Exercise 41, § 2

Let H be a hyperplane in a vector space E that does not contain 0 and let S be a convex set contained in H. a) Suppose that the intersection of S with each line in H is a compact segment. Let $a, b$ be two distinct points of E such that there exist two numbers $\lambda > 0, \mu > 0$ for which $b + \mu S \subset a + \lambda S$; show that if c is the point where the line joining a and b meets the hyperplane H' parallel to H which contains $a + \lambda S$, then $c \in b + \mu S$ and $b + \mu S$ is the image of $a + \lambda S$ by a homothety of centre c transforming a into b. (Reduce to the case where E is of dimension 2.) b) With the same hypotheses on S, let $a, b$ be two distinct points of E and suppose that there exists a point $c \in E$ and three numbers $\lambda > 0, \mu > 0, v \geq 0$ such that $(a + \lambda S) \cap (b + \mu S) = c + vS$. Show that if A (resp. B) is the cone with vertex $a$ (resp. $b$) generated by $a + \lambda S$ (resp. $b + \mu S$), and $H''$ the hyperplane parallel to H passing through c, then $H'' \cap A \cap B = \{ c \}$ (use a)). c) Suppose that H is the affine linear variety generated by S. Let C be the cone with vertex 0 generated by S. Show that the following two conditions are equivalent: $\alpha)$ E is a lattice for the order on E of which C is the set of elements $\geq 0$. $\beta)$ For any points $x, y$ of E and numbers $\lambda > 0, \mu > 0$ such that the set $(x + \lambda S) \cap (y + \mu S)$ is not empty, there exists $z \in E$ and $v \geq 0$ such that this set is $z + vS$. (To prove that $\alpha$ implies $\beta$), reduce to the case $y = 0$ and use the fact that if $(s_i)$ is a finite family of points of S and $(\lambda_i)$ is a family of real numbers such that $\sum_i \lambda_i s_i = 0$, then $\sum_i \lambda_i = 0$. To prove that $\beta$ implies $\alpha$ use $b$). When S satisfies the equivalent conditions $\alpha$ and $\beta$, we say that S is a simplex in E. When E is of finite dimension, the convex envelope of a finite set of points affinely independent in H and generating H is a simplex \* (the converse is also true : $cf.$ INT, II, 2nd ed., § 2, exerc. 7)). \*