The set $\mathscr{N}$ of hyperplanes H such that $H \in \mathscr{H}$ and $a \notin H$ is locally finite since it is contained in $\mathscr{H}$. Hence, the set U of points of E that do not belong to any of the hyperplanes of the set $\mathscr{N}$ is open. Since $a \in U$, there is a connected open neighbourhood of a contained in U. The remainder of the proposition is clear.

Given two points x and y of E, denote by R\{x, y\} the relation

" For any hyperplane $H \in \mathscr{H}$, either $x \in H$ and $y \in H$ or $x$ and $y$ are strictly on the same side of H."

Clearly, R is an equivalence relation on E.

This is clear since $\mathscr{H}$ is locally finite.

Let F be a facet and a a point of F. A hyperplane $H \in \mathscr{H}$ contains F if and only if $a \in H$; the set $\mathscr{F}$ of these hyperplanes is thus finite; their intersection is an affine subspace L of E, which we shall call the affine support of F; the dimension of L will be called the dimension of F.

If $\mathscr{N}$ is the set of hyperplanes $H \in \mathscr{H}$ not containing F, then

We shall prove that the closure of F is given by

It is clear that the right-hand side contains the left-hand side. Conversely, let $x \in L \cap \bigcap_{H \in \mathscr{N}} \overline{D_H(a)}$. The open segment with extremities a and x is contained in L and in each of the $D_H(a)$ for $H \in \mathscr{N}$, and hence in F. It follows that x is in the closure of F, hence the formula.

(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.

This follows from (iii).

(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.

a) For any $H \in \mathfrak{H}'$, there exists an open half-space $D_H$ bounded by $H$ such that $C = \bigcap_{H \in \mathfrak{H}'} D_H$.

b) *The set C does not meet any hyperplane belonging to $\mathfrak{H} - \mathfrak{H}'$. Under these conditions, C is a chamber defined by $\mathfrak{H}$ in E, and $D_H = D_H(C)$ for all $H \in \mathfrak{H}$.

Properties a) and b) show that C is a convex subset of U; hence there is a chamber $C'$ with $C \subset C'$. Since $C \subset D_H$, we have $D_H = D_H(C)$ for all H in $\mathfrak{H}'$, hence $C = D_{\mathfrak{H}'}(C) \supset D_{\mathfrak{H}}(C)$ since $\mathfrak{H}' \subset \mathfrak{H}$; we have $D_{\mathfrak{H}}(C) = C'$ by (6), hence $C \supset C'$. Finally therefore, $C = C'$.

If E reduces to a single point, this is clear. Otherwise, let $a \in E$ and let $H_1, \ldots, H_m$ be the hyperplanes of $\mathfrak{H}$ containing $a$. Since $\mathfrak{H}$ is locally finite, there exists a neighbourhood V of $a$ that does not meet any hyperplane of $\mathfrak{H}$ other than $H_1, \ldots, H_m$. Let D be a straight line passing through $a$ and not contained in any of the $H_i$; if $x \in D, x \neq a$, and $x$ is sufficiently close to $a$, the open segment $]ax[$ is contained in V and does not meet any of the $H_i$. Then $]ax[ \subset U$; since $]ax[$ is connected, it is contained in a chamber C, hence $a \in \overline{C}$.

(i) There exists a point a in $\Omega$ that does not belong to any of the hyperplanes of $\mathfrak{H}$ that do not contain L.

(ii) If L is a hyperplane and $L \notin \mathfrak{H}$, there exists a chamber that meets $\Omega$.

(iii) If L is a hyperplane and $L \in \mathfrak{H}$, there exists a point a in $\Omega$ that does not belong to any hyperplane $H \neq L$ of $\mathfrak{H}$.

Denote by $\mathfrak{N}$ the set of hyperplanes H with $H \in \mathfrak{H}$ and $L \not\subset H$, and by $\mathscr{L}$ the set of hyperplanes of the affine space L of the form $L \cap H$ with $H \in \mathfrak{N}$. It is clear that $\mathscr{L}$ is a locally finite set of hyperplanes in L, and Prop. 6 shows that $\Omega$ meets a chamber $\Gamma$ defined by $\mathscr{L}$ in L. If $a$ is a point of $\Gamma \cap \Omega$, then $a \notin H$ for all $H \in \mathfrak{N}$, hence (i).

