MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $P$ be a compact, convex subset of $\mathbb{R}^n$ (infinite-dimensional generalisations welcome, but not necessary). Let's say that disjoint subsets $W_1$, $W_2$ $\subset P$ are opposed if there exist parallel hyperplanes $H_1$, $H_2$ supporting $P$, such that $W_i \subset H_i \cap P$.

Let $F_1$ and $F_2$ be faces of $P$, such that their extreme points are pairwise opposed: i.e. $v_1, v_2 \in P$ are opposed whenever $v_i$ is an extreme point of $F_i$. Are $F_1$ and $F_2$ opposed?

I have a tentative proof when $P$ has affine dimension equal to 2, which I am struggling to generalise even to 3 dimensions. The converse is trivial. I'd also be interested to know if a proof of this requires some restriction on $P$ (e.g. letting $P$ be a polytope).

share|cite|improve this question
Are not the vertices of a tetrahedron (more generally a simplex) always opposed? Gerhard "Ask Me About System Design" Paseman, 2012.08.02 – Gerhard Paseman Aug 3 '12 at 0:02
@Gerhard: Notice that he defined subset opposition to require disjointness, whereas no two facets of a simplex are disjoint. – Joseph O'Rourke Aug 3 '12 at 0:59
That is right. If t&e poster had requested as a condition that the faces F_i be disjoint, there would be less of an issue. Also, I am not sure what dimension a facdt is, but I note that some edges of a simplex are opposed in pairs. Gerhard "Ask Me About System Design" Paseman, 2012.08.02 – Gerhard Paseman Aug 3 '12 at 2:56
up vote 0 down vote accepted


Let $P$ be the convex hull of two parabolic arcs, say $$\{\\,(x,0,z)\in \mathbb R^3\mid 1\ge z=x^2\\,\}$$ and $$\{\\,(0,y,z)\in \mathbb R^3\mid 0\le z=-y^2+\varepsilon\cdot y+1\\,\}.$$ Take $$F_1=\{\\,(x,0,1)\in \mathbb R^3\mid |x|\le 1\\,\}$$ and $$F_2=\{\\,(0,y,0)\in \mathbb R^3\mid 0\le -y^2+\varepsilon\cdot y+1\\,\}$$

You can approximate it by a polyhedra in such a way that $F_1$ and $F_2$ are still edges. This leads to a polyhedral example.

share|cite|improve this answer
I don't understand the definition of the first parabolic arc. What is $y$? – Will Sawin Aug 2 '12 at 20:15
@Will, $y=x$, now it is fixed. – Anton Petrunin Aug 2 '12 at 21:33
Thank you very much, no wonder I was having no luck. I'm guessing that the convex hull of the points (0,0,0), (0,1,0), (0,e,1+e), (1,0,1), (-1,0,1), for small e, would work for a similar reason? – Sabri Aug 3 '12 at 11:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.