Timeline for Critical points in $ZF$ without Choice
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2019 at 22:54 | comment | added | Thomas Benjamin | (cont.) in the previous comment it should read "Reinhardt Cardinals" | |
| Jun 23, 2016 at 7:19 | comment | added | Thomas Benjamin | (cont.) set Stack Overflow...we do not get nontriviality on the ordinals. Does the same result hold in $ZF$ or is this just another restatement of the open question "Can there exist Reinhardt Cardibals in $ZF$?". | |
| Jun 23, 2016 at 7:16 | comment | added | Thomas Benjamin | @AsafKaragila: Also, regarding your statement, "In particular, $j$:$V$$\rightarrow$$V$ without a critical point must be the identity.", always hold in models of $ZF$ where $AC$ fails? I ask this because Prof. Hamkins, in his answer to Andrea Nespola's mathoverflow question "$ZFGC^{-f}$$+$$BAFA$$+$$\exists$$\kappa$($\kappa$ is Reinhardt) and its implications" states that the nontrivial elementary embeddings $j$:$V$$\rightarrow$$V$ provided by $BAFA$ "have no critical points". He also states that in $ZFGC^{-f}$$+$$BAFA$ that "any $\Sigma_1$ elementary embedding must fix every wellfounded | |
| Jun 22, 2016 at 11:27 | comment | added | Thomas Benjamin | @NoahSchweber: Under what conditions would an external elementary embedding move an ordinal? Also, can one always adjoin(?) (by forcing) an external elementary embedding that moves an ordinal? | |
| Jun 22, 2016 at 6:47 | comment | added | Asaf Karagila♦ | No, this is not true. If it had a critical point, then is some $x$, e.g. the critical point, such that $j(x)$ had a greater rank than $x$. It's not for every $x$. And if there is some $x$ with $j(x)$ having a greater rank, then the ordinal of that rank is moved by $j$. In either case, the rank cannot decrease. | |
| Jun 22, 2016 at 6:38 | comment | added | Thomas Benjamin | @AsafKaragila: So in other words (assuming $j$ is internal), if $j$ in your theorem has a critical point, then either $rank(x)$$\lt$$rank(j(x))$ or $rank(j(x))$$\lt$$rank(x)$? | |
| Jun 21, 2016 at 17:11 | comment | added | Asaf Karagila♦ | Well, definable is a big word. I think that "a class in a NBG/MK model" would be enough. :-) But good point nonetheless. | |
| Jun 21, 2016 at 17:07 | comment | added | Noah Schweber | Note that there is a crucial assumption here that $j$ is an internal elementary embedding, that is, that $j$ is definable: otherwise the induction on rank can't go through if $V$ is ill-founded. External elementary embeddings have much more freedom - for example, Cohen's construction of a model of ZF with an automorphism of finite order. There the automorphism does not move any ordinals (exercise), but is still nontrivial. Very little can be said about external elementary embeddings or automorphisms (although not nothing - e.g. the motivation for Cohen's result was that choice must fail!). | |
| Jun 21, 2016 at 15:36 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |