Is this theory the complete theory of the real ordered field?

Question

We know that the real ordered field can be characterized up to isomorphism as a complete ordered field. However this is a second order characterization. That raises the following question. Consider the following theory. We take as axioms the axioms for ordered fields, and then add an axiom schema that states that every nonempty set that is definable without parameters that is bounded above has a least upper bound. Is that theory the complete first order theory of the real ordered field? And if it is not, can someone exhibit a model of that theory that is not elementarily equivalent to the real ordered field? I asked this question on math stack exchange, but I did not receive an answer.

Answer 1

It is not. Using set forcing, we can add 'undefinable' reals in a controlled manner, while keeping complexity of parameter-free definable sets low.

Specifically, let $M$ be a countable $ω$-model of ZFC\P, real $r$ be Cohen generic over $M$, and $ℝ_M(r)$ be the minimal field of reals containing $r$ and all reals in $M$ (in the initial revision, I called it $ℚ_M(r)$; it is a proper subset of $ℝ^{M[r]}$). Then, in $ℝ_M(r)$, every parameter-free definable bounded subset has the least upper bound, but $\sqrt{|r|}$ does not exist.

Because $r$ is transcendental over $M$, $\sqrt{|r|}∉ℝ_M(r)$.

Next, let $X$ be a bounded set of reals parameter-free definable in $ℝ_M(r)$. $ℝ_M(r)$ may not be closed under square roots, but it witnesses that the Cohen forcing is homogeneous, and therefore $\sup(X)∈M$, as required.

In more detail, by genericity, $\sup(X)$ can be determined with arbitrary given precision by a forcing condition (with the forcing relation definable in $M$). For Cohen forcing (modulo choice of representation), the conditions are rational $p<t$ (asserting $p<r<t$). Now, if $r$ is Cohen generic, then so is $qr$ for all nonzero $q∈ℚ$, and $ℝ_M(qr) = ℝ_M(r)$ (and same with $q+r$), and therefore all conditions lead to the same $\sup(X)$.

An interesting remaining question is whether there are computable examples.

Answer 2

Noah Schweber suggested on math stack exchange that $\mathbb Q^{alg}(r)$ could satisfy this weak least upper bound property for some fixed transcendental number $r$, say $r=\pi$. In this case, it would be a counterexample.

I think this may be true, but I don't think it is possible to prove with current technology.

Consider definable sets of the form $$ \exists y \exists z \textrm{ such that } f(x,y,z)=0, -1 \leq y \leq 1, -1 \leq z \leq 1 , \textrm{ and } \forall w, y+1 \neq w^2$$ for a complicated polynomial $f$.

Then $x,y,z$ must lie on a bounded subset of the surface $S$ defined by the equation $f$. The last condition forces $y \notin \mathbb Q^{alg}$, so $(x,y,z)$ must lie on the intersection of a rational curve with this bounded subset. Unless this rational curve has a very particular form, possible $(x,y,z)$ are dense on any intersection of a rational curve with a bounded subset, since almost any rational function in $\pi$ will not be a perfect square.

Does this set have a supremum?

It might be reasonable to guess that rational curves on $S$ are either dense in the real topology, or contained in some proper closed subset, and in either case there exists a supremum, since we could either ignore the last condition or replace it with an additional equation, and in either case solve it in the usual real closed fields way.

But we can't prove this for an arbitrary surface. If neither is true, the topological closure of the space of rational curves could be some weird transcendental thing, and the supremum would be a bizarre transcendental.

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This problem motivates an alternative proposal to find a "counterexample". If this theory is complete, it certainly proves $\forall x, x>0 \implies \exists y, y^2=x$. If so, then it proves this using finitely many axioms, and thus finitely many parameter-free definable sets. We can try to show this is impossible by showing that these finitely many parameter-free definable sets admit least upper bounds (or are unbounded) in $\mathbb Q(r_1,\dots, r_n)$, where $r_1,\dots,r_n$ are independent transcendentals.

The trick here is that definable sets like the one above have "true" least upper bounds in $\mathbb R$ that are insensitive to our choice of $r_1,\dots, r_n$, so we can simply choose $r_i$ to be the least upper bounds of the finitely many paramater-free definable sets, if any of them are transcendental.

The difficulty with this approach is that the "true" least upper bound of other parameter-free definable sets could potentially depend on our choice of $r_1,\dots,r_n$. It could theoretically be that no choice works because the least upper bound keeps jumping around to confound us.

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This technique does, at least, answer Tim Campion's question about whether the theory of an ordered field with the axiom scheme that every bounded set definable without parameters in the language of a \emph{field} (not the language of an ordered field) has an upper bound.

Every set definable without parameters in the language of a field is independent of the ordering, by definition, and thus independent of the choice of transcendentals $r_1,\dots, r_n$. In particular, if it contains any nonconstant rational function in $r_1,\dots, r_n$, then it is unbounded. If it does not contain any nonconstant rational function in $r_1,\dots, r_n$, then it is contained in $\mathbb Q^{alg}$ and independent of $r_1,\dots, r_n$. Therefore its least upper bound, if it exists, is independent of $r_1,\dots, r_n$.

So given any finite set of parameter-free definable sets in the language of a field, we can choose $n$ to be at least the the number of sets, and then choose $r_1,\dots, r_n$ to be the least upper bounds of these sets, if any are transcendental, or arbitrary transcendentals, otherwise, showing that these finite set having least upper bounds does not imply that all positive numbers are squares.

Answer 3

Edit: As Emil Jeřábek has pointed out in the comments, there is an error in the paper linked below, and the example does not work. I'll leave the answer here to document the error and add to the interesting list of attempted answers gathered here. An erratum has now appeared: Corrigendum to "[..]" RML, referencing this discussion.

