Monday, July 28, 2014

Towards simultaneous resolutions Part I

\def\REG{\sf {REG}}


Recall that in our discussions, $\kappa$ is a regular cardinal, and  $N$ is a $\kappa$-presentable elementary submodel of $H(\chi)$ for some sufficiently large regular $\chi$.

Previously we have fixed some progressive set of regular cardinals $A$ with $A\in N$ and $|A|<\kappa<\min(A)$.  In this post, we want to let $A$ vary while $N$ remains fixed.

Let us agree to call a set $A$ of regular cardinals $N$-admissible if it satisfies the following:

  • $A\in N$, and
  • $|A|<\kappa<\min(A)$.

Given a progressive set of regular cardinals $A$, we let  $\lambda(A)$ denote $\max\pcf(A)$, and let $\bar{f}^A$ be the $<_\chi$-least  minimally obedient universal sequence $\bar{f}^{A}$ for $\lambda(A)$. (Here $<_\chi$ is the well-ordering of $H(\chi)$ that is in the background whenever we talk about ``elementary submodels of $H(\chi)$''.)

Our technical lemma tells us the following:

Whenever $A$ is $N$-admissible, there is a club $C[A]\subseteq\kappa$ with the property that for any $\alpha<\beta$ in $C[A]$, the set

$b[A,\alpha,\beta]=\{a\in A: \sup(N_\alpha\cap a)< f^A_{sup(N\cap \lambda(A))}(a)\}$

sastifies the following:

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

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

Moving forward, let $\langle M_\alpha:\alpha<\kappa\rangle$ be an $\in$-increasing and continuous
chain of elementary submodels of $H(\chi)$  with $N$ (and $\langle N_\xi:\xi<\kappa\rangle$) in $M_0$, such that for each $\alpha<\kappa$:

  • $|M_\alpha|<\kappa$
  • $M_\alpha\cap\kappa$ is an initial segment of $\kappa$, and 
  • $\langle M_\beta:\beta<\alpha\rangle\in M_{\alpha+1}$.
Now define

$C:=\{\delta<\kappa:\delta=M_\delta\cap \kappa\}.$

We know $C$ is closed and unbounded in $\kappa$.  In our next post, we'll look at some properties of this club $C$.

[Warning:  the requirements on the $M_\alpha$ may be edited as I move through the proof in the next couple of posts.]

1 comment:

Bill Chen said...

There may be a typo in the definition of $b[A,\alpha,\beta]$, since as written it doesn't depend on $\beta$ (and seems to be the empty set). But it is not difficult to infer what is meant from the statement of the Technical Lemma.