Timeline for Can ZFC be interpreted in a set theory having finitely many ranks?
Current License: CC BY-SA 3.0
14 events
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Jun 8, 2018 at 8:26 | vote | accept | Zuhair Al-Johar | ||
May 12, 2017 at 16:14 | comment | added | Zuhair Al-Johar | @Joel David Hamkins. I think the proper answer to your suggestion is that $(HF,\in)$ has infinitely many ranks. | |
May 6, 2017 at 11:42 | answer | added | Zuhair Al-Johar | timeline score: 1 | |
May 5, 2017 at 18:53 | answer | added | Zuhair Al-Johar | timeline score: 0 | |
May 4, 2017 at 13:54 | review | Close votes | |||
May 4, 2017 at 22:10 | |||||
May 4, 2017 at 13:36 | comment | added | Zuhair Al-Johar | Yes, but in $(HF,\in)$ for every finite rank n there is a finite rank n+1, truly there is no infinite rank per se, but ranking can be spoken of as being "potentially infinite". | |
May 4, 2017 at 12:19 | comment | added | Joel David Hamkins | My point was that even finite set theory $(HF,\in)$, which is bi-interpretable with arithmetic, can interpret ZFC set theory, provided it satisfies Con(ZFC), since it can build a Henkin model of ZFC. So we don't need any infinite ranks at all. | |
May 4, 2017 at 12:06 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, we can certainly interpret ZFC in theories that are not about sets or that extends sets by other concepts that do not need to speak of any ranks at all, for example Mereology augmented with some labeling function sending totalities of atoms to atoms in David Lewis manner plus necessary axioms can definitely manage to interpret ZFC! and it is in some sense without ranks. But this is another story. The essence of the question here was about whether SET theory needs to have infinitely many ranks in order to interpret ZFC. | |
May 3, 2017 at 19:22 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, I don't think I can answer to your question, but wouldn't that be an interpretation using another language than that of set theory and choice? How we are to think of ranks under such different milieus? | |
May 3, 2017 at 19:09 | comment | added | Stella Biderman | @Zuhair you're right. | |
May 3, 2017 at 19:03 | comment | added | Zuhair Al-Johar | @StellaBiderman, I don't think that $\exists x (N < x \wedge \forall y < x (P(y)<x) \wedge \forall y (y < x \wedge \forall z \in y (z < x) \implies U(y)<x)$, where "<" signifies "strict subnumreousity" is a theorem of ZFC. | |
May 3, 2017 at 18:41 | comment | added | Joel David Hamkins | Apart from idea in the post, can't we interpret a model of ZFC inside any model of arithmetic having Con(ZFC), just by building the Henkin model there? This wouldn't need any higher ranks at all. | |
May 3, 2017 at 17:44 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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May 3, 2017 at 17:30 | history | asked | Zuhair Al-Johar | CC BY-SA 3.0 |