# Partitioning the club guessing

In a recent paper, I am making use of the following  fact.

Theorem (Shelah, 1997). Suppose that $\kappa$ is an accessible cardinal (i.e., there exists a cardinal $\theta<\kappa$ such that $2^\theta\ge\kappa)$. Then there exists a sequence $\langle g_\delta:C_\delta\rightarrow\omega\mid \delta\in E^{\kappa^+}_\kappa\rangle$ such that $C_\delta$ is a club of $\delta$ of order-type $\kappa$, and with the property that for every club $D\subseteq\kappa^+$, there exists some $\delta\in E^{\kappa^+}_\kappa$ such that $$\sup\{ \alpha\in\text{nacc}(C_\delta)\cap D\mid g_\delta(\alpha)=n\}=\delta$$ for all $n<\omega$.

[What is $\text{nacc}(C_\delta)$? It is simply $\{\alpha\in C_\delta\mid \sup(C_\delta\cap\alpha)<\alpha\}$. By convention, we shall write $C_\delta(j)$ for the unique element $\alpha\in C_\delta$ such that $\text{otp}(C_\delta\cap\alpha)=j$. Thus, in our case, $\text{nacc}(C_\delta)$ is nothing but $\{ C_\delta(j+1)\mid j<\kappa\}$.]

In this short blog post, we shall include a proof of the above result. The starting point of the proof is the following more elementary fact:

Claim 1. If $\kappa$ is a regular uncountable cardinal, then there exists a sequence $\langle C_\delta\mid\delta\in E^{\kappa^+}_\kappa\rangle$ such that $C_\delta$ is a club in $\delta$ of order-type $\kappa$, and for every club $D\subseteq \kappa^+$, there exists some $\delta\in E^{\kappa^+}_\kappa$ with $\sup(\text{nacc}(C_\delta)\cap D)=\delta.$

The Soukups have written up a friendly note with a proof of the preceding, and there is probably no point in repeating it here. Next, let $\langle C_\delta\mid \delta\in E^{\kappa^+}_\kappa\rangle$ be given by the preceding. Since $\kappa$ is accessible, let $\theta$ be the least cardinal to satisfy $2^\theta\ge\kappa$. In particular, $2^{<\theta}<\kappa$.
Let $\langle \eta_i\mid i<\kappa\rangle$ be a sequence of distinct elements in ${}^\theta2$. For every club $D\subseteq\kappa^+$ and $\delta\in E^{\kappa^+}_\kappa$, denote
$$D(\delta):=\{ \eta\in{}^{<\theta}2\mid \sup\{ j<\kappa\mid C_\delta(j+1)\in D, \eta\sqsubseteq \eta_{j+1}\}=\kappa\}.$$

Claim 2. There exists a club $D\subseteq\kappa^+$ such that for every club $E\subseteq D$, there exists $\delta\in E^{\kappa^+}_\kappa$ with $\sup(\text{nacc}(C_\delta)\cap E)=\delta$, and $E(\delta)=D(\delta)$.
Proof. Suppose not. We shall define sequences $\langle D_i\mid i<\kappa\rangle$ as follows.
Put $D_0:=\kappa^+$. Now suppose that $i<\kappa$ and $\langle D_j\mid j<i\rangle$ have already been defined. Write $\bar{D_i}:=\bigcap_{j<i}D_j$. By the indirect assumption, we may now find a club $D_i\subseteq\bar{D_i}$ such that for every $\delta\in E^{\kappa^+}_\kappa$, one of the following holds:

1. $\sup(\text{nacc}(C_\delta)\cap D_i)<\delta$;
2. $D_i(\delta)\neq \bar{D_i}(\delta)$.

This completes the description of the construction.

Put $D:=\bigcap_{i<\kappa}D_i$. Since $D$ is a club in $\kappa^+$, let us pick $\delta\in E^{\kappa^+}_\kappa$ such that $\sup(\text{nacc}(C_\delta)\cap D)=\delta$. In particular, $\sup(\text{nacc}(C_\delta)\cap D_i)=\delta$ for all $i<\kappa$. Consequently, $D_i(\delta)\neq\bar{D_i}(\delta)$ for every $i<\kappa$. So $\langle D_i(\delta)\mid i<\kappa\rangle$ is a strictly decreasing chain of subsets of ${}^{<\theta}2$, contradicting the fact that $2^{<\theta}<\kappa$. $\square$

Let $D$ be given by Fact 2. For every $\delta\in E^{\kappa^+}_\kappa$, if $D(\delta)$ contains an infinite antichain, fix such an antichain $\{ \rho_{\delta,n} \mid n<\omega\}$,  and let $$g_\delta(C_\delta(j)):=\max\{0,n\mid n<\omega, \rho_{\delta,n}\sqsubseteq \eta_{j}\}.$$If $D(\delta)$ does not contain an infinite antichain, let $g_\delta:C_\delta\rightarrow\{0\}$ be the constant function.

Claim 3. For every club $C\subseteq\kappa^+$, there exists some $\delta\in E^{\kappa^+}_\kappa$ such that $$\sup\{ \alpha\in\text{nacc}(C_\delta)\cap C\mid g_\delta(\alpha)=n\}=\delta$$ for all $n<\omega$.
Proof. Given a club $C$, put $E:=C\cap D$. Pick $\delta\in E^{\kappa^+}_\kappa$ with $\sup(\text{nacc}(C_\delta)\cap E)=\delta$, and $E(\delta)=D(\delta)$. Note that $E(\delta)$ contains an infinite antichain. Indeed, Erdos and Tarski proved that any poset that does not contain an infinite antichain is the union of finitely many updirected subposets. So if $E(\delta)$ does not contain an infinite antichain, then there exists a finite subset $J$ of $\kappa$ such that for every $\eta\in E(\delta)$, there exists $j\in J$ with $\eta\sqsubseteq\eta_j$, contradicting the fact that $\sup\{ j<\kappa\mid C_\delta(j+1)\in E\}=\kappa>2^{<\theta}$, and the definition of $E(\delta)$.
As $E(\delta)=D(\delta)$, we infer that $D(\delta)$ contains an infinite antichain, and hence $g_\delta$ is nonconstant, and satisfies in particular $$\sup\{ j<\kappa\mid C_\delta(j+1)\in E, \rho_{\delta,n}\sqsubseteq\eta_{j+1}\}=\kappa$$ for all $n<\omega$. So, $$\sup\{ \alpha\in\text{nacc}(C_\delta)\cap C\mid g_\delta(\alpha)=n\}=\delta$$ for all $n<\omega$. $\square$

We conclude this post with a few questions:

Question 1. Can we get more than $\omega$ many colors?
Answer. Often, yes. The simplest way is by using generalizations (due to Milner-Prikry and Todorcevic) to the above-mentioned theorem of Erdos and Tarski. Shelah has additional ways. Whether one can raise the number of colors all the way up to $\kappa$ appears to be open.

Question 2. What about the case $\kappa=\omega$?
Answer. This is a no go. E.g., under PFA. See here.

Question 3. What about the case $\kappa>\omega$ strongly inaccessible?
Answer. This appears to be the hardest problem in this particular vein. It seems conceivable that for $\kappa$ measurable, we get a negative answer.

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