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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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Noah's answer is the most direct, hence best, but let me point out another way to see this. A $<\kappa$-closed forcing adds no new $<\kappa$-sequences of ordinals. But Prikry forcing adds (not surprisingly) a Prikry sequence, which is a new $\omega$-sequence of ordinals. (If you take the natural name for the Prikry sequence, apply the proof that it wouldn't be new if the forcing were $<\kappa$-closed, and make the simplest choices wherever an arbitrary choice is needed in the proof, you'd recover Noah's counterexample.) – Andreas Blass Jun 15 '12 at 12:49
up vote 2 down vote accepted

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.

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 $<\kappa$. This forcing would then be $<\kappa$-closed (depending, I guess, on the filter $D$), but this is a very different forcing.

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Thank you for your answer. – um Haitham Jun 15 '12 at 11:04
You can accept an answer by clicking on the check mark next to the answer. – Noah Schweber Jun 21 '12 at 2:15

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