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 $⟨C_{\alpha }\mid \alpha \in S⟩$ 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. $C_{\alpha }{\subseteq }^{\ast }D$, (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 $1\Rightarrow 2\Rightarrow 3\Rightarrow 4\Rightarrow 5$. In particular, if ${\mathrm{♢}}_S$ holds, then $S$ carries a club-guessing sequence. However, in contrast to $\mathrm{♢}$, club-guessing sequences are often consequences of ZFC, hence allowing to prove ZFC results that were previously known to follow from $\mathrm{♢}$-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  $E_{\lambda }^{\kappa }=\{\alpha <\kappa \mid \text{cf}(\alpha )=\lambda \}$.

• if ${\lambda }^{+}<\kappa$, then $S$ carries a club-guessing sequence in the sense of (2) above.
• if ${\lambda }^{+}=\kappa >{\mathrm{\aleph }}_1$, 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 $\lambda ={\mathrm{\aleph }}_1,\kappa ={\mathrm{\aleph }}_3$.

Proof. Suppose that $S\subseteq E_{{\mathrm{\aleph }}_1}^{{\mathrm{\aleph }}_3}$ 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 $E\subseteq {\mathrm{\aleph }}_3$ such that $⟨C_{\alpha }\cap E\mid \alpha \in S⟩$ is a club-guessing sequence in the sense of (2) above. Suppose not. Then there exists a decreasing chain of club subsets of ${\mathrm{\aleph }}_3$, $\{E_i\mid i<{\mathrm{\aleph }}_2\}$, with $E_0={\mathrm{\aleph }}_3$, such that for every $i<{\mathrm{\aleph }}_2$, and every $\alpha \in S$, either $sup(C_{\alpha }\cap E_i)<\alpha$, or $C_{\alpha }\cap E_i⊈E_{i+1}$.
Consider the club $E:=\bigcap _{i<{\mathrm{\aleph }}_2}{E}_i$. 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 $i<{\mathrm{\aleph }}_2$. It now follows that for every $i<{\mathrm{\aleph }}_2$, $C_{\alpha }\cap E_i⊈E_{i+1}$. So $⟨C_{\alpha }\cap E_i\mid i<{\mathrm{\aleph }}_2⟩$ is a strictly decreasing chain of subset of $C_{\alpha }$ of length ${\mathrm{\aleph }}_2$, contradicting the fact that $\text{otp}(C_{\alpha })={\omega }_1$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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 $[{\omega }_1{]}^{{\mathrm{\aleph }}_0}$, one of the following holds:

1. there exists a club $D\subseteq {\omega }_1$ such that $[D{]}^{{\mathrm{\aleph }}_0}\subseteq \mathcal{I}$;
2. there exists a stationary subset $T\subseteq {\omega }_1$ such that $[T{]}^{{\mathrm{\aleph }}_0}\cap \mathcal{I}=\mathrm{\varnothing }$.

Theorem (Eisworth-Nyikos, 2009). PFA implies $({\star }_c)$.
In particular, under PFA, there exists no-club guessing sequence for $\kappa ={\mathrm{\aleph }}_1,\lambda ={\mathrm{\aleph }}_0$, even in the weakest sense of (5) above.
Proof of the second part. Suppose that $⟨C_{\alpha }\mid \alpha <{\omega }_1⟩$ is a club-guessing sequence, with $\text{otp}(C_{\alpha })\le \omega$ for every $\alpha <{\omega }_1$. Let $$\mathcal{I}=\{X\in [{\omega }_1{]}^{{\mathrm{\aleph }}_0}\mid \mathrm{\forall }\alpha <{\omega }_1(C_{\alpha }\cap X\text{ is finite})\}.$$
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 $C_{\alpha }\cap D\in [D{]}^{{\mathrm{\aleph }}_0}$, we infer that $[D{]}^{{\mathrm{\aleph }}_0}$ is not a subideal of $\mathcal{I}$.
2. If $X\subseteq {\omega }_1$ and $\text{otp}(X)={\omega }^2$, then $[X{]}^{{\mathrm{\aleph }}_0}\cap \mathcal{I}\ne \mathrm{\varnothing }$. 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 $f:\omega \to X$ such that $f(n)\notin \bigcup _{m\le n}{C}_{{\alpha }_m}$. As $\text{otp}(X)={\omega }^2$, such a function may indeed be constructed. Put $Y:=f“\omega$. Then $Y\in [X{]}^{{\mathrm{\aleph }}_0}\cap \mathcal{I}$. It follows that $[T{]}^{{\mathrm{\aleph }}_0}\cap \mathcal{I}\ne \mathrm{\varnothing }$ 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 $X_n{\subseteq }^{\ast }X$ 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 $\psi :\omega \to \bigcup _{n<\omega }{X}_n$ . Then for all $\alpha <{\omega }_1$, define $f_{\alpha }:\omega \to \omega$ 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 $f:\omega \to \omega$ such that $f_n{\le }^{\ast }f$ 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 $X\notin \mathcal{I}$. Then, we may find some $\alpha <{\omega }_1$ such that $A_{\alpha }\cap X$ is infinite. As $f_{\alpha }{\le }^{\ast }f$ 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 $(A_{\alpha }\cap X)\cap X_n\ne \mathrm{\varnothing }$. 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 $\psi (m)\notin X$, contradicting the choice of $\beta$ in $X$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$.

