Hamilton was an early graffiti artist, he left his mark on Broom bridge. I doubt very much that he was considering 4D rotations.

Nuggets like this are memorable and valuable.

@FedorPetrov If you don't mind, I might add my elaborations as an answer to the corresponding question on MSE, with appropriate attribution, of course.

@Ypbor dual cone. Unfortunately comments are not the best place for clarification.

@Ypbor Please replace span by convex cone generated by...

@Ypbor If $y\ge 0$ (that is, component wise) and $y \in Y$ then $-y \in C$ and so $\eta(-y) \le 0$, hence $\eta(y) \ge 0$.

@Ypbor The notations mean span and sum respectively. The dual of the intersections is the sum of the duals (see math.stackexchange.com/q/3356026/27978).

(Cont.) Now consider this as extending $\eta$ onto all of $\mathbb{R}^n$. In particular, we see that the extended $\eta$ is non positive on $(-\infty,0]^n$, non negative on $X$ and $\eta(x)>0$ for some $x \in X$.

@Ypbor It took me a while to unravel. $Y$ is the span of $X$ so any functional that is zero on $X$ must be zero on $Y$. Since $\eta \ge 0$ on $X$, since it is non zero, it must be $>0$ somewhere on $X$. Note that $\cap_K \{ y \in Y | e_k (y) \ge 0 \} \subset \{ y \in Y | \eta(y) \ge 0 \}$ and taking the dual gives $\operatorname{sp} \{ \eta \} \subset +_k \operatorname{sp} \{ e_k \}$, so we can write $\eta = \sum_k \lambda_k e_k$ with $\lambda_k \ge 0$.

Took me a while. Simple, but rather subtle.

@Youem I understand that, but regardless of the representation of $\eta$, if $x_2>0$ then $\eta(x)<0$, so I do not follow the last paragraph in the answer.

@FedorPetrov The last paragraph is a little glib for me. I wonder could you elaborate or point me somewhere please? If you take $n=2$, $X= \{ x\mid x_1+x_2 = 0, x_1>0 \}$ then $C=\{0\}$ and the functional $\eta:Y \to \mathbb{R}$ given by $\eta(x)= x_1-x_2$ separates $C$ and $X$. If the coordinate function $x \mapsto x_2$ is positive (on $Y$) then $\eta(x) < 0$. I presume I am missing something.

Minor point, (B) means that $J$ is quasi convex, not necessarily convex.

Well, for a mathematician, a few just means finite.

Very apropos name.