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Let $M$ be a model of $\sf ZFC$ in which $\kappa$ is a measurable cardinal, and $\cal U$ is a normal measure on $\kappa$. We can define the Prikry forcing (the most simple one) as the poset: $$\Bbb P=\left\{(p,A)\mid p\in[\kappa]^{<\omega}, A\in\mathcal U, \max p<\min A\right\}.$$ We also define the order, $(q,B)$ is stronger than $(p,A)$ if $q$ is an end-extension of $p$, $B\subseteq A$ and $q\setminus p\subseteq A$. The generic $G$ adds an $\omega$-sequence, $x_G=\bigcup\{q\mid\exists A:(q,A)\in G\}$ which we call a Prikry sequence. We can show that $x\subseteq^* A$, for all $A\in\cal U$, that is $x\setminus A$ is finite.

In the other direction, if $M$ is a model of $\sf ZFC$ in which $\kappa$ is measurable, and $M\subseteq V$, such that in $V$ we have some $x\in[\kappa]^\omega$ such that for some $\cal U$ in $M$ which is a normal measure on $\kappa$, $x\subseteq^*A$ for all $A\in\cal U$, then $x=x_G$ for some generic filter $G$ over the Prikry forcing defined from $\cal U$.

One conclusion from this last statement is that if $x$ is a Prirky sequence for $\kappa$, and $y\subseteq x$ is infinite then $y$ is also a Prikry sequence. This raises the following question.

Suppose that $x$ is a Prikry sequence for $\kappa$, and $y\subseteq x$ is infinite. Is $x$ generic over $y$? More generally, if $y,w\subseteq x$ are disjoint infinite subsets, are they pairwise generic? What about the weaker condition $y\cap w$ being finite?

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3 Answers 3

up vote 14 down vote accepted

There is a theorem of Gitik, Kanovei and Koepke that characterises the degrees of constructibility in $M[G]$ where $G$ is prikry generic over the model $M$: they are isomorphic to the $P(\omega)/Fin$ of $M[G]$.

That is they show (letting $M, G$ be as above with $x=G_C$ the actual prikry sequence):

$$∀Z ∈ M[x] ∃y ⊆ x, y ∈ M[x] \wedge M[Z] = M[y].$$

They identify a $\kappa^+$-c.c quotient p.o. $P/y$ needed to see that if $y\subseteq x$ is an infinite subset of the Prikry generic $x$, then $x=G_C$ is $M[y]$-generic for $P/y$ and $M[G_C]$ is also $M[y][G_C]$.

This in short, I think answers your questions; (at least the answer to Q1 is "Yes". I am not quite sure what you mean by "pairwise generic" for the case of $y,w$ disjoint - say the Evens and Odds in the sequence $x$ - but anyway, yes in the above sense: $w$ is generic for a quotient forcing over $M[y]$). (This includes your `weaker version'.) I don't know if the result is published, but there are slides with quite a bit of detail of Koepke at:

http://www.math.uni-bonn.de/people/koepke/Talks/Submodels_of_Prikry_generic_extensions.pdf

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Thank you for the answer and the reference Philip. When I say that $y,w$ are pairwise generic I mean that neither is definable from the other, so $y\notin M[w]$ and $w\notin M[y]$ either. –  Asaf Karagila Aug 6 '13 at 15:59
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Then "Yes" once more! –  Philip Welch Aug 6 '13 at 16:03
    
Then "Thanks" once more! :-) –  Asaf Karagila Aug 6 '13 at 18:42
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I saw Kanovei talk about this at Caltech a few years ago. There are two notes, but neither seems (yet) published, both accessible from Gitik's page: "Intermediate models of Prikry generic extensions" by Gitik, Kanovei, Koepke, and "A remark on subforcings of the Prikry forcing", by Gitik. In the latter, he shows that any nontrivial subfrocing of Prikry forcing is isomorphic to Prikry forcing with the same ultrafilter. In the former, they prove the result Philip described. –  Andres Caicedo Aug 6 '13 at 19:01
    
@Andres: That's great! I'll go take a look! –  Asaf Karagila Aug 6 '13 at 19:23

It is possible to define a version of Prikry forcing which gives a minimal extension of the universe. See the paper

"A minimal Prikry-type forcing for singularizing a measurable cardinal, J. Symbolic Logic Volume 78, Issue 1 (2013), 85-100."

by Peter Koepke, Karen Räsch, and Philipp Schlicht.

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There have been numerous generalizations of Prikry forcing, such as those using supercompactness-type measures or trees of conditions with measure-one branching, and these also support analogues of your question. The corresponding generalizations of the Mathias property and the maximality properties are part of the subject of my paper, Canonical seeds and Prikry trees, JSL 62(2):373-396, 1997, which grew out of one of the chapters of my dissertation.

Meanwhile, that paper also explains that there is a negative answer to your question in the version of Prikry forcing where one does not assume the measure is normal. For example, it will be negative if one uses the product of a normal measure with itself.

(Indeed, this question was the initial problem that Woodin had put to me for this part of my dissertation, and he was surprised that it came out negatively, since his instructions to me were to prove the positive result!)

Basically, if the measure $\cal U$ is a product measure $\cal V\times\cal V$ (which is never Rudin-Kiesler minimal and therefore not isomorphic to a normal measure), then there will be disjoint Prikry sequences $x$ and $y$ which construct each other. You can think of it like this: iterating the ultrapower embedding of $\cal U$ is just like iterating $\cal V$, but taking two steps at once. So we can build two disjoint $\cal U$ Prikry sequences over $M_\omega$, by taking pairs of the critical sequence $x=\{ \langle\kappa_{2n},\kappa_{2n+1}\rangle\mid n\in \omega\}$ and $y=\{\langle\kappa_{2n+1},\kappa_{2n+2}\rangle\mid n\in \omega\}$. These are disjoint and both are Prikry for $j_{0,\omega}(\cal U)$-Prikry forcing over $M_\omega$, but they easily construct each other by disassembling the pairs.

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I was amused by the remark about Woodin. Thank you very much for the additional remarks. The question itself was something I asked in an informal Prikry forcing mini-course we're having at HUJI over the summer. It's good to know that the normality of the measure is important. –  Asaf Karagila Aug 7 '13 at 2:22
    
The original question was related to generalized Prikry forcing issues in David Law's dissertation. –  Joel David Hamkins Aug 7 '13 at 3:01
    
Sorry, it was George Kafkoulis's dissertation, not David Law's. –  Joel David Hamkins May 20 at 3:33

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