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Let $A$ and $B$ be nonempty convex cones in an infinite-dimensional vector space $V$. Assume that $A$ and $B$ partition $V$, that is, $A\cap B=\varnothing$ and $A\cup B=V$. Can $A$ and $B$ be separated by a linear functional on $V$?

(I have tried to prove that "the partition assumption" implies that either of the cones has an internal point, but have failed.)

Thanks.

Mikhail
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  • You just wrote "vector space $V$". Do you have any topology on $V$? Otherwise, what is an "internal point"? – gerw Nov 20 '23 at 07:34
  • The vector space $V$ is not endowed a topology. For the definition of an internal point, please, see p. 3 here: https://www.johndcook.com/SeparationOfConvexSets.pdf This definition does not require a topology. If one of the cones has an internal point, then $A$ and $B$ can be separated by a linear functional (see Theorem 4 at p. 7 in the cited notes). – Mikhail Nov 20 '23 at 09:05
  • Ok, thank you. This definition is known to me as "algebraic interior". I will think about your question. – gerw Nov 20 '23 at 13:55

1 Answers1

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I think that we can modify Example 1 from your linked reference. Let $V = c_c$, the space of finite sequences. Now, let \begin{align} A &= \{ x \in c_c \mid \text{last coefficient of $x$ is positive}\},\\ B &= \{ x \in c_c \mid \text{last coefficient of $x$ is negative}\} \cup \{0\}. \end{align} Clearly, these sets satisfy your partition assumption. Since $A$ cannot be separated from $\{0\}$ (see your reference), we cannot separate $A$ and $B$.

(This is very strange...)

gerw
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