Is a ball a polyhedron? In the book Introduction to Linear Optimization by Bertsimas Dimitri, a polyhedron is defined as a set $ \lbrace x \in \mathbb{R^n} | Ax \geq b \rbrace $, where A is an m x n matrix and b is a vector in $\mathbb{R^m}$. What it means is that a polyhedron is the intersection of several halfspaces.
A ball can also be viewed as the intersection of infinitely many halfspaces. So I was wondering if a ball is also a polyhedron by that definition or by any other definition that you might use?
Thanks and regards!
 A: The usual definition of a polyhedron requires that either one intersects a finite number of half-spaces, or one takes the convex hull of a finite set of points.
See the book Convex Polytopes by Branko Grünbaum (either the first or second edition).
A: No a ball is not a polyhedron, even by this definition. In your definition the matrix $A$ is of size $m\times n$, where $m\in\mathbb{N}$ thus the matrix is finite. The integer $m$ is an upper bound on the number of halfspaces which intersect to form the polyhedron. 
The reason $m$ is an upper bound is because suppose $A$ has two rows identical. Then there are two hyperspaces which are parallel so at least one of them does not form any part of the polyhedron.
A: Case $n=1$
When $n=1$ the ball is a segment and it is indeed a polyhedron.
Case $n=2$
Assume that the circle is a polyhedron. Think of the condition $A\mathbb{x}\ge\mathbb{b}$ as a system of linear inequalities, each of them defining a line and an associated halfplane. Since there is only a finite number of lines, there must exist $(x_0,y_0)$ such that $x_0^2 + y_0^2 = 1$ which is not in any of these lines. In particular $(x_0,y_0)$ would satisfy all the inequalities with $``>"$ rather than $``="$. This is a contradiction because, by continuity, $(x_0,y_0)$ would belong in the interior of the circle.
Case $n>2$
Define $h\colon\mathbb{R}^2\to\mathbb{R}^n$ by $h(x,y)=(x,y,0,\dots,0)$. The pre-image by $h$ of the ball in $\mathbb{R}^n$ is the circle. And since the pre-image of a polyhedron by an affine transformation is a polyhendron, we conclude that the $n$-dimensional ball cannot be a polyhedron either.
A: polyhedron is not ball cause its a solid figure bounded by plane polygons or faces
