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A p.o. $\mathcal{P}$ is $(<\kappa)$-closure, if for every decreasing sequence of $<\kappa$ 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?

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You should mention that you’ve posted the same question to MathOverflow. – Brian M. Scott Jun 15 '12 at 10:44
Thank you for your comments, but really I was in hurry and I need the answer very quickly. – um Haitham Jun 15 '12 at 11:12
up vote 2 down vote accepted

I believe that in Prikry forcing the filter $\mathscr{D}$ is generally taken to be a normal ultrafilter on $\kappa$, though it doesn’t matter here: this partial order clearly isn’t countably closed, since for a fixed $A\in\mathscr{D}$ no strictly descending chain $\langle s_0,A\rangle > \langle s_1,A\rangle > \langle s_2,A\rangle >\ldots$ can be extended: $\bigcup_n s_n$ is infinite.

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