\(
\newcommand{\wsat}{\mathrm{wsat}}
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\newcommand{\REG}{\mathrm{REG}}
\newcommand{\Ch}{\mathrm{Ch}}
\newcommand{\reg}{\mathrm{reg}}
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\newcommand{\cf}{\mathrm{cf}}
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\newcommand{\nacc}{\mathrm{nacc}}
\newcommand{\acc}{\mathrm{acc}}
\newcommand{\Tr}{\mathrm{Tr}}
\)
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.
Almost 2Huge
Tuesday, January 8, 2019
Thursday, January 3, 2019
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.
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.
Friday, December 28, 2018
Transversals and Elementary Submodels
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\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).
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).
Thursday, December 27, 2018
$T^4_J(\mu)$ and some motivation
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\newcommand{\Ch}{\mathrm{Ch}}
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\newcommand{\nacc}{\mathrm{nacc}}
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\newcommand{\Tr}{\mathrm{Tr}}
\)
Last time we defined the cardinals: $$ T^4_J(\mu)=\min\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}. $$ $$ T^5_J(\mu)=\sup\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}. $$ The cardinal $T^4_J(\mu)$ appears in section 6 of Sh410, and there Shelah claims that $T_J(\mu)\leq T^4_J(\mu)$. Unfortunately, it seems that the proof only works if we replace the $\min$ with a $\sup$, and so we're also defining $T^5_J(\mu)$. In this post I wanted to briefly talk about why we want to relate $T_J(\mu)$ to pcftheoretic cardinals, and then discuss the (potentially wrong) definitions of the above defined cardinals.
I first want to recall that, assuming $J$ is a $\sigma$based ideal on a cardinal $\kappa$ such that $\kappa^{<\sigma}<\mu$, we know that $T_J(\mu)$ is actually the cardinality of any maximal $J$adt $\mathcal{F}$, which is a natural thing to be curious about. On the other hand, we have quite a few tools to deal with pcftheoretic objects, so situating $\mathcal{F}$ among pcftheoretic objects is useful insofar as this potentially allows us to put bounds on $\mathcal{F}$. Beyond that, let's suppose that $\kappa\leq \theta<\sigma$, so $J=[\kappa]^{<\theta}$ is a $\sigma$based ideal on $\kappa$. Let $$ AD_J(\mu)=\sup\{\mathcal{F}: (\forall A\neq B\in\mathcal{F})(A,B\in[\mu]^{\kappa} \wedge A\neq_J B )\}. $$ In other words, $AD_J(\mu)$ is the supremum of the cardinality of $\theta$almost disjoint families of subsets of $\mu$ with size $\kappa$. Note that given any such family $\mathcal{F}$, if we let $f_A: \kappa\to \mu$ be a faithful enumeration of $A$ for each $A\in \mathcal{F}$, then $\{f_A : A\in \mathcal{F}\}$ is a $J$adt. On the other hand, if $\mathcal{F}$ is a $J$adt, let $A_f=\{(i,f(i)) : i<\kappa\}$ be the graph of $f$ for each $f\in\mathcal{F}$. Then the family $\{A_f : f\in \mathcal{F}\}$ is a $\theta$almost disjoint family. Thus, $T_J(\mu)=AD_J(\mu)$ for the above $J$. Here we see that relating $AD_J(\mu)$ to pcftheoretic objects continues the trend in much of Shelah's work of connecting cardinals related to collections of sets (e.g. covering numbers) to cardinals related to functions (e.g. pseudo powers).
Now, back to those $T^4$ and $T^5$ numbers. One thing I want to note is that neither of the proofs given in section 6 of 410 work (as written) with the given definition of $T^4_J(\mu)$. Another thing I want to talk about is that given ideals $I\subseteq J$ on $\kappa$, we have that $T^2_I(\mu)\leq T^2_J(\mu)$. To see this, suppose that $\lambda$ is weakly $I$representable as witnessed by the block sequence $\langle E(i) : i<\kappa\rangle$, so $\lambda\leq\max\pcf(\bigcup_{i\in A} E(i))$ for every $A\in I^+$. Since any $J$positive set is $I$positive as well, it follows that $\lambda\leq\max\pcf(\bigcup_{i\in A} E(i))$ for any $A\in J^+$ as well and thus that $\lambda$ is weakly $J$representable. The inequality follows.
