To be precise, by a section of $C$ I mean the intersection of $C$ with a hyperplane. I want every proper, closed, convex subset of the plane, up to translation and rotation, to be appear as sectiond. So, for example, I want both the unit disc and the disc of radius $8.45$ to appear as sections.

It was recently brought to my attention that there is a closed convex cone in $\mathbb{R}^3$ such that the set of conic sections is "dense" amongst the proper, closed, convex sets of the plane (dense in exactly metric, I do not know). This astonished me and motivated the above question. I would be very surprised if the answer is yes, and am kind of expecting some argument which shows that such a $C$ obviously cannot exist. But as I've just learnt, convex subsets can be highly exotic.

If the answer is positive a bonus question would be if it can be arranged so that every closed, convex subset of the plane, up to translation and rotation, appears exactly once as a section.

  • $\begingroup$ No, half planes are closed and convex. And yes, $C$ would have to unbounded. $\endgroup$ Jun 10, 2022 at 21:17
  • 2
    $\begingroup$ As a side remark, the answer is no in case you remove the word "closed" from the question. A 3-dimensional convex set has at most $2^{\mathbb N}$ sections while there are $2^{2^{\mathbb N}}$ different "shapes". $\endgroup$
    – Ruy
    Jun 10, 2022 at 21:24
  • 2
    $\begingroup$ Aubrun is still active on MO. I placed a comment there with a link to your question here, also asked for a standard reference. mathoverflow.net/questions/423174/… $\endgroup$
    – Will Jagy
    Jun 10, 2022 at 21:42
  • 2
    $\begingroup$ Specifically, it looks to me that the only convex subsets of three-space containing a plane are Cartesian products of that plane and an interval. $\endgroup$ Jun 10, 2022 at 23:23
  • 2
    $\begingroup$ Let $X$ be the space of affine isometric embeddings from $\mathbf{R}^2$ to $\mathbf{R}^3$ and $Y$ the space of convex compact subsets of $\mathbf{R}^2$, equipped with Hausdorff distance. I can imagine an argument showing impossibility along the following lines: if there was such a convex set $C$, there would be a onto and locally Lipschitz map from an open subset of $X$ to $Y$ ; this is not possible since $X$ is finite-dimensional while the space $Y$ is infinite-dimensional. (Here I consider only sections though the interior on $C$; tangent planes produce at most countably many sets). $\endgroup$ Jun 11, 2022 at 10:38

1 Answer 1


I'm not answering the question, but clarify on the denseness result alluded to: there is a convex cone $C \subset \mathbf{R}^3$ with the following property. If you fix an affine hyperplane $H$ not containing zero, a compact convex set $K \subset H$ and $\varepsilon >0$, then there is a linear transformation $f : \mathbf{R}^3 \to \mathbf{R}^3$ such that the intersection $f(C) \cap H$ is at Hausdorff distance $< \varepsilon$ from $K$.

These linear maps $f$ induce projective transformations on $H$ ; so in other words there is a planar convex body $L$ whose orbit under the projective group is dense (you only want to consider projective maps which do not send points of $L$ to infinity). This can exist because projective transformations do not respect distance, so you could hide the projective image of any convex set you want in an arbitrary small neighbourhood. I wrote a note about this some years ago. http://math.univ-lyon1.fr/homes-www/aubrun/recherche/dense-projective-orbit.pdf

You cannot achieve this with $\varepsilon =0$ if $C$ is a cone. For example you could not see both a triangle and a square as sections (through the interior) of the same cone, since they are not projectively equivalent.

I would also be surprised if a convex set with your property exist.

  • $\begingroup$ Thank you for giving an answer here! I see, you link to a reference also. $\endgroup$
    – Will Jagy
    Jun 11, 2022 at 15:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .