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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.

pitariver
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1 Answers1

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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 $f:X\rightarrow\mathbb{R}$ such that $$\sup_{a\in A}f(a)\leq\inf_{b\in B}f(b).$$

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

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

Proof: Let $t\in f(A)$. So $f(a)=t$ for some $a\in A$. Because $f$ is non-zero, we find some $x\in X$ such that $f(x)>0$. By the "algebraic openness" property of $A$, there exists $\varepsilon>0$ such that $a+(-\varepsilon,\varepsilon)\cdot x\subset A$. 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 $a\in A$ and $b\in B$ that $$f(a)<\sup_{\alpha\in A}f(\alpha)\leq\inf_{\beta\in B}f(\beta)<f(b).$$

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 $s\in\mathbb{R}$ such that $\phi(a)<s\leq\phi(b)$ for all $a\in A$ and $b\in B$. But there also exists a continuous linear functional $\psi$ and a constant $t\in\mathbb{R}$ such that $\psi(b)<t\leq\psi(a)$ for all $b\in B$ and $a\in A$. Then $f:=\phi-\psi$ is a continuous linear functional such that $f(a)<s-t<f(b)$ holds for all $a\in A$ and $b\in B$.

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?

SmileyCraft
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  • Roughly speaking, scalar multiplication is continuous exactly by the "algebraically openness" property, and vector addition is continuous because the property is invariant under translation. – SmileyCraft Jan 03 '19 at 00:25
  • @SimleyCraft I a m still not sure it's necessarily true (in anycase I understand how to solve the problem with the $\phi - \psi$ argument). For example why should scalar mult be continues? let $x_0 \in X, \lambda_0 \in \mathbb{R}, \lambda_0 x_0 \in U$ open, you would want a neighborhood of $x_0$, V, for which there exists $\delta>0$, $\vert \lambda - \lambda_0 \vert < \delta \implies \lambda V \subset U$ – pitariver Jan 03 '19 at 06:03
  • @pitariver I must admit I underestimated the difficulty of proving this fact. I did already realize the problems you mentioned, but I had some ideas of proving this, however they did not work after all. I edited the post with a new answer. However I would still love to know whether the "algebraic openness" defines a linear topology. – SmileyCraft Jan 03 '19 at 15:11
  • Thanks for the link, though sadly, it still doesn't solve the problem: I asked about a strict separation, yet Theorem 4 in the paper only guarantees "weak" separation(which I already heard of). The counter example given immediately after the theorem also doesn't tell us much about this case. I am starting to wonder whether the statement is actually true, even for the "half strict" case. – pitariver Jan 03 '19 at 18:45
  • Also in regards to the "algebraically open" sets defining a TVS, I am not entirely sure it's true. Thanks for the attention so far! – pitariver Jan 03 '19 at 18:50
  • @pitariver Oh right, totally forgot about the strictness. I guess we are left with more questions than answers. – SmileyCraft Jan 03 '19 at 19:49
  • @pitariver I solved the strictness problem still using Theorem 4 from the link. At least the OP question is now answered. – SmileyCraft Jan 03 '19 at 20:18
  • @pitariver Apparently algebraic openness does not define a topological vector space. https://math.stackexchange.com/questions/3060986/do-algebraically-open-sets-define-a-vector-space-topology/3060987#3060987 – SmileyCraft Jan 03 '19 at 20:56