The convex hull of a subset $X$ of $\mathbb{R}^2$ is the smallest convex subset of $\mathbb{R}^2$ containing $X$. My question is, if $X$ is countable, then is the convex hull of $X$ necessarily a Borel set?

If not, does anyone know of a counterexample?

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    $\begingroup$ The convex hull is also the intersection of all half-spaces that contain it, by the supporting hyperplane theorem/ $\endgroup$ – logarithm May 28 at 20:42
  • $\begingroup$ @logarithm Isn’t that only true of $X$ is closed? $\endgroup$ – Keshav Srinivasan May 28 at 20:45
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    $\begingroup$ Suppose $E=\{x_1,x_2,\dots \}.$ Let $E_n$ be the convex hull of $\{x_1,\dots, x_n \}.$ Isn't it true that the convex hull of $E$ is $\cup E_n?.$ $\endgroup$ – zhw. May 28 at 21:00
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    $\begingroup$ By en.wikipedia.org/wiki/… the ch of $X$ is the union of the closed triangles whose vertices are in $X$. Of which there are only countably many. $\endgroup$ – kimchi lover May 28 at 21:04

Yes,it is Borel.

Define $\text{conv}_n(E)$ by $x\in \text{conv}_n(E)$ iff $\exists a_1,a_2,\dots,a_n\in E,\exists t_1,t_2,\dots,t_n\in[0,1], \Sigma t_i=1$ such that $x=\Sigma_{i=1}^n t_ia_i$.

It is easily to seen that $\text{conv}_n(E)$ is Borel, so $\text{conv}(E)$ is Borel since $\text{conv}_n(E)=\cup_{n=1}^\infty\text{conv}_n(E)$ (Try to prove this equality,please note that the union is vonvex and contains $E$, or refer J. van Mill's book-infinite-Dimensional Topology, Page 7,Lemma 1.2.2)


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