define the reals in a non-archimedean elementary extension of the real field.

Question

Can it be done? We have the real field $(\Bbb R,+,-,\times,0,1,<)$, of course $(0,1,-,<)$ are definable using the rest. We take an elementary non-archimedean extension. Can we define the original set of reals in it?

Answer 1

No, you cannot: You have the supremum property for bounded definable sets, since this is first-order and your extension is elementary. If $\mathbb R$ is definable, it would be bounded (by any infinite element of the field). So there would be a "least" infinite number.

Answer 2

The first-order theory of the real numbers coincides with the first-order theory of real closed fields (http://en.wikipedia.org/wiki/Real_closed_field). This theory is O-minimal (http://en.wikipedia.org/wiki/O-minimal_theory) which implies that the definable subsets of such a field, $L$ say, comprise finite unions of points and open intervals. A proper subfield of a real closed field $L$ cannot have this form. (Because if $K$ is a definable subfield, then $\pm 1, \pm 1/2, \pm 1/3, \ldots \in K$, so $K$ contains infinitely many members of the closed interval $[-1, 1]$ (i.e., $\{x \in L \mid -1 \le x \le 1\}$), hence by O-minimality, $K$ must contain $[-1, 1]$. But then $K$ contains $x$ or $1/x$ for any $x$, so $K$ is the whole field $L$).