Proposition 3

Let F be a facet and L its affine support.

(i) The set F is a convex open subset of the affine subspace L of E. (ii) The closure of F is the union of F and facets of dimension strictly smaller than that of F. (iii) In the topological space L, the set F is the interior of its closure.

Since every open half-space and every hyperplane are convex subsets of E, formula (2) shows that F is the intersection of a family of convex subsets, and hence is convex. On the other hand, let a be in F, and let U be a convex open neighbourhood of a in E that does not meet any of the hyperplanes in the set $\mathfrak{N}$ of $H \in \mathfrak{H}$ such that $a \notin H$. For any $H \in \mathfrak{N}$, we thus have $U \subset D_H(a)$, hence $L \cap U \subset F$, so that F is open in the topological space L.

Let b be a point of $\overline{F} - F$, belonging to a facet $F'$, and let $\mathfrak{N}'$ be the set of hyperplanes $H \in \mathfrak{N}$ passing through b; put $\mathfrak{N}'' = \mathfrak{N} - \mathfrak{N}'$. For any H in $\mathfrak{N}''$, we have $b \notin H$ and $b \in D_H(a)$, hence $b \in D_H(a)$ and $D_H(b) = D_H(a)$; by the definition of a facet, we thus have

whereas (3) implies that

hence $F' \subset \overline{F}$. We cannot have $\mathfrak{N}' = \varnothing$, for this would imply that $F = F'$ by (2) and (4), contrary to the hypothesis that $b \notin F$ and $b \in F'$. The support of $F'$ is the set $L' = L \cap \bigcap_{H \in \mathfrak{N}'} H$; we have $a \in L$, but $a \notin H$ for $H$ in $\mathfrak{N}'$, so $L' \neq L$ and finally $\dim L' < \dim L$, that is $\dim F' < \dim F$. This proves (ii).

Let H be in $\mathfrak{N}'$ and let D be the open half-space bounded by H and distinct from $D_H(a)$; we have $b \in H \cap L$, and it is immediate that $D \cap L$ is a half-space of L bounded by the hyperplane $H \cap L$ of L; consequently, every neighbourhood of b in L meets $D \cap L$, and since $D \cap L$ is disjoint from $\overline{F}$ by (3), we see that the point b of $\overline{F} - F$ cannot be in the interior of $\overline{F}$ in the topological space L. Since F is open in L, we have (iii). Q.E.D.