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I have a question about the definition of $K^c$ for sequences of measures in Mitchell's "The Covering Lemma" chapter in the Handbook. I will give the definition he gives here, but for reference, this is Definition 4.1 on page 1559 of the Handbook. Please note that I have fixed two typos in clause 2.

Either the core model $K$, or the countably complete core model $K^c$, are defined as $L[\mathcal{U}]$ where the sequence $\mathcal{U}$ is defined by recursion on $\gamma$ as follows. Assume that $\mathcal{U}\upharpoonright\gamma$ has already been defined:

  1. If there is a mouse $M=J_{\gamma'}[\mathcal{U}']$ such that $\mathcal{U}'\upharpoonright\gamma=\mathcal{U}\upharpoonright\gamma$, the projectum of $M$ is smaller than $\gamma$, and no measure in $\mathcal{U}'-\mathcal{U}$ is full in $M$, then set $\mathcal{U}\upharpoonright\gamma'=\mathcal{U}'$.

  2. If there is no mouse as in clause 1,and if $J_\gamma[\mathcal{U}\upharpoonright\gamma]\vDash\gamma=\kappa^{++}$ for some $\kappa<\gamma$ such that there is a $J_\gamma[\mathcal{U}\upharpoonright\gamma]$-ultrafilter $U$ on $\kappa$ with $i^U(\mathcal{U}\upharpoonright\gamma)\upharpoonright\gamma+1=\mathcal{U}\upharpoonright\gamma$, then set $\mathcal{U}_\gamma=U$, provided it satisfies an iterability condition depending on which model is being constructed:

a) For the model $K^c$, the ultrafilter $U$ is added to the sequence only if $U$ is countably complete and $\mathrm{cf}(\mathrm{crit}(U))=\omega_1$.

b) For the true core model $K$, the ultrafilter $U$ is added to the sequence only if $\mathrm{Ult}(L[\mathcal{W}],U)$ is well-founded for every iterable inner model $L[\mathcal{W}]$ such that $\mathcal{W}\upharpoonright\gamma=\mathcal{U}$.

My question is in clause 2a. Why must we have $\mathrm{cf}(\mathrm{crit}(U))=\omega_1$? Certainly, you cannot have it have countable cofinality, as the mouse would not be countable certified. Why can we not have $\mathrm{cf}(\mathrm{crit}(U))\geq\omega_2$?

Zeman's book defines $K^c$ below $0^{sword}$ and below $0^¶$, and in both cases he does not have this condition.

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    $\begingroup$ Mitchell wants the class of fixed points of an iteration of $K^c$ to be thick; see Proposition 4.8. $\endgroup$
    – Gabe Goldberg
    Commented Jul 20 at 22:14
  • $\begingroup$ @GabeGoldberg Thanks for your reply, Gabe. As a follow-up question, is there a reason this is not done in Zeman's book? Is it a matter of convention whether or not $K^c$ only has measurables of cofinality $\omega_1$? $\endgroup$
    – Connor W
    Commented Jul 21 at 2:47
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    $\begingroup$ I don't know, it's been 10 years since I read Zeman's book. Steel makes a similar move when he defines $K^c$ in The Core Model Iterability Problem; look at the conditions for adding an extender to the $K^c$ sequence (Case 1 on page 6). Again, this is to ensure that there is a thick set of ordinals that are not measurable; look at property (iv) on page 9. In the end, $K^c$ is constructed in order to extract $K$, so I guess the certification it is a matter of convention as long as it does not affect $K$. $\endgroup$
    – Gabe Goldberg
    Commented Jul 21 at 15:27

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