With this in mind, now note that if $A_0\subseteq A_1$, then the ideal $J+(\kappa\setminus A_0)$ contains the set $\kappa\setminus A_1$ since $\kappa\setminus A_1\subseteq\kappa\setminus A_0$. Thus, $J+(\kappa\setminus A_1)\subseteq J+(\kappa\setminus A_0)$ and so $T_{J+\kappa\setminus A_1}(\mu)\leq T_{J+\kappa\setminus A_0}$. But this means that given a sequence of sets $A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n$ with each $A_n\notin J$, the cardinal $$ \sup \{T^2_{J+(\kappa\setminus A_n)}:n<\omega\} $$ is just $T^2_{J+\kappa\setminus A_0}$. This means that $$ T^4_J(\mu)=\min\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}=\min\{T^2_{J+\kappa\setminus A}: A\notin J\}. $$ $$ T^5_J(\mu)=\sup\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}=\sup\{T^2_{J+\kappa\setminus A}: A\notin J\}. $$ I'm not entirely sure if this is a mistake on my part in not reading the definition correctly or getting an inequality reversed, or an oversight in Sh410. Either way, I don't know what the correct definition of $T^4_J(\mu)$ should actually be in order to make all of the claims in section 6 go through. With that said, my goal is to now repair the proof of the inequality $T_J(\mu)\leq T^4_J(\mu)$ in that section and use that to figure out what teh right version of $T^4_J(\mu)$ should be given the proof.
Last time we defined the cardinals: $$ T^4_J(\mu)=\min\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}. $$ $$ T^5_J(\mu)=\sup\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}. $$ The cardinal $T^4_J(\mu)$ appears in section 6 of Sh410, and there Shelah claims that $T_J(\mu)\leq T^4_J(\mu)$. Unfortunately, it seems that the proof only works if we replace the $\min$ with a $\sup$, and so we're also defining $T^5_J(\mu)$. In this post I wanted to briefly talk about why we want to relate $T_J(\mu)$ to pcftheoretic cardinals, and then discuss the (potentially wrong) definitions of the above defined cardinals.
I first want to recall that, assuming $J$ is a $\sigma$based ideal on a cardinal $\kappa$ such that $\kappa^{<\sigma}<\mu$, we know that $T_J(\mu)$ is actually the cardinality of any maximal $J$adt $\mathcal{F}$, which is a natural thing to be curious about. On the other hand, we have quite a few tools to deal with pcftheoretic objects, so situating $\mathcal{F}$ among pcftheoretic objects is useful insofar as this potentially allows us to put bounds on $\mathcal{F}$. Beyond that, let's suppose that $\kappa\leq \theta<\sigma$, so $J=[\kappa]^{<\theta}$ is a $\sigma$based ideal on $\kappa$. Let $$ AD_J(\mu)=\sup\{\mathcal{F}: (\forall A\neq B\in\mathcal{F})(A,B\in[\mu]^{\kappa} \wedge A\neq_J B )\}. $$ In other words, $AD_J(\mu)$ is the supremum of the cardinality of $\theta$almost disjoint families of subsets of $\mu$ with size $\kappa$. Note that given any such family $\mathcal{F}$, if we let $f_A: \kappa\to \mu$ be a faithful enumeration of $A$ for each $A\in \mathcal{F}$, then $\{f_A : A\in \mathcal{F}\}$ is a $J$adt. On the other hand, if $\mathcal{F}$ is a $J$adt, let $A_f=\{(i,f(i)) : i<\kappa\}$ be the graph of $f$ for each $f\in\mathcal{F}$. Then the family $\{A_f : f\in \mathcal{F}\}$ is a $\theta$almost disjoint family. Thus, $T_J(\mu)=AD_J(\mu)$ for the above $J$. Here we see that relating $AD_J(\mu)$ to pcftheoretic objects continues the trend in much of Shelah's work of connecting cardinals related to collections of sets (e.g. covering numbers) to cardinals related to functions (e.g. pseudo powers).
