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?

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    $\begingroup$ 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). $\endgroup$ Feb 11, 2019 at 9:47
  • $\begingroup$ By the definition that you posted, yes, a polyhedron can be an empty set. $\endgroup$
    – supinf
    Feb 11, 2019 at 10:42

1 Answer 1


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'.

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    $\begingroup$ 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. $\endgroup$ Feb 11, 2019 at 9:39

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