How is a halfspace an affine convex cone?

Question

Wikipedia says "An affine convex cone is the set resulting from applying an affine transformation to a convex cone" (https://en.wikipedia.org/wiki/Convex_cone)

It also says a halfspace is an affine convex cone. I am trying to understand why a halfspace is an affine convex cone.

A half-space can't be a cone (I think). If $C =\{x : Ax\leq b\}$ is the set of all $x$ in a half-space, $C$, I think $C$ is not a cone because it is not closed under positive scalar combinations. For example, if $Ax \leq b$ defines the half-space, then, in $R$, there exists an $\alpha : A\alpha x > b$. Thus $C$ is not closed under positive scalar combinations.

So, $C$ is a convex set, but I think it is not a convex cone.

So how is a halfspace an affine convex cone? Is the matrix $A$ the "affine transformation?" of a convex cone? If so, what is the convex cone it is transforming?

Answer 1

An (affine) half-space is an affine convex cone, because it can be obtained by translation of a half-space $S$ whose boundary is an $(n-1)$-dimensional subspace through the origin, which is a convex cone, since clearly if $x\in S$ then $\lambda x\in S$ for any $\lambda>0$. Formally, a half-space of the latter form can be written as $$ f(x)\ge 0 $$ where $f$ is an element of the dual space. Then, $f(\lambda x)=\lambda f(x)\ge 0$ for any $\lambda>0$ whenever $f(x)\ge 0$.