Mfrhtyj

Is the following statement true?

Let $X$ be a real linear space, $A,B subset X$ two disjoint convex sets with the following "algebraic openness" property: Every $x in A$ is an internal point of A, and so is every point of B and internal point of B. Then there exists a linear functional $f:X rightarrow mathbb{R}$ and $t in mathbb{R}$ such that for all $x in A, y in B$ we have
$$ f(x) < t < f(y)$$

A similar version (but in the context of TVS and one of the sets is open) is given in Wikipedia, Note that it only admits a "half strict" separation. I believe that it is still true when topological openness is replaced by the "algebraic openness" definition given here, but the question is if I assume both sets are "algebraically open", is it true that I can get a strict separation from both sides? Any help is appreciated.

IMPORTANT EDIT: A theorem related to the question stated in the OP can be found here. Theorem 4 states: If $A$ and $B$ are disjoint convex sets in $X$ and $A$ has an internal point, then $A$ and $B$ can be (weakly) separated. That is, there exists $f:Xrightarrowmathbb{R}$ such that $$sup_{ain A}f(a)leqinf_{bin B}f(b).$$

EDIT 2: We can use this theorem to answer the question in the OP. This follows from the following proposition.

Proposition: Let $f:Xtomathbb{R}$ be linear and non-zero. Then for any $A$ obeying the "algebraic openness" property, we have that $f(A)$ is open.

Proof: Let $tin f(A)$. So $f(a)=t$ for some $ain A$. Because $f$ is non-zero, we find some $xin X$ such that $f(x)>0$. By the "algebraic openness" property of $A$, there exists $varepsilon>0$ such that $a+(-varepsilon,varepsilon)cdot xsubset A$. Hence, $(t-varepsilon f(x),t+varepsilon f(x))subseteq f(A)$, so $f(A)$ is open.

Combining the two results, we have for all $ain A$ and $bin B$ that $$f(a)<sup_{alphain A}f(alpha)leqinf_{betain B}f(beta)<f(b).$$

ORIGINAL ANSWER: The set of all sets with your "algebraic openness" property makes $X$ a topological vector space. Hence, if $A$ and $B$ are disjoint, convex and open in this topology, then there exists a continuous linear functional $phi$ and a constant $sinmathbb{R}$ such that $phi(a)<sleqphi(b)$ for all $ain A$ and $bin B$. But there also exists a continuous linear functional $psi$ and a constant $tinmathbb{R}$ such that $psi(b)<tleqpsi(a)$ for all $bin B$ and $ain A$. Then $f:=phi-psi$ is a continuous linear functional such that $f(a)<s-t<f(b)$ holds for all $ain A$ and $bin B$.

EDIT 3: We now know that the "algebraic openness" property does not define a topological vector space. Do algebraically open sets define a vector space topology?