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How can we force the failure of $\square(\kappa)$ at an inaccessible $\kappa$, where $\square(\kappa)$ is defined as follows: There is a sequence $(C_i:i< \kappa)$ such that:

(1) $C_{i+1} = \{i\}$ and $C_i$ is closed and cofinal in $i$ if $i$ is a limit ordinal.

(2) If $i$ is a limit point of $C_j$, then $C_i = C_j \cap i$.

(3) There is no club $C$ (a subset of $\kappa$) such that for all limit points $i$ in $C$ the equality $C_i= C \cap i$ holds.

Update. By a recent result of Magidor and Vaananen ( On Löwenheim-Skolem-Tarski numbers for extensions of first order logic), it is consistent, relative to the existence of a supercompact cardinal, that $\square(\kappa)$ fails at the least strongly inaccessible cardinal.

Is it possible to reduce the consistency strength to a weakly compact cardinal or even a measurable cardinal?

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@Mohammad: There is no need to keep reposting your problem. People may be thinking about it. If it is difficult, it takes time. It may help if you mention what in the literature you have consulted already. – Andrés E. Caicedo Nov 30 '10 at 14:39
up vote 5 down vote accepted

The consistency strength of the failure of $\square(\kappa)$ for non-weakly Mahlo inaccessible cardinal $\kappa$ (in particular, for the first strongly inaccessible) is higher than weakly compact. For example it implies that $0^\#$ exists. We know that global square holds in $L$, i.e. there is a sequence $\langle C_\alpha | \alpha \text{ is singular limit ordinal} \rangle$ with the properties: $\forall \alpha,\, \text{otp } C_\alpha < \alpha$ and for every accumulation point $\beta$ of $C_\alpha$, $C_\alpha \cap \beta = C_\beta$.

In $V$, there is a club $D\subset \kappa$ through the singular cardinals below $\kappa$. If $0^\#$ doesn't exists, the covering theorem will imply that those cardinals are singular in $L$ as well, and therefore $C_\alpha$ is defined for every $\alpha \in D$.

Now we can define a $\square(\kappa)$ sequence, $\langle E_\alpha | \alpha < \kappa \rangle$ in $V$ as follow:

  • For $\alpha \in \text{acc } D$, $E_\alpha = C_\alpha \cap D$ (unless this set is bounded below $\alpha$. In this case, $\text{cf }\alpha = \omega$ and we can choose $E_\alpha$ to be arbitrary $\omega$ sequence cofinal in $\alpha$).
  • For $\alpha \notin \text{acc }D$, $E_\alpha = \alpha \setminus (\max D\cap \alpha)$

It's easy to see that $\langle E_\alpha | \alpha < \kappa\rangle$ is coherent (since the global square is coherent) and thread-less (since every thread meets $D$ club many times, but $\text{otp }E_\alpha < \alpha$ for every $\alpha \in D$).

Edit: As Mohammad noted, we can get better lower bound by using more appropriate inner model. We need to violate global square on the singular cardinals. By $\square$ On the singular cardinals (Theorem 1), we need at least many measurable cardinals of uncountable Mitchell order in order to get a model without global square on the singular cardinals.

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Thanks for your nice answer. The same argument applied to the Dodd-Jensen core model shows that there should be an inner model with a measurable cardinal. – Mohammad Golshani Nov 15 '13 at 3:14
Stable links $\gg$ links. – Asaf Karagila Nov 27 '13 at 19:51

In general, one cannot force the failure of $\square(\kappa)$ at a fixed cardinal $\kappa$. Indeed, if $\kappa$ is any regular uncountable cardinal which is not weakly compact in $L$, then there is a nontrivial $\square(\kappa)$ sequence which is moreover constructible. The fact that $\kappa$ is not weakly compact in $L$ cannot be destroyed by forcing. On the other hand, $\square(\kappa)$ always fails at a weakly compact cardinal.

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François, this is not the intended interpretation of the question. You could start with a supercompact and at the end just preserve its inaccessibility, and that would be fine. There is some literature of forcing $\lnot\square(\kappa)$ for $\kappa=\lambda^+$ a successor (it is harder than for $\lnot\square_{\lambda}$ and seems to involve serious large cardinals), but I do not recall any explicit approach to the question as asked. It seems very difficult, but there may be a trick hiding somewhere. This is a duplicate, by the way. – Andrés E. Caicedo Nov 30 '10 at 14:36
Oh, I see, the previous version was closed, ok. – Andrés E. Caicedo Nov 30 '10 at 14:38
@François : For example, PFA implies $\lnot\square(\kappa)$ for any $\kappa>\omega_1$, so one way of doing what is asked is to force PFA (or the P-ideal dichotomy, or even MRP) with a supercompact below your inaccessible. But the question seems rather meant to be in a context where there are no supercompacts below $\kappa$. – Andrés E. Caicedo Nov 30 '10 at 14:43
Andres, your interpretation makes some sense, but I couldn't see a good formulation along these lines that excludes the trivial forcing... (I closed the other question as a duplicate.) – François G. Dorais Nov 30 '10 at 14:43

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