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This might be silly, but I am not sure.

Let $A \subseteq \mathbb R^2$. Suppose that for any two points $x,y \in A$, I "add" the straight segment $[x,y]$ between them. Is the result convex?

That is, is $\cup_{(x,y)\in A^2} [x,y]$ convex?

Asaf Shachar
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  • What if $A$ is simply the vertices of a triangle (e.g. $A = { (0,0),(0,1),(1,1)}$)? – Minus One-Twelfth May 14 '20 at 10:18
  • Maybe if you do it twice. First step we get from some points to something like a "polygon". Second step we fill in the "polygon". – Vepir May 14 '20 at 10:33
  • @Vepir. Interesting question. I was actually implicitly thinking on $A$ as a continuous path in the plane. I have now asked about this here: https://math.stackexchange.com/questions/3674400/does-connecting-any-two-points-in-a-graph-result-in-a-convex-set – Asaf Shachar May 14 '20 at 10:35

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No. If you take three non-collinear points then the union of the lines joining them is the triangle with those points as vertices. This is not convex.

Kavi Rama Murthy
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  • Thank you. I was implicitly thinking on $A$ as a continuous path in the plane. Since I forgot to mention this, I (of course) accepted your answer. Do you have an idea if the answer changes if we begin with a graph of a function? I asked about this here: https://math.stackexchange.com/questions/3674400/does-connecting-any-two-points-in-a-graph-result-in-a-convex-set – Asaf Shachar May 14 '20 at 10:34