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).

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    $\begingroup$ Are not the vertices of a tetrahedron (more generally a simplex) always opposed? Gerhard "Ask Me About System Design" Paseman, 2012.08.02 $\endgroup$ Aug 3 '12 at 0:02
  • $\begingroup$ @Gerhard: Notice that he defined subset opposition to require disjointness, whereas no two facets of a simplex are disjoint. $\endgroup$ Aug 3 '12 at 0:59
  • $\begingroup$ 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 $\endgroup$ Aug 3 '12 at 2:56


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.

  • $\begingroup$ I don't understand the definition of the first parabolic arc. What is $y$? $\endgroup$
    – Will Sawin
    Aug 2 '12 at 20:15
  • $\begingroup$ @Will, $y=x$, now it is fixed. $\endgroup$ Aug 2 '12 at 21:33
  • $\begingroup$ 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? $\endgroup$
    – Sabri
    Aug 3 '12 at 11:10

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