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

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  • $\begingroup$ What is definition of a polytope ? $\endgroup$
    – HK Lee
    Aug 17 '17 at 7:55
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    $\begingroup$ 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
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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 $P$)

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