Timeline for What arithmetic is interpretable in Mayberry's Euclidean set theory?
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| Jun 10, 2018 at 14:01 | comment | added | Adam Epstein | @ColinMcLarty Granted, this seems orthogonal to your concerns, as the failure of Transitive Containment is due to missing Replacement Axioms rather than a missing Infinity Axiom. In your comment, are you removing the Axiom of Infinity or adding its negation? If the latter, I believe $\in$-induction yields Transitive Containment, this since Replacement is available (even provable by induction on finite cardinality). | |
| Jun 10, 2018 at 13:45 | comment | added | Adam Epstein | @ColinMcLarty I was thinking about something similar today, yet found myself back here following up something else. In any event, I believe I can exhibit a supertranstive model of Zermelo set theory which satisfies ∈-induction but not Transitive Containment: the model has infinite sets, yet every transitive set is is finite. As $\omega$ is missing from the model, the Axiom of Infinity is understood as the existence of an infinite set. The approach is directly inspired by Matthias' paper Slim Models of Zermelo Set Theory, so it is possible this is very familiar to him. | |
| Nov 14, 2017 at 19:43 | comment | added | Ali Enayat | It is worth pointing out the following paper of Pettigrew on a system related to Mayberry´s since besides its content, its list of references are useful: On Interpretations of Bounded Arithmetic and Bounded Set Theory, Notre Dame Journal of Formal Logic Volume 50, Number 2, 2009 | |
| Nov 14, 2017 at 11:41 | comment | added | Adam Epstein | I haven't had time to look through Mayberry's book, but several years ago I did read some of his work with Pettigrew, which I believe is related. My recollection is that Unbounded Separation is not assumed there (and there are related remarks on p. 384 of the book). I posted a comment about that here: mathoverflow.net/questions/121406/… | |
| Nov 13, 2017 at 14:51 | comment | added | Emil Jeřábek | (Also, I apparently have a time machine.) | |
| Nov 13, 2017 at 14:46 | comment | added | Colin McLarty | Putting these two comments together, since ZF-Infinity can interpret ZF-Infinity + TC, does it correctly follow that the question of which arithmetics are interpretable does not depend on whether or not we actually add TC as an axiom? Maybe the relation of TC to ∈-induction is a different question but am I right to believe that, in ZF without infinity but with the usual ∈-minimal element formulation of foundation, TC is equivalent to ∈-induction? | |
| Nov 13, 2017 at 14:46 | comment | added | Emil Jeřábek | @ColinMcLarty Yes, and yes. | |
| Nov 13, 2017 at 13:43 | comment | added | Joel David Hamkins | The theory "ZF without infinity" is ambiguous unless you specify how you are treating the axiom of foundation. For example, if you just use the usual $\in$-minimal element formulation, then you can't prove the $\in$-induction scheme, so many people like to have $\in$-induction added as a scheme. | |
| Nov 13, 2017 at 13:41 | comment | added | Emil Jeřábek | So, the theory is contained in $\mathrm{ZF_{Fin}}$, and contains $\mathrm{ZF}-\mathrm{Inf}$? Well then, it is mutually interpretable with PA, so your question boils down to characterization of arithmetical theories interpretable in PA. | |
| Nov 13, 2017 at 4:50 | history | asked | Colin McLarty | CC BY-SA 3.0 |