TITLE: Is a ball a polyhedron?
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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.