Timeline for How hard is it to get "absolutely" no amorphous sets?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2022 at 1:16 | vote | accept | Noah Schweber | ||
| Oct 28, 2022 at 3:31 | comment | added | Elliot Glazer | @AsafKaragila I see, nice. | |
| Oct 28, 2022 at 1:03 | comment | added | Asaf Karagila♦ | This paper: arxiv.org/abs/2112.14103 | |
| Oct 28, 2022 at 1:00 | comment | added | Asaf Karagila♦ | @ElliotGlazer: No, it does not. Indeed, not even DC is enough. Feferman's model, which is $L(\Bbb R)$ of the Cohen model satisfies DC, and by abstract nonsense results, the Cohen model is a generic extension of it (we add a set of reals, each real is in the model, and the set is in a generic extension). So even if DC holds, there can be a generic extension with an amorphous set. Alternatively, my paper with Jonathan Schilhan shows even higher DC won't do in a direct proof. | |
| Oct 26, 2022 at 12:29 | comment | added | Joel David Hamkins | Why ask about CTM instead of just asking about the theory ZF + "it is not forceable that there is an amorphous set"? The CTM aspect seems irrelevant. | |
| Oct 26, 2022 at 12:27 | answer | added | Elliot Glazer | timeline score: 7 | |
| Oct 25, 2022 at 8:07 | comment | added | Farmer S | @ElliotGlazer, why don't you write that as an answer? | |
| Oct 24, 2022 at 6:12 | comment | added | Noah Schweber | @AsafKaragila Indeed, that's quite lovely! | |
| Oct 24, 2022 at 5:56 | comment | added | Asaf Karagila♦ | Related? mathoverflow.net/a/412400/7206 | |
| Oct 24, 2022 at 3:26 | comment | added | Elliot Glazer | Incidentally, it would be interesting to see if just "Infinite = Dedekind-infinite" is sufficient to get "generically no amorphous sets." | |
| Oct 24, 2022 at 2:05 | comment | added | Elliot Glazer | If I'm not mistaken, an infinite set $X$ is a universe for any such $T$ iff $X$ is orderable and there is a bijection between $X$ and $X^2.$ The forward direction is by considering $T$ to be (a finite fragment of) PA and the backward direction by taking the E.M. model generated with $X$ as a set of order-indiscernibles. This would mean expansive models are precisely those satisfying choice by Tarski's characterization of choice. | |
| Oct 24, 2022 at 1:00 | history | asked | Noah Schweber | CC BY-SA 4.0 |