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I put the words in quotes because this question is NOT about topological affine spaces.

My definitions:
Let $A$ be a real affine space and let $K$ be a convex subset of $A$.
We define a point $x\in A$ to be an interior point of $K$ if every line $p$ that passes through $x$: $x\in p\subseteq A$, is in the interior of the segment $p\cap K$ (assuming standard topology on real line). Similarly, an exterior point is such a point $x\in A$ which is in the exterior of $p\cap K$ for every line $p$ that passes through $x$. A line is just a one-dimensional affine subspace of $A$.

The question:
Let $K$ be a convex subset of a real affine space $A$ such that the affine closure of $K$ is exactly $A$. If there exists an exterior point of $K$, does there exist an interior point of $K$ too?

My results:
Firstly, if $K$ doesn't span $A$ then this obviously doesn't have to hold: just let $K$ be a point. In case of finitely dimensional spaces, this is true: if a set has exterior and spans the whole space, then it has an interior too; this can be proven inductively. In that matter, I hve proven that in the case of my question, for every finitely dimensional affine subspace $\alpha$ of $A$ there is a translation $\beta$ of $\alpha$ such that $K\cap\beta$ has an interior with respect to $\beta$ (it is not a full interior point). Also, I fail to find a counter-example in the space of real polynomials, so maybe this holds for $\aleph_0$-dimensional spaces at least? On the other side, the existance of the exterior point might seem extra in these cases that I could prove, but it is important in a similar question where I want to prove the existance of an exterior point given the existance of an interior point. This is easily provable (and holds iff $K\neq A$) and that is where my question comes from. Third case would be an example of a convex set with neither exterior nor interior: all polynomials with the leading coefficient being positive (not negative nor $0$).

Comment:
This question extends this question. It is not exactly a duplicate, I investigated the exterior too.

cnikbesku
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1 Answers1

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Perhaps, the answer is a simple one. There exists a convex set with exterior and no interior which spans the whole space. Obtaining a convex set with exterior but no interior can be done in a very simple way: pick a convex set no neither interior nor exterior and intersect it with a "half-space" (pick a basis of $A$ and restrict only one coordinate to be $\geq$ than something). Then we just need to give some effort to make it still span the whole space.

Here is an example:
All real polynomials with no negative coefficients. This set is obviously convex in the space of all polynomials. The polynomial $-1$ is in the exterior, but no polynomial can be in its interior: if $n>\deg(p)$, then the line $p+t\cdot x^n$ is in the set iff $t\geq0$. The interior of $\{t\geq0\}$ is $\{t>0\}$, so $t=0$, i.e. $p$, is not in the interior.

cnikbesku
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