# Why is every nontrivial surface of a polyhedron an intersection of facets?

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?

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"polyeder"$=$"polyhedron" en.wiktionary.org/wiki/Polyeder – 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