Non-Archimedean non-standard models for R

Let $\langle R,0,1,+,\cdot,<\rangle$ 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.

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Dave, due to the high intersection of users between the MO and MSE communities it is considered impolite to post a question on both sites simultaneously. Please remember that for future reference. (Cross posted on MathOverflow mathoverflow.net/questions/94387) –  Asaf Karagila Apr 18 '12 at 15:50

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.

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Thanks for your reply, Henning, but I'm afraid that doesn't satisfy me, because this solution adds constants to the language. I try to find a model, in which there is an element that is greater than all constant terms in the model. So your model doesn't fit for me, as $c$ isn't greater than the constant term $c+1$. –  Dave Apr 18 '12 at 12:05
Well, you can always "forget" about the added constant (formally speaking, the structure that you get in the extended language is a model of the theory in reduced language). But in your question you seem to be asking if there is a model where $\{1, 1 + 1, 1 + 1 + 1, ... \}$ is not bounded. –  Levon Haykazyan Apr 18 '12 at 12:13
I guess reducing the language of the model does get me a satisfying model. Thanks for that! I'm sorry if my question was unclear. What I meant was, that it's obvious that there are models in which the set isn't bounded; I was thinking if that is true in every model, and the answer is no. –  Dave Apr 18 '12 at 12:18

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".

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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.

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