Proving that a strange model for real numbers exists using the compactness theorem.

Question

Problem statement:

*Let's look at the model $(\mathbb{R},+,\cdot,<,0,1)$. Prove that there exists a model of the theory $Th(\mathbb{R})$ in wich the following holds true: $$\exists a(a > 0 \wedge \forall r(r > 0 \Rightarrow r > a))$$ *

I found this problem in a textbook and it is expected to be solved using the compactness theorem.

But the problem i have with the above is that in the theory of real numbers the above statement can be proven to be incorrect, hence it is not correct in any model of $Th(\mathbb{R})$. I suspect that there is possibly an error in the problem statement, but just to check if there is something i overlooked, i decided to post the problem here. Does anyone have an idea how to make something out of this?

Thanks in advance.

Answer 1

Either you have made an error in transcribing the problem here, or the author made an error when writing the text.

The problem is almost certainly asking for a model with an element $a$ having the property that:

• $a > 0$
• For every positive real number $r$, you have $r > a$.

Note that $r > a$ is only required to hold for $r$ being a positive real number constant, not all positive elements of the model.

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Incidentally, the relevant subject here if you want to study models of this specific theory is the theory of real closed fields; the models you seek are precisely the real closed fields that contain $\mathbb{R}$ as a proper subfield.

Answer 2

The statement as quoted and the statement as you transcribed in your comments to Hurkyl are not equivalent. Your transcribed statement, $\exists a > 0 \forall r (r > 0 \implies r>a)$ is, as you noticed, obviously contradictory, instantiating $r$ to $a$ itself. But this is not what the problem is asking you to derive. The problem is hinting at the construction of the hyperreals, an application of the Compactness theorem used by Abraham Robinson to justify the introduction of infinitesimal elements. In this new model, $a$ will be a hyperreal, and so will not be in the scope of the restricted quantifier "for all positive real numbers". Here's a simple construction.

Expand the language of your theory by adding one new constant $c_r$ to each real number $r$, which is taken to denote that number, and take $Th(\mathbb{R})$ to be the set $\{\phi \; | \; \langle \mathbb{R}, 0, 1, \dots, r, \dots, +, \times, -, ^{-1}, < \rangle \models \phi\}$, i.e. the set of all true sentences of the expanded language.

Now, expand the language again by introducing a new constant $h$ and let $\Sigma = \{h > 0 \} \cup \{c_r > h \; | \; r \in \mathbb{R}, r>0\}$. I claim that $\Sigma$ is satisfiable. To see this, let $\Sigma_0$ be a finite subset of $\Sigma$. Then there will be finitely many constants occurring in $\Sigma_0$, say $c_{r_0}, \dots, c_{r_n}$. Without loss of generality, let $r_0$ be the least real number among $r_0, \dots, r_n$. Interpret $h$ as $\frac{r_0}{2}$. Then it's clear that $h>0$ (since $r_0 > 0$) and, moreover, $h < c_{r_i}$ for any $i \leq n$. So $\mathbb{R} \models \Sigma_0$, whence $\Sigma_0$ is satisfiable. But $\Sigma_0$ was an arbitrary finite subset of $\Sigma$, so $\Sigma$ is finitely satisfiable. Hence, by Compactness, $\Sigma$ is satisfible, and hence has a model, say $\mathbb{R}^*$. It's clear that we also have $h >0$ and, for every positive real number $r$, $c_r > h$, so $\mathbb{R}^*$ is the desired model.