Proposition 4

Let F be a facet, and let L be an affine subspace of E that is the intersection of hyperplanes belonging to $\mathfrak{H}$; denote by $\mathfrak{N}$ the set of hyperplanes $H \in \mathfrak{H}$ that do not contain L. The following conditions are equivalent:

(i) There exists a facet F' with support L that meets $\overline{F}$. (ii) There exists a facet F' with support L contained in $\overline{F}$. (iii) There exists a point x in $L \cap \overline{F}$ that does not belong to any of the hyperplanes of $\mathfrak{N}$.

If these conditions are satisfied, $L \cap D_{\mathfrak{N}}(F)$ is the unique facet with support $L$ contained in $\overline{F}$.

(i) $\Longrightarrow$ (ii): Since $\overline{F}$ is a union of facets (Prop. 3, (ii)), every facet that meets $\overline{F}$ meets a facet contained in $\overline{F}$, and so is equal to it.

(ii) $\Longrightarrow$ (iii): Every point $x$ of $F'$ satisfies (iii) since every hyperplane of $\mathfrak{H}$ containing $x$ contains $F'$, and hence $L$.

(iii) $\Longrightarrow$ (i): Let $x$ be a point satisfying (iii) and let $F'$ be the facet containing $x$; it is clear that $F'$ meets $\overline{F}$. Let $H \in \mathfrak{H}$; then $x \notin H$ if $H \in \mathfrak{N}$ and clearly $x \in H$ if $H \notin \mathfrak{N}$; consequently the support of $F'$ is the intersection of the hyperplanes of $\mathfrak{H} - \mathfrak{N}$, and is equal to $L$.

Finally, let $F'$ be a facet with support $L$ contained in $\overline{F}$, and let $x$ be a point of $F'$; since no hyperplane of $\mathfrak{N} \subset \mathfrak{H}$ passes through $x$, Prop. 1 shows that there exists a convex open neighbourhood $U$ of $x$ that does not meet any hyperplane of $\mathfrak{N}$. Since $x$ is in the closure of $F$, $U \cap F \neq \varnothing$; now $\mathfrak{N}$ is the set of hyperplanes $H \in \mathfrak{H}$ that do not contain $F'$, and for all $H$ in $\mathfrak{N}$ we have $D_H(x) = D_H(U) = D_H(U \cap F) = D_H(F)$ and formula (2) implies that

Q.E.D.