Shelah’s approachability ideal (part 2)

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 $S\subseteq E_{|\theta |}^{{\lambda }^{+}}$, there exists a sequence $⟨C_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ 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 $d_{\alpha }:\alpha \to \lambda$. Denote $a_{\alpha }^i:=d_{\alpha }^{-1}[i]$. Then for every $\alpha <{\lambda }^{+}$, $\{a_{\alpha }^i\mid i<\lambda \}$ 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:$${\mathrm{\Omega }}_{\alpha }:=\{i<\lambda \mid \mathrm{\forall }\beta \in a_{\alpha }^i(a_{\beta }^i=a_{\alpha }^i\cap \beta )\}.$$

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:

• $i\in {\mathrm{\Omega }}_{\alpha }$;
• $\text{otp}(E\cap S\cap \alpha )=\alpha$;
• $sup(E\cap S\cap a_{\alpha }^i)=\alpha$;
• $\text{otp}(E\cap S\cap a_{\alpha }^i)=\text{otp}(a_{\alpha }^i)$ is divisble by $\theta$.

Subclaim 1. There exists an $i^{\ast }<\lambda$, for which $E(i^{\ast })$ 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 $E_i(i)=\mathrm{\varnothing }$. 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 $\text{cf}(\alpha )\ne \text{cf}(\lambda )$, the set $D:=\{i<\lambda \mid sup(E\cap S\cap a_{\alpha }^i)=\alpha \}$ is co-bounded in $\lambda$.
Also, continuity entails that the set $D^{\prime }:=\{i\in D\mid \text{otp}(E\cap S\cap a_{\alpha }^i)=\text{otp}(a_{\alpha }^i)\}$ is a club in $\lambda$. Again, by continuity, we may pick $i\in D^{\prime }\cap {\mathrm{\Omega }}_{\alpha }$ such that $\text{otp}(a_{\alpha }^i)$ is divisible by $\theta$. Then, by $E_i\subseteq E$, we get that $\text{otp}(E_i\cap S\cap a_{\alpha }^i)=\text{otp}(a_{\alpha }^i)$. So $\alpha \in E_i(i)$, contradicting the choice of $E_i$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$

Let $i^{\ast }<\lambda$ be given by the previous claim.

Subclaim 2. There exists a club $E^{\ast }\subseteq {\lambda }^{+}$ such that for every club $D\subseteq {\lambda }^{+}$, the set $\{\alpha \in E^{\ast }(i^{\ast })\mid a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq D\}$ is nonempty.
Proof. Suppose not. Then there exists a decreasing and continuous sequence of clubs $⟨E_i\mid i<\lambda ⟩$ such that $E_0={\lambda }^{+}$, and for every $i<\lambda$, the set $\{\alpha \in E_i(i^{\ast })\mid a_{\alpha }^{i^{\ast }}\cap E_i\subseteq E_{i+1}\}$ is empty.
Put $E:=\bigcap _{i<\lambda }{E}_i$, and pick $\alpha \in E(i^{\ast })$. Then $\alpha \in E_i(i^{\ast })$ for all $i<\lambda$, and hence $⟨a_{\alpha }^{i^{\ast }}\cap E_{i+1}\mid i<\lambda ⟩$ must be a strictly-decreasing sequence of subsets of $a_{\alpha }^{i^{\ast }}$, contradicting the fact that $|a_{\alpha }^{i^{\ast }}|<\lambda$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$

Let $E^{\ast }\subseteq {\lambda }^{+}$ be given by the previous claim.

