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How I can prove that for the class $K$ of well-ordered structures there is no finite set of statements $T$ such that $\text{Mod} (T) = K$?


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This question would benefit from more details to be honest. In what context have you come across the question? What have you tried so far? – Simon Hayward Nov 27 '12 at 11:46
The (characteristic-classes) tag was not at all relevant. – Alex Kruckman Nov 28 '12 at 1:30

You ask how to prove that there is no finite set of set of statements $T$ such that $\text{Mod}(T) = K$. But the usual definition of elementary class allows the theory $T$ to be infinite. And, in fact, the class of well-ordered structures is not elementary, i.e. there is no (finite or infinite) first order theory $T$ in the language $\{<\}$ such that $\text{Mod}(T) = K$.

To prove this, suppose we had some theory $T$ picking out the well-orders. Consider the expanded language $\{<\}\cup\{c_0,c_1,c_2,\dots\}$, where the $c_i$ are constant symbols. Let $T' = T \cup \{c_{i+1} < c_i\,|\,i\in\mathbb{N}\}$. This theory expresses that the $c_i$ form an infinite descending chain.

Now every finite subset of $T'$ is consistent (there is a well-order with a descending chain of length $n$ for any $n$), so $T'$ is consistent by compactness. So there is a model $M\models T'$, which cannot be a well-order. But then also $M\models T$, contradicting our assumption on $T$.

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As an (not really better, but maybe a little curious) alternative to Alex Kruckman's proof, you can use the fact that a class of models is elementary if and only if it is closed under elementary equivalence and ultraproducts (as a side note, a class is a basic elementary class, that is, it is finitely axiomatizable iff both it and its complement have the property, see e.g. Keisler & Chang).

If you take $M_n=\{0,1,\ldots,n\}$ with the standard ordering, then the ultraproduct of $M_n$ (with respect to a non-principal ultrafilter) is not well-ordered (the sequence $a_n=[0,0,\ldots,0,1,2,3,4,\ldots]$ with $n+1$ zeroes at the beginning is an infinite descending sequence).

In fact, it is not even closed under elementary equivalence, because the ultrapower of $\bf N$ with standard ordering is elementarily equivalent to $\bf N$, but (if the ultrapower is with respect to a non-principal ultrafilter) it is not well-ordered (we have the same infinite descending sequence as above).

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