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.......Let T denote the first order theory of dense linear order with no infimum and no supremum. The only atomic formulae in the language of T are "$x \lt y$" and "$x=y$" (where "x" and "y" are variables). This theory T is axiomatizable (it has about 6 or 7 axioms). T has no finite models and is categorical in the smallest infinite cardinal. Therefore, by a theorem of Vaught, T is consistent, complete and decidable. One would hardly call T a trivial theory, since many of its theorems are used to prove results in more general theories of ordered sets. .......The first order set theory T* is obtained from T by substituting the symbol for set membership for the symbol "$\lt$" wherever "$\lt$" occurs (in all the formulae of T). The axioms of T* are the transformed axioms of T after the substitution. The language of T* is the language of first order ZF. Then (although there may be a gross error here) I claim that T* shares with T the properties of being consistent, complete and decidable. Every sentence of first order ZF can be proved or disproved in T* (including all the sentences that are undecidable). If anybody wants to know what T* has to say about the Continuum Hypothesis, they can even work it out for themselves-because T* is decidable. Nevertheless, T* seems like the most trivial set theory that it is possible to have. What can one do with it? Not even the weakest sub-theory of arithmetic appears to be interpretable in it. What can one say about it? The set-theoretic universe of T* is infinite, yet not one of the sets that it contains is specifically definable within it. It violates the axiom of foundation in the worst possible way. It negates almost all axioms of comprehension that (in other set theories) are used to prove the existence of sets satisfying a variety of conditions. Last, but not least T* thumbs its nose at Godel"s incompleteness theorem. Can it be be that all trivial set theories are not created equal?

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 Why not have a set theory with the axioms $\exists x\forall y:x=y$ and $\forall x:\neg(x\in x)$. That theory is much simpler and shares the "advantages" of your theory. – Michael Greinecker Oct 11 at 21:08
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 I don't see what makes this a "set theory" (except that it and set theory both use a language with a single binary relation). I like the idea of seeing what happens to the ZFC axioms in this theory, and it could be a good exercise for a logic class, but I don't see it as shedding any light on set theory or foundational issues. – Henry Cohn Oct 11 at 21:20
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 I agree with Henry. I wouldn't consider changing the notation for the primitive symbol as producing a different theory; $T^*$ is still the theory of dense linear orders without endpoints, not a set theory. – Andreas Blass Oct 11 at 21:46
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 Perhaps the OP would be interested in the answer at mathoverflow.net/questions/12584/…, which explains that a surprising number of the ZFC axioms are true in the structure $\langle\mathbb{R}^+,\lt\rangle$, including extensionality, union, weak power set, weak pairing, foundation, a form of AC and weak collection. All these axioms are theorems of the theory $T^*$ mentioned in the question. Nevertheless, my opinion is that this is not really a set theory, because of the extreme violations of the separation axiom. – Joel David Hamkins Oct 12 at 1:36
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 It is hard to tell what $T^*$ says about $CH$ since there are a ton of different first order formulas that you could say represent $CH$. This formulas are all equivalent under $ZFC$ but not under $T^*$. – Ramiro de la Vega Oct 12 at 9:52
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closed as not a real question by Ryan Budney, Michael Greinecker, Henry Cohn, David Roberts, Andreas Blass Oct 11 at 21:44

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