# Is the convex hull of a countable set a Borel set?

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?

• The convex hull is also the intersection of all half-spaces that contain it, by the supporting hyperplane theorem/ – logarithm May 28 at 20:42
• @logarithm Isn’t that only true of $X$ is closed? – Keshav Srinivasan May 28 at 20:45
• 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 $\cup E_n?.$ – zhw. May 28 at 21:00
• 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. – kimchi lover May 28 at 21:04

## 1 Answer

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)