0
$\begingroup$

I'm searching for a reference for the following simple fact about polytopes.

FACT:

If $\{x\;|\;Ax \leq b\}$ defines a polytope, and if $c\geq 0$ then $\{x\;|\;Ax\leq b+c\}$ also defines a polytope (i.e. a bounded polyhedron).

Although this fact is intuitively obvious, and not difficult to prove, I would like to justify it in at most one or two lines of text. So does this fact follows directly as a special case of some well known theorem?

Obs: It is clear that if $Ax\leq b$ is satisfied, then $Ax\leq b+c$ is satisfied. Therefore, the polytope defined by $Ax \leq b$ is included in the polyhedron defined by $Ax\leq b+c$. The question is about showing that $Ax\leq b+c$ defines a polytope, i.e., a bounded polyhedron.

  • 0
    @RodrigodeAzevedo this is clear. But this only shows that the polytope defined by $Ax \leq b$ is included in the *polyhedron* defined by $Ax\leq b+c$. The question is about showing that $Ax\leq b+c$ defines a *polytope*, i.e., a bounded polyhedron. – 2017-01-31
  • 0
    It is not clear what you are after. Since a reference to a theorem is accepted, you're not looking for an intuitive solution, yet a two line proof is unacceptable? – 2017-01-31

1 Answers 1

0

Just observe that the polyhedra $P=\{x \in \mathbb{R}^n\ |\ Ax\leq b\}$ and $Q=\{x \in \mathbb{R}^n\ |\ Ax\leq b+c\}$ have the same recession cone $$rec(P)=rec(Q)=\{x \in \mathbb{R}^n\ |\ Ax\leq 0\}$$. By hypothesis $rec(P)=\{\emptyset\}$ because $P$ is a polythope. The thesis follows.