How many faces does an $n$-dimensional slice through the center of $[-1,1]^m$ have?
I read that:
"...any $n$-dimensional slice of $[-1,1]^m$ is a body with at most $m$ pairs of faces."
How should I prove this? How do we define a slice here?
My thoughts:
I took the example of $m=n=3$, and it made sense, i.e. cutting a biscuit sort of a shape from a cube in $3D$. I don't know how to generalize though!
Could someone please provide hints or any visuals that might aid my imagination?
I would take "slice" to mean the intersection of a linear subspace of $\mathbf{R}^m$ with the cube. If $m=n$ then your subspace is all of $\mathbf{R}^m$ and your slice of the cube is the entire cube. (So I'm not sure what you mean by "biscuit sort of shape".) If you take $m=3$ and $n=2$, you are slicing the cube with a plane. If the plane is parallel to a pair of faces of the cube, the slice will be a square, but tilting the plane can also give rectangular and hexagonal slices.
The boundary of the slice will be formed from the intersection of the linear subspace and the boundary of the cube, which has $2m$ faces (of dimension $m-1$), so the slice won't have more than $2m$ faces. Since the cube is convex, its intersection with a linear subspace is also convex, so we don't have to worry that a face of the cube will contribute more than one face to the slice. Hence $2m$ really is the maximum. The hexagonal slice of the $3$-dimensional cube shows that this maximum is attainable. On the other hand, if $n=1$, the slice is a line segment, and the number of faces is always $2$ (the two endpoints of the line segment) no matter what $m$ is.
If you take the $4$-dimensional cube, its intersection with a plane will be a polygonal region with up to $8$ sides. Its intersection with a $3$-dimensional subspace will be a polytope with up to $8$ faces.