## Monday, July 07, 2014

### Proof of Lemma, Part (B)

$\DeclareMathOperator{\pp}{pp} \def\pcf{\rm{pcf}} \DeclareMathOperator{\cov}{cov} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq} \def\cof{\rm{cof}} \def\ran{\rm{ran}} \DeclareMathOperator{\ch}{Ch} \DeclareMathOperator{\PP}{pp} \DeclareMathOperator{\Sk}{Sk}$

Let's take a stab at proving our technical lemma.

Recall $N$ is $\kappa$-presentable as witnessed by $\langle N_\alpha:\alpha<\kappa\rangle$, and let us define
• $\gamma=\sup(N\cap\lambda)$
• $\gamma_\alpha= \sup(N_\alpha\cap\lambda)$, and
• for $\alpha<\beta<\kappa$, $b(\alpha,\beta)=\{a\in A:\sup(N_\alpha\cap a)\leq f_{\gamma_\beta}(a)\}$.
We are trying to produce a club $C\subseteq\kappa$ such that for any $\alpha<\beta$ in $C$ we have
• $b(\alpha,\beta)$ is a generator for $\lambda$ in $\pcf(A)$, and
• $\ch_N\restr b(\alpha,\beta)= f_\gamma\restr b(\alpha,\beta)$.
For the last statement to hold, we need only show

$b(\alpha,\beta)\subseteq \{a\in A: \sup(N\cap a)\leq f_\gamma(a)\}$.

Recall that in our last post we dealt with the set

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

so the last requirement is equivalent to demanding $b(\alpha,\beta)\subseteq B^*$.

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Observation 1:  For all sufficiently large $\alpha<\kappa$ we have

$f_\gamma(a)<\sup(N\cap a)$ if and only if $f_\gamma(a)<\sup(N_\alpha\cap a)$, and

therefore for all sufficiently large $\alpha<\kappa$ the set of $a\in A$ for which $\sup(N_\alpha\cap a)$ is less than or equal to $f_\gamma(a)$ is just the set $B^*$:

$(\forall^*\alpha<\kappa)[\{a\in A: \sup(N_\alpha\cap a)\leq f_\gamma(a)\}] = B^*$.
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Next, let us fix a club $E\subseteq \gamma$ for which

$f_\gamma = \sup\{f_\xi:\xi\in E\}$.

By minimality, we may assume that $E\subseteq \{\gamma_\alpha:\alpha<\kappa\}$, and that the least $\alpha$ for which $\gamma_\alpha\in E$ is sufficiently large'' in the sense of Observation 1.

Now let $C\subseteq\kappa$ be the club such that $E = \{\gamma_\alpha:\alpha\in C\}$, so for each $a\in A$, we have

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

We will show that $C$ is as required, so let us fix $\alpha<\beta$ in $C$

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Observation 2:  $b(\alpha,\beta)\subseteq B^*$
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Proof:

By definition,

$b(\alpha,\beta)=\{a\in A: \sup(N_\alpha\cap a)\leq f_{\gamma_\beta}(a)\}$

Since $\beta\in C$, we know $\gamma_\beta\in E$ and $f_{\gamma_\beta}(a)\leq f_\gamma(a)$,  so

$b(\alpha,\beta)\subseteq\{a\in A:\sup(N_\alpha\cap a)\leq f_\gamma(a)\}$.

But $\alpha\in C$ is sufficiently large'', so the set on the right is just $B^*$, and we are done.

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Observation 3: $b(\alpha,\beta)$ is a generator for $\lambda$.
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Since $b(\alpha,\beta)\subseteq B^*$, it suffices to prove

$B^*\setminus b(\alpha,\beta) \in J_{<\lambda}$.

Let $B\in N_0$ be a pcf generator for $\lambda$.  As in the previous post, we know that the sequence $\langle f_\xi\restr B:\xi<\lambda\rangle$ is cofinal in $\prod B$ modulo $J_{<\lambda}$.

The function $\ch_{N_\alpha}\restr B$ is in $N_{\alpha + 1}\cap\prod B$, and so there is a $\xi\in N_{\alpha+1}\cap\lambda$ with

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

Since $\alpha+1\leq\beta$, it follows that  $\xi\in N_{\beta}\cap\lambda$ hence $\xi<\gamma_\beta$ and

$\ch_{N_\alpha}\restr B \leq_{J_{<\lambda}} f_{\gamma_\beta}\restr B$

that is,

$\{a\in B: f_{\gamma_\beta}(a)< \sup(N_\alpha\cap a)\}\in J_{<\lambda}$.

Since $B^*$ is also a generator for $\lambda$, we know $B^*\setminus B\in J_{<\lambda}$, and so

$\{a\in B^*: f_{\gamma_\beta}(a)< \sup(N_\alpha\cap a)\}\in J_{<\lambda}$.

But the above set is precisely $B^*\setminus b(\alpha,\beta)$, so we are done.

This completes the proof of our Technical Lemma.