# 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

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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Archimedean real-closed ordered fields also exist, such as $R$ or any of its real-closed subfields.