7.4 Problems 99
We will show that X = K.
Assume that 3y e X 3 y $ K. Then, by Theorem 7.3.1, there is a hyperplane
separating y and K;
3a ^ 0 3 aTy < inf aTx
xEK
Let XQ = m{X£x(aTx)- xo is finite and 3xo £ X 3 aTxo = bo (because by
Weierstrass' Theorem: The continuous function aTx achieve its minimum over
any closed bounded set).
Hence, the hyperplane H = {x : aTx = bo} is a supporting hyperplane to X.
Since b 0 < aTy < mixeK aTx, H is disjoint from K. Let T = H D X. Then,
T is a bounded closed convex set of H, which can be regarded as a space
in Rn_1. T ^ 0, since XQ € T. Hence, by induction hypothesis, T contains
extreme points; and by the previous Lemma, these are the extreme points
of X. Thus, we have found extreme points of X not in K, Contradiction.
Therefore, X C K, and hence X = K (since K C X, i.e. K is closed and
bounded). •
Remark 7.4.4 Let us investigate the implications of this theorem for convex
polytopes. A convex polytope is a bounded polyhedron. Being the intersection of
closed halfspaces, a convex polytope is closed. Thus, any convex polyhedron is
the closed convex hull of its extreme points. It can be shown that any polytope
has at most a finite number of extreme points, and hence a convex polytope is
equal to the convex hull of a finite number of points. The converse can also be
established, yielding the following two equivalent characterizations.
Theorem 7.4.5 A convex polytope can be described either as a bounded in-
tersection of a finite number of closed half spaces, or as the convex hull of a
finite number of points.
Problems
7.1. Characterize (draw, give an example, list extreme points and halfspaces)
the following polytopes:
a) zero dimensional polytopes.
b) one dimensional polytopes.
c) two dimensional polytopes.
7.2. d-simplex
d-simplex is the convex hull of any d+1 independent points in R™ (n > d).
Standard d — simplex with d+1 vertices in Rd+1 is
d+l
Ad = {x£ Rd+1 : ]T Xi = 1; x{ > 0, i = 1,..., d + 1}.
Characterize A2 in R^3.