# 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.

Thank for your help.

• 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

## 2 Answers

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 $B$.

• or a compact from a closed one]---> didn't you mean [or a convex compact from a closed one] ? – Duchamp Gérard H. E. Jul 10 '17 at 7:49
• 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
• @DuchampGérardH.E. Sure, I thought of convex as a standing assumption, but added convexity now. – Dirk Jul 10 '17 at 10:26
• So without topology you also consider arbitrary linear functionals (no convexity assumption?). Are you sure that you do not have any topology (implicitly)? For example, do you really consider all linear functionals (i.e. the algebraic dual)? I can't imagine a situation where one really deals with this… (but this may be just me…). – Dirk Jul 10 '17 at 10:28
• @Ashkan [ And topology has nothing to do with convexity]--->I wouldn't be so affirmative. If you consider the finest locally convex topology (Bourbaki TVS Ch II § 4), algebraic and continuous dual coincide. – Duchamp Gérard H. E. Jul 10 '17 at 21:04

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.

• Thanks , I knew that. Actually If only B has interior then it's fine, they can be separated. But I I am really interested the case where there's no other assumptions on A and B, except they are convex and pointed cone.... – Red shoes Jul 10 '17 at 14:30