The order-type of clubs in a square sequence

Recall Jensen’s notion of square:

Definition (Jensen): For an infinite cardinal $\lambda$, $\square_\lambda$ asserts the existence of a sequence $\overrightarrow C=\left\langle C_\alpha\mid\alpha\in\text{acc}(\lambda^+)\right\rangle$ such that for every limit $\alpha<\lambda^+$:

• $C_\alpha$ is a club subset of $\alpha$ of order-type $\le\lambda$;
• if $\beta\in\text{acc}(C_\alpha)$, then $C_\beta=C_\alpha\cap\beta$.

Now, consider the set $S(\overrightarrow C):=\{ \alpha<\lambda^+\mid \text{otp}(C_\alpha)=\lambda\}$.

If $\lambda$ is a regular cardinal, then $S(\overrightarrow C)=\{\alpha<\lambda^+\mid\text{cf}(\alpha)=\lambda\}$, and in particular, it is stationary. Next, let us ask: what happens in the case that $\lambda$ is singular?
Solovay shows in Section 3 of this paper that a $\square_\lambda$-sequence $\overrightarrow C$ with a stationary $S(\overrightarrow C)$ exists in L for all singular cardinals $\lambda$. On the other hand, Jensen proved that if $\square_\lambda$ holds for a given singular caridnal $\lambda$, then this may be witnessed by a sequence $\overrightarrow C$ for which $S(\overrightarrow C)$ is empty(!) [a proof of the latter may be found in the lecture notes of Magidor’s talks at the Appalachian set theory]. Nevertheless, to the best of my knowledge, the following question has been overlooked:

Question: Suppose that $\square_\lambda$ holds for a given singular cardinal $\lambda$.
May this be witnessed by a sequence $\overrightarrow C$ for which $S(\overrightarrow C)$ is stationary?

I’m holding this peripheral question at the back of my head for a long time now, without much of success. I once approached Shelah after a seminar talk, and asked about this. After a short while, Shelah concluded that the answer is positive, and outlined an argument (unfortunately, no yellow notes were provided). On my way back home, I digested Shelah’s argument, and got myself convinced that the following should be true:

Proposition (Shelah). Suppose that $\square_\lambda$ holds for a given singular cardinal $\lambda$. Then this may be witnessed by a sequence $\overrightarrow C$ for which $S(\overrightarrow C)$ is cofinal in $\lambda^+$.

Once convinced, I came up with a completely different proof, which is, in fact, a very mild adaptation of existing arguments. Here goes:

Proof of Proposition. We commence with defining a poset $\mathbb P=\langle P_\lambda,\le\rangle$ as follows. $p\in P_\lambda$ iff all of the following holds:
(a) $\text{dom}(p)=(\beta+1)\cap\text{acc}(\lambda^+)$ for some $\beta\in\text{acc}(\lambda^+)$;
(b) $p(\alpha)$ is a club subset of $\alpha$ of order-type $\le\lambda$, for all $\alpha\in\text{dom}(p)$;
(c) if $\alpha\in\text{dom}(p)$ and $\beta\in\text{acc}(p(\alpha))$, then $p(\beta)=p(\alpha)\cap\beta$.
Then, let $\le\:=\ \subseteq\restriction (P_\lambda)^2$.

Evidently, $\mathbb P$ is a forcing notion that introduces a (generic) $\square_\lambda$-sequence.
Next, for every ordinal $\alpha<\lambda^+$, let $$D_\alpha:=\{ p\in P_\lambda\mid \exists\beta\in(\text{dom}(p)\setminus\alpha)(\text{otp}(p(\beta))=\lambda)\}.$$

Sublclaim.  $D_\alpha$ is dense in $\mathbb P$ for all $\alpha<\lambda^+$.

Proof of Subclaim. Recall that we work under the assumption that $\square_\lambda$ holds, so let us pick a witness $\langle C_\delta\mid\delta<\lambda^+\rangle$.

