Thursday, July 03, 2014

Proof of Lemma, Part (A)

\def\REG{\sf {REG}}

[WARNING:  Edits may follow as I notice typos or add clarifications]

Let $N$, $A$, $\lambda$, and $\bar{f}$ be as in the last post, and define

$B^*=\{a\in A: \sup(N\cap a) = f_{\sup(N\cap\lambda)}(a)\}$.

Our first step in proving our technical lemma is the following:
Goal 1:  $B^*$ is a generator for $\lambda$.

Let $\gamma$ denote $\sup(N\cap\lambda)$, and let $B\in N$ be a generator for $\lambda$. (Such a $B$ exists in $N$ as $\lambda$ and $A$ are in $N$).  We will show that $B$ and $B^*$ are equal modulo the ideal $J_{<\lambda}$, and this will suffice.
Stage 1: Some easy observations
Note that $\gamma<\lambda$ as $N$ has cardinality $\kappa<\min(A)$, and if we define


then the sequence $\langle \gamma_\alpha:\alpha<\lambda\rangle$ enumerates a club of $\gamma$ of order-type $\kappa$.  Note as well that each $N_\alpha$ is in $N$, and this means each $\gamma_\alpha$ is as well.

Since $\bar{f}$ is minimally obedient at cofinality $\kappa$, there is a club $E\subseteq\gamma$ of
order-type $\kappa$ such that

$f_\gamma(a) = \sup\{f_\xi(a):\xi\in E\}$ for each $a\in A$.

This remains true for any club $D\subseteq E$ (by minimality), so there is a club $C\subseteq \kappa$ such that

$f_\gamma(a)=\sup\{f_{\gamma_\alpha}(a):\alpha\in C\}$

Notice that $A\subseteq N$ (as $A\in N$, $|A|<\kappa$, and $\kappa+1\subseteq N$). Since we know each $\gamma_\alpha$ is in $N$ as well, it follows that for each $a\in A$,  $f_\gamma(a)$ is the supremum of a set of ordinals in $N\cap a$, and therefore

$f_\gamma(a)\leq \sup (N\cap a)$ for each $a\in A$, that is

$f_\gamma \leq \ch_N$.

Thus, $B^*$ may be defined equivalently by:

$B^*=\{a\in A: \sup(N\cap a)\leq f_{\gamma}(a)\}$.

Stage 2:  $B^*\setminus B\in J_{<\lambda}$

Note that $A\setminus B$ is in $N$.  Since $B$ is a generator for $\lambda$, we know

$\lambda\notin\pcf(A\setminus B)$

and therefore $\prod(A\setminus B)/J_{<\lambda}$ is $\lambda^+$-directed.

This means that the sequence $\langle f_\xi\restr(A\setminus B):\xi<\lambda\rangle$ is bounded mod $J_{<\lambda}$, and so there is a function $h\in \prod A$ such that for each $\xi<\lambda$,

$\{a\in (A\setminus B): h(a)\leq f_\xi(a)\}\in J_{<\lambda}$.

In particular,

$\{a\in (A\setminus B): h(a)\leq f_\gamma(a)\}\in J_{<\lambda}$.

Now comes the important point:  we can assume that the function $h$ is in $N$ because $N$ sees everything needed to describe such an $h$.

Once we have such an $h\in N\cap\prod A$, it follows that $h(a)\leq \sup(N\cap a)$ for all $a\in A$,
and so

$\{a\in (A\setminus B): \sup(N\cap a)\leq f_\gamma(a)\}\in J_{<\lambda}$

But this says exactly that $B^*\setminus B\in J_{<\lambda}$, as required.

Stage 3: $B\setminus B^*\in J_{<\lambda}$

Note that $a\in B\setminus B^*$ means $a\in B$ and $f_\gamma(a)<\sup(N\cap a)$.  Our discussion in
Stage 1 tells us that there is an $\alpha(a)<\kappa$ such that $f_\gamma(a)<\sup(N_{\alpha(a)}\cap a)$.
Since $|A|<\kappa$, if follows that there is a single $\alpha<\kappa$ such that

$B\setminus B^* = \{a\in B: f_\gamma(a)<\sup(N_\alpha\cap a)\}$.

We will show that the set on the right is in $J_{<\lambda}$.

To do this, we need to use the fact that $\bar{f}$ is a universal sequence for $\lambda$, as this implies (see Theorem 4.13 of Abraham-Magidor) that $\langle f_\xi\restr B:\xi<\lambda\rangle$ is cofinal in $\prod B/J_{<\lambda}$.

Since $\ch_{N_\alpha}\restr B\in N$, this implies that there is a $\xi\in N\cap\lambda$ such that

$\ch_{N_\alpha}\restr B <_{J_{<\lambda}} f_\xi\restr B$.

Since $\xi<\gamma$, we know $f_\xi <_{J_{<\lambda}} f_\gamma$, and therefore

$\ch_{N_\alpha}\restr B <_{J_{<\lambda}} f_\gamma\restr B$.

But this means

$\{a\in B:  f_\gamma(a)\leq \sup (N_\alpha\cap a)\}\in J_{<\lambda}$

Since $B\setminus B^*\subseteq\{a\in B: f_\gamma(a)\leq \sup(N_\alpha\cap a)\}$, we are done.

1 comment:

Bill Chen said...

Are we using that $\lambda$ is the maxpcf here (as assumed in the previous post)? In this case, the proof of Stage 2 is trivial.