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I am investigating convex analysis (unfortunately without a book) and currently I am trying to understand and clasify the algebraic structure of convex sets. Related questions that I have studied are, for example, this and this.

The question:
Let $A$ be a real affine space and let $K\subsetneq A$ be a convex set whose affine closure is $A$. Let its exterior be non-empty, i.e. let the interior of $A\setminus K$ be non-empty, i.e. let there exist a point $x\in A$ such that every one-dimensional affine subspace of $A$ which contains $x$ (a line through $x$) contain an interval which contains $x$ and is disjoint with $K$. Does, then, there exist a convex subset of $A\setminus K$, with non-empty interior, whose affine closure is $A$?

Discussion:
In the case of finite-dimensional spaces, this question is easy: every proper convex set has topological interior and exterior. In the case when $K$ has algebraic interior, we can apply a form of Hahn-Banach theorem to it to show that there exists a "half-space" which contains whole $K$, hence the complement of that half-space would be a convex set spanning the whole $A$. If we don't require that the interior of $L$ to be non-empty, then the reflection of $K$ over some exterior point $x$ is the set we want.
So, we can assume that $A$ has infinitely many dimensions and that the algebraic interior of $K$ is empty.
There are examples of sets $K$ when the answer to my question is affirmative: Take the space of polynomials and let $K$ be the set of polynomials with no negative coefficients. Now let $L$ be the set of all polynomials whose first coefficient is less than $-1$. This $L$ is the set I am asking for. So, if the answer to my question doesn't need any more branching into cases, then it could only be that for any such $K$ there exists such a set. Perhaps there is at least one set for which this doesn't hold?

cnikbesku
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  • You may want to specify what you mean by "space" for $A$. For instance if it is a LCS you can apply both forms of the geometric version of Hahn Banach. I think specifying is a good way to start because then you may see where Hahn Banach fails and maybe get counter examples – blamethelag Jan 09 '23 at 22:40
  • @blamethelag Good point, I forgot to focus on that, thanks. the space $A$ is just a real affine space. There are no norms, topologies nor metrics. Nothing like that, just a vector space and the affine space "generated" by it – cnikbesku Jan 09 '23 at 22:42
  • Also is it true that if there exists $L \subset A \setminus K$ convex such that $\overline{\operatorname{aff}} L = A$ then $\overline{\operatorname{aff}} \overline{\operatorname{co}} A \setminus K = A$? – blamethelag Jan 09 '23 at 23:02
  • @blamethelag can you please just tell what this "upperlined co" means? the other is, I suppose, the affine closure. – cnikbesku Jan 09 '23 at 23:05
  • It's the closed convex hull which you can replace by the convex hull. I was just mentioning that in fact your question reduces to: the affine hull of $A \setminus K$ is $A$. It can be deduced from the fact that $A \setminus K$ has no empty interior, according to the second link you gave. So the question boils down to: a set with non empty interior has affine closure that is the full space. – blamethelag Jan 09 '23 at 23:32
  • @blamethelag It is good, because the existance of an exterior point yields that $A\setminus K$ spans $A$, but I don't see why my question boils to it? Yes, I have just added another condition on the set I am looking for (for it to have algebraic interior), but even without it, I don't see, sorry. Why would this fact yield an existance of my $L$? – cnikbesku Jan 10 '23 at 00:00
  • Let us continue this discussion in chat. – blamethelag Jan 10 '23 at 13:44

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