Fix a condition $p\in P_\lambda$. Let $\beta>(\alpha\cup\max(\text{dom}(p)))$ be a large enough ordinal in $\lambda^+$, which is divisible by $\lambda$.
Let $E_\beta$ be some club subset of $\beta$ of order-type $\lambda$ such that $\min(E_\beta)>\text{dom}(p)$.
Define $q$ as follows:

• $\text{dom}(q):=(\beta+1)\cap\text{acc}(\lambda^+)$;
• for every $\delta\in\text{dom}(q)$, let
  $$q(\delta):=\begin{cases}
  p(\delta),&\delta\in\text{dom}(p)\\
  E_\beta\cap\delta,&\delta=\sup(E_\beta\cap\delta)\\
  C_\delta\setminus\sup(E_\beta\cap\delta),&\text{otherwise}
  \end{cases}.$$

Let us verify that $q$ is  a legitimate condition of $\mathbb P$.
Evidently, $q(\delta)$ is a club subset of $\delta$ of order-type $\le\lambda$, for all $\delta\in\text{dom}(q)$.
Next, suppose that $\delta\in \text{dom}(q)$ and $\gamma\in\text{acc}(q(\delta)$. We consider three cases:

• if $\delta\in\text{dom}(p)$, then $\gamma\in\text{dom}(p)$ and $q(\delta)\cap\gamma=p(\delta)\cap\gamma=p(\gamma)=q(\delta)$;
• if $\delta=\sup(E_\beta\cap\delta)$, then $q(\delta)=E_\beta\cap\delta$, and hence $\sup(E_\beta\cap\gamma)=\gamma$.
  So, $q(\beta)=E_\beta\cap\gamma=q(\delta)\cap\beta$.
• if $\delta\not\in\text{dom}(p)$ and $\delta>\sup(E_\beta\cap\delta)$, then $\delta>\gamma>\sup(E_\beta\cap\delta)$,and hence $\sup(E_\beta\cap\gamma)=\sup(E_\beta\cap\delta)<\gamma$.
  So $q(\gamma)=C_\gamma\setminus\sup(E_\beta\cap\delta)=C_\delta\cap\gamma\setminus\sup(E_\beta\cap\delta)=q(\delta)\cap\gamma$.

So, $q\in\mathbb P$ and $q\ge p$.
As $\beta>\alpha$ and $\text{otp}(q(\beta))=\lambda$, we conclude that $q\in D_\alpha$. This completes the of proof of the subclaim.

Next, let us recall the notion of strategic closure.

Definition. Given an ordinal $\alpha$, and a poset $\mathbb Q$, one defines a two-player game $G^{II}_\alpha(\mathbb Q)$ as follows:
Players $I$ and $II$ take turns to write down the entries of an increasing sequence of conditions in $\mathbb Q$, $\langle p_\beta\mid\beta<\alpha\rangle$, whereas $I$ plays at odd stages and $II$ plays at even stages (including all limit stages). If a play reaches a stage where player $II$ cannot move, then player $I$ wins, otherwise, player $II$ wins.

We say that the poset $\mathbb Q$ is $\alpha$-strategically closed iff player $II$ has a winning strategy in the game $G^{II}_\alpha(\mathbb Q)$.

Back to our business, let us point out the following:

Sublclaim (folklore). $\mathbb P$ is $\lambda^+$-strategically closed.

Proof of Subclaim. By a theorem of Velleman, if $\square_\kappa$ holds for a given infinite cardinal $\kappa$, then any poset which is $\sigma$-closed and $(\kappa+1)$-strategically closed, is, in fact, $\kappa^+$-strategically closed. As our $\mathbb P$ is clearly $\sigma$-closed, the hypothesis concerning $\square_\lambda$, reduces our task to showing that $II$ has a winning strategy in the game $G^{II}_{\lambda+1}(\mathbb P)$.
Here goes.

Let $C$ be some club subset of $\lambda$, with $\text{otp}(C)=\text{cf}(\lambda)$.
Suppose that $\alpha<\lambda+1$, and that $\langle p_\beta\mid\beta<\alpha\rangle$ is a sequence of conditions that has already been produced during the game. For all $\beta<\alpha$, let $\gamma_\beta$ be the unique limit ordinal such that $\text{dom}(p_\beta)=(\gamma_\beta+1)\cap\text{acc}(\lambda^+)$.
We now consider all cases in which player $II$ has to make a move.

