The set of separating hyperplanes is a convex cone

Question

Let $A, B ⊂ \mathbb{R}^n$ be two nonempty sets such that $A ∩ B = ∅$. $H(A, B) := \{(w, d) ∈ \mathbb{R}^{n+1} : \sup_{x\in A} \langle w,x\rangle ≤ d ≤ \inf_{y \in B} \langle w, y\rangle \}$

I can't understand how to prove that $H(A, B)$ is a convex cone.

Is my interpretation of the set correct? That $H(A, B)$ is just a set of all the hyperplanes which separates $A $ from $B$, including the origin?

Any help is very much appreciated

Answer 1

The set $H(A,B)$ is the set of all affine hyperplanes separating $A$ and $B$; not just those that pass through the origin.

To prove it's a convex cone, assume $(w_i,d_i)\in H(A,B)$ for each $i$, and take linear combination with nonnegative coefficients $\alpha_i$. The pair $$\left( \sum_i \alpha_i w_i, \sum_i \alpha_i d_i \right)$$ belongs to $H(A,B)$, because $$\left< \sum_i \alpha_i w_i, x\right> = \sum_i \alpha_i \left< w_i, x\right> \le \sum_i \alpha_i d_i$$ for all $x\in A$, and similarly for $B$.

Answer 2

Your interpretation is correct.

$H(A,B)$ could also be thought of as the intersection of half spaces that pass through the origin. Hence it is a convex cone.