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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4$\begingroup$ Why does a routine application of compactness not yield the model you want? $\endgroup$– Juris StepransCommented Apr 18, 2012 at 11:42
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$\begingroup$ @Juris Steprans: I'm sorry, I can't see what you're implying. Could you be more precise, please? $\endgroup$– DaveCommented Apr 18, 2012 at 11:51
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2$\begingroup$ 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) $\endgroup$– Asaf Karagila ♦Commented Apr 18, 2012 at 15:50
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2 Answers
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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.
answered Apr 18, 2012 at 12:09
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$\begingroup$ 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". $\endgroup$– DaveCommented Apr 18, 2012 at 12:13
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$\begingroup$ No, that doesn´t work. You first need to add a constant to the language. $\endgroup$ Commented Apr 18, 2012 at 12:15
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1$\begingroup$ 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! $\endgroup$– DaveCommented Apr 18, 2012 at 12:19
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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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$\begingroup$ 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". $\endgroup$ Commented Apr 21, 2016 at 20:44
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2$\begingroup$ @GeraldEdgar While I agree with the general sentiment, the compactness theorem for countable theories does not need the axiom of choice. $\endgroup$ Commented Apr 21, 2016 at 20:49
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1$\begingroup$ 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. $\endgroup$– ACLCommented Apr 22, 2016 at 6:11
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$\begingroup$ 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 $K$. $\endgroup$ Commented Apr 22, 2016 at 7:00
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$\begingroup$ And I disagree with ACL: $F(t)$ cannot be real closed. For instance, $t$ will never be a square in $F(t)$. $\endgroup$ Commented Apr 22, 2016 at 7:07