Non-Archimedean non-standard models for R Let $\langle R,0,1,+,\cdot,&lt\rangle$ be the standard model for R, and let S be a countable model of R (satisfying all true first-order statements in R). Is it true that the set 1,1+1,1+1+1,… is bounded in S? My intuition says "no", but I am yet to find a counter example. I read something about rational functions, but I cannot verify it is, indeed, a non-standard model of R.
 A: Let $T$ be the set of all true sentences about $\mathbb R$ and construct $T'$ by adding to $T$ a new constant $c$ together with the axioms $c>1$, $c>1+1$, $c>1+1+1$, ...
Every finite subset of $T$ has $\mathbb R$ as a model, so $T'$ is consistent by the compactness theorem, and has a countable model because $T'$ contains only countably many symbols. This shows that a countable $S$ can be non-Archimedean.
On the other hand, there must also be an Archimedean countable model. This follows directly from the downward Löwenheim-Skolem theorem, which produces a subset of $\mathbb R$ that is closed under the operations and is elementarily equivalent to $\mathbb R$ itself.
A: The (first-order) theory of real-closed fields is complete.  So any real-closed field that has the desired properties (countable, non-Archimedean) will do.  We can use devices from Model Theory. However, an algebraically natural approach is to start with the rational functions in $x$ with real algebraic coefficients, and the standard lexicographic ordering. Then we extend this to a real-closed field.  
This yields the field of Puiseux series with real algebraic coefficients.  It is real-closed, so elementarily equivalent to the field $\mathbb{R}$. And it is not Archimedean, since $x>1$, $x>1+1$, $x>1+1+1$, and so on.  To get infinitely many non-isomorphic such fields, we can add $n$ transcendentals to the base field, for some positive integer $n$, or countably many transcendentals, and again form the field of Puiseux series. 
Your question did not ask for countable Archimedean fields that are elementarily equivalent to $\mathbb{R}$. But they are easy to find.  The simplest is the field of real algebraic numbers. 
A: This is not what "non-standard model of ..." means.  
You are asking whether non-Archimedean real-closed ordered fields exist.  An example is to add some algebraically independent transcendental elements to the real algebraic numbers, or the Puiseux series field over the real algebraic numbers. 
Archimedean real-closed ordered fields also exist, such as $R$ or any of its real-closed subfields.
A nonstandard model is one that applies a construction on the positive integers (in this case, using them to build Z then Q and finally R) to the integers in a nonstandard model of arithmetic or set theory, but maintaining the set of sentences that are true in the model (transfer principle).  The integers in a nonstandard model of arithmetic are Archimedean when "seen from inside the model" and non-Archimedean "from the outside".
