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Consider a compact subset $A$ of $R^n$. Let me call $A$ special if any point $x$ that is on the boundary of $\textsf{conv.hull}(A)$ admits a unique representation as a convex combination of points in $A$.

My question is: Are "almost all" such compact subsets special?

If $A$ is finite, I think an argument by induction works: At the addition of a new point, the surface of the existing convex hull is $n-1$-dimensional, and so there is zero probability that the new point falls into this surface.

But to show the same result for (uncountably) infinite sets A (or at least A's generated by reasonably behaved parametric curves of one parameter), I am lost. From looking online, it seems that prevalence (Hunt, 1992) seems to be the right concept. I am however unfamiliar with the techniques, and cannot readily come up with a proof, but I'm hoping that maybe such a result is already known, if only one knows the right language?

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  • $\begingroup$ I don't understand. By "extremal" do you mean an extreme point? But an extreme point of the convex hull of $A$ is always a member of $A$, and can't be a nontrivial convex combination of any other points of $A$. Or do you just mean a boundary point? $\endgroup$
    – Robert Israel
    Commented Jun 22, 2018 at 21:57
  • $\begingroup$ I mean a point that solves $\arg\max_{x\in \textsf{conv.hull}(A)} \psi'x$ for some $\psi\in\mathbb{R}^n$. Sorry... I think boundary point is the right name. I will update the question right now. $\endgroup$
    – mimuller
    Commented Jun 22, 2018 at 22:56
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    $\begingroup$ The definitions of "shy" and "prevalent" require a complete metric linear space. Nonempty compact subsets of $\mathbb R^n$ do form a complete metric space with the Hausdorff metric, but it's not a linear space. $\endgroup$
    – Robert Israel
    Commented Jun 23, 2018 at 0:14
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    $\begingroup$ Your argument in the finite case is wrong. When you add a new point previously existing points could lose unique representation. (The "measure zero" conclusion should still be true, though.) $\endgroup$
    – Nik Weaver
    Commented Jun 23, 2018 at 15:43
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    $\begingroup$ The continuous functions from $[0,1]$ to $\mathbb R^n$ form a complete metric linear space, and maybe the corresponding set is prevalent there. $\endgroup$
    – Robert Israel
    Commented Jun 24, 2018 at 8:07

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