How do you prove that each face of a polytope is also a polytope? I think it may be through induction.

  • $\begingroup$ What is definition of a polytope ? $\endgroup$
    – HK Lee
    Aug 17 '17 at 7:55
  • 1
    $\begingroup$ The definition of polytope is the convex hull of finite set of points. @HKLee $\endgroup$ Aug 17 '17 at 11:24
  • $\begingroup$ I have one more question. What is the definition of face ? $\endgroup$
    – HK Lee
    Aug 17 '17 at 12:22
  • $\begingroup$ Let K be a subset of an n-dimensional Euclidean space, then the face, F, is the set x in K such that inner product <x, u> is equal to an alpha. @HKLee $\endgroup$ Aug 17 '17 at 12:55

If you know that "polytopes" and "bounded polyhedra" are the same objects, then what you are asking is easy: It is straightforward that every face of a polyhedron is a polyhedron. Hence, every face of a bounded polyhedron is a bounded polyhedron.

Here by polytope I mean (as you) the convex hull of finitely many points. By a (convex) polyhedron I mean a subset of $\mathbb R^d$ defined by finitely many affine-linear inequalities. By a face of a polytope or polyhedron $P$ I mean the part of $P$ where some linear functional is maximized. (This is almost what you say in "the face, F, is the set x in K such that inner product $<x, u>$ is equal to an alpha", except one needs to assume that alpha is the maximum value taken by $<x, u>$ on $P$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.