Now, back to those $T^4$ and $T^5$ numbers. One thing I want to note is that neither of the proofs given in section 6 of 410 work (as written) with the given definition of $T^4_J(\mu)$. Another thing I want to talk about is that given ideals $I\subseteq J$ on $\kappa$, we have that $T^2_I(\mu)\leq T^2_J(\mu)$. To see this, suppose that $\lambda$ is weakly $I$representable as witnessed by the block sequence $\langle E(i) : i<\kappa\rangle$, so $\lambda\leq\max\pcf(\bigcup_{i\in A} E(i))$ for every $A\in I^+$. Since any $J$positive set is $I$positive as well, it follows that $\lambda\leq\max\pcf(\bigcup_{i\in A} E(i))$ for any $A\in J^+$ as well and thus that $\lambda$ is weakly $J$representable. The inequality follows.
With this in mind, now note that if $A_0\subseteq A_1$, then the ideal $J+(\kappa\setminus A_0)$ contains the set $\kappa\setminus A_1$ since $\kappa\setminus A_1\subseteq\kappa\setminus A_0$. Thus, $J+(\kappa\setminus A_1)\subseteq J+(\kappa\setminus A_0)$ and so $T_{J+\kappa\setminus A_1}(\mu)\leq T_{J+\kappa\setminus A_0}$. But this means that given a sequence of sets $A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n$ with each $A_n\notin J$, the cardinal $$ \sup \{T^2_{J+(\kappa\setminus A_n)}:n<\omega\} $$ is just $T^2_{J+\kappa\setminus A_0}$. This means that $$ T^4_J(\mu)=\min\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}=\min\{T^2_{J+\kappa\setminus A}: A\notin J\}. $$ $$ T^5_J(\mu)=\sup\{\sup T^2_{J+(\kappa\setminus A_n)}: A_n\subseteq A_{n+1}\subseteq\kappa=\bigcup_{n<\omega}A_n, A_n\notin J\}=\sup\{T^2_{J+\kappa\setminus A}: A\notin J\}. $$ I'm not entirely sure if this is a mistake on my part in not reading the definition correctly or getting an inequality reversed, or an oversight in Sh410. Either way, I don't know what the correct definition of $T^4_J(\mu)$ should actually be in order to make all of the claims in section 6 go through. With that said, my goal is to now repair the proof of the inequality $T_J(\mu)\leq T^4_J(\mu)$ in that section and use that to figure out what teh right version of $T^4_J(\mu)$ should be given the proof.
Tuesday, December 25, 2018
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 $F\geqG$. 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 $$G\leq\kappa^{<\sigma}\times\mu<\mu\times\mu=\mu\leqF.$$
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 pcftheoretic 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.
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 $F\geqG$. 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 $$G\leq\kappa^{<\sigma}\times\mu<\mu\times\mu=\mu\leqF.$$
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 pcftheoretic 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.
Friday, December 21, 2018
Almost Disjoint Transversals
Definition: Suppose that $\kappa$ and $\mu$ are cardinals with $\kappa<\mu$, and $J$ is an ideal on $\kappa$. We say that $\mathcal{F}\subseteq {}^\kappa\mu$ is a $J$adt (almost disjoint transversal) if for every $f\neq g\in \mathcal{F}$, we have that $f\neq_J g$.
One reason for looking at almost disjoint transversals is that they show up in the context of cardinal arithmetic quite often, as they're used to prove the first GalvinHajnal formula and they appear in several of Shelah's papers (e.g. 186, 410, 506, and 589). In particular, we want to consider the following cardinal:
$T_J(\mu)=\sup\{F : F\subseteq{}^\kappa\mu\text{ is a }J\text{adt}\}$
I want to use the next few posts to work through a result of Shelah that, under reasonable assumptions on the ideal $J$, relates $T_J(\mu)$ to a pcftheoretic cardinal. This is all based on the arguments in section 6 of Shelah paper 410. In the remainder of this post we will define some auxiliary invariants, with the goal of showing all of these are equal in most cases. First, we need some definitions.
Definition: Given an ideal $J$ on a set $A$, we say that $A$ is $J$positive ($A\in J^+$) if $A$ is not in $J$.
Definition: Given a cardinal $\theta$, we say that an ideal $J$ is $\theta$based if for any $J$positive $A$, there is a $J$positive $B\subseteq A$ such that $B<\theta$.
At this point, we will assume that our ideal $J$ is $\theta$based where $\kappa^{<\theta}<\mu$. At the end of this series of posts, we will discuss some situations under which these assumptions are satisfied, as they are not terribly stringent. Finally, we need one more definition which is a bit adhoc, but will help us in defining the invariants I mentioned earlier.
