Timeline for Set theory inside arithmetics via the Ackermann yoga
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2020 at 20:23 | vote | accept | Mirco A. Mannucci | ||
| Jun 13, 2011 at 3:29 | comment | added | Ali Enayat | Mirco: $PA$ proves the negation of the axiom of infinity for the Ackermann interpretation, so as far as the Ackermann interpretation is concerned, every set is "doomed" to be finite from $PA$'s point of view. So there indeed is a severe limit to the "Yoga", unless - as in Ressayre's theorem - one moves to an external venue. | |
| Jun 12, 2011 at 20:41 | comment | added | Mirco A. Mannucci | But this is not exactly what i add in mind: my question is, can I add, via the Ackermann interpretation, a statement such as : There exist a number which is the code of an infinite set", or maybe even more, and get some Ackermann interpretation of ZF inside the arithmetical theory. In other words, are there limits for the Ackermann's yoga? | |
| Jun 12, 2011 at 20:41 | comment | added | Mirco A. Mannucci | Yes Ali, I see the point: you can take, say T= PA + CON(ZFC). Via the arithmetized completeness theorem, any model of T will be able to "construct" an inner model of ZFC. Same applies to stronger consistency statements, such as CON(ZFC + exists a Mahlo cardinal", etc. | |
| Jun 9, 2011 at 2:55 | comment | added | Ali Enayat | [continued from above} But such internal models, from the point of view of the ambient model of arithmetic, have nonstandard integers. | |
| Jun 9, 2011 at 2:54 | comment | added | Ali Enayat | @Mirco: in some sense, the answer is yes, since you can add all kinds of consistency statements (of recursive theories, such as $ZF$ plus such and such large cardinals, to $PA$. Since $PA$ "knows" about the completeness theorem (thanks to a theorem of Hilbert-Barnays), arithmetical models of such consistency statements can define an epsilon relation on themselves that turns them into models of set theory; moreover, the model of arithmetic has even a definable truth-predicate for such internal models of set theory. | |
| Jun 8, 2011 at 22:24 | comment | added | Mirco A. Mannucci | Ackermann (but perhaps weaker than PA), and enrich it with axioms stating the existence of Ackermann's infinite "sets"? In other words, can I bridge the real gap between arithmetics and ZF all the while remaining in an arithmetical theory? And if yes, can I push this "catching up" to put on equal footings not only arithmetics and ZF, but actually ZF "beefed up" with high infinity axioms? | |
| Jun 8, 2011 at 22:17 | comment | added | Mirco A. Mannucci | Yes, I see your point Ali: Ressayre's model is not directly related to the "internal set theory" of the ambient model M. It is from outside that we know what M does not, namely that certain "sets" are infinite. As far as M is concerned, its internal set theory is all finite. Ok, but my initial question is not about PA. Instead, it can be reformulated this way: the ackermann yoga tells us that the only real distinction between set theory and arithmetics is that ZF knows about actual infinity, whereas PA does not. Very well. Then, can I start from an arithmetical theory strong enough to verify | |
| Jun 8, 2011 at 21:07 | comment | added | Ali Enayat | @Mirco: Ressayre's result applies to all nonstandard model of $PA$, and all consistent extension $T$ of $ZF$, including those that assert the existence of large cardinals; so for better-or-worse, it does not lend itself as a tool of classification. Also, since the model produced in Ressayre's theorem is not "internal" to the ambient model of arithmetic, the theorem does not say much about the "internal set theory" of the model, rather, it shows it is so rich that, externally, we can carve out models of strong set theories from it. | |
| Jun 8, 2011 at 19:10 | comment | added | Mirco A. Mannucci | Indeed this results is a real pearl: totally fascinating! Do you know of any type of classification of nonstandard models of PA on the basis of how strong their internal set theory is? For instance, in certain models an Ackermann "ordinal" could be an inaccessible cardinal.. | |
| Jun 7, 2011 at 18:17 | comment | added | Ali Enayat | @Emil: Yes, you have the perfect explanation. | |
| Jun 7, 2011 at 17:54 | comment | added | Emil Jeřábek | Aha, I’ll answer myself. $M_k$ is not an end-extension of $A$, hence it only needs to satisfy existential consequences of $T$ to make it work. Now, in set theory, there is no MRDP theorem, and existential sentences have very little expressive power: they assert the existence of a finite configuration of points with prescribed elementhood relation on them. It is easy to see that every such sentence is decidable in ZF without infinity (or its negation), hence if $T$ is consistent, its existential consequences are indeed valid in $M_k$. Alright, this is magic. | |
| Jun 7, 2011 at 17:26 | comment | added | Emil Jeřábek | I don’t see why $T^A$ is consistent with $(M_k,\in_{Ack})$. Doesn’t this require $M_k$ to satisfy all $\Sigma_1$ consequences of $T$, or something like that? | |
| Jun 7, 2011 at 17:03 | history | answered | Ali Enayat | CC BY-SA 3.0 |