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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 '12 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 '12 at 11:51
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 – Asaf Karagila Apr 18 '12 at 15:50
up vote 6 down vote accepted

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 '12 at 12:13
No, that doesn´t work. You first need to add a constant to the language. – Ramiro de la Vega Apr 18 '12 at 12:15
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 '12 at 12:19

Take for $S$ the field $F(t)$ with $F$ being the algebraic closure of $\mathbb Q$ inside $\mathbb R$. Equip it with the unique ordering for which $t-x>0$ for every integer $x$, and take a maximal ordered algebraic extension of $F(t)$. The field $S$ that you get is real-closed, i.e., it has the same first-order theory as $\mathbb R$. It is countable, and has non-archimedean elements ($t$, for instance).

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I think it is more satisfying to write down a concrete model like this. Rather than depend on "compactness" which gives you a model out in "Axiom of Choice Land". – Gerald Edgar Apr 21 at 20:44
@GeraldEdgar While I agree with the general sentiment, the compactness theorem for countable theories does not need the axiom of choice. – Emil Jeřábek Apr 21 at 20:49
Is it so different from the other answer? If you add a constant $t$ to the language with the axioms that it be greater than any integer $x$, then $t$ is transcendental and any model contains $F(t)$ — the simplifying point being that $F(t)$ is already real closed. – ACL Apr 22 at 6:11
My construction is perhaps more explicit, and I can make it completely explicit (neither choice nor compactness involved): embed $F(t)$ into $F((s))$ with $s=1/t$ , and set $ K=\bigcup_n F((s^{1/n}))$. Equip $K$ with the ordering extending that of $F$ for which $s$ is positive and smaller than every positive rational number; this ordering extends that of $F(t)$, and makes $K$ a real closed field. Now you may take for $S$ the algebraic closure of $F(t)$ inside $K$. – Antoine Ducros Apr 22 at 7:00
And I disagree with ACL: $F(t)$ cannot be real closed. For instance, $t$ will never be a square in $F(t)$. – Antoine Ducros Apr 22 at 7:07

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