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One of the problems of infinitary logic is that it is possible for compactness to fail in a spectacular way: for example, one can concoct an inconsistent set of axioms whose proper subsets are all consistent. Nonetheless:

Question. Suppose we somehow managed to prove that a theory $\mathbb{T}$ (i.e. a set of axioms) in infinitary logic is consistent. What further assumptions do we need to show that $\mathbb{T}$ has a set model?


If we allow non-standard semantics then we can always construct a model of $\mathbb{T}$, provided $\mathbb{T}$ satisfies various ‘smallness’ conditions. For example, if $\mathbb{T}$ is a theory in $L_{\kappa \omega}$, we can construct a topos $\mathcal{E}$ containing a model of $\mathbb{T}$ that is generic in the sense that the only sentences in $L_{\kappa \omega}$ satisfied by the generic model are those that are intuitionistically provable from $\mathbb{T}$. (This was shown by Butz and Johnstone [1998].) Taking a localic boolean cover of $\mathcal{E}$ would then yield a boolean-valued model of $\mathbb{T}$, though we would lose genericity. (Of course, if $\mathcal{E}$ has a point then we can even get a set model.)

It should be possible to translate the above into set theory as the construction of a model of $\mathbb{T}$ in a forcing extension of the universe. This seems to suggest that the only obstruction to having a set model of $\mathbb{T}$ is the existence of $\kappa$-complete ultrafilters in certain $\kappa$-complete boolean algebras constructed from $\mathbb{T}$.


Addendum. I have found a model existence theorem for countable theories in certain countable fragments of $L_{\omega_1, \omega}$: see Theorem 5.1.7 in [Makkai and Reyes, First order categorical logic]. The proof seems to based on a remarkable result of Rasiowa and Sikorski concerning the existence of sufficiently nice ultrafilters.

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    What is the definition of consistency in infinitary logic? Isn't it the same as having a model?2012-12-10
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    Syntactic, of course. There are obvious infintary analogues of the usual rules of inference.2012-12-10
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    Could you clarify your opening statement? The set of axioms $\{0=1,0\neq1\}$ also has the property of being inconsistent, yet every proper subset is consistent.2012-12-14
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    Sure, but that's degenerate. In finitary first-order logic, a theory is consistent if and only if every finite subset is consistent.2012-12-14
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    One theorem along these lines is the Bawrise compactness theorem, http://en.wikipedia.org/wiki/Barwise_compactness_theorem2013-07-25
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    Are you looking for something along the lines of The Model Existence Theorem for $L_{\omega_1, \omega}$? See e.g. Marker's notes http://homepages.math.uic.edu/~marker/math512-F13/inf.pdf . It is not strictly about infinitary proof systems, but it can probably be converted to a suitable form.2014-05-07
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    @ZhenLin: I know this is an old question, and maybe the result you cite in your Addendum is quite similar, but I have found a very general proof in Keisler's Model Theory for Infinitary Logic. I could post it tomorrow if you are still interested.2014-07-18
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    Please feel free!2014-07-18

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