Timeline for Does the existence of the von Neumann hierarchy in models of Zermelo set theory with foundation imply that every set has ordinal rank?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 1, 2013 at 20:37 | vote | accept | Victoria Gitman | ||
| Sep 1, 2013 at 20:24 | answer | added | Sam Roberts | timeline score: 12 | |
| Sep 1, 2013 at 19:40 | answer | added | Thomas Forster | timeline score: 4 | |
| Sep 1, 2013 at 19:02 | answer | added | Adam Epstein | timeline score: 2 | |
| Sep 1, 2013 at 17:31 | comment | added | Adam Epstein | Granted, the model so obtained will not have $\omega$ as an element. For my own purposes this was more or less the point. Since you want to have $\omega$, the construction has to be modified. I think the general idea is sufficiently robust, but I'll have to make some claims I've not entirely checked. | |
| Sep 1, 2013 at 16:21 | answer | added | Christoph-Simon Senjak | timeline score: 3 | |
| Sep 1, 2013 at 15:41 | comment | added | Adam Epstein | The model I would propose is something I came up with in connection with my own question mathoverflow.net/questions/117910/… for which I also believe the answer is no. In a nutshell, take a model of finite set theory in which there is an infinite descending chain $x_{n+1}\in x_n$. This is possible, even assuming Foundation. Then formally adjoin $\omega$ levels of the cumulative hierarchy. | |
| Sep 1, 2013 at 15:30 | comment | added | Victoria Gitman | Yes, I also suspect that $A^*$ is stronger than $A$, but I have no idea how to construct a model of $T+A$ that does not satisfy $A^*$. | |
| Sep 1, 2013 at 15:28 | comment | added | Adam Epstein | I see that now, since you allow $V_{\omega+\omega}$ as a model. So you require that any von Neumann ordinal be the spine of a stage of the cumulative hierarchy, but you allow the existence of well-ordered sets longer than any von Neumann ordinal. I believe the answer is no. I will think a bit more and then try to justify this in an answer. | |
| Sep 1, 2013 at 15:18 | comment | added | Victoria Gitman | Adam, I actually do mean for the statements $A$ and $A^*$ to depend on the existence of the von Neumann ordinals. | |
| Sep 1, 2013 at 14:50 | comment | added | Adam Epstein | At first sight, the statement of $A$ appears depend on the existence of von Neumann ordinals, but I imagine you mean it to be taken as a assertion about recursions along arbitrarily long well-ordered sets? | |
| Sep 1, 2013 at 13:59 | history | asked | Victoria Gitman | CC BY-SA 3.0 |