In a previous post, we defined Shelah’s approachability ideal $I[\lambda]$. We remind the reader that a subset $S\subseteq\lambda$ is in $I[\lambda]$ iff there exists a collection $\{ \mathcal D_\alpha\mid\alpha<\lambda\}\subseteq\mathcal [\mathcal P(\lambda)]^{<\lambda}$ such that for club many $\delta\in S$, the union $\bigcup_{\alpha<\delta}\mathcal D_\alpha$ contains all the initial segments of some small cofinal subset, $A_\delta$, of $\delta$.
The notion “small” above stands for the assertion that $\text{otp}(A_\delta)<\delta$.
It is an interesting fact that in the proofs given in the previous post, the small witnessing sets $A_\delta$ were always of smallest possible size, i.e., $\text{otp}(A_\delta)=\text{cf}(\delta)<\delta$. This raises the question of whether one can get the witnessing sets to have higher order-type. In this post, we shall provide an alternative proof to the fact that $E^{\lambda^+}_{<\lambda}\in I[\lambda^+]$ for every regular cardinal $\lambda$, in which the witnessing sets would have as large order-type (below $\lambda$) as we like. As a bonus, the witnessing sets would also have the feature of club guessing.
Theorem (Shelah). For every regular uncountable cardinal $\lambda$, every non-zero limit ordinal $\theta<\lambda$, and every stationary set $S\subseteq E^{\lambda^+}_{|\theta|}$, there exists a sequence $\langle C_\alpha\mid \alpha<\lambda^+\rangle$ such that for every club $E\subseteq \lambda^+$, there exists an $\alpha\in S$ such that:
- $\sup(C_\alpha)=\alpha$, and $\text{otp}(C_\alpha)=\theta$;
- $C_\beta=C_\alpha\cap\beta$ for all $\beta\in C_\alpha$;
- $C_\alpha\subseteq S\cap E$.
Proof. For every ordinal $\alpha<\lambda^+$, fix an injection $d_\alpha:\alpha\rightarrow\lambda$. Denote $a_\alpha^i:=d_\alpha^{-1}[ i]$. Then for every $\alpha<\lambda^+$, $\{ a^i_\alpha\mid i<\lambda\}$ is an increasing and continuous chain in $[\alpha]^{<\lambda}$ that converges to $\alpha$. Next, for all $\alpha<\lambda^+$, we define a club in $\lambda$ as follows:$$\Omega_\alpha:=\{ i<\lambda\mid \forall\beta\in a^i_\alpha(a_\beta^i=a_\alpha^i\cap\beta)\}.$$
For a set $E\subseteq\lambda^+$ and $i<\lambda$, we define the set $E(i)$ as the set of all $\alpha\in S$ for which all of the following holds:
- $i\in\Omega_\alpha$;
- $\text{otp}(E\cap S\cap\alpha)=\alpha$;
- $\sup(E\cap S\cap a^i_\alpha)=\alpha$;
- $\text{otp}(E\cap S\cap a^i_\alpha)=\text{otp}(a^i_\alpha)$ is divisble by $\theta$.
Subclaim 1. There exists an $i^*<\lambda$, for which $E(i^*)$ is nonempty whenever $E$ is a club in $\lambda^+$.
Proof. Suppose not. Then for every $i<\lambda$, we may pick a club $E_i\subseteq\lambda^+$ for which $E_i(i)=\emptyset$. Put $E:=(\bigcap_{i<\lambda}E_i)\setminus\lambda$. Then $E$ is a club in $\lambda^+$, so let us pick some $\alpha\in E\cap S$ such that $\text{otp}(E\cap S\cap\alpha)=\alpha$.
As $\text{cf}(\alpha)\neq\text{cf}(\lambda)$, the set $D:=\{ i<\lambda\mid \sup(E\cap S\cap a^i_\alpha)=\alpha\}$ is co-bounded in $\lambda$.
Also, continuity entails that the set $D’:=\{ i\in D\mid \text{otp}(E\cap S\cap a^i_\alpha)=\text{otp}(a^i_\alpha)\}$ is a club in $\lambda$. Again, by continuity, we may pick $i\in D’\cap\Omega_\alpha$ such that $\text{otp}(a^i_\alpha)$ is divisible by $\theta$. Then, by $E_i\subseteq E$, we get that $\text{otp}(E_i\cap S\cap a^i_\alpha)=\text{otp}(a^i_\alpha)$. So $\alpha\in E_i(i)$, contradicting the choice of $E_i$. $\blacksquare$
Let $i^*<\lambda$ be given by the previous claim.
