By Gallier J.

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We do not know how the notion “algebraically open” relates to the concept of core. ). An important special case of separation is the case where A is convex and B = {a}, for some point, a, in A. 2. 17 (Minkowski) Let A be a nonempty, closed, and convex subset. Then, for every point a ∈ ∂A, there is a supporting hyperplane to A through a. 2. SUPPORTING HYPERPLANES AND MINKOWSKI’S PROPOSITION 35 Proof . Let d = dim A. , A has empty interior), then A is contained in some aﬃne subspace V of dimension d < dim X, and any hyperplane containing V is a supporting ◦ hyperplane for every a ∈ A.

13 (Farkas Lemma, Version II) Given any d × n real matrix, A, and any vector, z ∈ Rd , exactly one of the following alternatives occurs: 32 CHAPTER 3. SEPARATION AND SUPPORTING HYPERPLANES (a) The linear system, Ax = z, has a solution, x, such that x ≥ 0, or (b) There is some c ∈ Rd such that c z < 0 and c A ≥ 0. Proof . 10 and either z ∈ cone(A1 , . . , An ) or z ∈ / cone(A1 , . . , An ). One can show that Farkas II implies Farkas I. Here is another version of Farkas Lemma having to do with a system of inequalities, Ax ≤ z.

Ap })∗ can be written in matrix form as conv({a1 , . . , ap })∗ = {x ∈ Rn | A x ≤ 1}, where 1 denotes the vector of Rp with all coordinates equal to 1. We write P (A , 1) = {x ∈ Rn | A x ≤ 1}. There is a useful converse of this property as proved in the next proposition. 1. 4 Given any set of p points, {a1 , . . , ap }, in Rn with {a1 , . . , ap } = {0}, if A is the n × p matrix whose j th column is aj , then conv({a1 , . . , ap })∗ = P (A , 1), with P (A , 1) = {x ∈ Rn | A x ≤ 1}. Conversely, given any p × n matrix, A, not equal to the zero matrix, we have P (A, 1)∗ = conv({a1 , .