Are archimedean ordered fields axiomatizable?

Question

Let our first-order language $L$ be $\{+,-,*,0,1,<\}$. In that language, is the class of Archimedean ordered fields axiomatizable, and if it, is it finitely axiomatizable?

Answer 1

No, there cannot be a first-order axiomatization of archimedeanness.

For any proposed axiomatization (finite or not, recursive or not), if it has $\mathbb Q$ as a model, then a standard compactness argument shows that there's also a non-archimedean model.

(Temporarily add a new constant symbol $c$ to the language, and add new axioms $1+1+\cdots+1<c$ of every possible length. Every finite subset of the extended theory is consistent since it has $\mathbb Q$ with an appropriate choice for $c$ as a model. Thus the extended theory must have a model, which is non-archiemedean because its $c$ satisfies all the new axioms. If you now forget the new $c$ symbol, then the resulting structure is a non-archimedean model of the originally proposed theory).

Answer 2

Troposphere's answer is absolutely right, but let me add that there's a separate "coarse" argument we can use here as well.

Compactness (in its incarnation as the upwards Lowenheim-Skolem theorem) implies that any theory with an infinite model has models of arbitrarily large cardinality. But no Archimedean ordered field can have cardinality $>2^{\aleph_0}$. So any time you can find a cardinality bound (and aren't looking at finite structures), you know you've left the first-order realm.

Of course this requires you to prove that cardinality bound stated above; this isn't hard at all, but it does take a few steps, and Troposphere's answer avoids this work. But I think this type of argument is nonetheless worth pointing out.