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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.

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$\mathbb{R}^n$ 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.

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    @RobertIsrael: How do you define the intersection of *no* sets?2012-07-31
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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.

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    Wikipedia only defines ordinary half-spaces. The OP defined generalized half-spaces. It is any set of points $\vec x\in\mathbb R^n$ satisfying $\vec a\cdot\vec x=a_1x_1+a_2x_2+\cdots+a_nx_n\leq b$ , where $\vec a$ is allowed to be $\vec0$. This obviously does include $\mathbb R^n$ itself. Even with ordinary half-spaces, an empty intersection is all of $\mathbb R^n$ (just like the empty product is $1$).2018-10-24