# Can a polyhedron be an empty set?

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?

• 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). – Marc van Leeuwen Feb 11 at 9:47
• By the definition that you posted, yes, a polyhedron can be an empty set. – supinf Feb 11 at 10:42

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'.