An inconsistent form of club guessing

In this post, we shall present an answer (due to P. Larson) to a question by A. Primavesi concerning a certain strong form of club guessing.

We commence with recalling Shelah’s concept of club guessing.

Concept (Shelah). Given a regular uncountable cardinal $\kappa$ and a subset $S\subseteq\kappa$, we say that $\langle C_\alpha\mid\alpha\in S\rangle$ is a club-guessing sequence, if the following two holds:

• $C_\alpha$ is a subset of $\alpha$, for every $\alpha\in S$;
• for every club subset $D\subseteq\kappa$, there exists some nonzero $\alpha\in S$ with $\sup(C_\alpha)=\alpha$ such that $C_\alpha$ and $D$ are alike.

Now, here are a few concrete instances of being alike:

1. $C_\alpha=D\cap\alpha$;
2. $C_\alpha\subseteq D$;
3. $C_\alpha\subseteq^* D$, (so, $C_\alpha\setminus D$ is bounded below $\alpha$);
4. $\sup(\{ \beta\in C_\alpha\cap D\mid \sup(C_\alpha\cap\beta)<\beta\})=\alpha$, and $\text{otp}(C_\alpha)=\text{cf}(\alpha)$;
5. $\sup(C_\alpha\cap D)=\alpha$, and $\text{otp}(C_\alpha)=\text{cf}(\alpha)$.

Note that $1\Rightarrow2\Rightarrow3\Rightarrow4\Rightarrow5$. In particular, if $\diamondsuit_S$ holds, then $S$ carries a club-guessing sequence. However, in contrast to $\diamondsuit$, club-guessing sequences are often consequences of ZFC, hence allowing to prove ZFC results that were previously known to follow from $\diamondsuit$-type assumptions (here’s a recent example) :

Theorem (Shelah). Suppose that $\lambda<\kappa$ are given infinite regular cardinals, and $S$ is a stationary subset of  $E^{\kappa}_{\lambda}=\{\alpha<\kappa\mid \text{cf}(\alpha)=\lambda\}$.

• if $\lambda^+<\kappa$, then $S$ carries a club-guessing sequence in the sense of (2) above.
• if $\lambda^+=\kappa>\aleph_1$, then $S$ carries a club-guessing sequence in the sense of (4) above.

To give you a taste of the flavour of the proofs of these theorems, we provide here a proof of the instance $\lambda=\aleph_1, \kappa=\aleph_3$.

Proof. Suppose that $S\subseteq E^{\aleph_3}_{\aleph_1}$ is a given stationary set. For every $\alpha\in S$, pick a club subset $C_\alpha\subseteq\alpha$ of order-type $\omega_1$. We claim that there exists some club $E\subseteq\aleph_3$ such that $\langle C_\alpha\cap E\mid \alpha\in S\rangle$ is a club-guessing sequence in the sense of (2) above. Suppose not. Then there exists a decreasing chain of club subsets of $\aleph_3$, $\{ E_i\mid i<\aleph_2\}$, with $E_0=\aleph_3$, such that for every $i<\aleph_2$, and every $\alpha\in S$, either $\sup(C_\alpha\cap E_i)<\alpha$, or $C_\alpha\cap E_i\not\subseteq E_{i+1}$.
Consider the club $E:=\bigcap_{i<\aleph_2}E_i$. Let $\alpha\in S$ be an accumulation point of $E$ (the choice is possible, since the set of accumulation points of $E$ is a club, and $S$ is stationary). So $\sup(E\cap\alpha)=\alpha$, and hence $E\cap\alpha$ is a club in $\alpha$. As $\alpha\in S$, we get that $\text{cf}(\alpha)=\omega_1$, and so a standard close-off argument shows that $\sup(C_\alpha\cap E)=\alpha$. In particular, $\sup(C_\alpha\cap E_i)=\alpha$ for every $i<\aleph_2$. It now follows that for every $i<\aleph_2$, $C_\alpha\cap E_i\not\subseteq E_{i+1}$. So $\langle C_\alpha\cap E_i\mid i<\aleph_2\rangle$ is a strictly decreasing chain of subset of $C_\alpha$ of length $\aleph_2$, contradicting the fact that $\text{otp}(C_\alpha)=\omega_1$. $\square$

To see that not all instances of club-guessing are consequences of ZFC, we also mention the following. Let $(\star_c)$ denote the assertion that for every P-Ideal $\mathcal I$ over $[\omega_1]^{\aleph_0}$, one of the following holds:

1. there exists a club $D\subseteq\omega_1$ such that $[D]^{\aleph_0}\subseteq\mathcal I$;
2. there exists a stationary subset $T\subseteq\omega_1$ such that $[T]^{\aleph_0}\cap\mathcal I=\emptyset$.

Theorem (Eisworth-Nyikos, 2009). PFA implies $(\star_c)$.
In particular, under PFA, there exists no-club guessing sequence for $\kappa=\aleph_1, \lambda=\aleph_0$, even in the weakest sense of (5) above.
Proof of the second part. Suppose that $\langle C_\alpha\mid \alpha<\omega_1\rangle$ is a club-guessing sequence, with $\text{otp}(C_\alpha)\le\omega$ for every $\alpha<\omega_1$. Let $$\mathcal I=\{ X\in[\omega_1]^{\aleph_0}\mid \forall\alpha<\omega_1(C_\alpha\cap X\text{ is finite})\}.$$
It is clear that $\mathcal I$ is downward closed, as well as closed under finite unions. Next, we note the following:

1. If $D\subseteq\omega_1$ is a club, then there exists some limit $\alpha<\omega_1$ for which the element $C_\alpha$ of our club-guessing sequence satisfies $\sup(C_\alpha\cap D)=\alpha$.  As $C_\alpha\cap D\in[D]^{\aleph_0}$, we infer that $[D]^{\aleph_0}$ is not a subideal of $\mathcal I$.
2. If $X\subseteq\omega_1$ and $\text{otp}(X)=\omega^2$, then $[X]^{\aleph_0}\cap\mathcal I\neq\emptyset$. To see this, put $\alpha:=\sup(X)$, and let $\{ \alpha_m\mid m<\omega\}$ denote some enumeration of $\alpha+1$. Then recursively construct an injection $f:\omega\rightarrow X$ such that $f(n)\not\in \bigcup_{m\le n}C_{\alpha_m}$. As $\text{otp}(X)=\omega^2$, such a function may indeed be constructed. Put $Y:=f“\omega$. Then $Y\in[X]^{\aleph_0}\cap\mathcal I$. It follows that $[T]^{\aleph_0}\cap\mathcal I\neq\emptyset$ for every stationary subset $T\subseteq\omega_1$.

Thus, we have shown that the alternatives of $(\star_c)$ fails for $\mathcal I$, and this must mean that $\mathcal I$ is not a P-ideal. To get a contradiction, we prove that it is a P-ideal:

We need to show that for any given collection $\{ X_n\mid n<\omega\}\subseteq\mathcal I$, there exists some $X\in\mathcal I$ such that $X_n\subseteq^* X$ for all $n<\omega$. As $\mathcal I$ is downard closed, we may assume that $\{ X_n\mid n<\omega\}$ are mutually disjoint. Fix a bijection $\psi:\omega\rightarrow \bigcup_{n<\omega}X_n$ . Then for all $\alpha<\omega_1$, define $f_\alpha:\omega\rightarrow\omega$ by letting for all $n<\omega$: $$f_\alpha(n):=\min\{ m<\omega\mid X_n\cap A_\alpha\subseteq \psi“m\}.$$

As PFA implies that $\mathfrak b>\omega_1$, we may pick a function $f:\omega\rightarrow\omega$ such that $f_n\le^* f$ for all $n<\omega$. Now, let $X=\bigcup\{X_n\setminus \psi[f(n)]\mid n<\omega\}$. Then, for every $n<\omega$, $X_n\setminus X\subseteq \psi[f(n)]$ is finite. Now, let us assume towards a contradiction that $X\not\in\mathcal I$. Then, we may find some $\alpha<\omega_1$ such that $A_\alpha\cap X$ is infinite. As $f_\alpha\le^* f$ while $A_\alpha\cap X_n$ is finite for all $n<\omega$, let us pick a large enough $n<\omega$ such that $f_\alpha(n)<f(n)$ and $(A_\alpha\cap X)\cap X_n\neq\emptyset$. Pick $\beta\in A_\alpha\cap X\cap X_n$, and let $m:=\psi^{-1}(\beta)$. Since $\beta\in A_\alpha\cap X_n$, we get that $f_\alpha(n)>m$. In particular, $f(n)>m$, and so by definition of $X$, we get that $\psi(m)\not\in X$, contradicting the choice of $\beta$ in $X$. $\square$.

We remark that the assumption $\mathfrak b>\omega_1$ was not essential in the above proof, and also that Hirschorn established the consistency of a generalization of $(\star_c)$ together with CH.

We now arrive to the following strong form of club guessing.

Definition (Primavesi). For a regular uncountable cardinal $\kappa$, and a stationary subset $S\subseteq\kappa$, we say that $\text{Stat}_S$ holds, if there exists a sequence $\langle C_\alpha\mid\alpha\in S\rangle$ such that:

• $C_\alpha$ is a subset of $\alpha$;
• for every stationary subset $T\subseteq S$, there exists some $\alpha\in T$ with $\sup(C_\alpha)=\alpha$, for which $C_\alpha\subseteq^* T$.

So instead of guessing clubs, the sequence guesses stationary subset of $S$, but, more importantly, it does so within these sets.

I met Alex Primavesi at the Young Set Theory 2010 meeting, and he asked me about the consistency of his principle. It didn’t take long to realize that if $\kappa>\aleph_1$, then $\text{Stat}_S$ fails for all stationary subsets $S\subseteq\kappa$. Let’s first dispose of this observation!

Sketch of the proof. Suppose that $\langle C_\alpha\mid\alpha\in S\rangle$ witnesses $\text{Stat}_S$ for a stationary $S\subseteq\kappa$ with $\kappa>\aleph_1$. Without loss of generality, we may assume that $\sup(C_\alpha)=\alpha$ for all $\alpha\in S$. First, notice that for every stationary subset $T\subseteq S$, there exists a club $C\subseteq\kappa$ such that $\sup(C_\alpha\setminus T)<\alpha$ for every $\alpha\in S\cap C$. Next, pick distinct regular cardinals $\theta,\lambda$ smaller than $\kappa$ such that $S\cap E^\kappa_{\lambda}$ is stationary. Then, construct a decreasing chain of clubs $\{ D_i\mid i<\theta\}$ such that:

• $D_0:=\kappa$, and for all $i<\theta$:
• $D_{i+1}\subseteq\left\{ \alpha<\kappa\mid \alpha\not\in S\text{ or } \sup\left(C_\alpha\setminus S\cap E^\kappa_\lambda\cap D_i\right)<\alpha\right\}.$

Put $D:=\bigcap_{i<\theta}D_i$, and $T:=S\cap E^\kappa_\lambda\cap D$. Utilize the fact that $\theta\neq\lambda$, to argue that $\sup(C_\alpha\setminus T)<\alpha$ for every $\alpha\in T$. Consider $\alpha:=\min(T)$. As $\sup(C_\alpha\setminus T)<\alpha=\sup(C_\alpha)$, there must exist some $\beta\in C_\alpha\cap T$, contradicting the minimality of $\alpha$. $\square$

My proof uses the fact that $\kappa>\aleph_1$ in a crucial way (the existence of $\theta,\lambda$ follows from it), and hence the case $\kappa=\aleph_1$ requires a different handling. After asking people who know a great deal of the constructible universe whether $\text{Stat}_{\omega_1}$ holds in $L$, without getting definite answer, I considered another extreme – getting a model in which the nonstationary ideal on $\omega_1$ is dense, together with a variant of the standard club guessing property for $\omega_1$. I was under the impression that a $\mathbb P_{max}$-variation could produce such a model (and that $\text{Stat}_{\aleph_1}$ would hold there), so I wrote to Paul Larson. To my surprise, after a while, he sent me a very short and elegant disproof of Primavesi’s principle.

Proposition (Larson, 2011). $\text{Stat}_S$ fails for every stationary $S\subseteq\kappa$ (and every regular uncountable cardinal $\kappa$).
Proof. Suppose not, and let $\langle C_\alpha\mid\alpha\in S\rangle$ be a $\text{Stat}_S$-sequence. Recursively define a function $f:S\rightarrow \kappa+1$ by letting for all $\alpha\in S$:$$f(\alpha):=\begin{cases}\alpha,&\sup(C_\alpha\setminus f[\alpha])=\alpha\\\kappa,&\text{otherwise}\end{cases}$$ Put $T:=f[S]\cap\kappa$. Then $T\subseteq S$, and we consider two cases.

• If $T$ is stationary, then there must exist some $\alpha\in T$ such that $C_\alpha\subseteq^* T$. That is $\sup(C_\alpha\setminus T)<\alpha$. But then, $f(\alpha)\neq\alpha$, i.e., $\alpha\not\in T$. This is a contradiction.
• If $T$ is nonstationary, then there exists a club $C\subseteq\kappa$ which is disjoint from $T$. So $C\cap S$ is stationary, and we may find some $\alpha\in C\cap S$ such that $C_\alpha\subseteq^* C\cap S$. That is, $\sup(C_\alpha\setminus C\cap S)<\alpha$. It follows that $\sup(C_\alpha\setminus T)=\alpha$, and hence $f(\alpha)=\alpha$, i.e.,$\alpha\in T$. This contradicts the fact that $T\cap C=\emptyset$. $\square$

Open Problems

We saw that Primavesi’s principle is inconsistent. Still, there are other generalizations of club guessing which I find important, and their state is currently unclear. Let me mention the following two.

Quesion A (#10, in here). Suppose that $\lambda$ is an uncountable cardinal, and $S\subseteq E^{\lambda^+}_{<\lambda}$ is stationary.
Does there exist a sequence $\langle C_\alpha\mid \alpha\in S\rangle$ that guesses clubs in the following sense: for every club $D\subseteq\lambda^+$, there exists some $\alpha\in S$ with $\sup(C_\alpha)=\alpha$ such that $C_\alpha\subseteq S\cap D$.

Some partial answers:

• For $\lambda$ regular, the answer is yes.
• For $\lambda$ singular and $S$ such that $S\cap E^{\lambda^+}_{\neq\text{cf}(\lambda)}$ is stationary, the answer is yes.
• For $\lambda$ singular and $S\subseteq E^{\lambda^+}_{\text{cf}(\lambda)}$ that does not reflect, the answer is consistently no (from very large cardinals);
• For $\lambda$ singular and $S\subseteq E^{\lambda^+}_{\text{cf}(\lambda)}$ that reflects on some fixed cofinality club often (i.e., there is some uncountable regular cardinal $\theta<\lambda$ such that $\{ \alpha\in E^{\lambda^+}_\theta\mid S\cap\alpha\text{ is stationary }\}$ contains a club relative to $E^{\lambda^+}_\theta$), the answer is yes.
• For $\lambda$ singular and $S\subseteq E^{\lambda^+}_{\text{cf}(\lambda)}$ that reflects stationarily often, the answer is yes provided that $\square^*_\lambda$ holds.

So, the remaining case is that in which $\lambda$ is a singular cardinal, and $S\subseteq E^{\lambda^+}_{\text{cf}(\lambda)}$ merely reflects stationarily often. Notice that $\diamondsuit_S$ (and $\square^*_\lambda$) necessarily fails in any model that serves as a counterexample, however, that’s not very much of a problem, as this has already been established a couple of years ago.

Question B. Suppose that $\lambda$ is an uncountable regular cardinal, and that $S,T$ are disjoint stationary subsets of $E^{\lambda^+}_\lambda$.
Does there exist a sequence $\langle C_\alpha\mid \alpha\in S\rangle$ that guesses clubs in the following sense: for every club $D\subseteq\lambda^+$, there exists some $\alpha\in S$ with $\text{otp}(C_\alpha)=\lambda$ such that $$\sup(C_\alpha\cap D\cap T)=\alpha.$$

Note that by $\text{otp}(C_\alpha)=\lambda$, we get that $\sup(C_\alpha\cap\beta)<\beta$ for all $\beta\in C_\alpha\cap T$. Also notice that $\diamondsuit_S$ necessarily fails in any model that serves as a counterexample.