Let $I$ be a normal ideal on $P_{\kappa} (\lambda)$. Let $V$ denote our ground model. Now we force with the $I$-positive sets, and if $G$ is the resulting generic filter, it can be shown that $G$ is a $V$-ultrafilter on $P_{\kappa} (\lambda)$ extending the dual filter of $I$, which is normal if $I$ is normal.

We can build now the so called generic ultrapower, i.e. we construct inside $V[G]$, $Ult_{G} (V)$ the class of all functions in $V$ with domain $\kappa$ and the usual binary relation $\in^{\ast}$. We assume that the generic ultrapower is well-founded, so we identify it with its transitive collaps $M \cong Ult_{G} (V)$

My question now is: Is this $M$ closed under sequences from $V[G]$ of length $\lambda$, i.e. $M^{\lambda} \cap V[G] = M^{\lambda} \cap M$?

$M$ is closed under sequences from $V$ of length $\lambda$, this is clear to me but I don't have a good argument for the sequences of $V[G]$.

  • 2
    $\begingroup$ You want the functions used to build the ultrapower to have domain $P_\kappa(\lambda)$, not $\kappa$, since the measure lives on $P_\kappa(\lambda)$. $\endgroup$ – Joel David Hamkins Apr 20 '10 at 13:38
  • $\begingroup$ I added several tags. $\endgroup$ – Joel David Hamkins Apr 20 '10 at 14:40

If your ideal is normal, fine, precipitous and has the disjointing property (a consequence of saturation), then the answer is yes. As you likely know, you need more assumptions than you had stated, just in order to know that the ultrapower is well-founded. The difference in closure for the ultrapower that you mentioned appears to be related to the difference between having the disjointing property or not.

For a reference, I recommend Matt Foreman's chapter for the Handbook of Set Theory, which states the following theorem (it is Theorem 2.25 in the preliminary version I have here, but the published number may differ).

Theorem. Suppose I is a normal, fine, precipitous ideal on $Z\subset P(X)$, where $|X|=\lambda$. Let $G\subset P(Z)/I$ be generic, and $M$ the generic ultrapower of $V$ by $G$. Then $P(\lambda)\cap V\subset M$. Further, if $I$ has the disjointing property, then $M^\lambda\cap V[G]\subset M$.

Note that this theorem covers your case of $Z=P_\kappa(\lambda)$.

To prove the first part, you simply observe that $[id]$ represents $j " \lambda$, and then for any $A\subset\lambda$ you can get $j"A$ using the function $g(z)=z\cap A$. Now, from $j"\lambda$ and $j"A$ you can easily build $A$ in $M$.

For the second part, the part you were interested in, you use the disjointing property in order to know that a term for a $\lambda$-sequence of elements of $M$ can be transformed into a $\lambda$-sequence of terms in $M$. That is, if $\langle\dot a_\alpha :\alpha<\lambda\rangle$ is a $\lambda$-sequence of terms for objects in $M$, then disjointing allows us to find in $V$ a sequence of functions $\vec g = \langle g_\alpha: \alpha<\lambda\rangle$ such that $[g_\alpha]^G = \dot a_\alpha^G$. From this, it follows that the function $g(z) = \langle g_\alpha(z) | \alpha\in z\rangle$ represents $j(\vec g)(j"\lambda)$, which is $\langle j(g_\alpha)_\beta(j"\lambda) | \beta\in j"\lambda\rangle$, from which we can construct $\langle j(g_\alpha)(j"\lambda) | \alpha <\lambda\rangle$, which is the desired $\lambda$-sequence.

  • $\begingroup$ This is exactly what I was looking for. Thank you very much! $\endgroup$ – Stefan Hoffelner Apr 20 '10 at 15:00

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