failure of $\square(\kappa)$ at an inaccessible $\kappa$

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.

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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. – Andres Caicedo Nov 30 2010 at 14:39

1 Answer

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. – Andres Caicedo Nov 30 2010 at 14:36
Oh, I see, the previous version was closed, ok. – Andres Caicedo Nov 30 2010 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 $\kappa$. – Andres Caicedo Nov 30 2010 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 2010 at 14:43