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.
 A: 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.
