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John Mayberry published what he calls a Euclidean set theory in his book The Foundations of Mathematics in the Theory of Sets. It is ZF with the axiom of infinity replaced by an axiom saying "the whole is always larger than the part." In other words every set is asserted to be Dedekind finite: no proper subset of $S$ is in bijection with $S$.

The hereditarily finite sets of ZF form a model of this set theory. But, besides that this set theory obviously does not give a set which models arithmetic, it has no minimal definable infinite sequence of sets. Every definable infinite sequence of sets has definable subsequences that also contain (the interpretant of) 0 and are closed under (the interpretation of) successor.

Is there a known description of which theories of arithmetic are interpretable in this set theory? Of course I mean a description in more conventional terms than just saying "the ones interpretable here."

For that matter, this theory obviously implies the theory ZF-inf which is just ZF with the negation of the usual ZF axiom of infinity. See Non-standard models of finite set theory

But is there a known more specific relation between these set theories?

These questions seem not to be answered in the book, but left as open questions (p. 387).

Colin McLarty
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    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. – Emil Jeřábek Nov 13 '17 at 13:41
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    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. – Joel David Hamkins Nov 13 '17 at 13:43
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    @ColinMcLarty Yes, and yes. – Emil Jeřábek Nov 13 '17 at 14:46
  • 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? – Colin McLarty Nov 13 '17 at 14:46
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    (Also, I apparently have a time machine.) – Emil Jeřábek Nov 13 '17 at 14:51
  • 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:

    https://mathoverflow.net/questions/121406/where-in-ordinary-math-do-we-need-unbounded-separation-and-replacement/140984#140984

     – Adam Epstein Nov 14 '17 at 11:41
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    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 – Ali Enayat Nov 14 '17 at 19:43
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    @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. – Adam Epstein Jun 10 '18 at 13:45
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    @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).

     – Adam Epstein Jun 10 '18 at 14:01

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