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

Can any two disjoint nonempty convex sets in a vector space be separated by a hyperplane?

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

Yes, you are on the right track there. The substitute for the interior is now the concept of 'core'.

Let $X$ a vector space and $A\subseteq X.$ Then, the 'core' of $A,$ denoted as $cor(A),$ is defined as:

$$cor(A)=\{x\in A: \forall\;d\in X\;\exists \;\bar{t}>0: x+td\in A\;\forall\; t\in (0,\bar{t})\}.$$ Note that this is an algebraic definition, no topology involved.

Curious fact: If $A$ is convex, then $cor(A)=int(A)$ on any given vector space topology in $X$ such that $int(A)\neq \emptyset.$

The separation theorem in vector spaces can now be read as

$\textbf{Theorem:}$ Let $A,B\subseteq X$ vector space. Assume that $cor(A)\neq \emptyset.$ Then, $A$ and $B$ can be separated by a hyperplane if and only if $cor(A)\cap B=\emptyset.$

The details of this are in the book of Holmes. Hope this helps.

Answer 2

Sorry, I don't have enough points to comment, so I will post this here. Write a comment that you saw it and I will then delete it. This may be of help, I don't know if you are aware of it:

https://duckduckgo.com/?q=hahn+banach+for+hyperplanes&atb=v35-2a_&dbexp=b&ia=web