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Hi all! I have the following problem. Suppose I have a sequence of models $M_1,M_2,...$, all of which have the same countable domain (call it $D$). $x_1,x_2,...$ is a well-ordering of $D$. $\phi(x)$ is a formula. For each $n$, I have $M_n \models \phi(x_1),...,\phi(x_n)$. I wish to construct a model $M'$, ideally again having the same domain $D$, such that $M' \models \forall x (\phi(x))$.

You can see that what I want here is similar to an argument by compactness; but as far as I understand, the compactness theorem doesn't apply here. I've also done some fiddling with ultraproducts; but the problem I run into there is that the ultraproduct expands the universe. I don't have any objection to expanding the universe, but it keeps me from concluding $\forall x (\phi(x))$, because (at least in the approach I took, with the ultrafilter being the set of cofinite subsets), Los's theorem only gives me $\phi$ for objects of the form $(x_i,x_i, x_i, ...)$, modulo the equivalence relation.

Any ideas? Thank you!

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How are we excluding the situation where $\phi(x)$ says that $x$ has property blah and that there is a $y$ without property blah? –  Andres Caicedo Jan 19 '13 at 7:40
    
Andres: That's an excellent question, and it shows that what I'm asking for can't be done in general. In my specific problem, $\phi(x)$ has a form which excludes that case. But it seems clear that I haven't asked the right question, because I haven't included enough constraints to yield a solvable problem. I will see if I can repair my question; and in the meantime, thanks for your help! –  Nick Thomas Jan 19 '13 at 8:01
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Is there any other relation between the models $M_n$? If not, then your assumption is equivalent to "there is a model $M$ with "$\phi^M$ infinite". (Assuming your formula is first order.) Also: please tell us if $\phi$ mentions the well-order. –  Goldstern Jan 19 '13 at 15:24
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A good strategy would be using the Omitting Types Theorem to omit the type $\lnot\phi(x)$. Unfortunately, your hypotheses don't seem to imply that $\lnot\phi(x)$ is not isolated. –  François G. Dorais Jan 19 '13 at 15:39
    
Goldstern: My trouble is figuring out what other relations between the models might be relevant here. (Obviously I'll post if I figure that out.) Unfortunately, I do not understand the part in quotation marks. :-/ (Care to explain more?) $\phi$ does not mention the well-order. Francois: Thanks for the suggestion! I am going to play with it and see if it gets me anywhere. –  Nick Thomas Jan 19 '13 at 21:17
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