Subclaim 2. There exists a club $E^*\subseteq\lambda^+$ such that for every club $D\subseteq\lambda^+$, the set $\{ \alpha\in E^*(i^*)\mid a_\alpha^{i^*}\cap E^*\subseteq D\}$ is nonempty.
Proof. Suppose not. Then there exists a decreasing and continuous sequence of clubs $\langle E_i\mid i<\lambda\rangle$ such that $E_0=\lambda^+$, and for every $i<\lambda$, the set $\{ \alpha\in E_i(i^*)\mid a^{i^*}_\alpha\cap E_i\subseteq E_{i+1}\}$ is empty.
Put $E:=\bigcap_{i<\lambda}E_i$, and pick $\alpha\in E(i^*)$. Then $\alpha\in E_i(i^*)$ for all $i<\lambda$, and hence $\langle a^{i^*}_\alpha\cap E_{i+1}\mid i<\lambda\rangle$ must be a strictly-decreasing sequence of subsets of $a^{i^*}_\alpha$, contradicting the fact that $|a^{i^*}_\alpha|<\lambda$. $\blacksquare$
Let $E^*\subseteq\lambda^+$ be given by the previous claim.
Subclaim 3. There exists an ordinal $\tau^*<\lambda$ which is divisible by $\theta$ such that for every club $D\subseteq\lambda^+$, the set $\{ \alpha\in E^*(i^*)\mid a_\alpha^{i^*}\cap E^*\subseteq D\ \&\ \text{otp}(a_\alpha^{i^*})=\tau^*\}$ is nonempty.
Proof. Otherwise, for every ordinal $\tau<\lambda$ which is divisible by $\theta$, there exists a club $D_\tau\subseteq\lambda^+$ for which $\{ \alpha\in E^*(i^*)\mid a_\alpha^{i^*}\cap E^*\subseteq D_\tau\ \&\ \text{otp}(a_\alpha^{i^*})=\tau\}$ is empty. Put $D:=\bigcap\{ D_\tau\mid \tau<\lambda\text{ is divisible by }\theta\}$. By the choice of $E^*$, then, we may pick $\alpha\in E^*(i^*)$ such that $a_\alpha^{i^*}\cap E^*\subseteq D$. Put $\tau:=\text{otp}(a^{i^*}_\alpha)$. Then by $\alpha\in E^*(i^*)$, $\tau$ is divisible by $\theta$, contradicting the fact $a_\alpha^{i^*}\cap E^*\subseteq D_\tau$. $\blacksquare$
Let $\tau^*$ be given by the previous claim.
As $\tau^*$ is divisible by $\theta$, let us pick a cofinal subset $u\subseteq\tau^*$ of order-type $\theta$. Then, define for all $\alpha<\lambda^+$:
$$C_\alpha:=\{ \beta\in E^*\cap S\cap a_\alpha^{i^*}\mid \text{otp}(a_\beta^{i^*})\in u\}.$$
To see that $\langle C_\alpha\mid \alpha<\lambda^+\rangle$ works. Suppose that we are given a club $E\subseteq\lambda^+$.
Pick $\alpha\in E^*(i^*)$ such that $a_\alpha^{i^*}\cap E^*\subseteq E$ and $\text{otp}(a_\alpha^{i^*})=\tau^*$. Then:
- $\alpha\in S$;
- $\sup(E^*\cap S\cap a^{i^*}_\alpha)=\alpha$;
- $C_\alpha\subseteq a^{i^*}_\alpha\cap E^*\subseteq E$;
- $i^*\in\Omega_\alpha$, and hence for all $\beta\in a^{i^*}_\alpha$, we have $a^{i^*}_\beta=a^{i^*}_\alpha\cap\beta$, and $C_\beta=C_\alpha\cap\beta$;
- $\text{otp}(E^*\cap S\cap a^{i^*}_\alpha)=\tau^*$, and $\text{otp}(C_\alpha)=\text{otp}\{\beta\in E^*\cap S\cap a^{i^*}_\alpha\mid \text{otp}(E^*\cap S\cap a^{i^*}_\alpha\cap\beta)\in u\}=\text{otp}(u)=\theta$.
This completes the proof. $\square$
Corollary. For every uncountable cardinal $\lambda$ and every ordinal $\theta<\lambda$, there exists a sequence $\langle C_\alpha\mid \alpha<\lambda^+\rangle$ satisfying the following.