Assume now that L is a hyperplane; any hyperplane containing L is then equal to it, so we may distinguish two cases:

a) $L \notin \mathfrak{H}$: then $\mathfrak{N} = \mathfrak{H}$, and we have $a \notin H$ for all $H \in \mathfrak{H}$; thus $a$ belongs to a chamber defined by $\mathfrak{H}$ in E, hence (ii).

b) $L \in \mathfrak{H}$: then $\mathfrak{N} = \mathfrak{H} - \{L\}$, hence (iii).

By Prop. 7, (iii), there exists a point $a$ of $H$ that does not belong to any hyperplane $H' \neq H$ of $\mathfrak{H}$; by Prop. 6, there exists a chamber $C$ such that $a \in \overline{C}$; Prop. 4 then shows that $H$ is a wall of $C$.

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.

Let H and $H'$ be two non-parallel hyperplanes, $a \in H$ and $a' \in H'$; there exist two hyperplanes M and $M'$ of the vector space T such that $H = M + a$ and $H' = M' + a'$; since H and $H'$ are not parallel, we have $M \neq M'$ and hence $T = M + M'$; hence there exist $u \in M$ and $u' \in M'$ such that $a' - a = u - u'$, and the point $u + a = u' + a'$ belongs to $H \cap H'$.

The lemma being trivial when $L = H$, we assume that there exists a point $a$ in L with $a \notin H$. Put $\lambda = f'(a), \lambda' = -f(a)$ and

then $\lambda' \neq 0$ since $a \notin H$; moreover, since $H \neq H'$, there exists $b \in H$ such that $b \notin H'$, so that $f(b) = 0, f'(b) \neq 0$, and thus $g(b) = -f(a).f'(b)$ is non-zero. The set $L_1$ of points where the affine function $g \neq 0$ vanishes is then a hyperplane of E; we have $g(a) = 0$, so $a \in L_1$.

a) Assume that H and $H'$ are parallel: g and f both vanish at every point of $L_1 \cap H$, hence so does $f'$ since $\lambda' \neq 0$; thus every point of $L_1 \cap H$ belongs to $H'$; but since H and $H'$ are parallel and distinct, they are disjoint, so $L_1 \cap H = \varnothing$, and Lemma 1 shows that $L_1$ is parallel to H. Since $a \in L$ and $a \in L_1$, we thus have $L = L_1$.

b) Assume that H and $H'$ are not parallel: by Lemma 1, there is a point c in $H \cap H'$; we give E the vector space structure obtained by taking c as the origin. Then $H \cap H'$ is a vector subspace of $E$ of codimension 2, and since $a \notin H$, the vector subspace $M$ of $E$ generated by $H \cap H'$ and $a$ is a hyperplane; since $H \cap H' \subset L \cap L_1$ and $a \in L \cap L_1$, we have $M \subset L \cap L_1$, hence $M = L = L_1$. Q.E.D.

a) The hyperplanes H, H' and L are parallel.

b) The hyperplanes H and H' are not parallel, and $\overline{H \cap H'} \subset L$.

Then, L meets C.

Let $b$ (resp. $b'$) be a point of the face of C with support H (resp. $H'$); it is immediate that every point of the segment $[bb']$ distinct from $b$ and $b'$ belongs to C.

Introduce an affine function $f$ that vanishes at every point of H and is such that $f(x) > 0$ for $x$ in $D_H(C)$; similarly, introduce an affine function $f'$ having an analogous property with respect to $H'$. By applying Lemma 2, we can find numbers $\lambda$ and $\lambda'$ and an affine function $g$ having the properties stated in the lemma. We have $(\lambda, \lambda') \neq (0, 0)$, and for every point $x$ of $L \cap D_H(C) \cap D_{H'}(C)$, we have $f(x) > 0,\ f'(x) > 0$ and $\lambda.f(x) + \lambda'.f'(x) = 0$, so $\lambda \lambda' < 0$. On the other hand, we have $g(b) = \lambda'.f'(b)$ and $g(b') = \lambda.f(b')$, and since $f(b') > 0,\ f'(b) > 0$, we have $g(b)g(b') < 0$. The points $b$ and $b'$ are thus strictly on opposite sides of the hyperplane L, hence there exists a point c of L which belongs to $[bb']$, and is distinct from $b$ and $b'$, hence that belongs to C.