An ordered field $K$ with the ordering $<$ is Archimedean if for any $x \in K$ there exists $n \in \mathbb{N}$ (where $\mathbb{N}$ is the copy of the natural numbers in $K$) such that $|x|<n$.
It is an easy task to show that for every Archimedean ordered field $K$ there is a non-Archimedean ordered field $L$ such that they are elementarily equivalent, i.e. they satisfy exactly the same first-order sentences in the language of ordered fields $(+,\cdot,0,1,<)$. (Namely, one can take the first-order theory of $K$ and extend it by all sentences of the form $c> 1+ \ldots + 1$, where $c$ is a new constant symbol. Let us call this theory $T$. Now any finite subset of $T$ has an Archimedean model. Thus, by the Compactness Theorem, $T$ has a model, which, by construction, is non-Archimedean.)
I was wondering: Is there a single first-order sentence $\sigma$ in the language of ordered fields such that any model of $\sigma$ is a non-Archimedean ordered field?
I would expect the answer to be 'No.' but cannot find where this result has been proved.
After looking a bit further, I found the answer here:
Do ordered fields and archimedian ordered fields have the same first-order theory?
Indeed, there is a sentence which is false for all Archimedean ordered fields but true for some non-Archimedean. It is the conjuction of all axioms for ordered fields plus the following sentence:
$\exists a,b \ (0<a<b \wedge \forall x \ (x^2\leq a \vee b\leq x^2))$.