Let me add that Dmytro Taranovsky has given a correct example, but I would still be very interested to see a "forcing-free" and especially an "algebraically natural" example of a $\emptyset$-definably complete ordered field which is not definably complete.

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This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.

Answer 4

In my hasty first reading I did not notice the stipulation about parameter-freeness; I am leaving this answer "only for the record".

The answer to the question is in the positive, provided "real ordered field" is interpreted as "real closed field", and parameters are allowed. The details are worked out in a section 3 of a paper by Salehi and Zarza (a preprint of which can be found here); the paper was published as follows:

S. Salehi & M. Zarza, First-Order Continuous Induction and a Logical Study of Real Closed Fields, Bulletin of the Iranian Mathematical Society volume 46, pages225–243(2020).

Answer 5

This is a non-answer too long for a comment.

If $K$ is an ordered field that satisfies the least upper bound property for sets definable without parameters, then suprema of sets definable with parameters can be arbitrarily closely approximated in $K$.

Consequently, the completion of $K$ is a real-closed field (and $K$ itself is a real-closed field if it happens to be henselian).

To see this, let $X=\{x:K\models\phi(x,\vec a)\}$ be a nonempty bounded set. Applying an affine transformation if necessary, we may assume $0\in X\subseteq(-\infty,1)$. Let $\psi(w)$ denote the formula $$\begin{multline*} e>0\land\forall\vec u\:\bigl[\phi(0,\vec u)\land\forall x\:(\phi(x,\vec u)\to x<1)\\\to\exists z\:\bigl(\phi(z,\vec u)\land\forall x\:(\phi(x,\vec u)\to x<z+e)\bigr)\bigr]. \end{multline*}$$ In words, $\psi(e)$ says that for every set of the form $Y=\{x:K\models\phi(x,\vec u)\}$ for some $\vec u\in K$, if $0\in Y\subseteq(-\infty,1)$, then $\sup Y$ can be approximated within distance $e$, in the sense that there exists $z\in Y$ such that $Y\subseteq(-\infty,z+e)$.

Now, $E=\{e:K\models\psi(e)\}$ is definable without parameters, and clearly $1\in E\subseteq(0,+\infty)$, hence by assumption (or rather, by the equivalent greatest lower bound property), there exists $e_0=\inf E\ge0$.

If $e_0=0$, we are done. However, $e_0>0$ is impossible, because $$e\in E\implies e/2\in E.$$ Indeed, using the notation above, if $z\in Y\subseteq(-\infty,z+e)$, then either $z\in Y\subseteq(-\infty,z+e/2)$, or there exists $z'\ge z+e/2$ such that $z'\in Y\subseteq(-\infty,z'+e/2]$.

Answer 6

EDIT 2: This wrong proof at least shows that the upper bound property holds in $K(\pi)$, when $\phi$ is $\Sigma_1$. Indeed, if $\phi$ is $\Sigma_1$ then we have the inclusion of solution sets

$$ \phi(K) \subset \phi(K(\pi)) \subset \phi(\mathbb R). $$

Both $\phi(K)$ and $\phi(\mathbb{R})$ can be expressed in the same way as a finite union of points and intervals, so the supremum in $K$ is once again the supremum in $K(\pi)$.

EDIT: As Emil Jeřábek points out, this proof breaks by assuming the quantifier-free equivalence preserves the solution set in $K(\pi)$ as well.

Let $K$ denote the field of real algebraic numbers (real-closed, elementarily equivalent to $\mathbb{R}$). I may be missing something basic, but I think we can show that $K(\pi)$ is a counterexample to elementary equivalence in the following way:

First, note that $K$ satisfies the theory in question, since $\mathbb R$ does. Let $\phi(x)$ be a formula without parameters. Since $Th(K) = Th(\mathbb R) = RCF$ eliminates quantifiers, we can replace $\phi(x)$ with a quantifier-free formula whose solution set in $\mathbb{R}$ (and hence in all subfields) is the same. By model completeness of RCF the embedding $K \hookrightarrow \mathbb{R}$ is elementary, so we have

$$ \{x \in K: \phi(x) \} = \{x \in \mathbb{R}: \phi(x) \} \cap K. $$

The solution set of $\phi(x)$ in $\mathbb{R}$ is a finite union of points and intervals. Furthermore, all the points in this solution set are algebraic. If not, since rationals can separate the distinct points and intervals, and are definable without parameters, it follows that there is some formula $\psi$ without parameters whose unique solution in $\mathbb{R}$ is a transcendental number $s$. Additionally, $\psi$ is quantifier-free, since $\phi$ is. But this is clearly impossible.

Suppose that $\phi(x)$ defines a nonempty bounded subset of $K(\pi)$. Then $\phi(x)$ defines a bounded subset of $K$. Furthermore, the (nonempty) solution set of $\phi(x)$ in $\mathbb R$ consists of intervals and algebraic points, so its solution set in $K$ is nonempty as well. Since $K$ satisfies the theory, this means that $\phi(x)$ has a least upper bound $y$ in $K$. Furthermore, this $y$ must actually be the least upper bound of the set $\phi(x)$ defines in $\mathbb{R}$: By our argument above, the least upper bound is either an algebraic point, or the upper bound of an interval in which elements of $K$ are dense.

Therefore, there is some $y \in K$ which is the least upper bound of the set defined by $\phi(x)$, both in $K$ and in $\mathbb R$. But it follows that $y$ is also the least upper bound of the set defined by $\phi(x)$ in $K(\pi)$. This shows that $K(\pi)$, which is not real-closed, must satisfy this theory as well.