More on T_J numbers

\( \newcommand{\wsat}{\mathrm{wsat}} \newcommand{\GH}{\mathrm{GH}} \newcommand{\REG}{\mathrm{REG}} \newcommand{\Ch}{\mathrm{Ch}} \newcommand{\reg}{\mathrm{reg}} \newcommand{\rk}{\mathrm{rk}} \newcommand{\ot}{\mathrm{ot}} \newcommand{\pcf}{\mathrm{pcf}} \newcommand{\tcf}{\mathrm{tcf}} \newcommand{\cf}{\mathrm{cf}} \newcommand{\ran}{\mathrm{ran}} \newcommand{\dom}{\mathrm{dom}} \newcommand{\nacc}{\mathrm{nacc}} \newcommand{\acc}{\mathrm{acc}} \newcommand{\Tr}{\mathrm{Tr}} \)
The goal of this post is to relate pcf cardinals to the cardinal $T_J(\mu)$ that we defined in the previous post. Throughout, we will assume that $\kappa$ and $\mu$ are cardinals, and $J$ is a $\sigma$-based ideal on $\kappa$ such that $\kappa^{<\sigma}<\mu$. Recall the cardinals we're interested in:

$T_J(\mu)=\sup\{|F| : F\subseteq{}^\kappa\mu\text{ is a }J\text{-adt}\}$
$T^2_J(\mu)=\sup\{\lambda : \lambda\text{ is weakly }J\text{-representable}\}$
$T^3_J(\mu)=\sup\{\lambda : \lambda\text{ is }J\text{-representable}\}$

First, we establish a couple of facts about almost disjoint transversals and also show how to leverage the $\sigma$-based assumption.

Proposition: Suppose that $F$ is a $J$-adt. Then for any $f\in{}^\kappa\mu$, the set $\{g\in F : \neg(g\neq_J f)\}$ has cardinality at most $\kappa^{<\sigma}$.

Proof: Fix $F\subseteq{}^\kappa\mu$, and suppose that $f\in{}^\kappa\mu$. Note that if $g$ is in $F$ and $\neg(g\neq_J f)$, then the set $A^f_g=\{i\in \kappa : f(i)=g(i)\}\in J^+$. Since $J$ is $\sigma$-based, we may assume that $A^f_g\in[\kappa]^{<\sigma}$. Thus, there are only $\kappa^{<\sigma}$-many functions $g\in F$ such that $A^f_g\in J^+$ since $A^f_{g_1}=A^f_{g_2}\in J^+$ means that $g_1$ and $g_2$ agree on a $J$-positive set.

Proposition: The sup in $T_J(\mu)$ is achieved by a maximal family, and every maximal family has the same cardinality.

Proof: It suffices to show that, for any maximal $J$-adt $F$ and any $J$-adt family $G$, we have that $|F|\geq|G|$. With that in mind, fix a maximal $J$-adt $F\subseteq{}^\kappa\mu$ and a $J$-adt family $G$. For each $f\in G$, there is some $\Phi(f)$ such that $\neg(f\neq_J\Phi(f))$ since $F$ is maximal. By the previous lemma, we know that $|\Phi^{-1}[g]|\leq\kappa^{<\sigma}$ and thus that $|G|\leq\kappa^{<\sigma}\times |F|$. Since the constant functions form a $J$-adt, it follows that $$|G|\leq\kappa^{<\sigma}\times\mu<\mu\times\mu=\mu\leq|F|.$$

Next, we show that $T^2_J(\mu)$ and $T^3_J(\mu)$ are actually the same cardinal. Since this is mostly standard pcf tricks, I'm just going to sketch the argument and leave detail checking to less lazy folks.

Proposition: $T^2_J(\mu)=T^3_J(\mu)$

Proof: Since any $J$ representable cardinal is also weakly representable, it's clear that $T^3_J(\mu)\leq T^2_J(\mu)$. So let $\lambda$ be weakly representable as witnessed by $\langle E(i) : i< \kappa\rangle$. Since we'd like to show that $T^3_J(\mu)\geq \lambda$, we may replace $\lambda$ with the potentially larger cardinal $\lambda=\min\{\max\pcf(\bigcup_{i\in A}E(i)): A\in J^+\}$. Let $E=\bigcup_{i< \kappa}E(i)$, and note then that $\lambda\in \pcf(E)$ so we can fix a generator $B_\lambda[E]$ for $\lambda$. Now, let $E(i)=\{\lambda_{i,l} : l< n(i)\}$ and define $\lambda'_{i,l}$ as follows. If $\lambda_{i,l}\in B_\lambda[E]$, then we set $\lambda'_{i,l}=\lambda_{i,l}$. Otherwise, we let $u=\{(i,l): \lambda_{i,l}\notin B_\lambda[E]\}$ and set $$ I=\{v\subseteq u : \pcf(\{\lambda_i,l : (i,l)\in v\})\subseteq\lambda\}, $$
which is the pullback of $J_{< \lambda}[E\setminus B_\lambda[E]]$ to $u$. In particular, we know that $\prod_{(i,l)\in u}\lambda_{i,l}/I$ is $\lambda^+$-directed by some standard pcf theory since $J_{< \lambda}[E\setminus B_\lambda[E]]=J_{< \lambda^+}[E\setminus B_\lambda[E]]$. Using that, we can inductively define a $<_J$-increasing sequence $\langle f_\alpha : \alpha<\lambda\rangle$ with $<_J$-eub $f$. So defining $\lambda'_{i,l}=f(i,l)$ for $(i,l)\in u$, we get that $\tcf\prod_{(i,l)\in u}\lambda'_{i,l}/I=\lambda$. Set $E'(i)=\{\lambda'_{i,l}: l< n(i)\}$, then we claim that $\langle E'(i) : i<\kappa\rangle$ witnesses that $\lambda$ is $J$-representable.

To see this, one can check that $\max\pcf(\bigcup_{i<\kappa}E'(i)))=\lambda$. On the other hand, if $A\in J^+$, then $\lambda\in \pcf(\bigcup_{i\in A}E(i))$ again by construction. So the claim follows.

Proposition: $T^3_J(\mu)\leq T_J(\mu)$.

Proof: Suppose that $\lambda$ is $J$-representable as witnessed by $\langle E(i) : i< \kappa\rangle$, and let $E=\bigcup_{i< \kappa} E(i)$. Let $\langle f_\alpha : \alpha< \lambda$ be increasing and cofinal in $\prod E/J_{< \lambda}[E]$. Now let $Y={}^{< \omega}(\mu+1)$ and for each $\alpha< \lambda$, define $g_\alpha\in{}^\kappa Y$ by $$ g_\alpha(i)=\langle \lambda_{i,0},\ldots, \lambda_{i,n(i)}, f_\alpha(\lambda_{i,0}),\ldots f_\alpha(\lambda_{i,n(i)})\rangle. $$ Then one can check that $\{g_\alpha : \alpha<\lambda\}$ is a $J$-adt.

At this point, we know that $T^2_J(\mu)=T^3_J(\mu)\leq T_J(\mu)$. We now define two more pcf-theoretic cardinals in order to get a better picture of how $T_J(\mu)$ sits in relation to pcf theoretic objects. $$ T^4_J(\mu)=\min\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}. $$ $$ T^5_J(\mu)=\sup\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}. $$ I want to point out that in 410, Shelah only defines the cardinal $T^4_J(\mu)$, but we need to define $T^5_J(\mu)$ because one of the proofs in section 6 requires the sup on the outside. Our goal now will be to show that $$T^4_J(\mu)\leq T_J(\mu)\leq T^5_J(\mu).$$ But, because the definition of the $T^4_J(\mu)$ and $T^5_J(\mu)$ are a bit confusing, I'll use the next post to talk about their definitions and prove the first inequality.