Proposition 9

Let $C$ be a chamber and $\mathscr{M}$ the set of walls of $C$. Then $C = D_{\mathscr{M}}(C)$ and every subset $\mathscr{L}$ of $\mathfrak{H}$ such that $C = D_{\mathscr{L}}(C)$ contains $\mathscr{M}$. A subset $F$ of $\overline{C}$ is a facet if and only if it is a facet of $E$ relative to the family $\mathscr{M}$.

a) Let $\mathscr{L}$ be a subset of $\mathfrak{H}$ such that $C = D_{\mathscr{L}}(C)$. Consider a hyperplane $L$ belonging to $\mathfrak{H}$ but not to $\mathscr{L}$; let $\mathfrak{N}$ be the set of hyperplanes $H \neq L$ belonging to $\mathfrak{H}$. Then $\mathscr{L} \subset \mathfrak{N}$, hence $C = D_{\mathfrak{N}}(C)$, and $L$ does not meet $D_{\mathfrak{N}}(C)$. By the implication (i) $\Longrightarrow$ (iii) in Prop. 4, the hyperplane $L$ is not a wall of $C$. Consequently, every wall of $C$ belongs to $\mathscr{L}$.

b) We assume that $C = D_{\mathscr{L}}(C)$. Let $H$ be a hyperplane belonging to $\mathscr{L}$ that is not a wall of $C$, and put $\mathscr{L}' = \mathscr{L} - \{ H \}$. By the implication (iii) $\Longrightarrow$ (i) in Prop. 4, the convex set $D_{\mathscr{L}'}(C)$ does not meet $H$, so $D_{\mathscr{L}'}(C) \subset D_H(C)$ and $C = D_{\mathscr{L}'}(C)$. If $\mathfrak{F}$ is a finite subset of $\mathscr{L}$ that does not contain any wall of $C$, we conclude by induction on the cardinal of $\mathfrak{F}$ that $C = D_{\mathscr{L} - \mathfrak{F}}(C)$.

c) Let $a$ be a point of $C$; clearly, $C \subset D_{\mathfrak{M}}(a)$. Let $a'$ be a point of $D_{\mathfrak{M}}(a)$; since the closed segment $[aa']$ is compact, the set $\mathfrak{F}$ of hyperplanes $H \in \mathfrak{H}$ that meet $[aa']$ is finite. Since $a$ and $a'$ are strictly on the same side of every wall of $C$, no wall of $C$ belongs to $\mathfrak{F}$; by b), we have $C = D_{\mathfrak{H} - \mathfrak{F}}(C)$. Since $a' \in D_{\mathfrak{H} - \mathfrak{F}}(a)$, we have $a' \in C$. We have therefore proved that $D_{\mathfrak{M}}(a) \subset C$, which establishes the first part of the proposition.

d) To prove the last assertion of the proposition, it clearly suffices to show that a subset $F$ of $\overline{C}$ that is a facet of $E$ relative to $\mathscr{M}$ is a facet of $E$ relative to $\mathfrak{H}$, or that every hyperplane $H \in \mathfrak{H}$ that meets $F$ contains $F$. So let $H$ be a hyperplane that meets $F$ but does not contain it. Since $F$ is open in its affine support, it is not completely on one side of $H$. It follows that $\overline{C}$ is not completely on one side of $H$ and hence that the hyperplane $H$ does not belong to $\mathfrak{H}$, which completes the proof.