Timeline for Does choice always hold in a model of ZF with point-wise parameter-free definable sets?
Current License: CC BY-SA 4.0
13 events
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| Nov 22, 2019 at 18:00 | comment | added | Noah Schweber | It may also help to replace ordinal definability with definability-in-some-arbitrary-parameter. The point is that every set is trivially definable from a parameter (namely, use the set itself as a parameter), so "definability from arbitrary parameters" is definable (by "$x=x$"); but clearly there's no reason to believe that every "special case" of this property is also definable (since in a sense everything is a special case of parameter-definability!). | |
| Nov 22, 2019 at 17:57 | comment | added | Noah Schweber | In considering Andreas' comment, it may be helpful to first recall that subsets of computable sets need not be computable. (The analogy is: subset ~ subcollection and computable ~ definable. The collection of definable sets is always a sub-collection of the collection of ordinal-definable sets, but the definability of the latter in no way implies the definability of the former - just as a subset of a computable set need not be computable.) | |
| Nov 22, 2019 at 16:50 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 22, 2019 at 13:51 | comment | added | Andreas Blass | The facts that one property (parameter-free definability) implies another (ordinal definability) and that the latter is definable do not entail that the former is also definable. (I think you may be using "special case" and "kind of" ambiguously.) | |
| Nov 22, 2019 at 10:10 | comment | added | Zuhair Al-Johar | @AndreasBlass, I'm perplexed! If "parameter free definability" is a special case of "ordinal definability", then how come it is not definable? I can understand that 'definability' [which allows parameters of any kind] is not definable, but "parameter free definability" is ought to be definable, because its a kind of ordinal definability? | |
| Nov 21, 2019 at 4:28 | vote | accept | Zuhair Al-Johar | ||
| Nov 21, 2019 at 2:11 | history | edited | François G. Dorais | CC BY-SA 4.0 |
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| Nov 21, 2019 at 0:08 | answer | added | Noah Schweber | timeline score: 4 | |
| Nov 20, 2019 at 23:35 | review | Close votes | |||
| Nov 25, 2019 at 3:05 | |||||
| Nov 20, 2019 at 22:40 | comment | added | Andreas Blass | Although, as @Asaf pointed out, AC does follow from your scheme (by taking $\varphi(x)$ to be "$x$ is ordinal definable"), the idea in the last paragraph of the question won't work, because definability (unlike ordinal definability) isn't definable. | |
| Nov 20, 2019 at 21:47 | comment | added | Asaf Karagila♦ | If all sets are definable, then all sets are ordinal definable, then $V=\rm HOD$ holds, then global choice holds. | |
| Nov 20, 2019 at 21:40 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 20, 2019 at 20:51 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |