Finishing the proof of Theorem 6.1 from Sh410

\( \newcommand{\wsat}{\mathrm{wsat}} \newcommand{\GH}{\mathrm{GH}} \newcommand{\REG}{\mathrm{REG}} \newcommand{\Ch}{\mathrm{Ch}} \newcommand{\reg}{\mathrm{reg}} \newcommand{\rk}{\mathrm{rk}} \newcommand{\ot}{\mathrm{ot}} \newcommand{\pcf}{\mathrm{pcf}} \newcommand{\tcf}{\mathrm{tcf}} \newcommand{\cf}{\mathrm{cf}} \newcommand{\ran}{\mathrm{ran}} \newcommand{\dom}{\mathrm{dom}} \newcommand{\nacc}{\mathrm{nacc}} \newcommand{\acc}{\mathrm{acc}} \newcommand{\Tr}{\mathrm{Tr}} \)
Recall that we're in the middle of trying to show that $T_J(\mu)\leq T^2_J(\mu)$. Resuming where we left off, we showed that $\ran f^*\subseteq N^a$, and so we can define $m(i)=\min\{m<\omega : f^*(i)\in N^a_m\}$ for every $i<\kappa$. At the end of the last post, I remarked that using this we can obtain each $f^*(i)$ starting from $\kappa^{<\sigma}$, Skolem functions, functions $f^n$, and finitely many regular cardinals below $\mu$ at each stage $n$. What this means for us is that for each $i<\kappa$, we can find finite sets $E(i)\subseteq \bigcup_{l< m(i)}A_l$ and $Y(i)\subseteq\kappa^{<\sigma}+1$ such that $f^*(i)\in M^{E(i),Y(i)}_{m(i)}$ where

$M^{E,Y}_0=Sk^\mathfrak{A}(Y)$;
$M^{E,Y}_{n+1}=Sk^\mathfrak{A}(M^{E,Y}\cup f^n\upharpoonright(E\cap M^{E,Y}_n) ).$

Now, let $A\subseteq\kappa$ be such that $\pcf(\bigcup_{i\in A} E(i))\subseteq\lambda$, and set $E=\bigcup_{i\in A} E(i)$ and $Y=\bigcup_{i\in A}Y(i)$. We claim that $A\in J$. Along these lines, suppose otherwise and note that we may assume $|A|<\sigma$ and thus that $|E|,|Y|<\sigma$. We will show, by induction on $n<\omega$, that the following hold:

1) $M^{E,Y}_n\cap E\in M_{n+1}$;
2) $f^n\upharpoonright (M^{E,Y}_n\cap E)\in M_{n+1}$;
3) $M^{E,Y}_n\in M_{n+1}$.

For ease of notation, we will set $E_n=M^{E,Y}_n\cap E$. At stage $n=0$, we begin by noting that $M^{E,Y}_0\prec N^a_0$ by definition, and thus that $M^{E,Y}_0,E_0\in [N^a_0]^{<\sigma}$. Working inside $M_1$, let $\mathfrak{A}_0=(M_0, \in, <_\chi)$ and note that $$ Sk^{\mathfrak{A}_0}(\kappa^{<\sigma}\cup\{\kappa^{<\sigma}\})= Sk^{\mathfrak{A}}(\kappa^{<\sigma}\cup\{\kappa^{<\sigma}\})=N^a_0. $$ Thus, $N^a_0$ and $[N^a_0]^{<\sigma}$ are definable in $M_1$. As $|[N^a_0]^{<\sigma}|=\kappa^{<\sigma}<\mu\leq\lambda$ and $\lambda+1\subseteq M_1$, it follows that $[N^a_0]^{<\sigma}\subseteq M_0$ and thus that $ M^{E,Y}_0,E_0\in M_1$. By assumption, $\pcf(E_0)\subseteq\pcf(E)\subseteq\lambda\subseteq M_1$, and for every $\theta\in \pcf(E_0)$ and $\alpha<\theta$, the function $f^\theta_0$ is definable in $M_1$. From our earlier comment, it follows that $f^0\upharpoonright E_0\in M_1$. At stage $n<\omega$, assume that $M^{E,Y}_n, E_n, f^n\upharpoonright E_n\in M_{n+1}$. Working inside $M_{n+1}$, define $\mathfrak{A}_{n+1}=(M_{n+1}, \in, <_\chi)$ and note that $$ Sk^{\mathfrak{A}_{n+1}}(M^{E,Y}_n\cup f^n\upharpoonright E_n)= Sk^{\mathfrak{A}}(M^{E,Y}_n\cup f^n\upharpoonright E_n)=M^{E,Y}_{n+1}. $$ Therefore, $M^{E,Y}_{n+1}\in M_{n+2}$. Since $|M^{E,Y}_{n+1}|\leq\kappa^{<\sigma}$, we see that $[M^{E,Y}_{n+1}]^{<\sigma}\subseteq M_{n+2}$ and thus that $E_{n+1}\in M_{n+2}$. Again, since $\pcf(E_{n+1}\subseteq\pcf(E)\subseteq\lambda$ and $E_n\subseteq E$, our construction of $f^{n+1}$ gives us that $f^{n+1}\upharpoonright E_{n+1}\in M_{n+2}$. So we now have that $M^{E,Y}_n\in M_{n+1}$ for all $n<\omega$. Set $M^{E,Y}=\bigcup_{n<\omega}M^{E,Y}_n$ and note that $M^{E,Y}\in M_\omega$ since $M_n\subseteq M_\omega$ for each $n<\omega$. Now, $|M^{E,Y}|\leq\kappa^{<\sigma}$ and so $[M^{E,Y}]^{<\sigma}\subseteq M_{\omega}$. But $|A|<\sigma$ and so $$\ran(f^*\upharpoonright A)\in [M^{E,Y}]^{<\sigma}\subseteq M_\omega.$$ In an earlier post, we showed that our choice of $f^*$ tells us that $A$ must be in $J$, which is a contradiction. In particular, $A$ is in fact in $J$. Now, I want to take a second to point out that we used the assumption that $A\notin J$ in order to show that $|M^{E,Y}_n|=\sigma\leq\kappa^{<\sigma}$ (since $J$ is $\sigma$-based). This is what allowed is to show that $E_n\in M_{n+1}$. So while the fact that the contradiction is the contradiction is the conclusion we're trying to establish might seem horribly inelegant, I don't see a way to avoid this.

Regardless, we have shown that for any $A\subseteq \kappa$, if $\max\pcf(\bigcup_{i\in A}E(i))\leq\lambda$, then $A\in J$. That is, if we let $$ \theta=\min\{\max\pcf(\bigcup_{i\in A}E(i)): A\notin J\}, $$ then we see that $\theta$ is weakly $J$-representable and $\theta>\lambda$. Thus, $T^2_J(\mu)>\lambda$, which completes the proof as noted earlier.