Definition: We say that a cardinal $\lambda$ is weakly $J$representable if there is a block sequence $\langle E(i) : i<\kappa\rangle$ with $E(i)\in[(\kappa^{<\sigma,}\mu]\cap\mathrm{Reg}]^{<\omega}$ such that for any $J$positive set $A$, $\lambda\leq\max pcf\bigcup_{i\in A}E(i)$. We say that $\lambda$ is $J$representable if instead we ask $\lambda=\max pcf\bigcup_{i\in A}E(i)$ for every $J$positive set $A$.
$T^2_J(\mu)=\sup\{\lambda : \lambda\text{ is weakly }J\text{representable}\}$
$T^3_J(\mu)=\sup\{\lambda : \lambda\text{ is }J\text{representable}\}$
Note here that the above cardinals are defined in terms of a $\kappa$sequence of finite blocks, which may seem somewhat strange initially. It's worth pointing out that this "chunky pcf" appears in a number of other places in Shelah's work primarily when Skolem Hulls and towers of elementary submnodels are involved. In a later post, we will see how the chunkyness of the pcf gets used. In the next post however, I will discuss how $T_J(\mu)$, $T^2_J(\mu)$, and $T^3_J(\mu)$ are related and define one last related invariant.
One reason for looking at almost disjoint transversals is that they show up in the context of cardinal arithmetic quite often, as they're used to prove the first GalvinHajnal formula and they appear in several of Shelah's papers (e.g. 186, 410, 506, and 589). In particular, we want to consider the following cardinal:
$T_J(\mu)=\sup\{F : F\subseteq{}^\kappa\mu\text{ is a }J\text{adt}\}$
I want to use the next few posts to work through a result of Shelah that, under reasonable assumptions on the ideal $J$, relates $T_J(\mu)$ to a pcftheoretic cardinal. This is all based on the arguments in section 6 of Shelah paper 410. In the remainder of this post we will define some auxiliary invariants, with the goal of showing all of these are equal in most cases. First, we need some definitions.
Definition: Given an ideal $J$ on a set $A$, we say that $A$ is $J$positive ($A\in J^+$) if $A$ is not in $J$.
Definition: Given a cardinal $\theta$, we say that an ideal $J$ is $\theta$based if for any $J$positive $A$, there is a $J$positive $B\subseteq A$ such that $B<\theta$.
At this point, we will assume that our ideal $J$ is $\theta$based where $\kappa^{<\theta}<\mu$. At the end of this series of posts, we will discuss some situations under which these assumptions are satisfied, as they are not terribly stringent. Finally, we need one more definition which is a bit adhoc, but will help us in defining the invariants I mentioned earlier.
Definition: We say that a cardinal $\lambda$ is weakly $J$representable if there is a block sequence $\langle E(i) : i<\kappa\rangle$ with $E(i)\in[(\kappa^{<\sigma,}\mu]\cap\mathrm{Reg}]^{<\omega}$ such that for any $J$positive set $A$, $\lambda\leq\max pcf\bigcup_{i\in A}E(i)$. We say that $\lambda$ is $J$representable if instead we ask $\lambda=\max pcf\bigcup_{i\in A}E(i)$ for every $J$positive set $A$.
$T^2_J(\mu)=\sup\{\lambda : \lambda\text{ is weakly }J\text{representable}\}$
$T^3_J(\mu)=\sup\{\lambda : \lambda\text{ is }J\text{representable}\}$
Note here that the above cardinals are defined in terms of a $\kappa$sequence of finite blocks, which may seem somewhat strange initially. It's worth pointing out that this "chunky pcf" appears in a number of other places in Shelah's work primarily when Skolem Hulls and towers of elementary submnodels are involved. In a later post, we will see how the chunkyness of the pcf gets used. In the next post however, I will discuss how $T_J(\mu)$, $T^2_J(\mu)$, and $T^3_J(\mu)$ are related and define one last related invariant.
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Finishing the proof of Theorem 6.1 from Sh410
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Definition: Suppose that $\kappa$ and $\mu$ are cardinals with $\kappa<\mu$, and $J$ is an ideal on $\kappa$. We say that $\mathcal{F}...