Timeline for About external automorphism on non-well founded model of Finite ZF?
Current License: CC BY-SA 4.0
9 events
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| Aug 5, 2022 at 19:45 | comment | added | Noah Schweber | Incidentally, we see an analogue of the same phenomenon even in internal elementary embeddings of well-founded models: if $j:V\rightarrow M$ is elementary and internal, then $j(\kappa)=\kappa+\lambda$ implies that either $\lambda=0$ (i.e. $j(\kappa)=\kappa$) or $\lambda$ is "very large" (e.g. $>\omega$ for starters). | |
| Aug 5, 2022 at 19:45 | vote | accept | Zuhair Al-Johar | ||
| Aug 5, 2022 at 19:41 | comment | added | Noah Schweber | @ZuhairAl-Johar "but this way we'll have no automorphism at all, because every time we can take mod n, and violate any automorphism." No, that's incorrect: this trick only works for standard $n$s. Models with nontrivial automorphisms do exist, but they have to move ordinals by nonstandard amounts. The point is that in order to get (any version of) the analogue of the argument in this answer to work, you need the relevant mod property to be actually definable. | |
| Aug 5, 2022 at 19:40 | comment | added | Zuhair Al-Johar | @NoahSchweber, but this way we'll have no automorphism at all, because every time we can take mod n, and violate any automorphism. But on the other hand there must be rank shifting automorphisms on Finite ZF, becaue it has an infinite model. | |
| Aug 5, 2022 at 19:36 | comment | added | Noah Schweber | @ZuhairAl-Johar As my comment says, no we cannot: just think mod 3 instead of mod 2. (Basically, the same reason generalizes to give the fact in my original comment.) | |
| Aug 5, 2022 at 19:35 | comment | added | Zuhair Al-Johar | But is that the only reason? I mean can we have $j(V_{n+2})=V_n$ | |
| Aug 5, 2022 at 19:33 | comment | added | Noah Schweber | (Whoops, I missed that this was about finite set theory; my previous comment is a bit silly in that context (where "$\omega^M$" simply isn't a thing), but does hold for nonstandard models of ZF.) | |
| Aug 5, 2022 at 18:55 | comment | added | Noah Schweber | By the same argument, if $j:M\rightarrow M$ is an external automorphism and $\kappa,\lambda$ are $M$-ordinals with $j(\kappa)=\kappa+\lambda$ (this is equivalent to $j(V_\kappa)=V_{\kappa+\lambda}$) and $\lambda<\omega^M$ then $\lambda$ must be divisible by every standard natural number $n$. More generally, $j(\kappa)=\kappa+\lambda$ (with no size bound on $\lambda$) implies a lot of homogeneity for $\lambda$. | |
| Aug 5, 2022 at 18:25 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |