Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{T}$.

share|improve this question
2  
What is the definition of consistency in infinitary logic? Isn't it the same as having a model? –  Levon Haykazyan Dec 10 '12 at 10:55
1  
Syntactic, of course. There are obvious infintary analogues of the usual rules of inference. –  Zhen Lin Dec 10 '12 at 13:17
1  
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. –  Hurkyl Dec 14 '12 at 3:20
1  
Sure, but that's degenerate. In finitary first-order logic, a theory is consistent if and only if every finite subset is consistent. –  Zhen Lin Dec 14 '12 at 7:38
    
One theorem along these lines is the Bawrise compactness theorem, en.wikipedia.org/wiki/Barwise_compactness_theorem –  Carl Mummert Jul 25 '13 at 15:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.