failure of $\square(\kappa)$ at an inaccessible $\kappa$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:43:11Z http://mathoverflow.net/feeds/question/47779 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47779/failure-of-square-kappa-at-an-inaccessible-kappa failure of $\square(\kappa)$ at an inaccessible $\kappa$ Mohammad 2010-11-30T10:13:14Z 2011-03-10T06:33:10Z <p>How can we force the failure of $\square(\kappa)$ at an inaccessible $\kappa$, where $\square(\kappa)$ is defined as follows: There is a sequence $(C_i:i&lt; \kappa)$ such that:</p> <p>(1) <code>$C_{i+1} = \{i\}$</code> and $C_i$ is closed and cofinal in $i$ if $i$ is a limit ordinal.</p> <p>(2) If $i$ is a limit point of $C_j$, then $C_i = C_j \cap i$.</p> <p>(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.</p> http://mathoverflow.net/questions/47779/failure-of-square-kappa-at-an-inaccessible-kappa/47782#47782 Answer by François G. Dorais for failure of $\square(\kappa)$ at an inaccessible $\kappa$ François G. Dorais 2010-11-30T11:06:10Z 2010-11-30T13:07:25Z <p>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.</p>