# Is closed convex set with finite number of extreme points convex polyhedron

I have this simple question related to convex set and convex polyhedron. As the content in the title, it's basically my question:

Is closed convex set with finite number of extreme points convex polyhedron

If not, can you please give me some hint to find the counter-example of that statement? Thanks a lot. I really appreciate

• What is a polyhedron? – gerw Jan 23 '16 at 7:19
• Sorry, my bad. It should be convex polyhedron. In wikipedia, "convex polyhedron is any point set that is the intersection of a finite number of half-spaces" – le duc quang Jan 23 '16 at 7:24

• For any closed, convex $C$, consider $C \times \{0\}$: no extreme points.
• Of course "convex hull of a finite set" includes things that it seems to me would at best be described as degenerate polyhedra; for example do we really want to call a line segment in $\Bbb R^3$ a polyhedron? {Of course}^2 the definition "intersection of finitely many (closed) half-spaces" also includes degenerate polyhedra (for a few seconds I thought that that segment was not the intersection of finitely many half spaces, but of course it is...) – David C. Ullrich Jan 23 '16 at 13:08
• @DavidC.Ullrich: Polytopes are defined as convex hull of a finite set of points - and the only difference is that polyhedrons might be unbounded. And both definitions include "degenerate" sets. In fact, a line segment in $\mathbb{R}^3$ is a non-degenerate one-dimensional polyhedron, that is embedded in a three-dimensional space. – gerw Jan 23 '16 at 13:32