Transversals and Elementary Submodels

\( \newcommand{\wsat}{\mathrm{wsat}} \newcommand{\GH}{\mathrm{GH}} \newcommand{\REG}{\mathrm{REG}} \newcommand{\Ch}{\mathrm{Ch}} \newcommand{\reg}{\mathrm{reg}} \newcommand{\rk}{\mathrm{rk}} \newcommand{\ot}{\mathrm{ot}} \newcommand{\pcf}{\mathrm{pcf}} \newcommand{\tcf}{\mathrm{tcf}} \newcommand{\cf}{\mathrm{cf}} \newcommand{\ran}{\mathrm{ran}} \newcommand{\dom}{\mathrm{dom}} \newcommand{\nacc}{\mathrm{nacc}} \newcommand{\acc}{\mathrm{acc}} \newcommand{\Tr}{\mathrm{Tr}} \)
I want to use this post to prove some facts about the relationship between almost disjoint transversals and elementary submodels that will be useful later on. As usual, we assume that $J$ is an ideal on a cardinal $\kappa$, and that $J$ is $\sigma$-based where $\kappa^{<\sigma}<\mu$ and $\aleph_0\leq\sigma$. First, we need a proposition from a previous post:

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 $\kappa^{<\sigma}$.

Now let's suppose that $\chi$ is a sufficiently large regular cardinal, and let $\mathfrak{A}=(H(\chi), \in, <_\chi)$. Let $\lambda$ be a cardinal with $\mu\leq\lambda$, and fix $M\prec\mathfrak{A}$ such that $\lambda+1\subseteq M$ with $|M|=\lambda$ and $J\in M$. Suppose that $T_J(\mu)>\lambda$, and let $\mathcal{F}$ be a $J$-adt with $|\mathcal{F}|>\lambda$ such that $F\in M$. Note that if $T_J(\mu)>\lambda$, then $M$ knows this and can provide us with a witness.

Lemma: There is a function $f^*\in \mathcal{F}$ such that $f^*\neq_J g$ for each $g\in M\cap{}^\kappa\mu$.

Proof: This follows directly from the previous proposition. Since $\kappa^{<\sigma}<|M\cap{}^\kappa\mu|$, we know that $\{f\in\mathcal{F} : (\exists g\in {}^\kappa\mu\cap M)(\neg(f\neq_J g))\}$ has size $\lambda<|\mathcal{F}|$.

Lemma: Suppose that $A\subseteq \kappa$ is $J$-positive, and $B\in[\mu]^{\leq\kappa^{<\sigma}}$, then $|\{f\in\mathcal{F} : f^{-1}[B]\cap A\in J^+\}|\leq\kappa^{<\sigma}$.

Proof: First note that if $f\in {}^\kappa\mu$ is such that $f^{-1}[B]\cap A\in J^+$, then we can find $B_f\subseteq f^{-1}[B]\cap A$ such that $|B_f|<\sigma$ and $B_f\in J^+$. So then, there are $\kappa^{<\sigma}$-many possible options for the sets $B_f$. Now suppose that we are given a collection of more than $\kappa^{<\sigma}$ many $f\in {}^\kappa\mu$ with the property that $f^{-1}[B]\cap A\in J^+$. Then there will be a set of $<\kappa^{<\sigma}$ many such $f$ which share the same $B_f=:C$. On the other hand, for each such $f$, $f[C]\subseteq B$ and $|B|\leq\kappa^{<\sigma}$, which means that each $f\upharpoonright C$ can take on at most $\kappa^{<\sigma}$ many values. So there are more than $\kappa^{<\sigma}$-many $f$ in our collection which agree on $C\in J^+$. Thus, $|\{f\in\mathcal{F} : f^{-1}[B]\cap A\in J^+\}|\leq\kappa^{<\sigma}$.

Tying all of this together, suppose that we have an $f^*\in \mathcal{F}$ such that $f\neq_J g$ for each $g\in {}^\kappa\mu\cap M$. If $A\subseteq\kappa$ is such that $\ran(f\upharpoonright A)\in M$, $A\in J$.

To see this, suppose otherwise and let $B=\ran(f\upharpoonright A)$. Since $A, B\in M$, we know that $$\mathcal{F}_0:=\{f\in\mathcal{F} : f^{-1}[B]\cap A\in J^+\}\in M$$ since it's definable from $\mathcal{F}$, $A$, $B$, and $J$, all of which are in $M$. On the other hand, $|B|\leq\kappa\leq\kappa^{<\sigma}$, it follows from the above proposition that $\mathcal{F}_0|\leq\kappa^{<\sigma}$. Since $\mathcal{F}_0\in M$, $|\mathcal{F}_0|<\lambda$, and $\lambda+1\subseteq M$, we can find a function $g\in M$ which is a bijection from $|\mathcal{F}_0|$ to $\mathcal{F}_0$. As $|\mathcal{F}_0|\subseteq M$, this immediately gives us that $\mathcal{F}_0\subseteq M$ and so $f^*\upharpoonright A\in M$ (since $f^*\upharpoonright A\in \mathcal{F}_0$ as $(f^*\upharpoonright A)^{-1}[B]\cap A=A\in J^+$) which contradicts our choice of $f^*$.

With these propositions out of the way, we will need a few more facts about elementary submodels coming from pcf theory. That will follow in the next post, and then we can get to repairing the argument from section 6 of 410 (and slightly strengthening the result).