TITLE: define the reals in a non-archimedean elementary extension of the real field.
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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 $L$).