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The Upward Lowenheim-Skolem theory decrees that there must be a (non-standard) model of the naturals of cardinality the same as that of the standard model of the Reals.

For any combined theory of the Reals and the Naturals, a proof similar to that of Upward Lowenheim-Skolem means that there must be a model with any infinite cardinality of Naturals. But, is there a model where both the Reals and the Naturals have the same cardinality?

Equivalently, is the statement "There is a bijection between the reals and the naturals" satisfiable in some combined theory of naturals and reals where the standard Peano axoims and the axioms of the reals hold / can be encoded.

Or, is $\forall r, \exists i, \textrm{isInteger}(i) \land f(i) = r$ first-order provable in some minimal axiom scheme that can prove every thing that Peano axioms and the real axioms can prove?

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  • $\begingroup$ What do you call "the reals axioms"? $\endgroup$ – Régis Jun 15 '18 at 7:09
  • $\begingroup$ Didn't realise there was debate regarding that. I'm going to say, dense linear order without endpoints and.. umm... Aaaah, did not realise that the least-upper-bound property is not FOL axiomatizable. $\endgroup$ – nishantjr Jun 15 '18 at 7:23
  • $\begingroup$ You might like to look at real closed fields. $\endgroup$ – Régis Jun 15 '18 at 19:13
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    $\begingroup$ Anyway, as long as your language vocabulary is countable, whatever theory you consider, it will have countable models. $\endgroup$ – Régis Jun 15 '18 at 20:45
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    $\begingroup$ Questions like this are very hard to ask accurately and the answer often depends subtly on what you ask. How are the naturals and reals related? Certainly upward L-S guarantees there are models of PA with cardinality $\mathfrak c$. Those have elements that are not part of our standard reals. If that isn't a problem, we are done. If you want the PA model to be a subset of the real model, I don't know what the reals based on that sort of PA model look like. $\endgroup$ – Ross Millikan Jun 16 '18 at 3:21
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If you take the real-closed field axioms RCF as your theory of reals, then it is easy to get what you want, since the computable reals satisfy RCF, and the computable reals are definitely countable. If you add an axiom Sup that every upper-bounded definable sequence of reals (over your theory) has a supremum, then the new theory RCF+Sup is no longer satisfied by the computable reals, but is satisfied by the collection of reals that can be computed by some finite Turing jump, which is still countable.

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  • $\begingroup$ Sup is an axiom schema, with one axiom for each formula over PA+RCF that defines a sequence of reals. $\endgroup$ – user21820 Jun 16 '18 at 3:06

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