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I am cross-posting this question from MSE, where I asked it about $3$ months ago and I decided to ask it here as well.


This question of mine arises from Kanamori's the higher infinite, where he tries to prove the result attributed to Silver and Solovay, that if $\omega_1^{L[U]} = \omega_1$, then there is a $\Pi_2^1$ set without the perfect set property.


Here we are dealing with ZFC$^-$(ZFC minus the Powerset Axiom) premice.

To be more precise, assume that $M$ is a transitive model of ZFC$^-$ and that $U$ is some set in $V$. We say that $\langle M, \in, U\rangle$ is a ZFC$^-$ premouse (at $\kappa$) iff $U$ is an $M$-ultrafilter over $\kappa$ and that for some $\zeta$, $M = L_\zeta[U].$

Also we say that two premice $\langle M, \in, U\rangle$ and $\langle N, \in, W\rangle$ are comparable iff $\exists F \exists \zeta \exists \eta$ such that $M = L_\zeta[F]$ and that $N = L_\eta[F]$.

Now there is a lemma (called the Comparison lemma) which states that:

If $\langle M, \in, U\rangle$ and $\langle N, \in, W\rangle$ are iterable premice, then they have iterates which are comparable.

Now let $\langle M, \in, U\rangle$ and $\langle N, \in, W\rangle$ be iterable premice, define $\lt_{ip}$ in the following manner:

$\langle M, \in, U\rangle \lt_{ip}\langle N, \in, W\rangle$ iff there exists some $F$ and some $\zeta$ and $\eta$ such that $\langle L_\zeta[F], \in, F\cap L_\zeta[F]\rangle$ is an iterate of $\langle M, \in, U\rangle$ and $\langle L_\eta[F], \in, F\cap L_\eta[F]\rangle$ is an iterate of $\langle N, \in, W\rangle$, such that $\zeta \lt \eta$.

Now Kanamori says that this ordering is a well-defined well-ordering of iterable set premice. And he says that this is straightforward to check, using the comparison lemma.


I am still stuck on showing that this is well-defined. The best idea I had was to show that: if $\alpha$ and $\beta$ are the first indices of the iterations of $M$ and $N$ which are comparable, then all the higher iterates should be comparable in a "coherent" fashion.

So what I did was this: Let $\langle M_\alpha, \in, U_\alpha, \kappa_\alpha, i_{\alpha\beta} \rangle_{\alpha\le\beta\in\text{On}}$ and $\langle N_\alpha, \in, W_\alpha, \lambda_\alpha, j_{\alpha\beta} \rangle_{\alpha\le\beta\in\text{On}}$ be the iterations of $M$ and $N$, respectively. Then let $\alpha$ and $\beta$ be the first ordinals where $M_\alpha$ and $N_\beta$ are comparable. Let $F, \zeta, \eta$ be such that: $M_\alpha = L_\zeta[F]$ and $N_\beta = L_\eta[F]$. There are $2$ cases:

$(1)$ $\zeta = \eta$: At this point I know that the rest of their iterations should be the same. But I think for totality's sake I have to show that $M = N$. Which is not obvious to me at the moment. (*)

$(2)$ $\zeta \lt \eta$: We can see that $M_\alpha,U_\alpha \in N_\beta$ so that the iteration of $N$ can witness the iteration of $M$ inside it. So for $\delta \in \text{On}$, we can see that $M_{\alpha + \delta} \in N_{\beta + \delta}$, but I can't generalize this argument for all $\delta \ge \alpha$ and $\xi \ge \beta$, i.e. that if $M_\delta$ and $N_\xi$ are comparable via some $G$, then $M_\delta$ falls below $N_\xi$.(**)

(*) and (**) are the two points where I can't finish this argument. Now my question is that, can the above argument be completed? Or is there some other way to prove that $\lt_{ip}$ is well-defined?

Also I would really appreciate any hints or remarks concerning the well-order part.


EDIT I:

The material here can be found in Kanamori's "The Higher Infinite", page $273$, $2$nd edition.


EDIT II: As Yair Hayut kindly pointed out in the comments below, my definition of $\lt_{ip}$ was flawed. It is now fixed.

Also it was pointed out that it is reasonable to identify each premouse with it's iterates. In this light $(1)$ becomes:

$(1)^*$ In the case $\zeta = \eta$ we should find some premouse $\langle B, \in, O\rangle$ such that both $M$ and $N$ are iterates of $B$. In this case we insure totality.


EDIT III:

Also $(1)^*$ is evidently equivalent to:

$(1)^+$ We have to show that one of $M$ or $N$ is an iterate of the other one.

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    $\begingroup$ Note that case (1) can certainly happen for $M \neq N$. Just take $M$ to be any iterated ultrapower of $N$. Your order seems to be a pre-well-order in which a mouse is identified with its iterations. One can well-order each equivalence class by the critical point. $\endgroup$
    – Yair Hayut
    Feb 4, 2020 at 10:32
  • $\begingroup$ @YairHayut, I actually gave this point some thought before, but I couldn't come up with any concrete examples, since these mice seem to jump to different relative constructive hierarchies in each step. Are there such examples? $\endgroup$ Feb 4, 2020 at 11:45
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    $\begingroup$ I think that your citation of Kanamori is inaccurate - you might need to iterate $M$ and $N$ in order to get to the same $F$ in the definition of $<_{ip}$. Now, take $M = L_{\alpha}[U]$. Let $N = \mathrm{Ult}(M, U)$. $N$ is also a mouse (of the form $L_{\beta}[U']$ where $U'$ is the image of $U$ under the ultrapower map). In the comparison process of $M$ and $N$, we are going to take the ultrapower of $M$ once and halt. $\endgroup$
    – Yair Hayut
    Feb 4, 2020 at 12:08
  • $\begingroup$ @YairHayut, Oh, you are absolutlely right. I made a mistake when writing the definition and confused myself. :) Thanks for pointing that out. So the main idea is to identify all premice with their iterations and then look at the order in that way. But even then I still have doubts about (1). I can imagine two different mice being iterated and getting merged at some point. Unless we can prove any pair of such mice are exactly the iterations of some mouse strictly below both? $\endgroup$ Feb 4, 2020 at 12:33
  • $\begingroup$ While it is (sometimes) possible to reconstruct a (fine structural) mouse from its iterates, I think that the arguments for this are more sophisticated than what you need here. Note that in order to compare two mice $M$ and $N$ as in the question, you can simply hit the measures in both of them $\mu = (\max |M|, |N|)^+$ times. Now, it is clear that if there is an iteration that makes $M$ to be an initial segment of $N$ than also the $\mu$-steps iteration will do the same, which is what you need for (2). $\endgroup$
    – Yair Hayut
    Feb 6, 2020 at 8:41

1 Answer 1

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Note that in those types of mice, the comparison process is simple:

Lemma: Let $M = L_{\alpha}[U]$, $N = L_{\beta}[W]$ be two mice. Let $\mu$ be a regular cardinal $\mu \geq (\max(\alpha,\beta))^+$. Then $M_{\mu} = L_{\alpha'}[F], N_{\mu} = L_{\beta'}[F]$ where $F$ is the club filter on $\mu$, $M_{\mu}$ is the iteration of $M$ for $\mu$ many steps and $N_\mu$ is the iteration of $N$ for $\mu$ many steps.

Proof: Let $C_M = \{\kappa_{\alpha} \mid \alpha < \mu\}$ be the critical points of the iteration on $M$. Then $C_M$ is a club at $\mu$, and $j_{\mu}(\kappa) = \mu$.

Moreover for all $X \in M_{\mu}$, $$X \in j_{\mu}(U)\iff \exists \zeta < \mu,\ X = j_{\zeta, \mu}(X'), \text{ and } X' \in j_{\zeta}(U)$$ In turn, this is equivalent to $$\forall\zeta' \geq \zeta,\ \kappa_{\zeta'} \in X.$$ On the other hand, if $X \notin j_{\mu}(X)$ then $j_{\mu}(\kappa) \setminus X \in j_{\mu}(U)$ and thus $X$ is disjoint from a club. We conclude that $M_\mu = L_{\alpha'}[F]$ for some ordinal $\alpha'$. The same argument works for $N$.

Lemma: Let $M, N, \alpha', \beta'$ be as above. $M<_{ip} N$ iff $\alpha' < \beta'$.

Proof: First, the order of $\alpha', \beta'$ does not depend on the choice of $\mu$. Moreover, if there is some iteration $M_{\xi}, N_{\eta}$ of different lengths such that $M_{\xi}$ is an initial segment of $N_{\eta}$ than continue by iterating both $M_{\xi}, N_{\eta}$ by $\mu$ many steps where $\mu$ is regular cardinal above the size of both $M_{\xi}, N_{\eta}$. Then by the previous lemma, $M_{\xi + \mu} = M_{\mu}$ is either an initial segment of $N_{\eta + \mu}$ or the other way. The second case cannot occur, since $M_{\xi}$ is an initial segment of $N_{\eta}$ and this is preserved by the longer iteration.

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  • $\begingroup$ Many thanks for the answer. I think I got it, but for the sake of reassuring myself, I have a question. In the last step we probably should assume towards a contradiction that $N_\eta$ is an intial segment of $M_\xi$ and iterate them $\mu$ times, for a large enough $\mu$, to get a contradiction, right? Because assuming the other way around, we have nothing to prove, no? $\endgroup$ Feb 6, 2020 at 13:09
  • $\begingroup$ Also, I still have some doubts about $(1)^+$. Is it even true? If not, what conditions should we put on our mice for that to happen?(conditions like countability, etc... that don't break the mentioned theorems proof). $\endgroup$ Feb 6, 2020 at 13:44

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