A polyhedron is a convex hull of finitely many points. I would like to show that the intersection of finitely many half spaces is a polyhedron, provided the boundedness.
Is this statement true, in general? If so, can somebody give me a hint to approach it?
Thank you in advance for your help.
Best Answer
Not quite an answer, but a reference.
A classic reference on this is Rockafellar's "Convex Analysis".
Theorem 19.1, The following properties of a convex set $C$ are equivalent: (a) $C$ is polyhedral. (b) $C$ is closed and has only finitely many faces. (c) $C$ is finitely generated.
Note that Rockafellar uses polyhedral to mean the intersection of a finite collection of closed half spaces.
(b) is irrelevant to this question.
Finitely generated means the convex hull of a finite set of points and directions, that is, $C$ is finitely generated iff there are $a_1,...,a_m$ and some fixed $k \in \{0,...,m\}$ such that any point of $C$ can be written as $\sum_{i=1}^m \lambda_i a_i$ with $\lambda_i \ge 0$ and $\sum_{i=1}^k \lambda_i = 1$. (This is verbatim from the text, the case $k=0$ requires comment, but is not relevant here as $k=m$ in your question.)
Since you are presuming boundedness, this means there are no directions, and hence finitely generated reduces to the convex hull of a finite number of points.
I liked Rockafellar's comment that precedes Theorem 19.1 so much that I added it to my profile years ago: