Separating convex sets in Vector spaces

Question

This question just popped on my mind.

Let $A, B$ two disjoint, nonempty convex sets in the vector space $X$, can they be separated via a nonzero linear function in $X' = \{ f : X \to R ~ | \quad \text{f is linear} \} ?$ i.e., does there exist $f \in X' \setminus \{ 0\}$ such that

$$ f(a) \leq f(b) \quad \forall a\in A, ~ \forall b \in B $$

If not under what minimal condition one can separate them.

My Thought : Since $A \cap B = \emptyset $ using Zorn Lemma we can find two disjoint maximal convex sets, say $U, ~ V$ such that $ A \subseteq U, ~ B \subseteq V $ and through maximality of $U, V$ we can deduce that $U \cup V = X$ in other words $U,~ V$ make a convex partition of the space. Now from this, can we say that $U, ~V$ are two sides of a hyperplane ? i.e., $$ U \subseteq \{ x \in X ~ | \quad f(x) \leq \alpha \} , ~ V \subseteq \{ x \in X ~ | \quad f(x) \geq \alpha \} $$

for some $f \in X'$ and $\alpha \in \Bbb R$

Question #2: What if we assume $A, B$ are pointed cones with $A \cap B = \{0\}$

EDIT: I realized the answer of question # 1 is No generally see below link

https://math.stackexchange.com/q/929690/219176

But Still any answer regarding minimal conditions that guarantees separation is my main interest, and an answer for question #2.

Thank for your help.

Answer 1

I do not know a definitive "weakest" condition and I doubt there is one. Many results in the realm of Hahn-Banach do the trick, i.e. there is the general result for $A,B$ convex and $A$ open (both open giving strict separation) and there is also Eidelheit's theorem saying that you can separate a point from a closed convex set (or a compact convex from a closed convex one). The latter one also holds for convex $A,B$ such that the interior of $A$ is non-empty and does not intersect $B$.

Answer 2

You can always consider $X$ as a locally convex space, provided with the finest such topology, i.e., as the inductive limit of its finite dimensional subspaces, and then apply the Hahn-Banach spaces. Thus you can separate when $A$ is compact (a very strong constraint) and $B$ is open (very weak) for this structure. Not sure if these are the weakest conditions (or even what that means in this situation) but this might be of interest to you.