# Is there a closed, convex subset $C$ of $\mathbb{R}^3$, such that any proper, closed, convex shape in the plane is a section of $C$?

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.

• No, half planes are closed and convex. And yes, $C$ would have to unbounded. Jun 10, 2022 at 21:17
• 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".
– Ruy
Jun 10, 2022 at 21:24
• 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/… Jun 10, 2022 at 21:42
• 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. Jun 10, 2022 at 23:23
• 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). Jun 11, 2022 at 10:38

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.