5
$\begingroup$

A polyhedron is defined as the intersection of finitely many generalized halfspaces. That is, a polyhedron is any set of the form $ \{x \in R : Ax \leq\ b \} $

I would like to understand this further.

Given that $ Ax \leq\ b, Ax \geq\ b \leftrightarrow\ Ax=b$, a polyhedron can be a hyperplane. Can it be a single point?

Further, in the case that there is no intersection (for example two parallel lines in $R^2$), does the set describe an empty polyhedron, or does it simply fail to define a polyhedron? I.e. can a polyhedron be empty?

$\endgroup$
  • 1
    $\begingroup$ The formulation of this question could be improved. Does $R$ mean $\Bbb R^3$, is $A$ a linear form on this set, and why are these halfspaces "generalized"? Also I would hesitate to call a complete halfspace a polyhedron (but there is no universally agreed upon definition). $\endgroup$ – Marc van Leeuwen Feb 11 at 9:47
  • $\begingroup$ By the definition that you posted, yes, a polyhedron can be an empty set. $\endgroup$ – supinf Feb 11 at 10:42
7
$\begingroup$

It only depends on your convention. A set defined by linear inequalities can certainly be empty. Weather it can be called a polyhedron or not, depends on the convention you choose. It's a matter of terminology, not a matter of `correctness'.

$\endgroup$
  • 4
    $\begingroup$ Furthermore, the definition of a "polyhedron" quoted by the OP is already both broader and narrower than the "classical" geometric definition. It is broader because it does not require a polyhedron to be bounded, or to have a non-empty interior. One the other hand, it is also narrower, since it effectively only admits convex polyhedra. $\endgroup$ – Ilmari Karonen Feb 11 at 9:39

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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