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The questions in model theory I am trying to tackle is: Show that there is a countable model of $Th(\langle \mathbb{R};+,.,-,-,1,< \rangle) $ which is non archimedean.

Honestly i dont really know what to do, i read in another stack exchange post that Th(R) is the set of all true first order sentences in the standard language that hold in the model R. Next, you notice that the standard language is itself countable and thus, applying Lowenheim-Skolem, you conclude that a countable model exists.

So i suppose that deals with countability, but what about the non archimedean property?

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  • $\begingroup$ What is $Th$ ?; $\endgroup$ – zoli Mar 22 '15 at 13:41
  • $\begingroup$ The set of L sentences which are true in $\langle \mathbb{R};+,.,-,-,1,< \rangle) $ $\endgroup$ – ENAFMTH Mar 22 '15 at 13:46
  • $\begingroup$ Incidentally, this theory is well-studied, and its models are called "real closed fields". $\endgroup$ – Hurkyl Jan 8 '17 at 18:56
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We use Compactness and Lowenheim-Skolem.

Let $T$ be the set of sentences in our language $L$ that are true in the reals. Produce a new language $L'$ by adding the constant symbol $c$ to $L$. Produce a new theory $T'$ by adding to $T$ the axioms $1\lt c$, $1+1\lt c$, $1+1+1\lt c$, and so on. (We have left out some parentheses.)

Then any finite subset $F$ of $T'$ is consistent, since if the "largest" new axiom in $F$ is $n\lt c$, then we can produce a model of $F$ by letting the underlying set be the reals, and interpreting $c$ to be a large enough integer. So $T'$ is consistent, and therefore has a countable model by Lowenheim-Skolem. This is a countable non-Archimedean model of $T$.

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  • $\begingroup$ It should be $n<c$ in the beginning of your second paragraph. $\endgroup$ – Pece Mar 23 '15 at 9:26
  • $\begingroup$ @Pece: Thank you, fixed. $\endgroup$ – André Nicolas Mar 23 '15 at 13:45
  • $\begingroup$ I am sure this is the answer that the OP is looking for, but the field of Puiseux series over the algebraic reals provides a nice algebraic alternative. $\endgroup$ – Rob Arthan Mar 23 '15 at 20:32
  • $\begingroup$ Yes, there are more concrete (and therefore more illuminating) examples. But compactness gives an instantaneous solution. More interesting, from the point of view of Model Theory, are explicit characterizations of the definable subsets of the reals. $\endgroup$ – André Nicolas Mar 23 '15 at 20:42
  • $\begingroup$ @Everybody: Oops! I meant to write "I am sure that this is not the answer the OP is looking for ...". $\endgroup$ – Rob Arthan Mar 23 '15 at 23:19

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