I'm having some trouble with the following problem from Using the Borsuk-Ulam Theorem.
Let $P = \mathrm{conv}\{e_1, -e_1, \dots, e_d, -e_d\}$ be a crosspolytope. Show that if a subset $F \subset \{e_1, -e_1, \dots, e_d, -e_d\}$ forms the vertex set of a proper face of the crosspolytope then there is no $i \in \{0, 1, \dots , d\}$ with both $e_i, -e_i \in F$.
The definition of face that I am working with is as follows. A face of a convex polytope $P$ is either $P$ itself or an intersection $P \cap h$ where $h$ is a hyperplane that does not dissect $P$.
As a hint the book suggests considering $h = \{x: \left<a_F, x\right> = 1\}$ for $a_F = \Sigma_{v\in F}v$ as a defining hyperplane.
So let's proceed by way of contradiction assuming that $F$ contains an $e_i$ and $-e_i$ for some $i$. For a hyperplane $h$ as defined above I'm guessing I want to show that $P$ intersects both of the open halfspaces $h^+ = \{x: \left<a_F, x\right> > 1\}$ and $h^- = \{x: \left<a_F, x\right> < 1\}$. Since $e_i - e_i = \vec{0}$ and $a_F$ is just a sum of standard basis vectors there will be a zero in the $i$th component of $a_F$. This means $<a_F, \pm e_i> = 0$ so $\pm e_i \in h^-$.
This is the point in the proof where I get stuck. My question is twofold:
How do we know that $h$ as defined above is a defining hyperplane? Why was it chosen?
How should I proceed at trying to find an element of $P$ in $h^+$? An element in $h^+$ is not immediately obvious to me.
If $F$ contains both $e_i$ and $-e_i$ it contains their midpoint $0$, since each face is a convex set. But $0$ is an interior point of $P$, so the only face of $P$ which contains $0$ is $P$ itself.
To see this using the definition in your book, the hyperplanes through $0$ have the equation $a_1x_1+\cdots+a_dx_d=0$. If $a_j\ne0$ then $e_j$ and $-e_j$ are on opposite sides of the hyperplane, so it does dissect $P$ non-trivially.