For every club $E\subseteq\lambda^+$, every large enough regular cardinal $\chi$, and any $x\in H(\chi)$, there exists an elementary submodel $M\preceq (H(\chi),\in,\ldots)$ such that:
- $M$ has size $\lambda$, and $x\in M$;
- $\alpha:=M\cap\lambda^+$ is an ordinal, and $C_\alpha\subseteq E$;
- $C_\alpha$ is a club in $\alpha$ of type $\ge\theta$;
- $C_\alpha\cap\beta\in M$ for all $\beta<\alpha$.
Proof. If $\lambda$ is a regular cardinal, then we take the sequence $\overrightarrow C=\langle C_\alpha\mid \alpha<\lambda^+\rangle$ provided by the previous theorem, and let $D_\alpha$ denote the closure of $C_\alpha$. To see that $\langle D_\alpha\mid\alpha<\lambda^+\rangle$ works, suppose that we are given a club $E\subseteq\lambda^+$. Consider the club $D\subseteq\lambda^+$ of all ordinals of the form $M\cap\lambda^+$ for some elementary submodel $M$ of size $\lambda$ that contains $x$ and $\overrightarrow C$. By the choice of $\overrightarrow C$, we can find some $\alpha<\lambda^+$ such that $C_\alpha$ is cofinal subset of $\alpha$, of type $\theta$, $C_\alpha\subseteq D\cap E$, and $C_\beta=C_\alpha\cap\beta$ for all $\beta\in C_\alpha$. In particular, $\alpha\in D$, and so we can find a model $M$ with $x\in M$ and $M\cap\lambda^+=\alpha$ such that $C_\beta\in M$ for all $\beta<\alpha$. As $C_\alpha\subseteq E$ and $E$ is closed, we also get that $D_\alpha\subseteq E$. As $C_\alpha\cap\beta\in M$ for all $\beta<\alpha$, we get also that $D_\alpha\cap\beta\in M$ for all $\beta<\alpha$.
If $\lambda$ is singular, we let $\kappa:=|\theta|^+$. Then, by statement (2) of our previous post, there exists a sequence $\langle C_\alpha\mid \alpha\in E^{\lambda^+}_\kappa\rangle$, and a matrix $\overrightarrow D=\langle D_{\delta,i}\mid\delta<\lambda^+, i<\lambda\rangle$ of bounded subsets of $\lambda^+$, such that for every club $E\subseteq\lambda^+$, there exists $\alpha<\lambda^+$ such that $C_\alpha$ is a club in $\delta$ of type $\kappa$, $C_\alpha\subseteq E$ and for every $\beta<\alpha$, there exists $\delta<\alpha$ and $i<\lambda$ such that $C_\alpha\cap\beta=D_{\delta,i}$. To see that $\langle C_\alpha\mid\alpha<\lambda^+\rangle$ works, suppose that we are given a club $E\subseteq\lambda^+$. Consider the club $D\subseteq\lambda^+$ of all ordinals of the form $M\cap\lambda^+$ for some elementary submodel $M$ of size $\lambda$ containing $x$ and $\overrightarrow D$. By the choice of $\langle C_\alpha\mid\alpha<\lambda^+\rangle$, we can find some $\alpha<\lambda^+$ such that $C_\alpha$ is a club in $\alpha$, of type $\kappa>\theta$, $C_\alpha\subseteq D\cap E$, and any proper initial segment of $C_\alpha$ is an element of $\{ D_{\delta,i}\mid \delta<\alpha,i<\lambda\}$. In particular, $\alpha\in D$, and we may fix a model $M$ with $M\cap\lambda^+=\alpha$ such that $x,\overrightarrow D\in M$. As $\{ D_{\delta,i}\mid \delta<\alpha,i<\lambda\}\subseteq M$, our proof is complete. $\square$
We conclude with a couple of a questions concerning possible strengthenings of the above opening theorem:
- What happens if we require that $C_\alpha$ be closed for all $\alpha<\lambda^+$? Does this stronger statement follow from ZFC?
- What happens if we take $\lambda=\aleph_\omega$, $S=E^{\aleph_{\omega+1}}_\omega$ and $\theta$ arbitrarily large countable limit ordinal? Does this version follow from ZFC?
(note that here we cannot expect the statement to hold true for any stationary subset $S\subseteq E^{\aleph_{\omega+1}}_\omega$)
We mention that a negative answer to (2) entails a negative answer to Question 4 from this paper of Foreman and Todorcevic.
