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.

Transversals: A slight Strengthening of Theorem 6.1 from Sh410

\( \newcommand{\wsat}{\mathrm{wsat}} \newcommand{\GH}{\mathrm{GH}} \newcommand{\REG}{\mathrm{REG}} \newcommand{\Ch}{\mathrm{Ch}} \newcommand{\reg}{\mathrm{reg}} \newcommand{\rk}{\mathrm{rk}} \newcommand{\ot}{\mathrm{ot}} \newcommand{\pcf}{\mathrm{pcf}} \newcommand{\tcf}{\mathrm{tcf}} \newcommand{\cf}{\mathrm{cf}} \newcommand{\ran}{\mathrm{ran}} \newcommand{\dom}{\mathrm{dom}} \newcommand{\nacc}{\mathrm{nacc}} \newcommand{\acc}{\mathrm{acc}} \newcommand{\Tr}{\mathrm{Tr}} \)
Finally getting to the point of this series of posts, I want to repair the proof of Theorem 6.1 from Sh410 and also strengthen it slightly. I say repair here, because I'm not entirely sure that the proof goes through as written with the stated definitions. The core ideas are all correct though, so nothing I'm doing in this post (and subsequent ones, as this proof will take a bit) is terribly novel. Let's recall some definitions. $$T_J(\mu)=\sup\{|F| : F\subseteq{}^\kappa\mu\text{ is a }J\text{-adt}\}$$ $$T^2_J(\mu)=\sup\{\lambda : \lambda\text{ is weakly }J\text{-representable}\}$$ $$T^3_J(\mu)=\sup\{\lambda : \lambda\text{ is }J\text{-representable}\}$$ A small note: $T^2_J(\mu)$ is defined differently in Sh506 and Sh589, but all of these definitions coincide under our hyoptheses.

Theorem (Shelah): Suppose that Suppose that $\kappa$, $\sigma$, and $\mu$ are cardinals with $\kappa^{<\sigma}<\mu$. If $J$ is a $\sigma$-based ideal on $\kappa$, then $T^3_J(\mu)=T^2_J(\mu)=T_J(\mu)$.

Proof: Note that by previous posts, we only need to show that $T_J(\mu)\leq T^2_J(\mu)$. For this, it will suffice to show that for every cardinal $\lambda\geq\mu$, if $\lambda< T_J(\mu)$, then $\lambda< T^2_J(\mu)$. Otherwise, if $T^2_J(\mu)< T_J(\mu)$, then $\mu\leq T^2_J(\mu)< T_J(\mu)$ would give us that $T^2_J(\mu)< T^2_J(\mu)$. So suppose that $\mu\leq\lambda< T_J(\mu)$, and fix a sufficiently large regular cardinal $\chi$. Let $\mathfrak{A}=(H(\chi), \in, <_\chi)$, and inductively build a sequence $\langle M_\xi : \xi\leq\omega+1\rangle$ such that:

1) $M_\xi\prec \mathfrak{A}$ for each $\xi\leq\omega+1$;
2) $|M_\xi|=\lambda$ and $\lambda+1\subseteq M_\xi$ for each $\xi\leq\omega+1$;
3) $M_\xi\in M_\eta$ for each $\xi\leq\eta\leq\omega+1$ (so $M_\xi\subseteq M_\eta$);
4) $J\in M_0$.

Note that since $J,\lambda\in M_0$, by elementarity we can find a witness $F$ to the fact that $T_J(\mu) >\lambda$ inside $M_0$. By an earlier remark, there is an $f^*\in F$ such that $f^*\neq_J g$ for each $g\in{}^\kappa\mu\cap M_{\omega+1}$. We now inductively define objects $N^a_n$, $N^b_n$, $f^n$, and $A_n$ for $n<\omega$ as follows. First, we set $$N^a_0=Sk^\mathfrak{A}(\kappa^{<\sigma}+1)$$ and $$N^b_0=Sk^\mathfrak{A}(\kappa^{<\sigma}+1\cup\{f^*\}).$$ At stage $n\geq 0$, suppose that $N^a_n$ and $N^b_N$ have been defined, and set $$A_n=N^a_n\cap((\kappa^{<\sigma})^+, \mu]\cap\REG$$ In order to define $f^n$, we first let $\mathcal{B}_n=[A_n]^{<\sigma}\cap M_{n+1}$ and let $N\prec\mathfrak{A}$ be a $\kappa^{<\sigma}$-presentable elementary submodel containing $\Ch_{N^b_n}\upharpoonright A_n$, $\mathcal{B}_n$, and $A_n$. Note that since $|A_n|=\kappa^{<\sigma}$, it follows that $\mathcal{B}_n\subseteq N$. With that in mind, set $f^n=\Ch_N\upharpoonright A_n$. By some standard pcf theory, for every $B$ in $\mathcal{B}_n\cup\{A_n\}$ we have $$f^n\upharpoonright B=\sup\{f^{\lambda_0}_{\gamma_0},\ldots, f^{\lambda_0k}_{\gamma_k}\},$$ where $k<\omega$, $\lambda_i\in\pcf(B)\cap N$, $\gamma_i<\lambda_i$, and $\langle f^\lambda_\alpha :\alpha<\lambda\rangle$ is universal for $\lambda$. It's not terribly important precisely what the functions $f^\lambda_\gamma$ are. The important fact is that once $B$, $\{\lambda_i : i\leq k\}$, and $\{\gamma_i : i\leq k\}$ are inside an elementary submodel of $\mathfrak{A}$, $f^n\upharpoonright B$ will also be in said elementary submodel. Now that $f^n$ has been defined, we finally define $$N^a_{n+1}=Sk^\mathfrak{A}(N^a_n\cup\ran(f^n))$$ and $$N^b_{n+1}=Sk^\mathfrak{A}(N^b_n\cup\ran(f^n)).$$ Continuing in this manner through $n<\omega$, set $$N^a=\bigcup_{n<\omega}N^a_n\qquad \text{and}\qquad N^b=\bigcup_{n<\omega}N^b_n.$$ By construction, we've ensured that characteristic functions of $N^a$ and $N^b$ agree on $N^b\cap\mu\cap\REG$. Since $N^a$ and $N^b$ agree up to $\kappa^{<\sigma}$ and $N^a\subseteq N^b$, it follows from some standard arguments of Shelah that $N^a\cap\mu=N^b\cap\mu$. In particular, we know that $\ran(f^*)\subseteq N_a$ and so we can define $m(i)=\min\{m<\omega : f^*(i)\in N^a_m\}$ for every $i<\kappa$.

I'm going to stop here and continue the remainder of the proof in the next post. But I want to give an idea of how we're using these two towers of elementary submodels and how block pcf will come into play here. The main idea here is that we can eventually get the range of $f^*$ starting only with $\kappa^{<\sigma}$, Skolem functions, functions $f^n$, and finitely many regular cardinals below $\mu$ at each stage $n$. These finitely many regular cardinals will be our vehicle for bringing block pcf into the picture. Another important part is that the restrictions of $f^n$ to a set $B$ can be seen in $M_{n+1}$ provided that $\pcf(B)\subseteq \lambda$. We will leverage these ideas to finish the proof off in the next post.