Subclaim 3. There exists an ordinal ${\tau }^{\ast }<\lambda$ which is divisible by $\theta$ such that for every club $D\subseteq {\lambda }^{+}$, the set $\{\alpha \in E^{\ast }(i^{\ast })\mid a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq D\mathrm{\&}\text{otp}(a_{\alpha }^{i^{\ast }})={\tau }^{\ast }\}$ is nonempty.
Proof. Otherwise, for every ordinal $\tau <\lambda$ which is divisible by $\theta$, there exists a club $D_{\tau }\subseteq {\lambda }^{+}$ for which $\{\alpha \in E^{\ast }(i^{\ast })\mid a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq D_{\tau }\mathrm{\&}\text{otp}(a_{\alpha }^{i^{\ast }})=\tau \}$ is empty. Put $D:=\bigcap \{D_{\tau }\mid \tau <\lambda \text{ is divisible by }\theta \}$. By the choice of $E^{\ast }$, then, we may pick $\alpha \in E^{\ast }(i^{\ast })$ such that $a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq D$. Put $\tau :=\text{otp}(a_{\alpha }^{i^{\ast }})$. Then by $\alpha \in E^{\ast }(i^{\ast })$, $\tau$ is divisible by $\theta$, contradicting the fact $a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq D_{\tau }$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$

Let ${\tau }^{\ast }$ be given by the previous claim.

As ${\tau }^{\ast }$ is divisible by $\theta$, let us pick a cofinal subset $u\subseteq {\tau }^{\ast }$ of order-type $\theta$. Then, define for all $\alpha <{\lambda }^{+}$:
$$C_{\alpha }:=\{\beta \in E^{\ast }\cap S\cap a_{\alpha }^{i^{\ast }}\mid \text{otp}(a_{\beta }^{i^{\ast }})\in u\}.$$

To see that $⟨C_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ works. Suppose that we are given a club $E\subseteq {\lambda }^{+}$.
Pick $\alpha \in E^{\ast }(i^{\ast })$ such that $a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq E$ and $\text{otp}(a_{\alpha }^{i^{\ast }})={\tau }^{\ast }$. Then:

• $\alpha \in S$;
• $sup(E^{\ast }\cap S\cap a_{\alpha }^{i^{\ast }})=\alpha$;
• $C_{\alpha }\subseteq a_{\alpha }^{i^{\ast }}\cap E^{\ast }\subseteq E$;
• $i^{\ast }\in {\mathrm{\Omega }}_{\alpha }$, and hence for all $\beta \in a_{\alpha }^{i^{\ast }}$, we have $a_{\beta }^{i^{\ast }}=a_{\alpha }^{i^{\ast }}\cap \beta$, and $C_{\beta }=C_{\alpha }\cap \beta$;
• $\text{otp}(E^{\ast }\cap S\cap a_{\alpha }^{i^{\ast }})={\tau }^{\ast }$, and $\text{otp}(C_{\alpha })=\text{otp}\{\beta \in E^{\ast }\cap S\cap a_{\alpha }^{i^{\ast }}\mid \text{otp}(E^{\ast }\cap S\cap a_{\alpha }^{i^{\ast }}\cap \beta )\in u\}=\text{otp}(u)=\theta$.

This completes the proof. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Corollary. For every uncountable cardinal $\lambda$ and every ordinal $\theta <\lambda$, there exists a sequence $⟨C_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ 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 $M⪯(H(\chi ),\in ,\dots )$ 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 $\overrightarrow{C}=⟨C_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ provided by the previous theorem, and let $D_{\alpha }$ denote the closure of $C_{\alpha }$. To see that $⟨D_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ 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 $⟨C_{\alpha }\mid \alpha \in E_{\kappa }^{{\lambda }^{+}}⟩$, and a matrix $\overrightarrow{D}=⟨D_{\delta ,i}\mid \delta <{\lambda }^{+},i<\lambda ⟩$ 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 $⟨C_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ 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 $⟨C_{\alpha }\mid \alpha <{\lambda }^{+}⟩$, 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. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

We conclude with a couple of a questions concerning possible strengthenings of the above opening theorem:

1. What happens if we require that $C_{\alpha }$ be closed for all $\alpha <{\lambda }^{+}$? Does this stronger statement follow from ZFC?
2. What happens if we take $\lambda ={\mathrm{\aleph }}_{\omega }$, $S=E_{\omega }^{{\mathrm{\aleph }}_{\omega +1}}$ 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 $S\subseteq E_{\omega }^{{\mathrm{\aleph }}_{\omega +1}}$)

We mention that a negative answer to (2) entails a negative answer to Question 4 from this paper of Foreman and Todorcevic.