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Let $\mathfrak{R}= ⟨\mathbb{R},<,+,-,\cdot,0,1⟩$ be the standard model of $Th(\mathbb{R})$ in the language of ordered fields. I need to show that there exists a (non standard) model of $Th(\mathbb{R})$ with the same cardinality of $\mathbb{R}$ but not isomorphic to $\mathfrak{R}$.

My doubts are on how to prove the "not-isomorphic" part, so I'll just sketch my proof of existence of a model:

Let's expand the language adding a new constant $c_x$ for each $x \in \mathbb{R}$ and an additional new constant $c$. We consider the following set of formulas:

$$ \Gamma = Th(\mathbb{R}) \cup \{c_x \ne c_y\}_{x,y \in \mathbb{R}, x \ne y} \cup \{c_x < c\}_{x \in \mathbb{R}}$$

We can show $\Gamma$ is finitely satisfiable, thus, by Compactness Theorem, satisfiable, thus it has a model with cardinality $\kappa \ge |\mathbb{R}|$. By the Löwenheim-Skolem Downward theorem it has a model of cardinality $c = |\mathbb{R}|$, say $^*\mathfrak{R}$.

Now my doubts are how to show that $^*\mathfrak{R} \not \cong \mathfrak{R}$. I guess the lack of an isomorphism should be showed after shrinking back the structures to the original language, i.e. removing all the constant symbols. Moreover, as the two models have the same cardinality we cannot simply conclude there is not a bijection, so I think a possible proof is assume there is such an isomorphism and get a contradiction with the order relation, but all my reasonings in this direction seem at a dead point. How can one show there is not such an isomorphism?

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    $\begingroup$ One can just add the axioms $\;\;\; 1\hspace{-0.05 in}+\hspace{-0.05 in}1+...\hspace{-0.05 in}+1\hspace{-0.05 in}+\hspace{-0.05 in}1 \: < \: c \;\;\;$ instead. $\;\;\;\;\;\;\;\;$ $\endgroup$ – user57159 Jul 2 '15 at 10:09
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    $\begingroup$ Let $f : ^{\star}\mathfrak R \to \mathfrak R$ be a bijection. Then for some natural number $n$, we have $1 + 1 + ... +1 > f(c)$, whereas $1 + 1 + ... +1 < c$ (the addition is $n$-times). Thus $f$ does not preserve formulas. $\endgroup$ – Levon Haykazyan Jul 2 '15 at 11:37
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    $\begingroup$ The standard model is Archimedean, the model obtained from adding each of $c\gt 1+\cdots +1$ is not. $\endgroup$ – André Nicolas Jul 2 '15 at 11:56
  • $\begingroup$ @RickyDemer sure. Shouldn't make a big difference though :) $\endgroup$ – Manlio Jul 2 '15 at 13:58
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    $\begingroup$ That it is not Archimedean? This is the definition of Archimedean. And if two ordered fields are isomorphic, they are both Archimedean or both not Archimedean. $\endgroup$ – André Nicolas Jul 2 '15 at 14:06
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As has been pointed out in comments, we do not need to add a constant symbol for every element of $\mathbb{R}$. Let $L$ be the usual first-order language for ordered fields. Extend $L$ to $L'$ by adding a constant symbol $c$. Let $T$ be the set of all sentences of $L$ true in the reals, under the usual interpretation. Make 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.

Then, as pointed out in the OP, the theory $T'$ is consistent, and therefore has a model $M$ of cardinality equal to the cardinality of $\mathbb{R}$. The sentences $1\lt c$, $1+1\lt c$, $(1+1)+1\lt c$, and so on, are true in $M$. It follows that there is an element $c_M$ of $|M|$ (the underlying set of $M$) such that $1_M\lt_M c_M$, $(1+1)_M\lt_M c_M$, and so on. Now to be technical we should let $M'$ be $M$ restricted to the language $L$. Then $|M'|$ contains an object, namely $c_M$, such that the inequalities $1_{M'}\lt_{M'} 1_{M'}$, and so on, hold.

Finally, we show there is no order isomorphism of $\mathbb{R}$ and $M'$. For let $\phi: \mathbb{R}\to M'$ be such an isomorphism, and suppose that $\phi(K)=c_M$. Since the reals are an Archimedean ordered field, there is a positive integer $n$ such that the interpretation of $1+1+\cdots +1$ ($n$ times) in $\mathbb{R}$ is greater than $K$.

Note that $\phi$ takes the interpretation of $1+1+\cdots+1$ in $\mathbb{R}$ to its interpretation in $M'$. Since $\phi$ is order-preserving, it follows that the interpretation of $1+1+\cdots +1$ ($n$ times) in $M'$ is greater (in $M'$) than $c_M$. But this is not the case.

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Instead of compactness, you can try an ultrapower argument.

Let $\cal U$ be a nonprincipal ultrafilter on $\Bbb N$. Then $\Bbb{R^N}/\cal U$ is elementarily equivalent to the reals. And it is not hard to show it has the wanted cardinality, and that it is non-standard.

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If you are familiar with types, another way to state the answer is as follows: If two models are isomorphic, they must realize the same types. However the type (over the empty set) realized by $c$ in $^*R$ is not realized in $R$.

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  • $\begingroup$ Looks a very quick way to conclude! Unfortunately we didn't meet type theory in the course of logic and, after a fast check, looks like the theory behind requires more than few minutes to be read :) Looks interesting though! Hope I'll have the chance to read something about! $\endgroup$ – Manlio Jul 3 '15 at 16:02
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The problem with your method is that the interpretation of $\{c_x \ | \ x \in \mathbb{R}\}$ in a model of $\Gamma$ could be any uncountable set. For instance, you could interpret all $c_x$ as distinct elements of $[0;1]$ and $c$ as $2$ in $\mathbb{R}$.

Another solution using non elementary model theoritic and algebraic results, but provinding an archimedean model:

Let $\mathcal{B}$ be a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$ and let $x$ be an element of $\mathcal{B}$. $\mathbb{Q}(\mathcal{B}-\{x\})$ is an uncountable proper subfield of $\mathbb{R}$. It is uncountable because $\mathcal{B}$ is, and it is proper because it is the increasing union of fields that don't contain $x$. Its real closure $F$ is proper because it is algebraic over it so it can't contain $x$. And $F$ is uncountable too.

Since the theory of real closed fields is a complete theory which $\mathfrak{R}$ models, $\mathfrak{F} = (F,+,.,0,1,<)$ is elementary equivalent to $\mathfrak{R}$.

Let's assume there is an isomorphism $f: \mathfrak{R} \rightarrow \mathfrak{F}$. $f|_{\mathbb{Q}} = id_{\mathbb{Q}}$, so $f(x) = f(\sup(\{q \in \mathbb{Q} \ | \ q < x\})) = \sup(\{f(q) \ | \ q \in \mathbb{Q}, q < x\}) = \sup(\{q \in \mathbb{Q} \ | \ q < x\}) = x$, which is impossible because $x \notin F$.

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