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Find a prime model and a countable $\omega$-saturated model of $Th((\mathbb{Q},+,0))$.

Define a function from $\mathbb{N}$ to $\mathbb{N}$ such that, for each n, $Th((\mathbb{Q},<))$ has not more than $f(n)$ $n$-types.


I know for the first part that i have to find a model $\mathfrak{A}$ of $Th((\mathbb{Q},+,0))$ such that every other model embedds elementary in this model $\mathfrak{A}$ (here we can use Fraisse's Method). I think that $(\mathbb{Q},+,0)$ self i such a model but is this true? But what about a counatble-saturated model. I know the definition but it is very difficult to understand this definition. Can someone give a hint?

For the second part: i know that for each n holds: $(q_0,\cdots,q_{n-1})$ in $Th((\mathbb{Q},<))$ realized the same n-type as $(r_0,\cdots,r_{n-1})$ iff for all i$,j<n$: $q_i<q_j$ iff $r_i<r_j$ (this is a equivalence relation with finitly many equivalence classes). The statement is clear, but how to define such a function $f$ with this information?

Thank you for help. :)

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1 Answer 1

A prime model is not one into which every model embeds elementarily. It's one which embeds into every model elementarily.

The prime models in a countable language can be characterized as those which omit all non-isolated types. Dually, an $\omega$-saturated model is one which realizes all types over finitely many parameters.

Do you know about the quantifier elimination for $\text{Th}(\mathbb{Q},+,0)$? You can use this to understand all the types relative to this theory, and once you understand all the types, you should see what it means to omit or realize them.

For the second question, you're correct that the type of $(q_1,\dots,q_n)$ depends only on the order of $q_1,\dots,q_n$. So there's exactly one $n$-type for each way of ordering these $n$ elements. It's a counting problem: you have $n$ choices for which one comes first, $n-1$ choices for which comes next...

EDIT: as pointed out in the comment, there's also the possibility that some of the $q_i$ could be equal, i.e. $q_1 < q_4 = q_3 < q_2$. So the counting becomes a little more complicated.

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The $q_i$ could be equal, not? –  Lord_Farin Oct 23 '12 at 16:37
    
ah yes, of course –  Alex Kruckman Oct 23 '12 at 16:51

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