# Are faces of a compact, convex body “opposed” iff their extreme points are pairwise “opposed”?

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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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

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.
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