(1) if $\alpha=2\beta+2$ for some ordinal $\beta$, then define $p_\alpha$ as follows:
(-) $\text{dom}(p_\alpha):=\gamma_{2\beta+1}+\omega+1$;
(-) $p_\alpha\restriction\text{dom}(p_{2\beta+1})=p_{2\beta+1}$;
(-) $p_\alpha(\max(\text{dom}(p_\alpha)))=\{\gamma_{2\beta+1}+n\mid n<\omega\}$;
(2) if $\alpha$ is a limit ordinal, and $\sup(C\cap\alpha)<\alpha$, then let:
(-) $\text{dom}(p_\alpha):=((\sup_{\beta<\alpha}\gamma_\beta)+1)\cap\text{acc}(\lambda^+)$
(-) $p_\alpha\restriction\text{dom}(p_{\beta})=p_{\beta}$ for all $\beta<\alpha$;
(-) $p_\alpha(\max(\text{dom}(p_\alpha)))=\{\gamma_\beta\mid \sup(C\cap\alpha)<\beta<\alpha\}$.
(3) if $\alpha$ is a limit ordinal, and $\sup(C\cap\alpha)=\alpha$, then let:
(-) $\text{dom}(p_\alpha):=((\sup_{\beta<\alpha}\gamma_\beta)+1)\cap\text{acc}(\lambda^+)$
(-) $p_\alpha\restriction\text{dom}(p_{\beta})=p_{\beta}$ for all $\beta<\alpha$;
(-) $p_\alpha(\max(\text{dom}(p_\alpha)))=\{\gamma_\beta\mid \beta\in C\cap\alpha\}$.

Denote $d:=p_\alpha(\max(\text{dom}(p_\alpha)))$.
Note that in either of the cases, $\text{otp}(d)<\lambda$, and hence $p_\alpha$ is indeed a legitimate condition of $\mathbb P$. Isn’t that lovely?

Next, suppose that $\delta\in\text{acc}(d)$, and let us verify that $p(\delta)=d\cap\delta$.
Let $\beta<\alpha$ be such that $\delta=\gamma_\beta$. Then $\beta$ is a limit ordinal, and there are two cases to consider.
(2) If $\sup(C\cap\alpha)<\alpha$, then $\sup(C\cap\alpha)<\beta<\alpha$,
and hence $\sup(C\cap\alpha)=\sup(C\cap\beta)<\beta$, and
$$p_\alpha(\delta)=p_\beta(\delta)=\{\gamma_{\beta’}\mid \sup(C\cap\beta)<\beta'<\beta\}=d\cap\delta.$$
(3) If $\sup(C\cap\alpha)=\alpha$, then the fact that $\gamma_\beta\in\text{acc}(d)$ entails that $\sup(C\cap\beta)=\beta$, and hence
$$p_\alpha(\delta)=p_\beta(\delta)=\{\gamma_{\beta’}\mid \beta’\in C\cap\beta\}=d\cap\delta.$$

Thus, altogether, we conclude that $\mathbb P$ is indeed $(\lambda+1)$-strategically closed, and infer that $\mathbb P$ is $\lambda^+$-strategically closed. This completes the of proof of the subclaim.

To complete our proof, let $\langle p_\alpha\mid\alpha<\lambda^+\rangle$ be the increasing sequence of conditions produced throughout the game $G^{II}_{\lambda^+}(\mathbb P)$ in which player $I$ chooses at odd stage $\alpha$ a condition  $p_\alpha\in D_\alpha$, and player II obeys a winning a strategy. Nota bene that we didn’t pass to a generic extension — the existence of our sequence is a consequence of the preceding subclaims.

Put $C_\alpha:=p_{\alpha+1}(\alpha)$ for all limit $\alpha<\lambda^+$. Then, $\overrightarrow C=\left\langle C_\alpha\mid\alpha\in\text{acc}(\lambda^+)\right\rangle$ is a $\square_\lambda$-sequence, and $S(\overrightarrow C)$ is cofinal in $\lambda^+$. This completes the proof of the Proposition.

As you can see, the original question hasn’t been answered yet, hence I look forward to your inputs.

Further reading:

1. S. Todorcevic, Special square sequences, Proc. Amer. Math. Soc. 105(1), 199–205, 1989.
2. Y. Yoshinobu, Directive trees and games on posets, Proc. Amer. Math. Soc. 130(5), 1477–1485, 2002.