Monday, July 07, 2014

Resolutions of $Ch_N$

\def\REG{\sf {REG}}

Assume the following:

  •  $A$ is a set of regular cardinals, 
  • $|A|<\kappa=\cf(\kappa)<\min(A)$,
  • $N$ is $\kappa$-presentable as witnessed by $\langle N_\alpha:\alpha<\kappa\rangle$ with $A\in N_0$,
  • for each $\lambda\in\pcf(A)$, $\bar{f}^\lambda=\langle f_\xi^\lambda:\xi<\lambda\rangle$ is a minimally obedient universal sequence for $\lambda$,
  • $\bar{F}=\{\bar{f}^\lambda:\lambda\in \pcf(A)\}\in N_0$.

Let us declare

$A_0:= A$,

$\lambda_0 := \max\pcf(A)$, and

$\gamma_0:=\sup(N\cap \lambda_0)$

Our technical lemma implies (among other things) that there is a set $B_{\lambda_0}\in N$ such that

  • $B_{\lambda_0}$ is a generator for $\lambda_0$ in $\pcf(A)$, and
  • $\ch_N\restr B_{\lambda_0} = f^{\lambda_0}_{\gamma_0}\restr B_{\lambda_0}$

Since $B_{\lambda_0}\in N$ [THIS IS CRITICAL!], we know the set 

$A_1:= A_0\setminus B_{\lambda_0}$

is in $N$ as well.  Furthermore, since $\lambda_0$ is $\max\pcf(A_0)$ and $B_{\lambda_0}$ is a generator for $\lambda_0$, we know


If we set

$\gamma_1 : = \sup(N\cap\lambda_1)$

then we are essentially back in the same situation described at the start of this post (we may have to lose an initial segment of the sequence $\langle N_\alpha:\alpha<\kappa\rangle$ to guarantee $A_1\in N_0$, but this is trivial.)

Any iteration of this procedure must stop after finitely many steps (i.e., eventually $A_i=\emptyset$) as the sequence of $\lambda_i$ is decreasing.

Thus, we end up with a sequence $\lambda_0>\lambda_1>\dots>\lambda_n$ of elements of $\pcf(A)$, and
corresponding generators $\{B_{\lambda_i}:i\leq n\}$ from $N$ such that

(*)   $A=\bigcup_{i\leq n}B_{\lambda_i}$


(**) $\ch_N\restr B_{\lambda_i} = f^{\lambda_i}_{\gamma_i}\restr B_{\lambda_i}$ for each $i\leq n$.

By the obedience of $\bar{F}$, we know

$f_{\gamma_i}^{\lambda_i}\leq \ch_N\restr A$ for each $i\leq n$,

and therefore

(***)  $\ch_N\restr A = \max\{f^{\lambda_0}_{\gamma_0},\dots, f^{\lambda_n}_{\gamma_n}\}.$

We call the sequence $\langle b_{\lambda_i}:i\leq n\rangle$ an $\bar{F}$-resolution of $\ch_N\restr A$.  More formally, 
an $\bar{F}$-resolution of $\ch_N\restr A$ is a sequence $\langle B_i:i\leq n\rangle$ so that, letting $\lambda_i = \max\pcf(B_i)$, we have
  • the sets $B_i$ are pairwise disjoint,
  • each $B_i\in N$
  • $B_i$ is a generator for $\lambda_i$ in $\pcf(A)$,
  • $\lambda _{i+1}<\lambda_i$ for $i<n$, and
  • $\ch_{N}\restr B_i  =  f^{\lambda_i}_{\sup(N\cap\lambda_i)}\restr B_i$ for each $i\leq n$.
Our technical lemma implies that these resolutions exist; this is the conclusion of Corollary 5.9 of Abraham-Magidor.  

Why the extra bells and whistles in presentation?

Note that any resolution of $\ch_N$ is actually an element of $N$, as it is a finite sequence of elements of $N$.  What if instead of a single $A\in N$ we were given a collection $\langle A_\alpha:\alpha<\theta\rangle$ where each $A_\alpha\in N$ is a set of regular cardinals satisfying


Certainly we can build a resolution for each $A_\alpha$ using the above procedure, and each of these resolutions is in $N$.  However,  we would like to have some way of guaranteeing that the entire sequence of resolutions is an element of $N$.  We will see that this can be done, assuming something a little stronger than $\kappa$-presentability for $N$.

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