8
$\begingroup$

I have cross-posted this question on MO. This is the link:https://mathoverflow.net/questions/351894/lt-ip-is-a-well-defined-well-ordering-of-iterable-set-premice


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:

It kindly was pointed out to me at MO in the comments by Yair Hayut that the definition of $\lt_{ip}$ here is a little bit flawed. I fixed it.

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.

$\endgroup$
7
  • 4
    $\begingroup$ @GEdgar "Premouse" is a technical term, and "premice" is plural for it. (Should probably be spelled "pre-mouse" and "pre-mice" to avoid this confusion.) $\endgroup$ Commented Oct 30, 2019 at 13:00
  • 5
    $\begingroup$ @GEdgar: In logic this is called a false premise. $\endgroup$ Commented Oct 30, 2019 at 14:31
  • 5
    $\begingroup$ @AsafKaragila And it's an easy mistake to repeat: it's an iterable false premise. $\endgroup$ Commented Oct 30, 2019 at 15:04
  • 2
    $\begingroup$ For that matter, irritable premice. $\endgroup$ Commented Feb 3, 2020 at 12:10
  • 2
    $\begingroup$ @WillJagy: The correct pun here is "a terrible premice"... (It does work better when considering a terrible mouse, though. But then again, a mouse is just a terrible premouse. So all mice are terrible by definition...) $\endgroup$ Commented Feb 4, 2020 at 12:50

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.