Timeline for What is the consistency strength of definable axiomatization of ZFC?
Current License: CC BY-SA 4.0
12 events
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| Jun 23, 2018 at 12:32 | comment | added | Joel David Hamkins | OK, I have edited. | |
| Jun 23, 2018 at 12:31 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jun 23, 2018 at 12:03 | comment | added | Zuhair Al-Johar | by the way it is not "new", I've explicitly written that infinity to be stated in a 'definable manner' to mean that, and I explicity also said that it stipulates a set of all finite von Neumann's, anyhow. | |
| Jun 23, 2018 at 11:52 | comment | added | Joel David Hamkins | Yes, your new infinity axiom is not true in my integer structure. And indeed, once you get infinity by your axiom, then you get the empty set and also many finite ordinals and also $P(\omega)$ and so on. | |
| Jun 23, 2018 at 11:39 | comment | added | Zuhair Al-Johar | the point is that the set that infinity states its existence is "parameter free definable", the formula "y is a finite von Neumann ordinal" is parameter free, so it doesn't matter whether it gives usual von Neumann's or not, we can separate on it to get the empty set. | |
| Jun 23, 2018 at 11:23 | comment | added | Zuhair Al-Johar | I'll write it up, but pretty much it is the usual definition of being a transitive set of a transitive sets that is either empty or a successor and every non empty element of it is a successor. | |
| Jun 23, 2018 at 11:20 | comment | added | Joel David Hamkins | You can't separate on a set unless it is definable. In my model, every "set" contains all the ordinals, since there aren't any. But you are right, it doesn't have a set whose elements are exactly the finite ordinals. Tell me how you want to define "von Neumann ordinal" since the various usual formulations are likely no longer equivalent in your theory. | |
| Jun 23, 2018 at 11:13 | comment | added | Zuhair Al-Johar | why? the infinity axiom states explicitly the existence of a set of all finite von Neumann ordinals, now 'x is a finite von Neumann ordinal" is parameter free formula, simply separate on it to get the empty set. | |
| Jun 23, 2018 at 11:11 | comment | added | Joel David Hamkins | No, it doesn't. It doesn't even prove the existence of the empty set. | |
| Jun 23, 2018 at 11:10 | comment | added | Zuhair Al-Johar | but this theory proves existence of finite von Neumann ordinals? | |
| Jun 23, 2018 at 10:58 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Jun 23, 2018 at 10:51 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |