# Separating convex sets in Vector spaces

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.

• Very nice answer to question number one was given by fedja many years ago mathoverflow.net/questions/37551/… – Paata Ivanishvili Jul 10 '17 at 17:44
• @Paata Ivanisvili, thank you very much, I already knew a counter example for part 1 but not part two, that was nice, and I think a similar idea maybe leads to a counter example for part two. – Red shoes Jul 10 '17 at 19:00

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 $B$.
• First Thanks for your answer. But My question is not that trivial, that can be answered by referring Hahn-Banach Separation theorems! I never assumed topology on $X$ so being open and closed doesn't make any sense here. – Red shoes Jul 10 '17 at 7:51
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.