Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the geometry of (convex) polyhedra used for linear optimization, one has the lemma:

Consider the inequality $Ax \leq b$ where $A^+ x \leq b^+$ (the non-implicit inequalities of $Ax \leq b$) doesn't have redundant equations.

Then $F$ is a facet of $P$ iff $F= \{x \in P \mid a_{i_0}^T x = \beta_{i_0} \}$ for an $i_0 \in I_+$.

(A surface, I am not sure if this is the right translation, can be written as $S=P \cap H$ for the polyhedron $P$ and a hyperplane $H = \{ x \mid c^T x =d \}$ with $c^T x \leq d$ redundant concerning $Ax \leq b$ and the intersection of $P$ and $H$ nonempty. A facet is maximal concering the set-inclusion of nontrivial surfaces of $P$).

This should imply that

Every non-trivial surface is the intersetion of facets.

How can I assert this?

share|cite|improve this question
"polyeder"$=$"polyhedron" – Lee Mosher Jul 28 '12 at 14:57
@Lee wasn't aware of this, I changed it. I hope the rest is okay. – Suedklee Jul 28 '12 at 15:04
@Suedklee: a) You didn't, you left it in the title. b) The plural of "polyhedron" is "polyhedra". – joriki Jul 28 '12 at 17:08
What's the relationship between $f$ and $F$? Are they supposed to denote the same thing? – joriki Jul 28 '12 at 17:10
@joriki yes it should be just $F$. – Suedklee Jul 28 '12 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.