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I want to prove the existence and uniqueness of the convex set described below, which is the convex hull. My thinking is that I'm to generate a set containing all the convex sets containing $A$ and take their intersection. Then pointing out that the intersection will also be convex. How could I formalize the set containing all such convex sets containing $A$?

Thanks in advance

If $A\subset\mathbb{R^n}$ is compact, then show that $\exists$ a unique convex subset $B$ of $\mathbb{R^n}$ such that $A\subset B$ and $B$ lies in any compact convex subset of $\mathbb{R^n}$ containing $A$.

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  • $\begingroup$ I think the definition in wiki is clear enough to solve your problem. en.wikipedia.org/wiki/Convex_hull $\endgroup$
    – vita nova
    Oct 10, 2017 at 3:28
  • $\begingroup$ The statement is a little disingenuous; there is a unique convex subset $B$ (called the convex hull) such that $A \subset B$ and $B \subset C$ for any convex $C$ containing $A$ (that is, not just compact ones). What is also truu in finite dimensions is that the convex hull of a compact set is also compact. one route to proving this is Carathéodory's theorem. $\endgroup$
    – copper.hat
    Oct 10, 2017 at 5:17

1 Answer 1

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To formalize: let $\mathscr C$ be the collection of sets (so it is a subset in the power set of $\mathbb R^n$):

$$\mathscr C:= \{ B\subset \mathbb R^n : B \text{ is convex and } A\subset B\}.$$

Then the set you want is

$$ A^h := \bigcap _{B\in \mathscr C} B.$$

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  • $\begingroup$ This makes sense. So I guess it would be enough to just say prior to describing the collection that since $A$ is compact, it's bounded, which guarantees that such convex sets contain it $\endgroup$ Oct 10, 2017 at 3:30
  • $\begingroup$ Indeed one can find a convex hull for any subset (indeed, $\mathbb R^n$ is convex and so $\mathscr C$ is always nonempty. The point is that when $A$ is compact, the convex hull is also compact (is that your question? @johnfowles) $\endgroup$
    – user99914
    Oct 10, 2017 at 3:33
  • $\begingroup$ Well, proving that the convex hull is compact is what I want to prove next, but I wanted to make sure that in proving the first part of the question, I was correctly defining the set which contained every convex set containing $A$ before I took the intersection of that set $\endgroup$ Oct 10, 2017 at 3:46

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