Separating hyperplanes for a convex cone

Question

Let $W$ be a fixed matrix. Define $$\operatorname{pos}W \triangleq \{t \mid Wy =t , y≥ 0\}.$$ It is called the positive hull of $W$. It represents the set of right-hand sides that can be obtained by a non-negative combination of the columns of $W$. The positive hull is easily seen to be a convex cone.

Let $p$ be a point not in the set $\operatorname{pos}W$. Then, there exists a hyperplane $H \triangleq \{x\mid\sigma^Tx =0\}$ that separates $p$ and $\operatorname{pos}W$.

How can we prove that the number of possible separating hyperplanes (separating $p$ and $\operatorname{pos}W$) is finite based on the fact that $\operatorname{pos}W$ is finitely generated?

Answer 1

Perhaps the argument is the following. Observe that $\operatorname{pos}W$ is a polyhedral cone (every finitely generated cone is polyhedral). That it is finitely generated can be seen from the fact that it is positively spanned by the finite number of columns of $W$.

Thus we can write $$\operatorname{pos}W=\{t\mid Wy=t,y\geq 0\}=\{t\mid Dt\leq 0\}$$ for some matrix $D\in \mathbb{R}^{m\times n}$, $n$ being the dimension of $t$. The last expression illustrates that the number of hyperplanes separating $\operatorname{pos}W$ is finite ($m$).

Answer 2

How can we prove that the number of possible separating hyperplanes (separating $p$ and $\operatorname{pos}W$) is finite based on the fact that $\operatorname{pos}W$ is finitely generated?

In general, it is not. There may be infinitely many separating hyperplanes.

Let $\sigma_{1}$, $\sigma_{2}$ define two distinct separating hyperplanes. Then it's straightforward to see that, for $0 \leq \lambda \leq 1$, $\lambda \sigma_{1} + (1 - \lambda) \sigma_{2}$ also defines a separating hyperplane.