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](http://jdh.hamkins.org/the-hierarchy-of-geometric-constructibility-can-we-go-back/).
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 $\mathbb{Q}(\sqrt[3]{2})$?



Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?
There are three answers. Throughout let $qcl(F)$ be the quadratic closure of a field $F$ inside $\mathbb{C}$.


**Part 1:** Yes there is a quadratically closed field strictly between $qcl(\mathbb{Q})$ and $qcl(\mathbb{Q}(2^{1/3}))$. First, find an $S\_4$-extension $K/\mathbb{Q}$ containing $\mathbb{Q}(2^{1/3})$. (My friend Darrin Doud tells me that the Galois closure of adjoining a root of $x^6 + 3x^4 + 3x^2 + 3$ will be such an $S\_4$ extension.) Note that $K$ is a subfield of $qcl(\mathbb{Q}(2^{1/3}))$, being constructible from $2^{1/3}$ by a sequence of square roots.


Now, $K$ has a unique subfield $\mathbb{Q}(\sqrt{-3})$ quadratic over $\mathbb{Q}$. Then $K/\mathbb{Q}(\sqrt{-3})$ is an $A\_4$-extension, and hence there exists a non-Galois subfield $F$ with $[F:\mathbb{Q}(\sqrt{-3})]=4$. Note, in particular, that the Galois closure of $F$ requires a degree $3$ extension. We know that $F$ is not contained in $qcl(\mathbb{Q})$ because $qcl(\mathbb{Q})$ is Galois but contains no cubic extension of $\mathbb{Q}$. Also, $qcl(F)$ is proper in $qcl(\mathbb{Q}(2^{1/3}))$, since $qcl(F)$ is formed only using power-of-2 extensions of $\mathbb{Q}$.


**Part 2:** The order is not dense. Just use Zorn's lemma (or transfinite induction, if you want to avoid the axiom of choice) to construct a quadratically closed subfield of $qcl(\mathbb{Q}(2^{1/3}))$ that is maximal with respect to not containing $2^{1/3}$.


**Part 3:** If by "geometrical figure" you mean a *finite* set of points, then I don't know the answer to the first question anymore. (Edited to add: I've now also answered this question in the negative. I've put it as a separate answer.)


