Consider the hierarchy of relative geometric constructibility by
straightedge and compass. Namely, given a geometric figure $B$, a
set of points in the plane, we define that geometric figure $A$ is
*constructible from* $B$, written as $$A\leq B,$$ if from points in
$B$ using straightedge and compass we may construct every point in
$A$.

This is a partial preorder on geometric figures, considered as sets of points in the Euclidean plane. The order gives rise to a corresponding strict order notion $A<B$, which holds when from $B$ we may construct $A$ but not conversely. That is, $$A<B\qquad\text{ if and only if }\qquad A\leq B\quad\text{ and }\quad B\not\leq A.$$

**Question.** Is the hierarchy of relative geometric constructibility a
dense order? That is, if $A<B$ for figures in the plane, is there a figure $C$
such that $A<C<B$?

I would also be interested to know whether the answer depends on considering only finite sets of points or infinite sets.

[**Update.** The main question is now answered by the maximality argument of Pace Nielsen below, but remains open in the case of finite figures. That is, if $A$ and $B$ are finite geometric figures with $A<B$, must there be some figure $C$ strictly between?]

Let me make a few observations. First, relative constructibility is symmetric on line segments. That is, if a line segment $AB$ is constructible from $CD$, then $CD$ is constructible from $AB$. Thus, the relation of relative constructibility on line segments is an equivalence relation, with no occurences of the strict order. The argument is given on my blog post. For this reason, there will be no counterexamples to density using segments only.

The symmetry claim is not true, however, for triangles. For example, from a unit length and $\sqrt[3]{2}$, we can construct a unit isosceles right triangle, but from such a triangle, we cannot construct the length $\sqrt[3]{2}$. This instance is also discussed on my blog post.

Since this case has the feeling of being perhaps a minimal extension, it might be considered as a natural candidate counterexample, and several people (including David Madore and Nikolay Nikolov) have proposed various arguments that make definite progress. But the matter is currently still open.

A constructively closed set $S$ of points in the plane is determined by any two of them and the set of points on the corresponding line $\ell$. The reason is that if point $p\in S$, then the distances between $p$ and two points on $S\cap\ell$ will be a distance witnessed on $S\cap\ell$, and so we can construct $p$ via circles from points on $S\cap\ell$.

Therefore, a constructively closed set of points in the plane is determined by a unit segment and the real field of distances realized by that set. (one can also simply view the points as complex numbers, as determined by a unit segment from the set.) For this reason, the question is equivalent to the following:

**Question.** If $K< L$ are quadratically closed field
extensions of the rationals, must there be a quadratically closed
field $F$ strictly between them?
$$K< F< L$$
Such a field would correspond to a strictly intermediate set of points between $K$ and $L$ in the hierarchy of relative constructibility.

Regarding the natural candidate counterexample, the question would be: is there a quadratically closed field strictly between the quadratic closure of $\mathbb{Q}$ and the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$?

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