# Halfspace as Polyhedra?

The definition of a polyhedron is any $x$ such that $Ax \geq b, A \in \mathbb{R}^{m\times n}, b \in \mathbb{R}^n$. A half space is defined as $a'x \geq b \text{ where } a,x \in \mathbb{R}^n ,b\in \mathbb{R}$.

So, can a halfspace be called a polyhedron where $m=1$?

(It won't be a polytope though, right?)

• As far as I know, "polytope" is the general concept for arbitary dimensions, and "polyhedron" and "polygon" are its specializations to 3 and 2 dimensions. "Polyhedra" is the plural of "polyhedron"; "a polyhedra" is a grammatical error. – hmakholm left over Monica Oct 13 '12 at 18:32
• @HenningMakholm, I fixed the plural error but I don't think polyhedron are only 3 dimensions. My book explicitly states polyhedrons of n dimensions. – Inquest Oct 13 '12 at 18:36
• I don't think that's standard terminology in geometry. For that matter, I don't think allowing a "polyhedron" to be unbounded is standard in geometry, nor is it standard to require of a general "polyhedron" that it is convex. But these conventions may well be common in other areas of mathematics; for example I suppose they are more or less what one wants to speak about in optimization theory. – hmakholm left over Monica Oct 13 '12 at 18:48
• I am reading a book on Simplex Method. – Inquest Oct 13 '12 at 19:00
• x @Inquest: Yes, that is optimization theory. – hmakholm left over Monica Oct 13 '12 at 19:01