Wednesday, June 12, 2013

Final Inequality 2

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

Picking up right where we left off, suppose $X\in [\lambda]^{<\theta}$.  We will find an acceptable model $M$ with $X\subseteq M\cap\lambda$.

Start by fixing a continuous increasing $\in$-chain  $\langle N_i:i<\theta\rangle$  of elementary submodels of $\mathfrak{A}$ satisfying:

• $\langle N_j:j\leq i\rangle\in N_{i+1}$,
• $|N_i|<\theta$
• $N_i\cap\theta$ is an initial segment of $\theta$
• $\{X,p\}\in N_0$

This is certainly possible as $\theta$ is a regular cardinal.   Note that our requirement that $N_i\cap\theta$ is an initial segment of $\theta$ has a lot of power:  for example, it implies that $X\subseteq N_0$, and that $N_j\subseteq N_i$ whenever $j<i<\theta$.

Given $i<\theta$, let us consider $\mathfrak{a}:= N_i\cap\lambda\cap\REG$.  We know the following facts:
• $|\mathfrak{a}|=|N_i|<\theta$, and
• $\mathfrak{a}$ is cofinal in $\lambda$ (as $\cf(\lambda)<\theta$).
This means that $\prod\mathfrak{a}$ is "covered" by our family $\mathcal{F}$:  if $g\in\prod\mathfrak{a}$ then there is an $f\in\mathcal{F}$ and $\mu<\lambda$ such that $g(\kappa)<f(\kappa)$ whenever $\kappa\in\mathfrak{a}\cap (\mu,\lambda)$.

We apply the observation in the previous paragraph to the characteristic function of $N_i$ on $\mathfrak{a}$. This gives us an ordinals $\alpha_i<\lambda(1)$ and $\mu_i<\lambda$ such that
$$\kappa\in\mathfrak{a}\cap (\mu_i,\lambda)\Longrightarrow \sup(N_i\cap \kappa)<f_{\alpha_i}(\kappa).$$
Moreover, we can choose $\alpha_i$ and $\mu_i$ in $N_{i+1}$, as the above argument can be done in the model $N_{i+1}$.  This is important, because it guarantees that the function $i\mapsto\alpha_i$ is one-to-one.

Since $\cf(\lambda)<\theta=\cf(\theta)<\lambda$, there is a $\mu<\lambda$ such that
$$|\{i<\theta:\mu_i\leq\mu\}|=\theta.$$
Fix such a $\mu$ (without loss of generality greater than $\cf(\lambda)$), and define
$$I^*:=\{i<\theta:\mu_i\leq\mu\}$$.

Now look at the set $A^*=\{\alpha_i:i\in I^*\}$.   This set is in $[\lambda(1)]^\theta$ because we made sure that the function $i\mapsto\alpha_i$ is one-to-one.  By the definition of $\mathcal{P}$, there is a set $B^*\in\mathcal{P}$ such that $A^*\cap B^*$ is infinite.

Let $\alpha$ be the supremum of the first $\omega$ elements of $A^*\cap B^*$, and find $I\subseteq I^*$
of order-type $\omega$ so that

• $\{\alpha_i:i\in I\}$ is strictly increasing,
• each $\alpha_i$ is in $A^*\cap B^*$, and
• $\sup\{\alpha_i:i\in I\}=\alpha$
Now let $B=B^*\cap \alpha$, and define

$M=\Sk_\mathfrak{A}(B\cup p\cup\mu+1)$

Clearly $M$ is an acceptable model, so we need to prove that $X\subseteq M$.

To be continued...