$(<\kappa)$-closure for Prikry forcing - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:51:33Z http://mathoverflow.net/feeds/question/99691 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99691/kappa-closure-for-prikry-forcing $(<\kappa)$-closure for Prikry forcing um Haitham 2012-06-15T10:36:14Z 2012-06-15T10:45:38Z <p>A p.o. $\mathcal{P}$ is $(&lt;\kappa)$-closure, if for every decreasing sequence of $&lt;\kappa$ conditions in the forcing $p_0 \geq p_1 \geq \cdots$, there is a condition that is below all of them. Prikry forcing $\mathcal{P}$ is the set of pairs $(s,A)$ where s is a finite subset of a fixed measurable cardinal $\kappa$, and $A$ is an element of a fixed filter $D$ on $\kappa$. A condition $(s,A)$ is stronger than $(t, B)$ if $t$ is an initial segment of $s$ and $A \cup (s-t) \subseteq B$. Why doesn't Prikry forcing have this property? Could someone help me out with this?</p> http://mathoverflow.net/questions/99691/kappa-closure-for-prikry-forcing/99692#99692 Answer by Noah S for $(<\kappa)$-closure for Prikry forcing Noah S 2012-06-15T10:45:38Z 2012-06-15T10:45:38Z <p>Consider the conditions $(s_i, \kappa)$ where $s_i={0, . . . , i}$ for $i\in\omega$. This is a sequence of $\omega$-many conditions, and clearly $(s_{i+1}, \kappa)\le (s_i, \kappa)$, but there is obviously no single condition below all the $(s_i, \kappa)$: were $(p, A)$ such a condition, then $p$ would have to be infinite, which contradicts the definition of condition.</p> <p>Note that we could alter the forcing to consider conditions of the form $(p, A)$, with $A$ as above, and $p$ a subset of $\kappa$ of size $&lt;\kappa$. This forcing would then be $&lt;\kappa$-closed (depending, I guess, on the filter $D$), but this is a very different forcing.</p>