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Will Sawin
Will Sawin
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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 $$ \exists y \exists z \textrm{ such that } f(x,y,z)=0, -1 \leq y \leq 1, -1 \leq z \leq 1 , \textrm{ and } \forall w, y \neq w^2$$$$ \exists y \exists z \textrm{ such that } f(x,y,z)=0, -1 \leq y \leq 1, -1 \leq z \leq 1 , \textrm{ and } \forall w, y+1 \neq w^2$$ 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.

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.

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.

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 $$ \exists y \exists z \textrm{ such that } f(x,y,z)=0, -1 \leq y \leq 1, -1 \leq z \leq 1 , \textrm{ and } \forall w, y \neq w^2$$ 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.

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.

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.

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.

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.

Source Link
Will Sawin
Will Sawin
  • 148.4k
  • 9
  • 324
  • 563

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 $$ \exists y \exists z \textrm{ such that } f(x,y,z)=0, -1 \leq y \leq 1, -1 \leq z \leq 1 , \textrm{ and } \forall w, y \neq w^2$$ 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.

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.