Skip to main content
14 events
when toggle format what by license comment
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
edited body
May 3, 2017 at 17:30 history asked Zuhair Al-Johar CC BY-SA 3.0