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 $⟨C_{\alpha }\mid \alpha \in S⟩$ 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 $C_{\alpha }{\subseteq }^{\ast }T$.

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 $\kappa >{\mathrm{\aleph }}_1$, 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 $⟨C_{\alpha }\mid \alpha \in S⟩$ witnesses ${\text{Stat}}_S$ for a stationary $S\subseteq \kappa$ with $\kappa >{\mathrm{\aleph }}_1$. 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 $S\cap E_{\lambda }^{\kappa }$ is stationary. Then, construct a decreasing chain of clubs $\{D_i\mid i<\theta \}$ such that:

• $D_0:=\kappa$, and for all $i<\theta$:
• $D_{i+1}\subseteq \{\alpha <\kappa \mid \alpha \notin S\text{ or }sup(C_{\alpha }\setminus S\cap E_{\lambda }^{\kappa }\cap D_i)<\alpha \}.$

Put $D:=\bigcap _{i<\theta }{D}_i$, and $T:=S\cap E_{\lambda }^{\kappa }\cap D$. Utilize the fact that $\theta \ne \lambda$, 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 $\alpha$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

My proof uses the fact that $\kappa >{\mathrm{\aleph }}_1$ in a crucial way (the existence of $\theta ,\lambda$ follows from it), and hence the case $\kappa ={\mathrm{\aleph }}_1$ 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 ${\text{Stat}}_{{\mathrm{\aleph }}_1}$ 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 $⟨C_{\alpha }\mid \alpha \in S⟩$ be a ${\text{Stat}}_S$-sequence. Recursively define a function $f:S\to \kappa +1$ by letting for all $\alpha \in S$:$$f(\alpha ):=\{\begin{array}{ll}\alpha ,& sup(C_{\alpha }\setminus f[\alpha ])=\alpha \\ \kappa ,& \text{otherwise}\end{array}$$ 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 $C_{\alpha }{\subseteq }^{\ast }T$. That is $sup(C_{\alpha }\setminus T)<\alpha$. But then, $f(\alpha )\ne \alpha$, i.e., $\alpha \notin T$. 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 $C_{\alpha }{\subseteq }^{\ast }C\cap S$. 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 $T\cap C=\mathrm{\varnothing }$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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 $⟨C_{\alpha }\mid \alpha \in S⟩$ 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 $S\cap E_{\ne \text{cf}(\lambda )}^{{\lambda }^{+}}$ is stationary, the answer is yes.
• For $\lambda$ singular and $S\subseteq E_{\text{cf}(\lambda )}^{{\lambda }^{+}}$ that does not reflect, the answer is consistently no (from very large cardinals);
• For $\lambda$ singular and $S\subseteq E_{\text{cf}(\lambda )}^{{\lambda }^{+}}$ that reflects on some fixed cofinality club often (i.e., there is some uncountable regular cardinal $\theta <\lambda$ such that $\{\alpha \in E_{\theta }^{{\lambda }^{+}}\mid S\cap \alpha \text{ is stationary }\}$ contains a club relative to $E_{\theta }^{{\lambda }^{+}}$), the answer is yes.
• For $\lambda$ singular and $S\subseteq E_{\text{cf}(\lambda )}^{{\lambda }^{+}}$ that reflects stationarily often, the answer is yes provided that ${\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}}_{\lambda }^{\ast }$ holds.

So, the remaining case is that in which $\lambda$ is a singular cardinal, and $S\subseteq E_{\text{cf}(\lambda )}^{{\lambda }^{+}}$ merely reflects stationarily often. Notice that ${\mathrm{♢}}_S$ (and ${\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}}_{\lambda }^{\ast }$) 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 $⟨C_{\alpha }\mid \alpha \in S⟩$ 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 ${\mathrm{♢}}_S$ necessarily fails in any model that serves as a counterexample.