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.

 

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12 Responses to The order-type of clubs in a square sequence

  1. Peter says:

    Sam just pointed me to your post and the fact that you’re actually posing an open question!

    I think that is an awesome idea!

    I’m wondering if other people have had the same problem as I did. Namely, I didn’t notice this idea until Sam pointed it out :(

    Do you plan another post? Maybe more targeted at a Polymath-esque response? Maybe announce it on the mailing list?

       0 likes

    • saf says:

      Hi Peter,
      Just notice that the “open problems” tag has been assigned to this post. In fact, it is most likely to be assigned to the majority of my future posts. The point is that once I get myself into writing down a clean proof of a (new or existing) theorem, then, often, along the way I would become familier with some of the surrounding open problems, and would use the opportunity to populate it.

      Note, however, that experience shows that people do not solve (“real”) open problems just a few days after becoming aware of it. Such problems requires an honest attention. Hence, my modest hopes boils down to the following scenario: A person becomes interested in a subject, he’ll find a post with some related proofs and some open problems, and then, may decide to dedicate some time to solve it.

      If you insist that I announce some open problems, then I invite anyone to have a look at this paper of mine, where I collect 23 open problems.

         0 likes

      • Peter says:

        Hm… the tag is at the bottom of the post, thinly printed.

        I guess what I’m trying to say is that a short sentence before jumping into the definitions could have helped me realize that there’s an open invitation in the post.

        This is not a criticism, just an observation — we talked about similar ideas in the hangout, other people also want to discuss open questions.

           0 likes

        • saf says:

          It looks like you overlooked the sentence “the following question has been overlooked” ;).
          Anyway, you are of course right, and here are the changes I made:
          1. highlighted the sentence with “the following question has been overlooked”, and assigned a different color to “As you can see, the original question hasn’t been answered yet…”
          2. modified functions.php so that the assigned categories will appear at the top of the page. (and moved “open problems” from tag to category).

             0 likes

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  4. saf says:

    I was thrilled to learn today from Peter Komjath that he had solved this problem over 25 years ago (in the following paper)!

       0 likes

    • Peter says:

      Awesome! and frustrating at the same time. We need better search methods…

         0 likes

      • saf says:

        Have you noticed the useless review? “the main result of the paper is a complete answer (under certain set-theoretic hypotheses) to the analogous problem..”.
        No wonder that none of my searches yielded this paper..

           0 likes

    • saf says:

      Komjath’s method yields a $\square_\lambda$-sequence $\langle C_\alpha\mid\alpha< \lambda^+\rangle$ such that stationarily many $\alpha<\lambda^+$ has $\text{otp}(C_\alpha)=\lambda$. However, at the same time, there exists a club $E\subseteq\lambda^+$ such that for every $\alpha<\lambda$: $$\text{otp}(C_\alpha)=\lambda\Rightarrow\text{otp}(C_\alpha\cap E)=\text{cf}(\lambda).$$ So, here is a corrected version of the original question: Suppose that $\square_\lambda$ holds for a given singular cardinal $\lambda$. Does it follow that there exists a $\square_\lambda$-sequence $\langle C_\alpha\mid \alpha<\lambda^+\rangle$ with the additional property that for every club $E\subseteq\lambda^+$, there exists some $\alpha<\lambda^+$ for which $\text{otp}(C_\alpha\cap E)=\lambda$?

         1 likes

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