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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? – Levon Haykazyan Dec 10 '12 at 10:55
Syntactic, of course. There are obvious infintary analogues of the usual rules of inference. – Zhen Lin Dec 10 '12 at 13:17
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
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, – Carl Mummert Jul 25 '13 at 15:19

The proof is from Keisler's Model Theory for Infinitary Logic and the result is due to Makkai which is closely related to earlier work of Henkin and Smullyan:

Fix a language $L$. Now let $C$ be a countable set of new constant symbols, and let $M$ be the language formed by adding each $c \in C$ to $L$. Then, we can make the infinitary logic $M_{\omega_1 \omega}$ corresponding to $M$.

Notational convention: For a given formula $\varphi$ of $M_{\omega_1 \omega}$, the formula $\varphi \neg$ is defined inductively as follows. (This is called 'moving the negation inside' and I'm not really sure why it is necessary since it shows us that $\varphi \neg $ is logically equivalent to $\neg \varphi$)

1) $(\neg \varphi) \neg $ is $\varphi$

2) $(\bigwedge_{\varphi \in \Phi}\varphi)\neg$ is $\bigvee_{\varphi \in \Phi} \neg \varphi$

3) $(\bigvee_{\varphi \in \Phi}\varphi)\neg$ is $\bigwedge_{\varphi \in \Phi} \neg \varphi$

4) $(\forall x\varphi)\neg$ is $\exists x \neg \varphi$

5) $(\exists x\varphi)\neg$ is $\forall x \neg \varphi$

Definition: Let S be a set of countable sets of sentences of $M_{\omega_1\omega}$. $S$ is said to be a consistency property iff for each $s \in S$, all of the following hold.

(C1) (consistency rule) Either $\varphi \not \in s$ or $(\neg \varphi) \not \in s$

(C2) ($\neg$ - rule) If $(\neg \varphi)\in s$ then $s \cup \{\varphi \neg\} \in S$

(C3) ($\bigwedge$ - rule) If ($\bigwedge \Phi \in s$ then for all $\phi \in \Phi$, $s \cup \{ \varphi \} \in S$

(C4) ($\forall$ - rule) If ($\forall x \varphi (x)) \in S$, then for all $c \in C$, $s \cup \{\varphi (c)\} \in S$

(C5) ($\bigvee$ - rule) If $(\bigvee \Phi) \in s$, then for some $\varphi \in \Phi$, $s \cup \{ \varphi \} \in S$

(C6) ($\exists$ - rule) If $(\exists x \varphi(x)) \in s$, then for some $c \in C$, $s \cup \{\varphi(c)\} \in S$

Now, by the term $Basic$ $terms$, we mean either a constant symbol or a term of the form $F(c_1,...,c_n)$ where $c_1,...,c_n \in C$ and $F$ is a function symbol of $L$.

(C7) (Equality Rules) Let $t$ be a basic term and $c,d\in C$. If $(c = d) \in S$, then $s \cup \{d =c\} \in S$. If $c = t$, $\varphi(t) \in s$, then $s \cup \{\varphi \} \in S$. For some $e \in C$, $s\cup \{e = t\} \in S$.

Theorem: (Model Existence Theorem). If $S$ is a consistency property and $s_0 \in S$, then $s_0$ has a model. The proof is not very long, but it is a little involved and there are some definition things I have left out. This is essentially a mini-chapter (5 pages) in the book.

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