## Non-Archimedean non-standard models for R

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Let $\langle \mathbb{R}, 0, 1, +, \cdot, <\rangle$ 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 $1,1+1,1+1+1,\ldots$ 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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Why does a routine application of compactness not yield the model you want? – Juris Steprans Apr 18 2012 at 11:42
@Juris Steprans: I'm sorry, I can't see what you're implying. Could you be more precise, please? – Dave Apr 18 2012 at 11:51

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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 math.SE math.stackexchange.com/q/133418/622) – Asaf Karagila Apr 18 2012 at 15:50

## 1 Answer

If $S$ is the set of real algebraic numbers then $1, 1+1, 1+1+1, \dots$ is unbounded in $S$. On the other hand, by compactness of first order logic (as Juris points out), there are models $S$ for which $1, 1+1, 1+1+1, \dots$ is bounded.

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Thanks for your answer. I'm afraid I don't understand how compactness comes in here. I'm guessing I should add the statements $\exists{x} x>1$, $\exists{x} x>1+1$ and so on to my set, and show that there are models that satisfy that set. But I think it only gives me that there are greater elements than any element of the type $1+1+\ldots+1$; I need one element that "covers them all". – Dave Apr 18 2012 at 12:13
No, that doesn´t work. You first need to add a constant to the language. – Ramiro de la Vega Apr 18 2012 at 12:15

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Yes, indeed. I think that a good solution for that would include adding another constant c to the language, then the theorems c>1, c>1+1 and so on, find a model and "forget" c (this is a model of the reduced language). Thanks! – Dave Apr 18 2012 at 12:19
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