Is $\mathbb{R}^n$ a polyhedron? A polyhedron is defined as the intersection of a finite collection of generalized halfspaces. In my book, a generalized halfspace is defined as a set where we allow the vector of coefficients $(a_1, a_2,\ldots, a_n )$ in the definition of a halfspace to vanish (or = 0). 
I said yes. $\mathbb{R}^n$ is a polyhedron because a generalized halfspace can be all of $\mathbb{R}^n$. But I feel there's something wrong with that logic. Please help. Thanks.
 A: $\mathbb{R}^n$ is not a polyhedron. You mentioned that a generalized halfspace can be all of $\mathbb{R}^n$, but that is not that case. By definition, a halfspace can only emcompass half of $\mathbb{R}^n$. I'm not sure what definition you're given for a generalized halfspace, but it's just the direct analogue of the three dimensional case. 
Since a polyhedron is by definition the intersection of a finite collection of halfspaces, if $\mathbb{R}^n$ is a polyhedron then there must be at least one halfspace that describes it. However, as noted above, $\mathbb{R}^n$ cannot be described by a single halfspace. Additionally, whenever another halfspace is added to your collection of halfspaces that describe $\mathbb{R}^n$, the number of possible points described cannot increase (that is, at best you retain the same number of points in the intersection). Since a single halfspace does not describe $\mathbb{R}^n$ and the addition of other halfspaces cannot add points to the description of $\mathbb{R}^n$, $\mathbb{R}^n$ cannot be described by the intersection of a finite number of halfspaces. Thus, $\mathbb{R}^n$ is not a polyhedron.
A: Geometry isn't exactly my area of expertise, either. So let's start with the basics: What is the definition of a "generalized half-space"? Wikipedia's article on half-spaces gives the following definition:

In geometry, a half-space is either of the two parts into which a plane divides the three-dimensional Euclidean space. More generally, a half-space is either of the two parts into which a hyperplane divides an affine space. 

From this definition, I would say that $\mathbb{R}^n$ is not a generalized halfspace itself. Now, can we take the intersection of a finite collection of generalized halfspaces to get $\mathbb{R}^n$? Let's assume that we can do so. Name the collection $\mathbb{H}$. Now consider one of these generalized halfspaces $H \in \mathbb{H}$. Now we must have $\mathbb{R}^n \subseteq H$ because by our assumption $\mathbb{R}^n = \bigcap \mathbb{H}$. However, this is a contradiction.
