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 $\mathbb{T}$.


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.

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
3  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.