M_\alpha\cap\kappa$,
it follows that

$\alpha = M_\alpha\cap\kappa\in C[A]$

All we needed here was for $A$ to be a member of $M_\alpha$, so this holds for all sufficiently large $\alpha\in C$.

In particular, if $\alpha<\beta$ in $C$ and $A\in M_\alpha$ is $N$-admissible, then $b[A,\alpha,\beta]$
is a generator for $\lambda(A)$  as in our previous posts.  Also note that $b[A,\alpha,\beta]$ is definable from $A$ (which gives us $\lambda(A)$ and $\bar{f}^A$) together with $N_\alpha$ and $N_\beta$, so we have the following:

Assume $\alpha<\beta<\gamma$ in $C$, and $A\in M_\alpha$ is $N$-admissible.  Then

(1) \(b[A,\alpha,\beta]\) is a generator for $\lambda(A)$ in $\pcf(A)$,

(2) $a\in b[A,\alpha,\beta]\Longrightarrow \sup(N\cap a) = f^A_{\sup(N\cap\lambda(A))}(a)$, and

(3)  $b[A,\alpha,\beta]\in M_\gamma$.

As a transition to our next post, note that $A\setminus b[A,\alpha,\beta]$ is also $N$-admissible (it's going to be an element of $N_{\beta+1}$), and $A\setminus b[A,\alpha,\beta]$ is in $M_\gamma$, so we are set up to iterate things.

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