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!
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.
polyhedron is not ball cause its a solid figure bounded by plane polygons or faces