Prove that the following two definitions of the convex hull are equivalent.

Question

I was wondering if a topology expert could help me solve this proof, as I have no idea but want to understand these concepts. This is not for homework.

Let X be a point set, not necessarily finite, in R^d. Prove the following two definitions of the convex hull are equivalent.

The set of all points that are convex combinations of all points in X.
The intersection of all convex sets that include X.

http://en.wikipedia.org/wiki/Convex_hull

http://en.wikipedia.org/wiki/Convex_combination

I can visualize in my head why this is true, but cannot convert it to mathematical terms. I would really appreciate any expert's help on this, as it is very simple but I have taking it away from just geometry, to a general proof.

The intersection of a family of convex sets is convex (show that, if you don't already have a theorem for it). The set of convex combinations of points of $X$ is a convex set containing $X$ (show that too, if not a theorem you can use). That's it. — Daniel Fischer, Sep 10 '13 at 19:09
Well, okay, you may need to explicitly mention that a convex combination of points of $X$ is contained in every convex set containing $X$. — Daniel Fischer, Sep 10 '13 at 19:13
@copper.hat But it's a convex set containing $X$, so it contains the first. — Daniel Fischer, Sep 10 '13 at 19:16
@DanielFischer: Without noting that the first set is contained in all convex sets containing $X$, your initial comment just gives one inclusion. — copper.hat, Sep 10 '13 at 19:20
As an aside, there is no topology involved here. Any vector space will do. — copper.hat, Sep 10 '13 at 19:25
This is special case of many similar situations, where we have "top-down" and "bottom-up" description of some object. See this answer. — Martin Sleziak, Feb 14 '15 at 18:16

score 2 · Accepted Answer · edited Feb 14 '15 at 17:29

There is a standard procedure when showing equalities like these: Let $A$ be the set, and define $\text{Conv}(A)$ as the set of all convex combinations of $A.$ You want to show that $\text{Conv}(A)=\bigcap\mathcal C$ where $\mathcal C=\{C\mid C\text{ is a convex superset of }A\}.$ So you should show that $\text{Conv}(A)$ is itself convex and contains $A,$ that tells you that $\bigcap\mathcal C\subseteq \text{Conv}(A).$ Then show that $\text{Conv}(A)$ is in each convex superset of $A,$ that tells you that $\text{Conv}(A)\subseteq\bigcap\mathcal C.$

The second inclusion is harder than the first. Let $B$ be a convex set containing $A.$ Use induction: Assume that for $n\in\mathbb N$ all convex combinations $\sum_{i=0}^n t_i x_i,\ t_i\ge 0,\ \sum t_i=1$ are in $B$, then try to express $\sum_{i=1}^{n+1}s_i x_i$ as $tx_{n+1}+(1-t)\sum_{i=0}^n t_i x_i$ for some $t\in[0,1]$.

By $B$ I presume you mean any convex set containing $A$? – copper.hat Sep 10 '13 at 19:30 — copper.hat, Sep 10 '13 at 19:30
@copper.hat: Yes, that's right. – Stefan Hamcke Sep 10 '13 at 19:31 — Stefan Hamcke, Sep 10 '13 at 19:31

score 0 · Answer 2 · answered Sep 10 '13 at 19:19

I'm not an expert, but still hope to be capable of answering.

The intersection of all convex supersets is the same as the smallest convex superset (why is it convex?). So it is enough to see that the set from the first definition is convex (obvious), "superset" (obvious) and "the smallest" (for example by induction on the number of nontrivial elements in the convex combination).

Prove that the following two definitions of the convex hull are equivalent.

2 Answers2