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Let $A \subset \mathbb{R}^n$. Suppose for all $x,y \in A$ with $x \not = y$, and for all $0 < t < 1$, we have $tx + (1-t)y \not \in A$. I like to say that $A$ is an "anti-convex" set. For instance, a sphere is anti-convex (I think?). Is there a former name for such a set property?

Spencer Kraisler
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    I don't think that concept has a name, so you are free to invent one - define it for your reader. You are right about the sphere - and for the boundary of any convex set as long as the boundary contains no segment (so is always curved in an appropriate sense). – Ethan Bolker Feb 03 '23 at 00:25
  • Hah I think you indirectly answer my question. Maybe I am wrong, but I think a set is anti-convex if and only if it's segment-free + boundary of a convex set – Spencer Kraisler Feb 03 '23 at 00:28
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    An arc of a circle satisfies your condition but it isn't the boundary of any convex subset of the plane. – Daniel Schepler Feb 03 '23 at 00:31
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    Hmm, it's reminiscent of, although definitely not the same as, an "extreme point". I think your first project is to find non-trivial examples. – JonathanZ Feb 03 '23 at 00:51
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    If $A$ is an anti-convex set, let $H$ be its convex hull. If $a \in A$ is in the interior of $H$, so there is some ball about $a$ completely contained in the interior $H$. If $a_1, a_2$ are points in the ball on opposite sides of $a$, Then there must be some points $a_1', a_2' \in A$ with $a_1 \in \overline{aa_1'}, a_2 \in \overline{aa_2'}$, but that would mean $a \in \overline{a_1'a_2'}$, contradicting that $A$ is anti-convex. Thus $A$ can only contain boundary points of $H$. Thus a set is anti-convex if and only if it is segment-free and a subset of the boundary of a convex set. – Paul Sinclair Feb 03 '23 at 23:20
  • @PaulSinclair Very neat. If you make that an answer, I'll accept it! – Spencer Kraisler Feb 05 '23 at 22:52
  • Actually, I think the argument is flawed. For example, consider $A$ to be a half-circle and a point $a$ in the interior of the concavity. This isn't anti-convex, but it illustrates the problem: If $a_1$ is between $a$ and the half-circle center, then the ray from $a$ thru $a_1$ does not intersect any other point of $A$, contrary to my claim. I still think the claim is true, but the proof is a bit harder than what I gave. – Paul Sinclair Feb 06 '23 at 03:30
  • @PaulSinclair: I don't think that your claim is true. Take the three corners of a triangle together with one point in its interior. Then, $A$ consists of four points, is anti-convex, but the interior point is not a boundary point of $H$. – gerw Feb 07 '23 at 12:35
  • @gerw - You are right. There are anti-convex sets that are not subsets of the boundary of any convex set. – Paul Sinclair Feb 07 '23 at 13:00
  • Maybe one has to change the definition of "anti-convex" to: "No point in $A$ can be written as a (proper) convex definition of points in $A$". Then, the statement might continue to hold. – gerw Feb 08 '23 at 07:55
  • If I’m not mistaken, that is implied by the definition, no? – Spencer Kraisler Feb 08 '23 at 17:02

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