Timeline for Does $ZFC^-$ plus $\mathcal{P}^\gamma$ have a countable $L_\alpha$ model if it has a countable model?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2022 at 2:17 | comment | added | Frode Alfson Bjørdal | @NoahSchweber A shorter answer is to say that S may only have non-standard models, which are not in $L$. So, the world may be as seen from the minimal transitive model of S. | |
| Feb 26, 2022 at 18:27 | comment | added | Frode Alfson Bjørdal | @NoahSchweber ".. it's consistent with 𝖹𝖥𝖢 that 𝖹𝖥𝖢 has a countable model but no well-founded model, and so in particular no level of 𝐿 satisfying 𝖹𝖥𝖢. Does this answer your question?" Yes | |
| Feb 26, 2022 at 6:33 | comment | added | Frode Alfson Bjørdal | At any rate, your comment on ill founded models suggests the question: Is there a countable ordinal $\alpha$ so that $L_\alpha$ is a model of S if S has a countable well founded model? | |
| Feb 26, 2022 at 6:02 | comment | added | Frode Alfson Bjørdal | @NoahSchweber I identified $ZFC$ with the set of its theorems. The notation I use is common, e.g. in the metalogic of modal logic. | |
| Feb 26, 2022 at 5:59 | comment | added | Noah Schweber | "$S\cup\{\mathcal{P}(A)\vert A\in ZFC\}$" makes no sense whatsoever. What is the powerset of an axiom? | |
| Feb 26, 2022 at 5:58 | comment | added | Noah Schweber | No, I am saying that it is consistent with $\mathsf{ZFC}$ that $\mathsf{ZFC}$ has illfounded models (hence ill-founded countable models by dLS) but no well-founded models of any cardinality. | |
| Feb 26, 2022 at 5:58 | comment | added | Frode Alfson Bjørdal | @NoahSchweber I would interpret $S$ with $\gamma=1$ as $S=S\cup\{\mathcal{P}(A)|A\in ZFC\}$ | |
| Feb 26, 2022 at 5:53 | comment | added | Frode Alfson Bjørdal | @NoahSchweber I don't understand. Do you say that the fact that there are unfounded models of ZFC rules out the existence of well founded models of ZFC? | |
| Feb 26, 2022 at 4:55 | comment | added | Noah Schweber | Even taking $\gamma=1$ (so not changing $\mathsf{ZFC}$ at all), it's consistent with $\mathsf{ZFC}$ that $\mathsf{ZFC}$ has a countable model but no well-founded model, and so in particular no level of $L$ satisfying $\mathsf{ZFC}$. Does this answer your question? | |
| Feb 26, 2022 at 2:43 | history | edited | Frode Alfson Bjørdal | CC BY-SA 4.0 |
added 2 characters in body
|
| Feb 26, 2022 at 2:43 | comment | added | Frode Alfson Bjørdal | @NoahSchweber My understanding was that the appropriate $\kappa$ for your $L_\kappa$ was uncountable, to the level $\beth_\gamma$, and so $L_\kappa$ would be a standard model. But I do not ask whether there is a countable ordinal $\alpha$ so that $L\alpha$ is a model of S if S has a standard model. | |
| Feb 26, 2022 at 2:25 | comment | added | Noah Schweber | (Ignoring the issue of precisely phrasing $\mathcal{P}^\gamma$) Yes, as with your earlier question: take the Mostowski collapse of a large enough countable elementary submodel of $L_\kappa$, for $\kappa$ appropriate, and apply Condensation. | |
| Feb 26, 2022 at 2:07 | history | asked | Frode Alfson Bjørdal | CC BY-SA 4.0 |