Timeline for Do all countable $\omega$-standard models of ZF with an amorphous set have the same inclusion relation up to isomorphism?
Current License: CC BY-SA 3.0
19 events
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| Jul 16, 2021 at 9:53 | answer | added | Harry West | timeline score: 2 | |
| Nov 21, 2017 at 12:33 | comment | added | Joel David Hamkins | I think it is explained in Chang-Keisler. In the general result, it is an invariant for the theory, not the isomorphism type, but for some kinds of Boolean algebras, it is an isomorphism invariant. I am confused about whether it goes transfinite, but since you have an increasing sequence of ideals, it would seem natural to quotient by the union of them. | |
| Nov 21, 2017 at 12:25 | comment | added | Asaf Karagila♦ | What do you do at limits? Or is there a proof that this process can only go at most $\omega$ steps or something? | |
| Nov 21, 2017 at 12:24 | comment | added | Joel David Hamkins | Yes, that is the right idea. We need to understand what happens in the iterated quotient process. What survives and for how long? | |
| Nov 21, 2017 at 12:21 | comment | added | Asaf Karagila♦ | So, if there is an amorphous set, then essentially $A^A$ gives you a sort of atom at the limit step? (Or maybe $A^\omega$? Or $A^{<\omega}$?) But now, if you have countably many "sufficiently different amorphous sets", you can mix and match them to get different kind of structures, no? | |
| Nov 21, 2017 at 12:17 | comment | added | Joel David Hamkins | My idea is to understand these invariants in connections with a model of ZF + an amorphous set. If we can show that the invariants are the same, then they are all isomorphic. OK, it's not a Boolean algebra, but only a relatively complemented distributed lattice, and in this case these are the Ersov invariants, but the idea is the same. | |
| Nov 21, 2017 at 12:16 | comment | added | Joel David Hamkins | Yes, you iterate this, and the number of steps it takes, plus the number of atoms left at the last step, is an isomorphism invariant (the Tarski invariants). | |
| Nov 21, 2017 at 11:56 | comment | added | Asaf Karagila♦ | So you basically consider what happens with inclusion modulo finite sets in the universe, and there there will be an atom again which is $A\times A$. That's the gist I understand from this, is that right? | |
| Nov 21, 2017 at 11:51 | comment | added | Joel David Hamkins | The atoms are the singletons. If $A$ is amorphous, then in the quotient by the ideal of finite sets, $A$ becomes an atom, since it is not finite, but every subset is either finite or finitely different from $A$. What happens with $A\times A$? Every subset is either finite or cofinite on every section, and uniformly on cofinitely many sections. I think this shows that $A\times A$ lasts for two steps in the Tarskian quotient process, which is what I meant by saying $A\times A$ is amorphous after one step. | |
| Nov 21, 2017 at 11:38 | comment | added | Asaf Karagila♦ | Okay, I'm not entirely sure that I understand that last part completely. What are the atoms here? And how would $A\times A$ become amorphous in the ideal generated by the atoms? | |
| Nov 21, 2017 at 11:35 | comment | added | Joel David Hamkins | Well, if you add one amorphous set $A$, then of course there are infinitely many disjoint copies of it $\{n\}\times A$. And also $A\times A$ is not amorphous, but it becomes amorphous in the quotient of the Boolean algebra by the ideal generated by the atoms. This is related to the Tarski invariants for a Boolean algebra. | |
| Nov 21, 2017 at 11:24 | comment | added | Asaf Karagila♦ | (If at some point you want to move this to somewhere with more than 600 characters, do feel free. :-)) | |
| Nov 21, 2017 at 11:20 | comment | added | Asaf Karagila♦ | Right, right. But start with a standard model (just to omit this $\omega$ from the notation everywhere). And add a single amorphous set to it. That gives you an "almost minimal" infinite cardinal, which I suspect is what you mean by this fact. That you get an $(\omega,\omega)$-gap in the cardinals with the lower chain being $\omega$ itself. But, I mean, what else do you get there which "bothers" the containment structure? | |
| Nov 21, 2017 at 11:12 | comment | added | Joel David Hamkins | If you add an amorphous set to an $\omega$-standard model of ZFC, then the subset relation is definitely not isomorphic to the original model, since the property that all subsets are either finite or co-finite is revealed by the subset relation. | |
| Nov 21, 2017 at 9:40 | comment | added | Asaf Karagila♦ | Well, what happens when you have the one amorphous set? | |
| Nov 20, 2017 at 12:13 | comment | added | Joel David Hamkins | That is a good idea, and I don't know what happens to the subset relation when you do this. It is a good candidate model for another isomorphism type. | |
| Nov 20, 2017 at 10:52 | comment | added | Asaf Karagila♦ | Joel, what happens when you add a proper class of "pairwise-generic" amorphous sets compared to adding just a single one? | |
| Apr 17, 2017 at 18:07 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
forgot the $\omega$-standard hypothesis
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| Apr 17, 2017 at 0:37 | history | asked | Joel David Hamkins | CC BY-SA 3.0 |