Spencinar #1, part 2

We just saw that the nonstationary ideal cannot be $\kappa$-saturated. A more refined question is whether it can even be $\kappa^+$ saturated. Sometimes this is just called saturated. Gitik and Shelah proved that the answer is no, if $\kappa>\omega_1$.

Theorem: If $\kappa,\theta$ are regular with $\theta^+<\kappa$, then $NS_\kappa \upharpoonright\mathrm{cof}(\theta)$ is not $\kappa^+$-saturated.

They start off with a principle that is inconsistent. Let $\kappa, \theta$ be regular, $\kappa>\omega_2, \theta^+$. The strong club guessing principle $\lozenge^*_\textrm{club}(\kappa,\theta)$ states that there exists a sequence $\langle S_\alpha : \alpha\in\kappa\cap \mathrm{cof}(\theta)\rangle$ such that

1. $S_\alpha\subseteq \alpha$,
2. $\sup S_\alpha =\alpha$,
3. $|S_\alpha|=\theta$,
4. For all $\beta\in S_\alpha$, $\mathrm{cf}(\beta)>\theta$ and if $\theta=\aleph_0$ then $\mathrm{cf}(\beta)>\omega_1$ (Spencer remarks that the "and if" part is an ad hoc condition used to handle a special case),
5. For all $C\subseteq \kappa$ club, $\{\alpha: \exists \beta<\alpha \, S_\alpha\setminus \beta\subseteq C\}$ contains a club intersected with $\mathrm{cof}(\theta)$. (Such an $\alpha$ is said to guess $C$)

Lemma 1: $\lozenge^*_\textrm{club}(\kappa,\theta)$ is false.

Proof: Otherwise there is such a sequence $\langle S_\alpha : \alpha\in\kappa\cap \mathrm{cof}(\theta)\rangle$.

Case 1: $\theta>\aleph_0$.

We will build a decreasing sequence of clubs $\langle E_n: n<\omega \rangle$ by

$$E_0=\kappa,$$

$$E_{n+1} \subseteq E'_n \textrm{ such that every point of }E_{n+1}\cap \mathrm{cof}(\theta) \textrm{ guesses } E'_n.$$

Let $E=\bigcap_n E_n$, and $\delta =\min(E\cap \mathrm{cof}(\theta))$. Since $\theta>\aleph_0$, there is a $\beta\in S_\delta \subseteq E$ and $S_\delta\setminus \beta \subseteq E$. Therefore $\beta \in E'_n$ for all $n$ and hence in $E'$, and $\mathrm{cf}(\beta)>\theta$ by (4). Thus we can find a point in $E\cap \mathrm{cof}(\theta))$ less than $\beta$, contradicting the minimality of $\delta$. 

Case 2: $\theta=\aleph_\omega$. 

This case is similar to Case 1, and is handled using the ad hoc condition. $\Box$

Now that we've defined this inconsistent principle, we will derive it from the assumption of a saturated nonstationary ideal. This would be enough to prove the theorem.

Lemma 2: Suppose $\kappa,\theta$ are regular cardinals with $\theta^+<\kappa$. If $NS_\kappa\upharpoonright \mathrm{cof}(\theta)$ is $\kappa^+$-saturated, then $\lozenge^*_\textrm{club}(\kappa,\theta)$ holds.

Proof: Consider the principle $\lozenge'_\textrm{club}(S)$ which is like $\lozenge^*_\textrm{club}(\kappa,\theta)$ above but weakening (5) to (5'):

• for all closed unbounded $C\subseteq \kappa$, $\{\alpha: \exists \beta<\alpha \, S_\alpha\setminus \beta\subseteq C\}$ is stationary.

It's a famous result of Shelah that this club-guessing theorem holds for all stationary $S\subseteq \kappa\cap\mathrm{cof}(\theta)$.

From the saturation assumption we can first get a local version of the strong club guessing principle.

Claim: Let $S\subseteq \kappa\cap\mathrm{cof}(\theta)$ be stationary and $\langle S_\alpha: \alpha\in S\rangle$ witness $\lozenge'_\textrm{club}(S)$. Then there is $S^*\subseteq S$ stationary such that for all $C\subseteq \kappa$ club, $\{\alpha: \exists \beta<\alpha \, S_\alpha\setminus \beta\subseteq C\}$ contains a club intersected with $S^*$.

Proof of the Claim: Suppose not. We will define clubs $\langle C_\alpha: \alpha<\kappa^+\rangle$ so that if $\xi<\zeta$ then for some $\gamma<\kappa$, $C_\zeta\setminus \gamma \subseteq C_\xi$ (so the sequence is "almost decreasing"). We will simultaneously define stationary sets $\langle A_\alpha:\alpha<\kappa^+\rangle$ which are subsets of $S$ and whose pairwise intersections are nonstationary. This will yield a contradiction to saturation.

For any club $D$, let $N(D)=\{\alpha: \exists \beta<\alpha \, S_\alpha\setminus \beta\subseteq D\}$ be the set of points that guess $D$.

By the contradictory assumption, pick $C_0$ club so that $S-N(C_0)$ is stationary, and let $A_0=S-N(C_0)$. Suppose $\langle C_\beta:\beta<\alpha\rangle$ and $\langle A_\beta:\beta<\alpha$ have already been constructed. Let $C=\Delta_{\beta<\alpha} C_\alpha$ (technically, the diagonal intersections are taken using some fixed surjection from $\kappa$ to $\alpha$ in the background).

Again using the contradictory assumption, there is $C_\alpha$ club so that $N(C)-N(C_\alpha)$ is stationary. Let $A_\alpha = N(C)-N(C_\alpha)$.

These $A_\alpha$ are have nonstationary (in fact, bounded) pairwise intersection: if $\beta<\alpha$, then coboundedly many points of $A_\alpha$ guess the diagonal intersection of the $C_\xi, \xi<\alpha$, so coboundedly many points of $A_\alpha$ guess $C_\beta$. But $A_\beta$ was taken to be disjoint from $C_\beta$. The claim is proven.

Find a maximal collection $\langle T_\alpha:\alpha<\kappa\rangle$ of pairwise intersection nonstationary sets so that each $T_\alpha$ has the property of $S^*$ in the claim. The sequence has size $\kappa$ by saturation. Gluing the pieces together, we have proved the Lemma, so we're done.