6
$\begingroup$

This question is about the "old-fashioned" coherent sequences, in the style of Mitchell

Mitchell, W. (1974). Sets constructible from sequences of ultrafilters. Journal of Symbolic Logic, 39(1), 57-66. doi:10.2307/2272343

Recall that a coherent sequence of measures of length $\ell$ is given by a function $o^{\mathcal{U}}\colon \ell\to \mathrm{Ord}$ and a sequence $$\mathcal{U}=\langle U_\alpha^\beta; \alpha<\ell, \beta<o^{\mathcal{U}}(\alpha)\rangle$$ where each $U_\alpha^\beta$ is a normal measure on $\alpha$ and, if $j_\alpha^\beta$ is the associated embedding, $$j_\alpha^\beta(\mathcal{U})\upharpoonright (\alpha+1,0) = \mathcal{U} \upharpoonright (\alpha,\beta)$$ (the restrictions are taken in the sense of the lexicographic order).

Now let's suppose that $V=L[\mathcal{U}]$ believes that $\mathcal{U}$ is a coherent sequence of normal measures, and look at some initial segment $\mathcal{U}\upharpoonright (\delta,0)$ of $\mathcal{U}$. Can we conclude that $\mathcal{U}\upharpoonright (\delta,0)$ is coherent in $L[\mathcal{U}\upharpoonright (\delta,0)]$ (that $\mathcal{U}\upharpoonright (\delta,0)$ is coherent in $V$ follows from the definition, as pointed out in the comments)?

Typically I would want to argue for this by saying that all relevant functions with domain $\gamma<\delta$ in $L[\mathcal{U}]$ are constructed before the next measurable after $\gamma$ appears on the $\mathcal{U}$ sequence, so the ultrapowers of $L[\mathcal{U}]$ and $L[\mathcal{U}\upharpoonright(\delta,0)]$ match well enough for the coherence property to be satisfied, but from what I can tell, $L[\mathcal{U}]$ does not have any of the nice condensation properties that would allow an argument like that to work (and unless I'm mistaken, reals such as $0^\sharp$ will definitely only appear after a measurable stage).

I can also imagine an argument that takes $L[\mathcal{U}]$ and iterates the first measure on $\mathcal{U}$ at or after $\delta$ out of the universe. I think that the limit model of this iteration should just be $L[\mathcal{U}\upharpoonright (\delta,0)]$, and elementarity should show that this initial segment of $\mathcal{U}$ is coherent, but I am not confident.

$\endgroup$
5
  • 2
    $\begingroup$ You asked why an initial segment $\mathcal W$ of a coherent sequence $\mathcal U$ is coherent. I feel like I'm missing something, because I don't see why that doesn't follow from the definition. Granted the coherence of $\mathcal W$, assuming $V = L[\mathcal U]$ did not reach $o(\kappa) = \kappa^{++}$, you get coherence of $\mathcal W$ in $L[\mathcal W]$ from Mitchell's Corollary 2.11 in "Beginning inner model theory." $\endgroup$ May 7, 2021 at 0:21
  • $\begingroup$ Also the limit model of the iteration you mention is closed under sharps, so it is not $L[A]$ for any set $A$. It's more like $K^\text{DJ}[\mathcal U\restriction \delta]$, I think. $\endgroup$ May 7, 2021 at 2:37
  • $\begingroup$ @GabeGoldberg You're right, coherence of $\mathcal{W}$ in $V$ follows from the definitions, so the actual question is about $L[\mathcal{W}]$. I didn't think to look at the Handbook, so Corollary 2.11 is useful, although my application would definitely need $o(\kappa)=\kappa^{++}$ (in fact, I'm most interested in this question when the initial segment $\mathcal{W}$ is $\mathcal{U}\upharpoonright (\kappa+1)$, where $\kappa$ has $o(\kappa)=\kappa^{++}$. Do you know if anything can be said in this case? $\endgroup$ May 7, 2021 at 16:34
  • $\begingroup$ Not sure. In "Beginning inner model theory," page 1477, right before Proposition 3.11, Mitchell seems to imply that a coherent sequence $\mathcal U$ need not be coherent in $L[\mathcal U]$. It actually seems to me like the proof of Corollary 2.11 might give you a $\mu$-measurable rather than just $o(\kappa) = \kappa^{++}$, if that's useful. $\endgroup$ May 7, 2021 at 18:43
  • $\begingroup$ Benny Siskind pointed out to me that on page 12 of Steel's "Introduction to itreated ultrapowers" (math.cmu.edu/users/jcumming/Appalachian/steel_cmu_2015_files/…), it is stated that whether $\mathcal U$ is coherent in $L[\mathcal U]$ is a longstanding open question. $\endgroup$ May 8, 2021 at 5:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.