Timeline for Can ZFC commit cardinality errors?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2019 at 21:39 | vote | accept | Zuhair Al-Johar | ||
| Sep 24, 2019 at 18:09 | comment | added | Zuhair Al-Johar | OK, I've removed the 'answered' status on the question, so that you can edit it. | |
| Sep 24, 2019 at 16:20 | comment | added | Greg Kirmayer | I am changing the answer to the modified questions because it is wrong. | |
| Sep 24, 2019 at 3:47 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Sep 21, 2019 at 6:36 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Sep 21, 2019 at 6:31 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Sep 20, 2019 at 17:55 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Sep 20, 2019 at 17:49 | vote | accept | Zuhair Al-Johar | ||
| Sep 24, 2019 at 17:05 | |||||
| Sep 20, 2019 at 16:43 | answer | added | Randall Holmes | timeline score: 6 | |
| Sep 20, 2019 at 16:18 | answer | added | Greg Kirmayer | timeline score: 2 | |
| Sep 20, 2019 at 9:19 | comment | added | Zuhair Al-Johar | @MonroeEskew, I meant statements similar in kind to the continuum hypothesis, I mean naturally looking statements that ZFC + V=L cannot settle, but "ZFC + No cardinality error is committed" can answer. Are there examples of such statements, since the latter theory is more minimal than the former. | |
| Sep 20, 2019 at 9:06 | comment | added | Monroe Eskew | If $\alpha$ is uncountable and $L_\alpha$ satisfies ZFC, then it will have non-definable reals, and thus countable sets without a definable bijection with $\omega$. | |
| Sep 20, 2019 at 9:03 | comment | added | Zuhair Al-Johar | @MonroeEskew, from those comments I'm realizing that ZFC + ZFC doesn't commit cardinality errors. To be an even more restrictive condition than ZFC + V=L. Are there natural mathematically looking statements that this theory can settle and that ZFC + V=L cannot? | |
| Sep 20, 2019 at 8:57 | comment | added | Monroe Eskew | It's a rather unusual one. I mostly like big ones. | |
| Sep 20, 2019 at 8:57 | comment | added | Zuhair Al-Johar | @MonroeEskew would you consider that model to be the standard model of ZFC? | |
| Sep 20, 2019 at 8:28 | comment | added | Monroe Eskew | Yes but more specifically, the least $L_\alpha$ satisfying ZFC. One can prove that every set is definable without parameters in that model. | |
| Sep 20, 2019 at 8:16 | comment | added | Zuhair Al-Johar | @do you mean a model of ZFC + V=L? | |
| Sep 20, 2019 at 8:15 | comment | added | Monroe Eskew | It is consistent. It holds in the minimal model. | |
| Sep 20, 2019 at 8:14 | comment | added | Zuhair Al-Johar | @MonroeEskew, can there be a definable bijection between the set of all reals and some ordinal, i.e. can there be a definable well ordering on the reals? would stipulating ZFC not committing cardinality error of second kind, be inconsistent? | |
| Sep 20, 2019 at 8:09 | comment | added | Monroe Eskew | It would require that whenever sets are in bijection, then there is a definable bijection. | |
| Sep 20, 2019 at 7:59 | comment | added | Zuhair Al-Johar | @MonroeEskew, yes this is part of it, but I'm not really sure if it burns down to that only. For example if I add the axiom that ZFC do not commit cardinality error of second type, would that enforce all sets in ZFC to be definable? | |
| Sep 20, 2019 at 7:08 | comment | added | Monroe Eskew | It sounds like you’re asking if it is consistent with ZFC for two sets to be in bijection, without there existing a definable bijection. Do I read that right? If so, please ask your questions in this more straightforward way. Anyway, the answer can be found by looking at a model with a countable set of reals that is not definable. For example, add omega many Cohen reals. See Jech’s book. | |
| Sep 20, 2019 at 0:35 | review | Close votes | |||
| Sep 27, 2019 at 3:05 | |||||
| Sep 19, 2019 at 23:18 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |