Hahn-Banach Theorem for convex polytopes and their supporting hyperplanes

Question

A polytope in $\mathbb R^n$ is the convex hull of a nonempty finite set in $\mathbb R^n$.

Let $C$ be a polytope in $\mathbb R^n$. We shall say that a hyperplane $H\subseteq \mathbb R^n$

$\bullet$ weakly supports $C$ if the set $C\setminus H$ is connected;

$\bullet$ strongly supports $C$ if $C\setminus H$ is connected and $\dim(C\cap H)=\dim(C)-1$.

Problem. Let $A,B$ be two disjoint polytopes in $\mathbb R^n$. Is there a hyperplane $H$ in $\mathsf R^n$ such that

1. $H$ weakly supports both $A$ and $B$;
2. $H$ strongly supports $A$ or $B$;
3. $H$ separates $A$ and $B$ in the sense that for any points $a\in A$ and $b\in B$ the segment $[a,b]$ intersects the hyperplane $H$?

Answer 1

Edited to give a more symmetric and easily visualized solution.

Counterexample in $\mathbb{R}^3$. $A$ and $B$ will be two tetrahedrons. $A$ has vertices at $(1,-1,0)$, $(-1,-1,0)$, $(0, -1, .1)$ and $(0,-.1,0)$. $B$ has vertices at $(0, 1, 1)$, $(0,1, -1)$, $(.1, 1, 0)$ and $(0, .1, 0)$. $A$ is wide and flat, $B$ is tall and thin.

$A$ is nearly contained in the $xy$ plane and $B$ is nearly contained in the $yz$ plane. Both have an outer side which is parallel to the $xz$ plane; their hyperplanes lie to the same side of $A$ and $B$. The hyperplanes of the other three sides of $A$ are nearly parallel to the $xy$ plane and slice through the middle of $B$, while the hyperplanes of the other three sides of $B$ are nearly parallel to the $yz$ plane and slice through the middle of $A$.

Answer 2

This is very similar to Nik Weaver's answer but may be easier to visualise. Fix a small $\epsilon>0$. Let $A$ have vertices $(\pm 1,0,-1)$ and $(0,\epsilon^2,-1)$ and $(0,0,\epsilon-1)$. This means that $A$ is close to a line segment $A'$ parallel to the $x$-axis, the bottom face is horizontal, and the other three faces are close to the plane $y=0$. Let $B$ be the image of $A$ under $(x,y,z)\mapsto (y,x,-z)$, so $B$ is close to a line segment $B'$ parallel to the $y$-axis, the top face is horizontal, and the other three faces are close to the plane $x=0$. Both $A$ and $B$ lie above the horizontal face of $A$ and below the horizontal face of $B$. The plane $y=0$ cuts the line segment $B'$ at $(0,0,1)$, and similarly, any of the three nearly-vertical supporting planes of $A$ cuts $B$. By symmetry, any of the three nearly-vertical supporting planes of $B$ cuts $A$. By plotting you can check that $\epsilon=1/2$ is (just) small enough, but smaller $\epsilon$ makes the picture clearer.

Maple code is as follows:

with(plots): with(plottools):
e := 1/3;
A := [[-1,0,-1],[1,0,-1],[0,e^2,-1],[0,0,e-1]];
B := map(u -> [u[2],u[1],-u[3]],A);
xp := (u,v) -> [u[2]*v[3]-u[3]*v[2],u[3]*v[1]-u[1]*v[3],u[1]*v[2]-u[2]*v[1]]:
dp := (u,v) -> u[1]*v[1]+u[2]*v[2]+u[3]*v[3]:
dsc := proc(u)
 local u0,v0,w0;
 u0 := (u[1] +~ u[2] +~ u[3])/~3.;
 v0 := u[1] -~ u0;
 v0 := v0 /~ sqrt(dp(v0,v0));
 w0 := u[2] -~ u0;
 w0 := w0 -~ dp(w0,v0) *~ v0;
 w0 := w0 /~ sqrt(dp(w0,w0));
 plot3d(u0 +~ (r * cos(t)) *~ v0 +~ (r * sin(t)) *~ w0,r=0..2.5,t=0..2*Pi,style=wireframe,colour=gray,scaling=constrained,axes=none);
end:
display(
 polygon([A[1],A[2],A[3]]),
 polygon([A[1],A[2],A[4]]),
 polygon([A[1],A[3],A[4]]),
 polygon([A[2],A[3],A[4]]),
 polygon([B[1],B[2],B[3]]),
 polygon([B[1],B[2],B[4]]),
 polygon([B[1],B[3],B[4]]),
 polygon([B[2],B[3],B[4]]),
 dsc([A[2],A[3],A[4]]),
 scaling=constrained,axes=none
);

Here is a picture with $\epsilon=1/2$ (but it is much easier to see if you use Maple or similar to make a plot that you can rotate with your mouse).

Answer 3

Another visualization with Maple $\ge$ 2015 is as follows.

with(PolyhedralSets):
ps1 := PolyhedralSet([[1, -1, 0], [-1, -1, 0], [0, -1, 1/10], [0, -1/10, 0]]):
ps2 := PolyhedralSet([[0, 1, 1], [0, 1, -1], [1/10, 1, 0], [0, 1/10, 0]]):
Plot([ps1, ps2], scaling = constrained);