Timeline for Can uncountable sets be proved to exist in this variant of ZFC with definability restrictions?
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11 events
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| Mar 27 at 21:30 | comment | added | Zuhair Al-Johar | Related to this answer is a short work I've done before at: sites.google.com/site/zuhairaljohar/parameter-free-z-zf | |
| Mar 27 at 20:47 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Mar 27 at 19:44 | comment | added | Zuhair Al-Johar | I think a note should be added that this , albeit related, yet it doesn't answer this question at least as it stands because its not obvious how one can recover the construction given in the referenced paper in this theory. | |
| Mar 27 at 15:41 | comment | added | Zuhair Al-Johar | Schindler & Schlicht are speaking of ZFC formalized with biconditional forms of pairing, union and power set. That doesn't apply to the implicational (conditional) forms, actually if you weaken just one of them to conditional form you cannot get the equivalence with ZFC. Here, in this theory I'm not sure how this would relate to those issues, since all axioms of pairing, union and powerset are not the full forms, the parameters are restricted to just parameter free definable sets. So, it is not clear to me if this is weaker than the biconditional parameter free ZFC. Hence the question. | |
| Mar 27 at 15:34 | comment | added | Joel David Hamkins | Schindler & Schlicht say the observation goes back at least to Lévy in 1976, whom they cite. | |
| Mar 27 at 15:30 | comment | added | Zuhair Al-Johar | This can be seen in Kanamori's article: Levy and set theory. Annals of Pure and Applied Logic 140 (2006) 233–252. See page 247 | |
| Mar 27 at 15:20 | comment | added | Zuhair Al-Johar | Parameter-free ZFC is not truly parameter free as its name connotes. You have unleashed parameters in pairing, union, and powerset axioms. Only Separation and Replacement is parameter free. Here you don't have that. The parameter free implicational form of ZFC for example is weaker than ZFC, because you don't have the full axioms of pairing, union and powerset; consult Levy for that. So, I don't think you can carry on the construction (in the article your referenced) that proves the equivalence with ZFC, I mean in this system of course. | |
| Mar 27 at 13:55 | comment | added | Joel David Hamkins | Using a definable set as a parameter is the same as not using any parameter, since it is definable. You don't need it as a parameter. But in any case, parameter-free ZFC would be even weaker than definable-parameters, and still fully powerful, so anything in between is also equally powerful. | |
| Mar 27 at 13:51 | comment | added | Zuhair Al-Johar | But, this is not parameter free ZFC. | |
| Mar 27 at 13:21 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Mar 27 at 13:15 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |